This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 4 Energy Transfer by Heat, Work, and Mass Chapter 4 ENERGY TRANSFER BY HEAT, WORK, AND MASS
Heat Transfer and Work 41C Energy can cross the boundaries of a closed system in two forms: heat and work. 42C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other forms are work. 43C An adiabatic process is a process during which there is no heat transfer. A system that does not exchange any heat with its surroundings is an adiabatic system. 44C It is a work interaction. 45C It is a work interaction since the electrons are crossing the system boundary, thus doing electrical work. 46C It is a heat interaction since it is due to the temperature difference between the sun and the room. 47C This is neither a heat nor a work interaction since no energy is crossing the system boundary. This is simply the conversion of one form of internal energy (chemical energy) to another form (sensible energy). 48C Point functions depend on the state only whereas the path functions depend on the path followed during a process. Properties of substances are point functions, heat and work are path functions. 49C The caloric theory is based on the assumption that heat is a fluidlike substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric. Boundary Work 410C It represents the boundary work for quasiequilibrium processes. 411C Yes. 412C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case. 413C 1 kPa m 3 = 1 k(N / m 2 ) m 3 = 1 kN m = 1 kJ 41 Chapter 4 Energy Transfer by Heat, Work, and Mass 414 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A4 through A6) P1 = 200 kPa 3 v1 = v g @ 200 kPa = 0.8857 m /kg Sat.vapor P2 = 200 kPa 3 v 2 = 1.3162 m /kg T2 = 300 o C Analysis The boundary work is determined from its definition to be Wb,out = P dV = P (V 2  V1 ) = mP(v 2  v1 )
1 P
(kPa) 1 200 2 2 V 1kJ = (5kg)(200kPa)(1.3162  0.8857) m 3 /kg 1kPa m 3 = 430.5 kJ Discussion The positive sign indicates that work is done by the system (work output). 415 Refrigerant134a in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A11 through A13) P1 = 800 kPa 3 v1 = v f @ 800 kPa = 0.0008454 m /kg Sat.liquid P2 = 800kPa 3 v 2 = 0.02846m /kg o T2 = 50 C Analysis The boundary work is determined from its definition to be m= and Wb, out = P dV = P (V 2  V1 ) = mP(v 2  v1 )
1 P
(kPa) 1 800 2 V1 0.2 m 3 = = 236.6 kg v1 0.0008454 m 3 / kg v 2 1kJ = (236.6kg)(800kPa)(0.02846  0.0008454)m 3 /kg 1kPa m 3 = 5227kJ Discussion The positive sign indicates that work is done by the system (work output). 42 Chapter 4 Energy Transfer by Heat, Work, and Mass 416 Problem 415 is reconsidered. The effect of pressure on the work done. as the pressure varies from 400 kPa to 1200 kPa is to be investigated. The work done is to be plotted versus the pressure. "Knowns" Vol_1L=200"[L]" x_1=0 "saturated liquid state" P=800"[kPa]" T_2=50"[C]" "Solution" Vol_1=Vol_1L*convert(L,m^3)"[m^3]" "The work is the boundary work done by the R134a during the constant pressure process." W_boundary=P*(Vol_2Vol_1)"[kJ]" "The mass is:" Vol_1=m*v_1"[m^3]" v_1=volume(R134a,P=P,x=x_1)"[m^3/kg]" Vol_2=m*v_2"[m^3]" v_2=volume(R134a,P=P,T=T_2)"[m^3/kg]" "Plot information:" v[1]=v_1 v[2]=v_2 P[1]=P P[2]=P T[1]=temperature(R134a,P=P,x=x_1) T[2]=T_2 P [kPa] 400 514.3 628.6 742.9 857.1 971.4 1086 1200 Wboundary [kJ] 7334 7073 6833 6607 6392 6186 5986 5790 43 Chapter 4 Energy Transfer by Heat, Work, and Mass 150 125 100 75 R134a T [C] 50 25 0 25 50 10 4 10 3 10 2 2 1
800 kPa 10 1 v [m /kg] 3 10 5 R134a 10 4 P [kPa] 2
10
3 1
31.31C 10 2 10 1 10 4 10 3 10 2 10 1 v [m /kg] 3 44 Chapter 4 Energy Transfer by Heat, Work, and Mass 7250 P = 800 kPa
6800 W boundary [kJ] 6350 5900 5450 5000 50 60 70 80 90 100 110 120 130 T[2] [C] 7500 T = 100 C
2
7150 W boundary [kJ] 6800 6450 6100 5750 400 500 600 700 800 900 1000 1100 1200 P [kPa] 45 Chapter 4 Energy Transfer by Heat, Work, and Mass 417E Superheated water vapor in a cylinder is cooled at constant pressure until 70% of it condenses. The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A4E through A6E) P1 = 60 psia 3 v1 = 9.399ft /lbm T1 = 500 o F P2 = 60 psia v 2 = v f + x 2 v fg x 2 = 0 .3 = 0.017378 + 0.3(7.177  0.017378) = 2.165ft 3 /lbm Analysis The boundary work is determined from its definition to be Wb, out = P dV = P(V 2  V1 ) = mP (v 2  v1 )
1 P
(psia) 60 2 1 v 2 1 Btu = (12 lbm)(60 psia)(2.165  9.399)ft 3 /lbm 5.4039 psia ft 3 = 963.8 Btu Discussion The negative sign indicates that work is done on the system (work input). 46 Chapter 4 Energy Transfer by Heat, Work, and Mass 418 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasiequilibrium. 2 Air is an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A1). Analysis The boundary work is determined from its definition to be
P
2 T = 12C 1 Wb ,out = 2 1 P dV = P1V1 ln V2 P = mRT ln 1 V1 P2 150 kPa = (2.4 kg)(0.287 kJ/kg K)(285 K)ln 600 kPa = 272 kJ
Discussion The negative sign indicates that work is done on the system (work input). V 419 Nitrogen gas in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasiequilibrium. 2 Nitrogen is an ideal gas. Analysis The boundary work is determined from its definition to be Wb, out =
P
2 2 1 P dV = P1V1 ln V2 P = P1V1 ln 1 V1 P2 150kPa 1kJ = (150kPa)(0.2m 3 ) ln 800kPa 3 1kPa m = 50.2kJ T = 300 K 1 V Discussion The negative sign indicates that work is done on the system (work input). 47 Chapter 4 Energy Transfer by Heat, Work, and Mass 420 A gas in a cylinder is compressed to a specified volume in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined by plotting the process on a PV diagram and also by integration. Assumptions The process is quasiequilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a PV diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, P = aV1 + b = ( 1200 kPa / m 3 )(0.42 m 3 ) + (600 kPa) = 96 kPa 1 P2 = aV2 + b = ( 1200 kPa / m 3 )(0.12 m 3 ) + (600 kPa) = 456 kPa and
P2 P
(kPa) 2 P = aV + b 1 Wb, out = Area = P1 + P2 (V 2  V1 ) 2 1kJ (96 + 456)kPa = (0.12  0.42)m 3 1kPa m 3 2 = 82.8kJ P1 0.1 0.42 V (m3) (b) The boundary work can also be determined by integration to be Wb ,out = V 22  V12 + b(V2  V1 ) 1 1 2 (0.12 2  0.42 2 )m 6 = (1200 kPa/m 3 ) + (600 kPa)(0.12  0.42)m 3 2 = 82.8 kJ 2 P dV = 2 (aV + b)dV = a GAS P = aV+ b Discussion The negative sign indicates that work is done on the system (work input). 48 Chapter 4 Energy Transfer by Heat, Work, and Mass 421E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a PV diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, At state 1:
P P = aV1 + b 1 15 psia = (5 psia / ft 3 )(7 ft 3 ) + b b = 20 psia (psia) 100 1 P = aV + b 2 At state 2: P2 = aV2 + b V2 = 24 ft
3 15 100 psia = (5 psia / ft 3 )V2 + ( 20 psia) and, 7 V (ft3) Wb ,out = Area = P1 + P2 (100 + 15)psia 1Btu (V 2  V1 ) = (24  7)ft 3 5.4039psia ft 3 2 2 = 181 Btu
Discussion The positive sign indicates that work is done by the system (work output). 422 [Also solved by EES on enclosed CD] A gas in a cylinder expands polytropically to a specified volume. The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Analysis The boundary work for this polytropic process can be determined directly from V P2 = P1 1 V 2
and, 0.03 m 3 = (150 kPa) 0.2 m 3 P2V 2  P1V1 1 n n 1.3 = 12.74 kPa P
(kPa) 1 150 PV Wb ,out = 2 1 P dV = (12.74 0.2  150 0.03) kPa m 3 = 1  1.3 = 6.51 kJ 1 kJ 1 kPa m 3 2 0.03 0.2 (m3) V Discussion The positive sign indicates that work is done by the system (work output). 49 Chapter 4 Energy Transfer by Heat, Work, and Mass 423 Problem 422 is reconsidered. The process described in the problem is to be plotted on a PV diagram, and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted. Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is required since the expression for boundary work depends on whether n=1 or n<>1" If n<>1 then BoundWork:=(P[2]*V[2]P[1]*V[1])/(1n)"Use Equation 322 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 320 when n=1" endif end "Inputs from the diagram window" {n=1.3 P[1] = 150 "kPa" V[1] = 0.03 "m^3" V[2] = 0.2 "m^3" Gas$='AIR'} "System: The gas enclosed in the pistoncylinder device." "Process: Polytropic expansion or compression, P*V^n = C" P[2]*V[2]^n=P[1]*V[1]^n "n = 1.3" "Polytropic exponent" "Input Data" W_b = BoundWork(P[1],V[1],P[2],V[2],n)"[kJ]" "If we modify this problem and specify the mass, then we can calculate the final temperature of the fluid for compression or expansion" m[1] = m[2]"[kg]" "Conservation of mass for the closed system" "Let's solve the problem for m[1] = 0.05 kg" m[1] = 0.05 "[kg]" "Find the temperatures from the pressure and specific volume." T[1]=temperature(gas$,P=P[1],v=V[1]/m[1])"[K]" T[2]=temperature(gas$,P=P[2],v=V[2]/m[2])"[K]"
3 n 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 P1 [kPa] 150 150 150 150 150 150 150 150 150 150 P2 [kPa] 150 102.6 70.23 48.06 32.88 22.5 15.4 10.53 7.208 4.932 T1 [K] 313.6 313.6 313.6 313.6 313.6 313.6 313.6 313.6 313.6 313.6 T2 [K] 2090 1430 978.8 669.7 458.3 313.6 214.6 146.8 100.5 68.74 V1 [m ] 0.03 0.03 0.03 0.03 0.03 0.03 0.03 0.03 0.03 0.03 410 Chapter 4 Energy Transfer by Heat, Work, and Mass
Boundary W ork vs n
30 25 20 15 W b [kJ] 10 5 0 0.0 0.4 0.8 1.2 1.6 2.0 n Pressures vs n
175 156 136 117 P[1] P ressure 97 78 58 39 19 0 0.0 P[2] 0.4 0.8 1.2 1.6 2.0 n Tem perature vs n
2250 1800 1350 T[2] Tem perature [K] 900 T[1]
450 0 0.0 0.4 0.8 1.2 1.6 2.0 n 411 Chapter 4 Energy Transfer by Heat, Work, and Mass 424 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1 The process is quasiequilibrium. 2 Nitrogen is an ideal gas. Analysis The boundary work for this polytropic process can be determined from Wb, out = P2V 2  P1V1 mR(T2  T1 ) = 1 n 1 n (2kg)(0.2968kJ/kg K)(360  300)K = 1  1.4 = 89.0 kJ
P
2 PVn=C 1 2 1 P dV = V Discussion The negative sign indicates that work is done on the system (work input). 425 [Also solved by EES on enclosed CD] A gas whose equation of state is v ( P + 10 / v 2 ) = Ru T expands in a cylinder isothermally to a specified volume. The unit of the quantity 10 and the boundary work done during this process are to be determined. Assumptions The process is quasiequilibrium. Analysis (a) The term 10 / v 2 must have pressure units since it is added to P. Thus the quantity 10 must have the unit kPam6/kmol2. (b) The boundary work for this process can be determined from
T = 300 K P NRu T 10 N 2 RT R T 10 10 =  P= u  2 = u  V V / N (V / N ) 2 v V2 v and V
2 4 Wb ,out = 1 V 1 dV = NRu T ln 2 + 10 N 2 V  V 1 1 V1 1 2 4 m3 = (0.5 kmol)(8.314 kJ/kmol K)(300 K)ln 2 m3 1 1 1 kJ + (10 kPa m 6 /kmol 2 )(0.5kmol) 2 4 m 3  2 m 3 1 kPa m 3 = 863 kJ 2 P dV = 2 NRu T 10 N 2 V  V2 Discussion The positive sign indicates that work is done by the system (work output). 412 Chapter 4 Energy Transfer by Heat, Work, and Mass 426 Problem 425 is reconsidered. Using the integration feature, the work done is to be calculated and compared, and the process is to be plotted on a PV diagram. "Input Data" N=0.5"[kmol]" v1_bar=2/N"[m^3/kmol]" v2_bar=4/N"[m^3/kmol]" T=300"[K]" R_u=8.314"[kJ/kmolK]" "The equation of state is:" v_bar*(P+10/v_bar^2)=R_u*T "P is in kPa" "using the EES integral function, the boundary work, W_bEES, is" W_b_EES=N*integral(P,v_bar, v1_bar, v2_bar,0.01)"[kJ]" "We can show that W_bhand= integral of Pdv_bar is (one should solve for P=F(v_bar) and do the integral 'by hand' for practice)." W_b_hand = N*(R_u*T*ln(v2_bar/v1_bar) +10*(1/v2_bar1/v1_bar))"[kJ]" "To plot P vs v_bar, define P_plot =f(v_bar_plot, T) as" {v_bar_plot*(P_plot+10/v_bar_plot^2)=R_u*T} " P=P_plot and v_bar=v_bar_plot just to generate the parametric table for plotting purposes. To plot P vs v_bar for a new temperature or v_bar_plot range, remove the '{' and '}' from the above equation, and reset the v_bar_plot values in the Parametric Table. Then press F3 or select Solve Table from the Calculate menu. Next select New Plot Window under the Plot menu to plot the new data." Pplot 622.9 560.7 509.8 467.3 431.4 400.6 373.9 350.5 329.9 311.6 vplot 4 4.444 4.889 5.333 5.778 6.222 6.667 7.111 7.556 8 650 600 550 500 450 P vs v bar 1 T = 300 K P plot [kPa] 400 350 300 250 200 150 100 50 0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 2 Area = W boundary v plot [m^3/kmol] 413 Chapter 4 Energy Transfer by Heat, Work, and Mass 427 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as P = aV 2 . The boundary work done during this process is to be determined. Assumptions The process is quasiequilibrium. Analysis The boundary work done during this process is determined from
P
2 P = aV2 Wb ,out 1 a 1 = P dV = 2 dV =  a V  V 1 1 V 1 2 1 1 1 kJ = (8 kPa m 6 )  3 0.1 m 0.3 m 3 1 kPa m 3 = 53.3 kJ 2 2 0.1 1 (m3) V 0.3 Discussion The negative sign indicates that work is done on the system (work input). 428E Hydrogen gas in a cylinder equipped with a spring is heated. The gas expands and compresses the spring until its volume doubles. The final pressure, the boundary work done by the gas, and the work done against the spring are to be determined, and a PV diagram is to be drawn. Assumptions 1 The process is quasiequilibrium. 2 Hydrogen is an ideal gas. Analysis (a) When the volume doubles, the spring force and the final pressure of H2 becomes Fs = kx 2 = k P2 = P1 + 15ft 3 V = (15,000lbf/ft) = 75,000lbf A 3ft 2 1ft 2 144in 2 = 188.3 psia P
2 1 Fs 75,000lbf = (14.7psia) + A 3ft 2 (b) The pressure of H2 changes linearly with volume during this process, and thus the process curve on a PV diagram will be a straight line. Then the boundary work during this process is simply the area under the process curve, which is a trapezoid. Thus, 15 30 V (ft3) Wb ,out = Area = = P1 + P2 (V2  V1 ) 2 1 Btu (188.3 + 14.7)psia (30  15)ft 3 5.40395 psia ft 3 = 281.7 Btu 2 (c) If there were no spring, we would have a constant pressure process at P = 14.7 psia. The work done during this process would be Wb ,out, no spring = PdV = P(V
1 2 2  V1 ) = 40.8 Btu 1 Btu = (14.7 psia)(30  15) ft 3 5.40395 psia ft 3 Thus, Wspring = Wb  Wb, no spring = 281.7  40.8 = 240.9 Btu Discussion The positive sign for boundary work indicates that work is done by the system (work output). 414 Chapter 4 Energy Transfer by Heat, Work, and Mass 429 Water in a cylinder equipped with a spring is heated and evaporated. The vapor expands until it compresses the spring 20 cm. The final pressure and temperature, and the boundary work done are to be determined, and the process is to be shown on a PV diagram. Assumptions The process is quasiequilibrium. Analysis (a) The final pressure is determined from P3 = P2 + (100 kN/m)(0.2 m) 1kPa Fs kx = P2 + = (150kPa) + 1kN/m 2 A A 0.1m 2 = 350 kPa P The specific and total volumes at the three states are T1 = 25 o C 3 v1 v f @ 25o C = 0.001003m /kg P1 = 150kPa V1 = mv1 = (50kg)(0.001003m 3 /kg) = 0.05m 3 V 2 = 0.2m 3 V3 = V 2 + x 23 A p = (0.2m ) + (0.2m)(0.1m ) = 0.22m
3 2 3 1 3 2 v3 = V3 0.22m 3 = = 0.0044 m 3 /kg 50kg m v At 350 kPa, vf = 0.0010 m3/kg and vg = 0.5243 m3/kg. Noting that vf < v3 < vg , the final state is a saturated mixture and thus the final temperature is T3 = [email protected] kPa = 138.88 o C
(b) The pressure remains constant during process 12 and changes linearly (a straight line) during process 23. Then the boundary work during this process is simply the total area under the process curve, Wb ,out = Area = P1 (V2  V1 ) + P2 + P3 (V3  V 2 ) 2 (150 + 350)kPa 1 kJ (0.22  0.2)m 3 = (150 kPa)(0.2  0.05)m 3 + 3 2 1 kPa m = 27.5 kJ
Discussion The positive sign indicates that work is done by the system (work output). 415 Chapter 4 Energy Transfer by Heat, Work, and Mass 430 Problem 429 is reconsidered. The effect of the spring constant on the final pressure in the cylinder and the boundary work done as the spring constant varies from 50 kN/m to 500 kN/m is to be investigated. The final pressure and the boundary work are to be plotted against the spring constant. P[3]=P[2]+(Spring_const)*(V[3]  V[2]) "P[3] is a linear function of V[3]" "where Spring_const = k/A^2, the actual spring constant divided by the piston face area squared" "Input Data" P[1]=150"[kPa]" m=50"[kg]" T[1]=25"[C]" P[2]=P[1]"[kPa]" V[2]=0.2"[m^3]" A=0.1"[m^2]" k=100"[kN/m]" DELTAx=20"[cm]" Spring_const=k/A^2"[kN/m^5]" V[1]=m*spvol[1] spvol[1]=volume(Steam,P=P[1],T=T[1]) V[2]=m*spvol[2] V[3]=V[2]+A*DELTAx*convert(cm,m) V[3]=m*spvol[3] "The temperature at state 2 is:" T[2]=temperature(Steam,P=P[2],v=spvol[2]) "The temperature at state 3 is:" T[3]=temperature(Steam,P=P[3],v=spvol[3]) Wnet_other = 0"[kJ]" W_out=Wnet_other + W_b12+W_b23"[kJ]" W_b12=P[1]*(V[2]V[1])"[kJ]" "W_b23 = integral of P[3]*dV[3] for Deltax = 20 cm and is given by:" W_b23=P[2]*(V[3]V[2])+Spring_const/2*(V[3]V[2])^2 "[kJ]" k [kN/m] 50 100 150 200 250 300 350 400 450 500 P3 [kPa] 250 350 450 550 650 750 850 950 1050 1150 Wout [kJ] 26.48 27.48 28.48 29.48 30.48 31.48 32.48 33.48 34.48 35.48 416 Chapter 4 Energy Transfer by Heat, Work, and Mass
10 5 Steam 10 4 P [kPa] 10 3 3
138.9C 10 2 1 2 111.4C 10 1
25C 10 0 10 4 10 3 10 2 10 1 10 0 10 1 10 2 v [m /kg] Final Pressure vs. spring constant
1200 3 1000 800 P[3]
600 400 200 50 100 150 200 250 300 350 400 450 500 k [kN/m] Total W ork done vs. spring constant
36 34 W out [kJ] 32 30 28 26 50 100 150 200 250 300 350 400 450 500 k [kN/m] 417 Chapter 4 Energy Transfer by Heat, Work, and Mass 431 Refrigerant134a in a cylinder equipped with a set of stops is heated and evaporated. The vapor expands until the piston hits the stops. The final temperature, and the boundary work done are to be determined, and the process is to be shown on a PV diagram. Assumptions The process is quasiequilibrium. Analysis (a) This a constant pressure process. Initially the system contains a saturated mixture, and thus the pressure is P2 = P = Psat@8 o C = 217.04 kPa 1 The specific volume of the refrigerant at the final state is v2 = V2 0.4 m 3 = = 0.04 m 3 / kg m 10 kg
P At 217.04 kPa (or 8 C), vf = 0.0007569 m 3/kg and vg = 0.0919 m3/kg. Noting that vf < v2 < vg , the final state is a saturated mixture and thus the final temperature is T2 = [email protected] kPa = 8 o C (b) The total initial volume is 1 2 V1 = m f v f + mg vg = 8 0.0007569 + 2 0.0919 = 0.19 m 3 Thus, Wb ,out = PdV = P(V
1 2 v
2  V1 ) 1 kJ = (217.04 kPa)(0.4  0.19)m 3 1 kPa m 3 = 45.6 kJ Discussion The positive sign indicates that work is done by the system (work output). 418 Chapter 4 Energy Transfer by Heat, Work, and Mass 432 Saturated refrigerant134a vapor in a cylinder is allowed to expand isothermally by gradually decreasing the pressure inside to 500 kPa. The boundary work done during this process is to be determined by using property data from the refrigerant tables, and by treating the refrigerant vapor as an ideal gas. Assumptions The process is quasiequilibrium. Analysis From the refrigerant tables, the specific volume of the refrigerant at various pressures at 50C are determined to be P, MPa 1.320 1.200 1.0 0.9 0.8 0.7 0.6 0.5 v, m3 / kg 0.01505 0.01712 0.02171 0.02472 0.02846 0.03324 0.03958 0.04842
P
(MPa) 1 1.32 T = 50C 0.5 2 V Plotting these on a PV diagram and finding the area under the process curve, the boundary work during this isothermal process is determined to be Wb = 262.9 kJ (b) Treating the refrigerant as an ideal gas, the boundary work for this isothermal process can be determined from Wb ,out = PdV = P V ln V
1 1 1 2 V2
1 = mRT ln P1 P2 1320 kPa 500 kPa = (10 kg)(0.08149 kJ/kg K)(323 K)ln = 255.5 kJ
which is sufficiently close to the experimental value. Discussion The positive sign indicates that work is done by the system (work output). 419 Chapter 4 Energy Transfer by Heat, Work, and Mass 433 Problem 432 is reconsidered. Using the integration feature, the work done is to be calculated and compared to the result obtained by the ideal gas assumption. Also, the process is to be plotted on a Pv diagram. "Let's plot the percent error as a function of P[2]" "Knowns" m=10"[kg]" T[1]=50"[C]" x[1]=1.0 "saturated vapor" P[2]=500"[kPa]" "Remove the {} when not using the solve table feature of EES." "Solution" "The process is isothermal:" T[2]=T[1]"[C]" Vol[1]=m*v[1]"[m^3]" v[1]=volume(R134a,T=T[1],x=x[1])"[m^3/kg]" Vol[2]=m*v[2]"[m^3]" v[2]=volume(R134a,P=P[2],T=T[2])"[m^3/kg]" "The boundary work is the integral of PdV over the isothermal process." "The ideal gas result is:" W_idealgas=P[1]*Vol[1]*ln(Vol[2]/Vol[1])"[kJ]" P[1]=pressure(R134a,T=T[1],x=x[1])"[kPa]" "The work using the experimental values is found by numerically integrating PdV over the process. P_int and V_int are integration functions" P_int=pressure(R134a,T=T[1],v=v_int)"[kPa]" m*v_int=Vol_int W_exp=integral(P_int,Vol_int,Vol[1],Vol[2],0.005)"[kJ]" PercentError=(W_expW_idealgas)/W_exp*100 420 Chapter 4 Energy Transfer by Heat, Work, and Mass PercentError [%] 0.2994 1.914 3.43 4.875 6.266 7.62 8.95 10.27 11.61 12.97
14 12 P2 [kPa] 1300 1200 1100 1000 900 800 700 600 500 400 Wexp [kJ] 3.857 26.42 50.55 76.65 105.2 136.9 172.6 213.6 262 321 Widealgas [kJ 3.845 25.92 48.82 72.91 98.62 126.5 157.2 191.7 231.6 279.3 PercentError [%] 10 8 6 4 2 0 400 500 600 700 800 900 1000 1100 1200 1300 P[2], kPa Percent error in the boundary w ork w hen treating R134a as an ideal gas during an isothermal expansion process. 10 5 R134a 10 4 P [kPa] 1 10 3 50C 2 10 2
0.05 0.1 0.2 0.5 10 1 10 4 10 3 10 2
3 10 1 v [m /kg] 421 Chapter 4 Energy Transfer by Heat, Work, and Mass 434 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this process is to be determined using the experimental data. Assumptions The process is quasiequilibrium. Analysis Plotting the given data on a PV diagram on a graph paper and evaluating the area under the process curve, the work done is determined to be 0.25 kJ. Other Forms of Work 435C The work done is the same, but the power is different. 436C The work done is the same, but the power is different. 437 A car is accelerated from rest to 100 km/h. The work needed to achieve this is to be determined. Analysis The work needed to accelerate a body the change in kinetic energy of the body, Wa = 100,000m 2 1kJ 1 1 2  0 m V2  V12 = (800kg) 3600s 1000kg m 2 /s 2 2 2 ( ) = 308.6kJ 438 A car is accelerated from 20 to 70 km/h on an uphill road. The work needed to achieve this is to be determined. Analysis The total work required is the sum of the changes in potential and kinetic energies, 70,000m 2 20,000m 2 1kJ 1 1 2 2  W a = m V2  V1 = (2000kg) 3600s 3600s 1000kg m 2 /s 2 2 2 ( ) = 347.2kJ and, 1kJ W g = mg (z 2  z1 ) = (2000kg)(9.807 m/s 2 )(40m) 1000kg m 2 /s 2 Thus, W total = W a + W g = 347.2 + 784.6 = 1131.8kJ = 784.6kJ 439E A engine of a car develops 450 hp at 3000 rpm. The torque transmitted through the shaft is to be determined. Analysis The torque is determined from 422 Chapter 4 Energy Transfer by Heat, Work, and Mass T= & W sh 550 lbf ft/s 450 hp = 787.8 lbf ft = & 2 (3000/60 )/s 2n 1 hp 440 A linear spring is elongated by 20 cm from its rest position. The work done is to be determined. Analysis The spring work can be determined from Wspring = 1 1 2 k x 2  x12 = (70 kN/m)(0.2 2  0)m 2 = 1.4 kN m = 1.4 kJ 2 2 ( ) 441 The engine of a car develops 75 kW of power. The acceleration time of this car from rest to 85 km/h on a level road is to be determined. Analysis The work needed to accelerate a body is the change in its kinetic energy, 85,000 m 2 1 kJ 1 1 2 2  0 Wa = m V2  V1 = (1500 kg) 3600 s 1000 kg m 2 /s 2 2 2 ( ) = 418.1 kJ Thus the time required is t = W a 418.1 kJ = = 5.57 s & 75 kJ/s Wa This answer is not realistic because part of the power will be used against the air drag, friction, and rolling resistance. 423 Chapter 4 Energy Transfer by Heat, Work, and Mass 442 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000/20 = 50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is Load = (50 chairs)(250 kg/chair) = 12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is 1 kJ = 24,525 kJ W g = mg ( z 2  z1 ) = (12,500 kg)(9.81 m/s 2 )(200 m) 2 2 1000 kg m /s At 10 km/h, it will take t = distance 1 km = = 0.1 h = 360 s velocity 10 km / h to do this work. Thus the power needed is & Wg = Wg t = 24,525 kJ = 68.1 kW 360 s The velocity of the lift during steady operation, and the acceleration during start up are 1 m/s V = (10 km/h) = 2.778 m/s 3.6 km/h V 2.778 m/s  0 = = 0.556 m/s 2 5s t During acceleration, the power needed is a= 1 kJ/kg 1 1 2 & W a = m V2  V12 / t = (12,500 kg) (2.778 m/s) 2  0 1000 m 2 /s 2 2 2 ( ) ( ) /(5 s) = 9.6 kW Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be h=
and 200 m 1 1 2 1 at sin = at 2 = (0.556 m/s 2 )(5 s) 2 (0.2) = 1.39 m 2 2 1000 m 2 /(5 s) = 34.1 kW 1 kJ/kg & W g = mg (z 2  z1 ) / t = (12,500 kg)(9.81 m/s 2 )(1.39 m) 1000 kg m 2 /s 2 Thus, & & & W total = W a + W g = 9.6 + 34.1 = 43.7 kW 424 Chapter 4 Energy Transfer by Heat, Work, and Mass 443 A car is to climb a hill in 10 s. The power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, & & & W total = W a + W g & (a) Wa = 0 since the velocity is constant. Also, the vertical rise is h = (100 m)(sin 30) = 50 m. Thus, 1 kJ & W g = mg (z 2  z1 ) / t = (2000 kg)(9.81 m/s 2 )(50 m) 1000 kg m 2 /s 2 and /(10 s) = 98.1 kW & & & W total = W a + W g = 0 + 98.1 = 98.1 kW
(b) The power needed to accelerate is 1kJ 1 1 2 & Wa = m V2  V12 / t = (2000kg) (30m/s )2  0 1000kg m 2 /s 2 2 2 and ( ) ( ) /(10s) = 90kW & & & W total = W a + W g = 90 + 98.1 = 188.1 kW
(c) The power needed to decelerate is 1 1 2 & Wa = m V2  V12 / t = (2000kg) (5m/s )2  (35m/s )2 2 2 and ( ) ( 1 ) 1000kgkJm /s 2 2 /(10s) = 120kW & & & W total = W a + W g = 120 + 98.1 = 21.9 kW (breaking power) 425 Chapter 4 Energy Transfer by Heat, Work, and Mass 444 A damaged car is being towed by a truck. The extra power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, & & & W total = W a + W g (a) Zero. & (b) Wa = 0 . Thus, z & & W total = W g = mg ( z 2  z1 ) / t = mg = mgV z = mgV sin 30 o t 50,000m 1kJ/kg (0.5) = 81.7kW = (1200kg)(9.8m/s 2 ) 3600s 1000m 2 /s 2 & = 0 . Thus, (c) W
g 1 2 & & W total = W a = m V2  V12 / t 2 90,000m 2 1kJ/kg 1  0 = (1200kg) 3600s 1000m 2 /s 2 2 ( ) /(12s) = 31.25kW 426 Chapter 4 Energy Transfer by Heat, Work, and Mass Conservation of Mass 445C Mass flow rate is the amount of mass flowing through a crosssection per unit time whereas the volume flow rate is the amount of volume flowing through a crosssection per unit time. 446C The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteadyflow process. 447C Flow through a control volume is steady when it involves no changes with time at any specified position. 448C No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (unless the density is constant). To be steady, the mass flow rate through the device must remain constant. 449E A garden hose is used to fill a water bucket. The volume and mass flow rates of water, the filling time, and the discharge velocity are to be determined. Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing. Properties We take the density of water to be 62.4 lbm/ft 3. Analysis (a) The volume and mass flow rates of water are & V = AV = (D 2 / 4)V = [ (1 / 12 ft) 2 / 4](8 ft/s) = 0.04363 ft 3 /s & & m = V = (62.4 lbm/ft 3 )(0.04363 ft 3 /s) = 2.72 lbm/s (b) The time it takes to fill a 20gallon bucket is t = 1 ft 3 20 gal V = 61.3 s = & 0.04363 ft 3 /s 7.4804 gal V (c) The average discharge velocity of water at the nozzle exit is Ve = & & V V 0.04363 ft 3 /s = = = 32 ft/s 2 Ae De / 4 [ (0.5 / 12 ft) 2 / 4] Discussion Note that for a given flow rate, the average velocity is inversely proportional to the square of the velocity. Therefore, when the diameter is reduced by half, the velocity quadruples. 427 Chapter 4 Energy Transfer by Heat, Work, and Mass 450 Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 2.21 kg/m3 at the inlet, and 0.762 kg/m3 at the exit. Analysis (a) The mass flow rate of air is determined from the inlet conditions to be & m = 1 A1V1 = (2.21 kg/m 3 )(0.008 m 2 )(30 m/s ) = 0.530 kg/s & & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit area of the nozzle is determined to be
V1 = 30 m/s A1 = 80 cm 2 AIR V2 = 180 m/s & A2 = m = 2 A2 V2 & 0.530 kg/s m = = 0.00387 m 2 = 38.7 cm 2 2 V2 (0.762 kg/ m 3 )(180 m/s) 451 Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent increase in the velocity of air as it flows through the drier is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 1.20 kg/m3 at the inlet, and 1.05 kg/m3 at the exit. & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . Then, & & m1 = m 2 1 AV1 = 2 AV2 V2 1.20 kg/m = 1 = = 1.14 V1 2 1.05 kg/m 3
3 V2 V1 (or, and increase of 14%) Therefore, the air velocity increases 14% as it flows through the hair drier. 428 Chapter 4 Energy Transfer by Heat, Work, and Mass 452E The ducts of an airconditioning system pass through an open area. The inlet velocity and the mass flow rate of air are to be determined. Assumptions Flow through the air conditioning duct is steady. Properties The density of air is given to be 0.078 lbm/ft 3 at the inlet. Analysis The inlet velocity of air and the mass flow rate through the duct are V1 = & & V1 V1 450 ft 3 /min = = = 825 ft/min = 13.8 ft/s A1 D 2 / 4 (10/12 ft )2 / 4
450 ft3/min AIR D = 10 in & & m = 1V1 = (0.078 lbm/ft 3 )( 450 ft 3 / min) = 35.1 lbm/min = 0.585lbm/s 453 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until the density rises to a specified level. The mass of air that entered the tank is to be determined. Properties The density of air is given to be 1.18 kg/m3 at the beginning, and 7.20 kg/m3 at the end. Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. The mass balance for this system can be expressed as Mass balance: Substituting, m in  m out = msystem m i = m 2  m1 = 2V  1 V mi = ( 2  1 )V = [(7.20  1.18) kg/m 3 ](1 m 3 ) = 6.02 kg
Therefore, 6.02 kg of mass entered the tank.
V1 = 1 m3 1 =1.18 kg/m3 429 Chapter 4 Energy Transfer by Heat, Work, and Mass 454 The ventilating fan of the bathroom of a building runs continuously. The mass of air "vented out" per day is to be determined. Assumptions Flow through the fan is steady. Properties The density of air in the building is given to be 1.20 kg/m3. Analysis The mass flow rate of air vented out is & & m air = Vair = (1.20 kg/m 3 )(0.030 m 3 /s) = 0.036 kg/s Then the mass of air vented out in 24 h becomes & m = m air t = (0.036 kg/s)(24 3600 s) = 3110 kg Discussion Note that more than 3 tons of air is vented out by a bathroom fan in one day. 455E Chickens are to be cooled by chilled water in an immersion chiller. The mass flow rate of chicken through the chiller is to be determined. Assumptions Chickens are dropped into the chiller steadily. Properties The average mass of a chicken is 4.5 lbm. Analysis Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of & m chicken = (500 chicken/h)(4.5 lbm/chicken) = 2250 kg/h = 0.625 lbm/s Note that chicken can be treated conveniently as a "flowing fluid" in calculations. 430 Chapter 4 Energy Transfer by Heat, Work, and Mass 456 A desktop computer is to be cooled by a fan at a high elevation where the air density is low. The mass flow rate of air through the fan and the diameter of the casing for a given velocity are to be determined. Assumptions Flow through the fan is steady. Properties The density of air at a high elevation is given to be 0.7 kg/m3. Analysis The mass flow rate of air is & & mair = Vair = (0.7 kg/m 3 )(0.34 m 3 /min ) = 0.238 kg/min = 0.0040 kg/s
If the mean velocity is 110 m/min, the diameter of the casing is D & V = AV = V 4
2 D= & 4V = V 4(0.34 m 3 /min) = 0.063 m (110 m/min) Therefore, the diameter of the casing must be at least 6.3 cm to ensure that the mean velocity does not exceed 110 m/min. Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations. 431 Chapter 4 Energy Transfer by Heat, Work, and Mass Flow Work and Energy Transfer by Mass 457C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass. 458C Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids at rest do not possess any flow energy. 459C Flowing fluids possess flow energy in addition to the forms of energy a fluid at rest possesses. The total energy of a fluid at rest consists of internal, kinetic, and potential energies. The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies. 460E Steam is leaving a pressure cooker at a specified pressure. The velocity, flow rate, the total and flow energies, and the rate of energy transfer by mass are to be determined. Assumptions 1 The flow is steady, and the initial startup period is disregarded. 2 The kinetic and potential energies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at 30 psia. Properties The properties of saturated liquid water and water vapor at 30 psia are vf = 0.017004 ft 3/lbm, vg = 13.748 ft 3/lbm, ug = 1088.0 Btu/lbm, and hg = 1164.3 Btu/lbm (Table A5E). Analysis (a) Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity are m= & m= Vliquid vf = 0.13368 ft 3 0.4 gal 1 gal 0.017004 ft 3 /lbm = 3.145 lbm m 3.145 lbm = = 0.0699 lbm/min = 1.165 10  3 lbm/s 45 min t & mv g (1.165 10 3 lbm/s)(13.748 ft 3 /lbm) 144 in 2 & m = = V= 1 ft 2 g Ac Ac 0.15 in 2 H2O Sat. vapor P = 30 psia Q = 15.4 ft/s (b) Noting that h = u + Pv and that the kinetic and potential energies are disregarded, the flow and total energies of the exiting steam are eflow = Pv = h  u = 1164.3  1088.0 = 76.3 Btu/lbm = h + ke + pe h = 1164.3 Btu/lbm
Note that the kinetic energy in this case is ke = V2/2 = (15.4 ft/s)2 = 237 ft2/s2 = 0.0095 Btu/lbm, which is very small compared to enthalpy. (c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass, & & E mass = m = (1.165 10 3 lbm/s)(1164.3 Btu/lbm) = 1.356 Btu/s
Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside (which is hfg) since it relates directly to the amount of energy supplied to the cooker. 432 Chapter 4 Energy Transfer by Heat, Work, and Mass 461 Refrigerant134a enters a compressor as a saturated vapor at a specified pressure, and leaves as superheated vapor at a specified rate. The rates of energy transfer by mass into and out of the compressor are to be determined. Assumptions 1 The flow of the refrigerant through the compressor is steady. 2 The kinetic and potential energies are negligible, and thus they are not considered. Properties The enthalpy of refrigerant134a at the inlet and the exit are (Tables A12 and A13) h1 = h g @0.14 MPa = 236.04 kJ/kg P2 = 0.8 MPa h2 = 284.39 kJ/kg T2 = 50C Analysis Noting that the total energy of a flowing fluid is equal to its enthalpy when the kinetic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass into and out of the compressor are & & & E mass, in = m in = mh1 = (0.04 kg/s)(236.04 kJ/kg) = 9.442 kJ/s = 9.44 kW & & & E mass, out = m out = mh2 = (0.04 kg/s)(284.39 kJ/kg) = 11.38 kJ/s = 11.38 kW
Discussion The numerical values of the energy entering or leaving a device by mass alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity here is the difference between the outgoing and incoming energy flow rates, which is & & & E mass = E mass, out  E mass, in = 11.38  9.44 = 1.94 kW
This quantity represents the rate of energy transfer to the refrigerant in the compressor.
2 0.8 MPa 50C R134a compressor 1
0.14 MPa 433 Chapter 4 Energy Transfer by Heat, Work, and Mass 462 Warm air in a house is forced to leave by the infiltrating cold outside air at a specified rate. The net energy loss due to mass transfer is to be determined. Assumptions 1 The flow of the air into and out of the house through the cracks is steady. 2 The kinetic and potential energies are negligible. 3 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kPam3/kgK (Table A1). The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kgC (Table A2). Analysis The density of air at the indoor conditions and its mass flow rate are = P 101.325 kPa = = 1.189 kg/m 3 RT (0.287 kPa m 3 /kg K)(24 + 273)K & & m = V = (1.189 kg/m 3 )(150 m 3 /h) = 178.35 kg/h = 0.0495 kg/s Noting that the total energy of a flowing fluid is equal to its enthalpy when the kinetic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass into and out of the house by air re & & & E mass, in = m in = mh1 & & & E mass, out = m out = mh2 The net energy loss by air infiltration is equal to the difference between the outgoing and incoming energy flow rates, which is & & & & & E mass = E mass, out  E mass, in = m(h2  h1 ) = mC p (T2  T1 ) = (0.0495 kg/s)(1.005 kJ/kg C)(24  5)C = 0.945 kJ/s = 0.945 kW This quantity represents the rate of energy transfer to the refrigerant in the compressor. Discussion The rate of energy loss by infiltration will be less in reality since some air will leave the house before it is fully heated to 24C. Cold air 5C Warm air
24C Warm air 24C 434 Chapter 4 Energy Transfer by Heat, Work, and Mass Review Problems 463 The weight of the cabin of an elevator is balanced by a counterweight. The power needed when the fully loaded cabin is rising , and when the empty cabin is descending at a constant speed are to be determined. Assumptions 1 The weight of the cables is negligible. 2 The guide rails and pulleys are frictionless. 3 Air drag is negligible. Analysis (a) When the cabin is fully loaded, half of the weight is balanced by the counterweight. The power required to raise the cabin at a constant speed of 2 m/s is 1N & mgz = mgV = (400kg ) 9.8m/s 2 (2m/s ) W= 1kg m/s 2 t ( ) 1kW = 7.84kW 1000 N m/s If no counterweight is used, the mass would double to 800 kg and the power would be 27.84 = 15.68 kW. (b) When the empty cabin is descending (and the counterweight is ascending) there is mass imbalance of 400150 = 250 kg. The power required to raise this mass at a constant speed of 2 m/s is 1N & mgz = mgV = (250 kg ) 9.81 m/s 2 (2 m/s ) W= 1 kg m/s 2 t ( ) 1 kW = 4.9 kW 1000 N m/s If a friction force of 1200 N develops between the cabin and the guide rails, we will need F z 1 kW & Wfriction = friction = Ffriction V = (1200 N )(2 m/s ) 1000 N m/s = 2.4 kW t of additional power to combat friction which always acts in the opposite direction to motion. Therefore, the total power needed in this case is & & & W total = W + W friction = 4.9 + 2.4 = 7.3kW
Counter weight Cabin 435 Chapter 4 Energy Transfer by Heat, Work, and Mass 464 A cylinder equipped with an external spring is initially filled with air at a specified state. Heat is transferred to the air, and both the temperature and pressure rise. The total boundary work done by the air, and the amount of work done against the spring are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The process is quasiequilibrium. 2 The spring is a linear spring. Analysis (a) The pressure of the gas changes linearly with volume during this process, and thus the process curve on a PV diagram will be a straight line. Then the boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, Wb ,out = Area = P1 + P2 (V2  V1 ) 2 Air 200 kPa 0.2 m3 1 kJ (200 + 800)kPa = (0.5  0.2)m 3 1 kPa m 3 2 = 150 kJ
(b) If there were no spring, we would have a constant pressure process at P = 200 kPa. The work done during this process is Q P
(kPa) 800 1 2 Wb,out, no spring = PdV = P(V
2 1 2  V1 ) 1 kJ = (200 kPa )(0.5  0.2)m 3 /kg 1 kPa m 3 = 60 kJ Thus, 200 Wspring = Wb  Wb, no spring = 150  60 = 90 kJ
0.2 0.5 V (m3) 436 Chapter 4 Energy Transfer by Heat, Work, and Mass 465 A cylinder equipped with a set of stops for the piston is initially filled with saturated liquidvapor mixture of water a specified pressure. Heat is transferred to the water until the volume increases by 20%. The initial and final temperature, the mass of the liquid when the piston starts moving, and the work done during the process are to be determined, and the process is to be shown on a Pv diagram. Assumptions The process is quasiequilibrium. Analysis (a) Initially the system is a saturated mixture at 100 kPa pressure, and thus the initial temperature is T1 = Tsat @100 kPa = 99.63 o C
The total initial volume is V1 = m f v f + m g v g = 2 0.001043 + 3 1.6940 = 5.084 m 3
Then the total and specific volumes at the final state are
H2O 5 kg V3 = 1.2V1 = 1.2 5.084 = 6.101 m 3 v3 =
Thus, V3 6.101 m 3 = = 1.220 m 3 /kg m 5 kg o T3 = 259.0 C 3 v3 = 1.220 m /kg (b) When the piston first starts moving, P2 = 200 kPa and V2 = V1 = 5.084 m3. The specific volume at this state is P3 = 200 kPa P
2 3 v2 = V2 5.084 m 3 = = 1.017 m 3 /kg m 5 kg 1 v which is greater than vg = 0.8857 m3/kg at 200 kPa. Thus no liquid is left in the cylinder when the piston starts moving. (c) No work is done during process 12 since V1 = V2. The pressure remains constant during process 23 and the work done during this process is Wb = 1 kJ P dV = P2 (V3  V2 ) = (200 kPa )(6.101  5.084)m 3 1 kPa m 3 2 3 = 203 kJ 437 Chapter 4 Energy Transfer by Heat, Work, and Mass 466E A spherical balloon is initially filled with air at a specified state. The pressure inside is proportional to the square of the diameter. Heat is transferred to the air until the volume doubles. The work done is to be determined. Assumptions 1 Air is an ideal gas. 2 The process is quasiequilibrium. Properties The gas constant of air is R = 0.06855 Btu/lbm.R (Table A1E). Analysis The dependence of pressure on volume can be expressed as 1 6V D = V = D 3 6 13 6V P = kD 2 = k P D2 or, 6 k 23 23 = P1V1  2 3 = P2V 2 2 3
23 AIR 10 lbm 30 psia 800 R Also, and Thus, P2 V 2 = = 2 2 3 = 1.587 P1 V1 PV P1V1 P2V2 = T2 = 2 2 T1 = 1.587 2 (800R ) = 2539R P1V1 T1 T2 Wb = 3k 6 3 6V 53 53 k dV = V2  V1 = ( P2V2  P1V1 ) 1 1 5 5 3 3 = mR(T2  T1 ) = (10 lbm)(0.06855 Btu/lbm R )(2539  800)R = 715 Btu 5 5 2 P dV = 2 23 23 ( ) 438 Chapter 4 Energy Transfer by Heat, Work, and Mass 467E Problem 466E is reconsidered. Using the integration feature, the work done is to be determined and compared to the 'hand calculated' result. N=2 m=10"[lbm]" P_1=30"[psia]" T_1=800"[R]" V_2=2*V_1 R=1545"[ftlbf/lbmolR]"/molarmass(air)"[ftlbf/lbmR]" P_1*Convert(psia,lbf/ft^2)*V_1=m*R*T_1 V_1=4*pi*(D_1/2)^3/3"[ft^3]" C=P_1/D_1^N (D_1/D_2)^3=V_1/V_2 P_2=C*D_2^N"[psia]" P_2*Convert(psia,lbf/ft^2)*V_2=m*R*T_2 P=C*D^N*Convert(psia,lbf/ft^2)"[ft^2]" V=4*pi*(D/2)^3/3"[ft^3]" W_boundary_EES=integral(P,V,V_1,V_2)*convert(ftlbf,Btu)"[Btu]" W_boundary_HAND=pi*C/(2*(N+3))*(D_2^(N+3)D_1^(N+3))*convert(ftlbf,Btu)*convert(ft^2,in^2)"[Btu]" N 0 0.3333 0.6667 1 1.333 1.667 2 2.333 2.667 3 850 Wboundary [Btu] 548.3 572.5 598.1 625 653.5 683.7 715.5 749.2 784.8 822.5 800 750 W boundary [Btu] 700 650 600 550 500 0 0.5 1 1.5 2 2.5 3 N 439 Chapter 4 Energy Transfer by Heat, Work, and Mass 468 A water tank open to the atmosphere is initially filled with water. The tank discharges to the atmosphere through a long pipe connected to a valve. The initial discharge velocity from the tank and the time required to empty the tank are to be determined. Assumptions 1 The flow is uniform and incompressible. 2 The draining pipe is horizontal. 3 The tank is considered to be empty when the water level drops to the center of the valve. Analysis (a) Substituting the known quantities, the discharge velocity can be expressed as V= 2 gz 2 gz = = 0.1212 gz 1.5 + fL / D 1.5 + 0.015(100 m)/(0.10 m) Then the initial discharge velocity becomes V1 = 0.1212 gz1 = 0.1212(9.81 m/s 2 )(2 m) = 1.54 m/s
where z is the water height relative to the center of the orifice at that time. (b) The flow rate of water from the tank can be obtained by multiplying the discharge velocity by the pipe crosssectional area, D & V = Apipe V2 = 4
2 z D0 D 0.1212 gz Then the amount of water that flows through the pipe during a differential time interval dt is D & dV = Vdt = 4
2 0.1212 gz dt (1) which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, dV = Atan k (dz ) = 
2 D 0 dz 4 (2) where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, D 2 4 0.1212 gz dt = 
2 D 0 D2 dz dt =  0 4 D2 dz 0.1212 gz = 2 D0 D 2 z 1 2 dz 0.1212 g The last relation can be integrated easily since the variables are separated. Letting tf be the discharge time and integrating it from t = 0 when z = z1 to t = tf when z = 0 (completely drained tank) gives tf t =0 dt =  D 2 0.1212 g 2 D0 0 z = z1 z 1 / 2 dz t f =  2 D0 1 0 z2
1 2 z1 D 2 0.1212 g = 2 2D0 1 D 2 0.1212 g z12 Simplifying and substituting the values given, the draining time is determined to be tf = 2 2 D0 D 2 z1 2(10 m) 2 = 0.1212 g (0.1 m) 2 2m = 25,940 s = 7.21 h 0.1212(9.81 m/s 2 ) Discussion The draining time can be shortened considerably by installing a pump in the pipe. 440 Chapter 4 Energy Transfer by Heat, Work, and Mass 469 Milk is transported from Texas to California in a cylindrical tank. The amount of milk in the tank is to be determined. Assumptions Milk is mostly water, and thus the properties of water can be used for milk. Properties The density of milk is the same as that of water, = 1000 kg/m3. Analysis Noting that the thickness of insulation is 0.05 m on all sides, the volume and mass of the milk in a full tank is determined to be Vmilk = (Di2 / 4) Li = [ (1.9 m) 2 / 4](6.9 m) = 19.56 m 3 m milk = Vmilk = (1000 kg/m 3 )(19.56 m 3 ) = 19,560 kg
The volume of the milk in gallons is 264.17 gal V milk = (19.56 m 3 ) = 5167 gal 1m3 Di L 441 Chapter 4 Energy Transfer by Heat, Work, and Mass 470 The rate of accumulation of water in a pool and the rate of discharge are given. The rate supply of water to the pool is to be determined. Assumptions 1 Water is supplied and discharged steadily. 2 The rate of evaporation of water is negligible. 3 No water is supplied or removed through other means. Analysis The conservation of mass principle applied to the pool requires that the rate of increase in the amount of water in the pool be equal to the difference between the rate of supply of water and the rate of discharge. That is, dm pool dt & & = mi  m e & mi = dm pool dt & + me & Vi = dV pool dt & + Ve since the density of water is constant and thus the conservation of mass is equivalent to conservation of volume. The rate of discharge of water is & Ve = Ae Ve = (D 2 /4) Ve = [ (0.05 m) 2 /4](5 m/s) = 0.00982 m 3 /s The rate of accumulation of water in the pool is equal to the crosssection of the pool times the rate at which the water level rises, dV pool dt = Across section Vlevel = (3 m 4 m)(0.015 m/min) = 0.18 m 3 /min = 0.00300 m 3 /s Substituting, the rate at which water is supplied to the pool is determined to be & Vi = dV pool dt & + Ve = 0.003 + 0.00982 = 0.01282 m 3 /s Therefore, water is supplied at a rate of 0.01282 m 3/s = 12.82 L/s. 471 A fluid is flowing in a circular pipe. A relation is to be obtained for the average fluid velocity in therms of V(r), R, and r. Analysis Choosing a circular ring of area dA = 2rdr as our differential area, the mass flow rate through a crosssectional area can be expressed as & m= V (r )dA = A R 0 V (r )2 r dr
R dr r Setting this equal to and solving for Vav, Vav = 2 R
2 V(r )r dr
R 0 442 Chapter 4 Energy Transfer by Heat, Work, and Mass 472 Air is accelerated in a nozzle. The density of air at the nozzle exit is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 4.18 kg/m3 at the inlet. & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . Then, & & m1 = m2 1 A1 V1 = 2 A2 V2 A V 120 m/s 2 = 1 1 1 = 2 (4.18 kg/m 3 ) = 2.64 kg/m 3 A2 V2 380 m/s
Discussion Note that the density of air decreases considerably despite a decrease in the crosssectional area of the nozzle. 1
AIR 2 443 Chapter 4 Energy Transfer by Heat, Work, and Mass 473 A long roll of large 1Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The mass flow rate of the plate is to be determined. Assumptions The plate moves through the bath steadily. Properties The density of steel plate is given to be = 7854 kg/m . Analysis The mass flow rate of the sheet metal through the oil bath is
3 Steel plate 10 m/min & & m = V = wtV = (7854 kg/m 3 )(1 m)(0.005 m)(10 m/min) = 393 kg/min = 6.55 kg/s Therefore, steel plate can be treated conveniently as a "flowing fluid" in calculations. 474 The air in a hospital room is to be replaced every 20 minutes. The minimum diameter of the duct is to be determined if the air velocity is not to exceed a certain value. Assumptions 1 The volume occupied by the furniture etc in the room is negligible. 2 The incoming conditioned air does not mix with the air in the room. Analysis The volume of the room is V = (6 m)(5 m)(4 m) = 120 m 3 To empty this air in 20 min, the volume flow rate must be
3 & V = 120 m = 0.10 m 3 /s V= t 20 60 s HospitalRoom
654 m 10 bulbs If the mean velocity is 5 m/s, the diameter of the duct is D & V = AV = V 4
2 D= & 4V = V 4(0.10 m 3 /s) = 0.16 m (5 m/s) Therefore, the diameter of the duct must be at least 0.16 m to ensure that the air in the room is exchanged completely within 20 min while the mean velocity does not exceed 5 m/s. Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations. 444 Chapter 4 Energy Transfer by Heat, Work, and Mass 475E A study quantifies the cost and benefits of enhancing IAQ by increasing the building ventilation. The net monetary benefit of installing an enhanced IAQ system to the employer per year is to be determined. Assumptions The analysis in the report is applicable to this work place. Analysis The report states that enhancing IAQ increases the productivity of a person by $90 per year, and decreases the cost of the respiratory illnesses by $39 a year while increasing the annual energy consumption by $6 and the equipment cost by about $4 a year. The net monetary benefit of installing an enhanced IAQ system to the employer per year is determined by adding the benefits and subtracting the costs to be Net benefit = Total benefits total cost = (90+39) (6+4) = $119/year (per person) The total benefit is determined by multiplying the benefit per person by the number of employees, Total net benefit = No. of employees Net benefit per person = 120$119/year = $14,280/year Discussion Note that the unseen savings in productivity and reduced illnesses can be very significant when they are properly quantified. 476 ... 478 Design and Essay Problems 445 Chapter 5 The First Law of Thermodynamics Chapter 5 THE FIRST LAW OF THERMODYNAMICS
Closed System Energy Balance: General Systems 51C No. This is the case for adiabatic systems only. 52C Warmer. Because energy is added to the room air in the form of electrical work. 53C Warmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied to this room to run the refrigerator, which is eventually dissipated to the room as waste heat. 54C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport. 55 Water is heated in a pan on top of a range while being stirred. The energy of the water at the end of the process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + W pw,in  Qout = U = U 2  U 1 30 kJ + 0.5 kJ  5 kJ = U 2  10 kJ U 2 = 35.5 kJ
Therefore, the final internal energy of the system is 35.5 kJ. 51 Chapter 5 The First Law of Thermodynamics 56E Water is heated in a cylinder on top of a range. The change in the energy of the water during this process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 123 4 4
Change in internal, kinetic, potential, etc. energies Qin  Wb,out  Qout = U = U 2  U 1 65 Btu  5 Btu  8 Btu = U U = U 2  U 1 = 52 Btu
Therefore, the energy content of the system increases by 52 Btu during this process. 57 A classroom is to be airconditioned using window airconditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from & & & & Qcooling = Qlights + Qpeople + Qheat gain where & Qlights = 10 100 W = 1 kW & Qpeople = 40 360 kJ / h = 4 kW & Qheat gain = 15,000 kJ / h = 4.17 kW Substituting, & Qcooling = 1 + 4 + 4.17 = 9.17 kW Thus the number of airconditioning units required is 9.17 kW = 1.83 2 units 5 kw / unit 15,000 kJ/h Room
40 people 10 bulbs Qcool 52 Chapter 5 The First Law of Thermodynamics 58 An industrial facility is to replace its 40W standard fluorescent lamps by their 35W high efficiency counterparts. The amount of energy and money that will be saved a year as well as the simple payback period are to be determined. Analysis The reduction in the total electric power consumed by the lighting as a result of switching to the high efficiency fluorescent is Wattage reduction = (Wattage reduction per lamp)(Number of lamps) = (40  34 W/lamp)(700 lamps) = 4200 W Then using the relations given earlier, the energy and cost savings associated with the replacement of the high efficiency fluorescent lamps are determined to be Energy Savings = (Total wattage reduction)(Ballast factor)(Operating hours) = (4.2 kW)(1.1)(2800 h/year) = 12,936 kWh/year Cost Savings = (Energy savings)(Unit electricity cost) = (12,936 kWh/year)($0.08/kWh) = $1035 The implementation cost of this measure is simply the extra cost of the energy efficient fluorescent bulbs relative to standard ones, and is determined to be Implementation Cost = (Cost difference of lamps)(Number of lamps) = [($2.26$1.77)/lamp](700 lamps) = $343 This gives a simple payback period of Simple payback period = Implementation cost $343 = = 0.33 year (4.0 months) Annual cost savings $1035 / year Discussion Note that if all the lamps were burned out today and are replaced by highefficiency lamps instead of the conventional ones, the savings from electricity cost would pay for the cost differential in about 4 months. The electricity saved will also help the environment by reducing the amount of CO2, CO, NOx, etc. associated with the generation of electricity in a power plant. 53 Chapter 5 The First Law of Thermodynamics 59 The lighting energy consumption of a storage room is to be reduced by installing motion sensors. The amount of energy and money that will be saved as well as the simple payback period are to be determined. Assumptions The electrical energy consumed by the ballasts is negligible. Analysis The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become Energy Savings = (Number of lamps)(Lamp wattage)(Reduction of annual operating hours) = (24 lamps)(60 W/lamp )(3285 hour s/year) = 4730 kWh/year Cost Savings = (Energy Savings)(Unit cost of energy) = (4730 kWh/year)($0.08/kWh) = $378/year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor, Implementation Cost = Material + Labor = $32 + $40 = $72 This gives a simple payback period of Implementation cost $72 = = 0.19 year (2.3 months) Annual cost savings $378 / year Simple payback period = Therefore, the motion sensor will pay for itself in about 2 months. 510 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined. Analysis The total electric power consumed by the lights in the classrooms and faculty offices is & E lighting, classroom = (Power consumed per lamp) (No. of lamps) = (200 12 110 W) = 264,000 = 264 kW & E lighting, offices = (Power consumed per lamp) (No. of lamps) = (400 6 110 W) = 264,000 = 264 kW & & & E lighting, total = E lighting, classroom + E lighting, offices = 264 + 264 = 528 kW Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is Unoccupied hours = (4 hours/day)(240 days/year) = 960 h/yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are & Energy savings = ( E lighting, classroom )( Unoccupied hours) = (528 kW)(960 h/yr) = 506,880 kWh Cost savings = (Energy savings)(Unit cost of energy) = (506,880 kWh)($0.082/kWh) = $41,564/yr
Discussion Note that simple conservation measures can result in significant energy and cost savings. 54 Chapter 5 The First Law of Thermodynamics 511 The radiator of a steam heating system is initially filled with superheated steam. The valves are closed, and steam is allowed to cool until the pressure drops to a specified value by transferring heat to the room. The amount of heat transfer is to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis We take the radiator as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout = U = m(u 2  u1 ) Qout = m(u1  u 2 ) (since W = KE = PE = 0) STEAM
V = const. Q Using data from the steam tables (Tables A4 through A6), some properties are determined to be P1 = 300kPa v1 = 0.7964m 3 /kg T1 = 250 o C u1 = 2728.7kJ/kg P2 = 100kPa v f = 0.001043, v g = 1.6940m 3 /kg u f = 417.36, u fg = 2088.7kJ/kg Noting that v1 = v2 and vf < v2 < vg , the mass and the final internal energy becomes m= V1 0.020 m 3 = = 0.0251 kg v1 0.7964 m 3 /kg v 2  v f 0.7964  0.001043 x2 = = = 0.470 v fg 1.6940  0.001043 P 1 u 2 = u f + x 2 u fg = 417.36 + (0.470 2088.7) = 1399.0 kJ/kg
Substituting, 2 v Qout = m(u1  u 2 ) = (0.0251 kg)(2728.7  1399.0) kJ/kg = 33.4 kJ 55 Chapter 5 The First Law of Thermodynamics 512 A rigid tank is initially filled with superheated R134a. Heat is transferred to the tank until the pressure inside rises to a specified value. The mass of the refrigerant and the amount of heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = U = m(u 2  u1 ) (since W = KE = PE = 0) R134a 200 kPa Using data from the refrigerant tables (Tables A11 through A13), the properties of R134a are determined to be P1 = 200kPa v f = 0.0007532, v g = 0.0993m 3 /kg x1 = 0.4 u g = 221.43kJ/kg u f = 36.69, v1 = v f + x1 v fg = 0.0007532 + [0.4 (0.0993  0.0007532 )] = 0.04017 m 3 /kg u1 = u f + x1u fg = 36.69 + [0.4 (221.43  36.69 )] = 110.59kJ/kg P2 = 800kPa u 2 = 349.82 kJ/kg (Superheated vapor) (v 2 = v1 ) Then the mass of the refrigerant is determined to be V 0.5m 3 = 12.45kg m= 1 = v1 0.04017 m 3 /kg (b) Then the heat transfer to the tank becomes P 2 1 Qin = m(u 2  u1 ) = (12.45 kg)(349.82  110.59) kJ/kg = 2978 kJ v 56 Chapter 5 The First Law of Thermodynamics 513E A rigid tank is initially filled with saturated R134a vapor. Heat is transferred from the refrigerant until the pressure inside drops to a specified value. The final temperature, the mass of the refrigerant that has condensed, and the amount of heat transfer are to be determined. Also, the process is to be shown on a Pv diagram. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout = U = m(u 2  u1 ) Qout = m(u1  u 2 ) (since W = KE = PE = 0) Using data from the refrigerant tables (Tables A11 through A13), the properties of R134a are determined to be P1 = 120psia v1 = v g @120 psia = 0.3941ft 3 /lbm sat. vapor u1 = u g @120 psia = 105.06Btu/lbm P2 = 30psia v f = 0.01209, v g = 1.5408ft 3 /lbm (v 2 = v1 ) u f = 16.24, u g = 95.40Btu/lbm The final state is saturated mixture. Thus, T2 = Tsat @ 30 psia = 15.38 F (b) The total mass and the amount of refrigerant that has condensed are m= x2 = V1 20 ft 3 = = 50.8 lbm v1 0.3941 ft 3 / lbm v2  v f v fg 0.3941  0.01209 = 0.250 = 1.5408  0.01209 R134a 120 psia Sat. vapor P
1 m f = (1  x 2 )m = (1  0.250)(50.8 lbm) = 38.1 lbm Also, u 2 = u f + x 2 u fg = 16.24 + [0.250 (95.40  16.24 )] = 36.03 Btu/lbm (c) Substituting, Qout = m(u1  u 2 ) = (50.8 lbm)(105.06  36.03) Btu/lbm = 3507 Btu
2 v 514 An insulated rigid tank is initially filled with a saturated liquidvapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a Pv diagram. 57 Chapter 5 The First Law of Thermodynamics Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is wellinsulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies H2O V = const. (since Q = KE = PE = 0) We We,in = U = m(u 2  u1 ) VIt = m(u 2  u1 ) The properties of water are (Tables A4 through A6) P1 = 100kPa v f = 0.001043, v g = 1.6940m 3 /kg x1 = 0.25 u f = 417.36, u fg = 2088.7 kJ/kg v1 = v f + x1v fg = 0.001043 + [0.25 (1.6940  0.001043)] = 0.42428m 3 /kg T
2 u1 = u f + x1u fg = 417.36 + (0.25 2088.7 ) = 939.5kJ/kg 1 v 2 = v1 = 0.42428m 3 /kg u 2 = u g @ 0.42428 m 3 /kg = 2556.7 kJ/kg sat.vapor Substituting, 1000VA (110 V)(8A )t = (5kg)(2556.7  939.5) kJ/kg 1kJ/s t = 9189s 153.2min v 58 Chapter 5 The First Law of Thermodynamics 515 Problem 514 is reconsidered. The effect of the initial mass of water on the length of time required to completely vaporize the liquid as the initial mass varies from 1 kg to 10 kg is to be investigated. The vaporization time is to be plotted against the initial mass. PROCEDURE P2X2(v[1]:P[2],x[2]) If v[1] > V_CRIT(Steam) then P[2]=pressure(Steam,v=v[1],x=1) x[2]=1 else P[2]=pressure(Steam,v=v[1],x=0) x[2]=0 EndIf End "Knowns" {m=5"[kg]"} P[1]=100"[kPa]" y=0.75 "moisture" Volts=110"[V]" I=8"[amps]" "Solution" "Conservation of Energy for the closed tank:" E_dot_inE_dot_out=DELTAE_dot E_dot_in=W_dot_ele"[kW]" W_dot_ele=Volts*I*CONVERT(J/s,kW)"[kW]" E_dot_out=0"[kW]" DELTAE_dot=m*(u[2]u[1])/DELTAt_s"[kW]" DELTAt_min=DELTAt_s*convert(s,min)"[min]" "The quality at state 1 is:" x[1]=1y u[1]=INTENERGY(Steam,P=P[1], x=x[1])"[kJ/kg]" v[1]=volume(Steam,P=P[1], x=x[1])"[m^3/kg]" T[1]=temperature(Steam,P=P[1], x=x[1])"[C]" "Check to see if state 2 is on the saturated liquid line or saturated vapor line:" Call P2X2(v[1]:P[2],x[2]) u[2]=INTENERGY(Steam,P=P[2], x=x[2])"[kJ/kg]" v[2]=volume(Steam,P=P[2], x=x[2])"[m^3/kg]" T[2]=temperature(Steam,P=P[2], x=x[2])"[C]" tmin [min] 30.63 61.26 91.89 122.5 153.2 183.8 214.4 245 275.7 306.3 m [kg] 1 2 3 4 5 6 7 8 9 10 59 Chapter 5 The First Law of Thermodynamics
Steam 700 600 500
T [C] 400 300 200
437.9 kPa 2
100 kPa 100 0 10 3 1
0.05 0.1 0.2 0.5 10 2 10 1 10 0
v [m /kg]
3 10 1 10 2 10 3 350 300 250 200 150 100 50 0 1 t min [m in] 2 3 4 5 6 7 8 9 10 m [kg] 510 Chapter 5 The First Law of Thermodynamics 516 One part of an insulated tank contains compressed liquid while the other side is evacuated. The partition is then removed, and water is allowed to expand into the entire tank. The final temperature and the volume of the tank are to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Evacuated 0 = U = m(u 2  u1 ) u1 = u 2 (since W = Q = KE = PE = 0) Partition The properties of water are (Tables A4 through A6) P1 = 600kPa v1 v f @60o C = 0.001017 m 3 /kg T1 = 60 o C u1 u f @ 60o C = 251.11kJ/kg H2O We now assume the final state in the tank is saturated liquidvapor mixture and determine quality. This assumption will be verified if we get a quality between 0 and 1. P2 = 10kPa v f = 0.00101, v g = 14.67 m 3 /kg (u 2 = u1 ) u f = 191.82, u fg = 2246.1 kJ/kg x2 =
Thus, T2 = =Tsat @ 10 kPa = 45.81 C v 2 = v f + x 2 v fg = 0.0010 + [0.0264 (14.67  0.00101)] = 0.388m 3 /kg and, V = mv2 =(2.5 kg)(0.388 m 3/kg) = 0.97 m3 u2  u f u fg = 251.11  191.82 = 0.0264 2246.1 511 Chapter 5 The First Law of Thermodynamics 517 Problem 516 is reconsidered. The effect of the initial pressure of water on the final temperature in the tank as the initial pressure varies from 100 kPa to 600 kPa is to be investigated. The final temperature is to be plotted against the initial pressure. "Knowns" m=2.5"[kg]" {P[1]=600"[kPa]"} T[1]=60"[C]" P[2]=10"[kPa]" "Solution" "Conservation of Energy for the closed tank:" E_inE_out=DELTAE E_in=0"[kJ]" E_out=0"[kJ]" DELTAE=m*(u[2]u[1])"[kJ]" u[1]=INTENERGY(Steam,P=P[1], T=T[1])"[kJ/kg]" v[1]=volume(Steam,P=P[1], T=T[1])"[m^3/kg]" T[2]=temperature(Steam,P=P[2], u=u[2])"[kJ/kg]" T_2=T[2]"[C]" v[2]=volume(Steam,P=P[2], u=u[2])"[m^3/kg]" V_total=m*v[2]"[m^3]" P1 [kPa] 100 200 300 400 500 600 T2 [C] 45.79 45.79 45.79 45.79 45.79 45.79 700 600 500 Steam T [C] 400 300 200
600 kPa 100 0 10 4 1
10 kPa 2
0.05 0.1 0.2 0.5 10 3 10 2 10 1
3 10 0 10 1 10 2 10 3 v [m /kg] 512 Chapter 5 The First Law of Thermodynamics 50 40 30 T 2 [C] 20 10 0 100 200 300 400 500 600 P[1] [kPa] 513 Chapter 5 The First Law of Thermodynamics 518 A cylinder is initially filled with R134a at a specified state. The refrigerant is cooled at constant pressure. The amount of heat loss is to be determined, and the process is to be shown on a Tv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout  Wb,out = U = m(u 2  u1 )  Qout = m(h2  h1 ) (since KE = PE = 0) R134a 800 kPa Q since U + Wb = H during a constant pressure quasiequilibrium process. The properties of R134a are (Tables A11 through A13) P1 = 800kPa h1 = 294.98kJ/kg o T1 = 60 C P1 = 800kPa h2 = h f @ 20o C = 77.26kJ/kg T1 = 20 o C Substituting, Qout =  (5 kg)(77.26  294.98) kJ/kg = 1089 kJ T
1 2 v 514 Chapter 5 The First Law of Thermodynamics 519E A cylinder contains water initially at a specified state. The water is heated at constant pressure. The final temperature of the water is to be determined, and the process is to be shown on a Tv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb ,out = U = m(u 2  u1 ) Qin = m(h2  h1 ) (since KE = PE = 0) H2O 120 psia Q since U + Wb = H during a constant pressure quasiequilibrium process. The properties of water are (Tables A11E through A13E) V 2ft 3 = 4ft 3 /lbm v1 = 1 = m 0.5lbm P1 = 120psia h1 = 1216.9Btu/lbm 3 v1 = 4ft /lbm Substituting, T
2 1 v 200 Btu = (0.5lbm)(h2  1216.9)Btu/lbm h2 = 1616.9 Btu/lbm
Then, o T2 = 1161.4 F h2 = 1616.9 Btu/lbm P2 = 120 psia 515 Chapter 5 The First Law of Thermodynamics 520 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddlewheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The cylinder is wellinsulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in + W pw,in  W b,out = U (since Q = KE = PE = 0)
H2O P = const. We,in + W pw,in = m(h2  h1 ) (VIt ) + W pw,in = m(h2  h1 ) since U + Wb = H during a constant pressure quasiequilibrium process. The properties of water are (Tables A4 through A6) We Wpw P1 = 150kPa h1 = h f @150 kPa = 467.11kJ/kg 3 sat.liquid v1 = v f @150 kPa = 0.0010528m /kg P2 = 150kPa h2 = h f + x 2 h fg = 467.11 + (0.5 2226.5) = 1580.36kJ/kg x 2 = 0.5 m=
Substituting, VIt + (300kJ) = (4.75kg)(1580.36  467.11)kJ/kg VIt = 4988kJ V= 1000VA 4988kJ = 230.9V (8A)(45 60s) 1kJ/s 1 2 V1 0.005m 3 = = 4.75kg v1 0.0010528m 3 /kg
P v 516 Chapter 5 The First Law of Thermodynamics 521 A cylinder is initially filled with steam at a specified state. The steam is cooled at constant pressure. The mass of the steam, the final temperature, and the amount of heat transfer are to be determined, and the process is to be shown on a Tv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout  Wb ,out = U = m(u 2  u1 )  Qout = m(h2  h1 ) (since KE = PE = 0) since U + Wb = H during a constant pressure quasiequilibrium process. The properties of water are (Tables A4 through A6) P1 = 1MPa v1 = 0.2825m 3 /kg o T2 = 350 C h1 = 3157.7 kJ/kg m= V1 1.5m 3 = = 5.31kg v1 0.2825m 3 /kg
T H2O 1 MPa 350C Q 1 2 (b) The final temperature is determined from P2 = 1MPa T2 = Tsat @1MPa = 179.91 o C sat.vapor h2 = [email protected] = 2778.1kJ/kg (c) Substituting, the energy balance gives Qout =  (5.31 kg)(2778.1  3157.7) kJ/kg = 2016 kJ v 517 Chapter 5 The First Law of Thermodynamics 522 [Also solved by EES on enclosed CD] A cylinder equipped with an external spring is initially filled with steam at a specified state. Heat is transferred to the steam, and both the temperature and pressure rise. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. 4 The spring is a linear spring. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting that the spring is not part of the system (it is external), the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb ,out = U = m(u 2  u1 ) Qin = m(u 2  u1 ) + Wb,out (since KE = PE = 0) H2O 200 kPa 200C Q The properties of steam are (Tables A4 through A6)
P P1 = 200kPa v1 = 1.0803m 3 /kg T1 = 200 o C u1 = 2654.4kJ/kg m= V1 0.5m = = 0.463kg v1 1.0803m 3 /kg V2 0.6m 3 = = 1.296m 3 /kg m 0.463kg
3 2 1 v2 = v o T2 = 1131 C v 2 = 1.296m 3 /kg u 2 = 4321.9kJ/kg (b) The pressure of the gas changes linearly with volume, and thus the process curve on a PV diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, P + P2 Wb = Area = 1 (V2  V1 ) = (200 + 500)kPa (0.6  0.5)m 3 1kJ 3 = 35kJ 1kPa m 2 2 (c) From the energy balance we have P2 = 500kPa Qin = (0.463 kg)(4321.9  2654.4)kJ/kg + 35 kJ = 807 kJ 518 Chapter 5 The First Law of Thermodynamics 523 Problem 522 is reconsidered. The effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer as the initial temperature varies from 150C to 250C is to be investigated. The final results are to be plotted against the initial temperature. "The process is given by:" "P[2]=P[1]+k*x*A/A, and as the spring moves 'x' amount, the volume changes by V[2]V[1]." P[2]=P[1]+(Spring_const)*(V[2]  V[1]) "P[2] is a linear function of V[2]" "where Spring_const = k/A, the actual spring constant divided by the piston face area" "Conservation of mass for the closed system is:" m[2]=m[1] "The conservation of energy for the closed system is" "E_in  E_out = DeltaE, neglect DeltaKE and DeltaPE for the system" Q_in  W_out = m[1]*(u[2]u[1]) DELTAU=m[1]*(u[2]u[1]) "Input Data" P[1]=200"kPa" V[1]=0.5"m^3" T[1]=200"C" P[2]=500"kPa" V[2]=0.6"m^3" m[1]=V[1]/spvol[1] spvol[1]=volume(Steam,T=T[1], P=P[1]) u[1]=intenergy(Steam, T=T[1], P=P[1]) spvol[2]=V[2]/m[2] "The final temperature is:" T[2]=temperature(Steam,P=P[2],v=spvol[2]) u[2]=intenergy(Steam, P=P[2], T=T[2]) Wnet_other = 0 W_out=Wnet_other + W_b "W_b = integral of P[2]*dV[2] for 0.5<V[2]<0.6 and is given by:" W_b=P[1]*(V[2]V[1])+Spring_const/2*(V[2]V[1])^2 "Compare this with the result given in Example 313" Qin [kJ] 778.2 793.2 808 822.7 837.1 T1 [C] 150 175 200 225 250 T2 [C] 975 1054 1131 1209 1285 Wout [kJ] 35 35 35 35 35 519 Chapter 5 The First Law of Thermodynamics
10 5 Steam 10 4 1132 C
10 3 200 C P [kPa] 2 1 10 2 10 1 Area = W b
10 2 10 1 10 0 10 1 10 0 10 3 v [m /kg] 3 840 830 820 Q in [kJ] 810 800 790 780 770 150 170 190 210 230 250 T[1] [C] 520 Chapter 5 The First Law of Thermodynamics
1300 1250 1200 T[2] [C] 1150 1100 1050 1000 950 150 170 190 210 230 250 T[1] [C] 50 40 W out [kJ] 30 20 10 0 150 170 190 210 230 250 T[1] [C] 521 Chapter 5 The First Law of Thermodynamics 524 A cylinder equipped with a set of stops for the piston to rest on is initially filled with saturated water vapor at a specified pressure. Heat is transferred to water until the volume doubles. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies 300 kPa Qin  Wb ,out = U = m(u 3  u1 ) Qin = m(u 3  u1 ) + Wb,out (since KE = PE = 0) H2O
200 kPa Sat. Vapor The properties of steam are (Tables A4 through A6) P1 = 200kPa v1 = v g @ 200kPa = 0.8857m 3 /kg sat.vapor u1 = u g @ 200kPa = 2529.5kJ/kg V 0.5m = 0.5645kg m= 1 = v1 0.8857 m 3 /kg v3 = V3 1m = = 1.7715m 3 /kg m 0.5645kg
v
3 3 P 2 3 1 T3 = 878.9 o C 3 v3 = 1.7715m /kg u 3 = 3813.8kJ/kg (b) The work done during process 12 is zero (since V = const) and the work done during the constant pressure process 23 is P3 = 300kPa Wb ,out = 1 kJ P dV = P (V3  V2 ) = (300 kPa)(1.0  0.5)m 3 1 kPa m 3 2 3 = 150 kJ (c) Heat transfer is determined from the energy balance, Qin = m(u 3  u1 ) + Wb ,out = (0.5645 kg)(3813.8  2529.5) kJ/kg + 150 kJ = 875.0 kJ 522 Chapter 5 The First Law of Thermodynamics Closed System Energy Analysis: Ideal Gases 525C No, it isn't. This is because the first law relation Q  W = U reduces to W = 0 in this case since the system is adiabatic (Q = 0) and U = 0 for the isothermal processes of ideal gases. Therefore, this adiabatic system cannot receive any net work at constant temperature. 526E The air in a rigid tank is heated until its pressure doubles. The volume of the tank and the amount of heat transfer are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, pe ke 0 . 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. Properties The gas constant of air is R = 0.3704 psia.ft 3/lbm.R (Table A1E). Analysis (a) The volume of the tank can be determined from the ideal gas relation,
Air 20 lbm 50 psia 80F V= Q (b) We take the air in the tank as our system. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass mRT1 (20 lbm)(0.3704 psia ft 3 /lbm R)(540 R) = = 80.0 ft 3 P1 50 psia = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = U Qin = m(u 2  u1 ) mC v (T2  T1 ) The final temperature of air is PV P2V 1 = T1 T2 T2 = P2 T1 = 2 (540 R) = 1080 R P 1 The internal energies are (Table A17E) u1 = u@ 540 R = 92.04 Btu / lbm u2 = u@ 1080 R = 186.93 Btu / lbm Substituting, Qin = (20 lbm)(186.93  92.04)Btu/lbm = 1898 Btu Alternative solutions The specific heat of air at the average temperature of Tave = (540+1080)/2= 810 R = 350F is, from Table A2Eb, Cv,ave = 0.175 Btu/lbm.R. Substituting, Qin = (20 lbm)( 0.175 Btu/lbm.R)(1080  540) R = 1890 Btu Discussion Both approaches resulted in almost the same solution in this case. 523 Chapter 5 The First Law of Thermodynamics 527 The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K. The final pressure in the tank and the amount of heat transfer are to be determined. Assumptions 1 Hydrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 240C and 1.30 MPa. 2 The tank is stationary, and thus the kinetic and potential energy changes are negligible, ke pe 0 . Properties The gas constant of hydrogen is R = 4.124 kPa.m3/kg.K (Table A1). The constant volume specific heat of hydrogen at the average temperature of 400 K is , Cv,ave = 10.352 kJ/kg.K (Table A2). Analysis (a) The final pressure of hydrogen can be determined from the ideal gas relation, PV PV 1 = 2 T1 T2 P2 = T2 300 K P = (250 kPa) = 150 kPa 1 T1 500 K (b) We take the hydrogen in the tank as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout = U Qout =  U =  m(u 2  u1 ) mC v (T1  T2 ) where H2 250 kPa 500 K Q m= P1V (250 kPa)(3.0 m 3 ) = = 0.3637 kg RT1 (4.124 kPa m 3 /kg K)(500 K) Substituting into the energy balance, Qout = (0.3637 kg)(10.352 kJ/kgK)(500  300)K = 753.0 kJ 524 Chapter 5 The First Law of Thermodynamics 528 A resistance heater is to raise the air temperature in the room from 7 to 23C within 20 min. The required power rating of the resistance heater is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. 4 Heat losses from the room are negligible. 5 The room is airtight so that no air leaks in and out during the process. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Also, Cv = 0.718 kJ/kg.K for air at room temperature (Table A2). Analysis We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constantvolume closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in = U mC v, ave (T2  T1 ) (since Q = KE = PE = 0) or, & We,in t = mCv , ave (T2  T1 ) The mass of air is V = 4 5 6 = 120 m3 m= PV (100 kPa)(120 m3 ) 1 = = 149.3 kg RT1 (0.287 kPa m3 / kg K)(280 K) We 456 m3 7C AIR Substituting, the power rating of the heater becomes (149.3 kg)(0.718 kJ/kg o C)(23  7) o C & We,in = = 1.91 kW 15 60 s
Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more proper to be conservative and to use H instead of using U in heating and airconditioning applications. 525 Chapter 5 The First Law of Thermodynamics 529 A room is heated by a radiator, and the warm air is distributed by a fan. Heat is lost from the room. The time it takes for the air temperature to rise to 20C is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. 4 The local atmospheric pressure is 100 kPa. 5 The room is airtight so that no air leaks in and out during the process. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Also, Cv = 0.718 kJ/kg.K for air at room temperature (Table A2). Analysis We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constantvolume closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + W fan,in  Qout = U mC v , ave (T2  T1 ) (since KE = PE = 0) or, & & & (Qin + Wfan, in  Qout )t = mC v, ave (T2  T1 ) The mass of air is V = 4 5 7 = 140 m3 PV (100 kPa)(140 m3 ) m= 1 = = 172.4 kg RT1 (0.287 kPa m3 / kg K)(283 K) Using the Cv value at room temperature,
Steam Wpw 5,000 kJ/h ROOM
4m 5m 7m 10,000 kJ/h [(10,000  5,000)/3600 kJ/s + 0.1 kJ/s]t = (172.4kg)(0.718 kJ/kgo C)(20  10) o C
It yields t = 831 s Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more proper to be conservative and to using H instead of use U in heating and airconditioning applications. 526 Chapter 5 The First Law of Thermodynamics 530 A student living in a room turns her 150W fan on in the morning. The temperature in the room when she comes back 10 h later is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and airconditioning applications. 4 All the doors and windows are tightly closed, and heat transfer through the walls and the windows is disregarded. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Also, Cv = 0.718 kJ/kg.K for air at room temperature (Table A2). Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be expressed as Ein  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in = U We,in = m(u2  u1 ) mCv (T2  T1 ) The mass of air is V = 4 6 6 = 144 m3 PV (100 kPa)(144 m3 ) 1 = = 174.2 kg RT1 (0.287 kPa m3 / kg K)(288 K) The electrical work done by the fan is m= & We = We t = (0.15 kJ / s)(10 3600 s) = 5400 kJ Substituting and using the Cv value at room temperature, 5400 kJ = (174.2 kg)(0.718 kJ/kgC)(T2  15)C T2 = 58.2C ROOM
4m 6m 6m Fan 527 Chapter 5 The First Law of Thermodynamics 531E A paddle wheel in an oxygen tank is rotated until the pressure inside rises to 20 psia while some heat is lost to the surroundings. The paddle wheel work done is to be determined. Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 181F and 736 psia. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 The energy stored in the paddle wheel is negligible. 4 This is a rigid tank and thus its volume remains constant. Properties The gas constant and molar mass of oxygen are R = 0.3353 psia.ft 3/lbm.R and M = 32 lbm/lbmol (Table A1E). The specific heat of oxygen at the average temperature of Tave = (735+540)/2= 638 R is C v,ave = 0.160 Btu/lbm.R (Table A2E). Analysis We take the oxygen in the tank as our system. This is a closed system since no mass enters or leaves. The energy balance for this system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 123 4 4
Change in internal, kinetic, potential, etc. energies W pw,in  Qout = U W pw,in = Qout + m(u 2  u1 ) Qout + mC v (T2  T1 )
The final temperature and the mass of oxygen are PV PV 1 = 2 T1 T2 m= Substituting, Wpw,in = (20 Btu) + (0.812 lbm)(0.160 Btu/lbm.R)(735  540) R = 45.3 Btu T2 = P2 20 psia T1 = (540 R) = 735 R P 14.7 psia 1 O2 14.7 psia 80F 20 Btu PV (14.7 psia)(10 ft 3 ) 1 = = 0.812 lbm RT1 (0.3353 psia ft 3 / lbmol R)(540 R) Discussion Note that a fan actually causes the internal temperature of a confined space to rise. In fact, a 100W fan supplies a room with as much energy as a 100W resistance heater. 528 Chapter 5 The First Law of Thermodynamics 532 One part of an insulated rigid tank contains an ideal gas while the other side is evacuated. The final temperature and pressure in the tank are to be determined when the partition is removed. Assumptions 1 The kinetic and potential energy changes are negligible, ke pe 0 . 2 The tank is insulated and thus heat transfer is negligible. Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 0 = U = m( u2  u1 ) u2 = u1 Therefore, T2 = T1 = 50C Since u = u(T) for an ideal gas. Then, P1V1 P2V 2 V 1 = P2 = 1 P1 = (800 kPa ) = 400kPa T1 T2 V2 2 IDEAL GAS 800 kPa 50C Evacuated 529 Chapter 5 The First Law of Thermodynamics 533 A cylinder equipped with a set of stops for the piston to rest on is initially filled with helium gas at a specified state. The amount of heat that must be transferred to raise the piston is to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 There are no work interactions involved. 4 The thermal energy stored in the cylinder itself is negligible. Properties The specific heat of helium at room temperature is Cv = 3.1156 kJ/kg.K (Table A2). Analysis We take the helium gas in the cylinder as the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this constant volume closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 500 kPa Qin = U = m(u2  u1 ) Qin = m(u2  u1 ) = mCv (T2  T1 ) The final temperature of helium can be determined from the ideal gas relation to be P1V P2V P 500 kPa (298 K) = 1490 K = T2 = 2 T1 = 100 kPa T1 T2 P1 Substituting into the energy balance relation gives Qin = (0.5 kg)(3.1156 kJ/kgK)(1490  298)K = 1857 kJ
He 100 kPa 25C Q 530 Chapter 5 The First Law of Thermodynamics 534 An insulated cylinder is initially filled with air at a specified state. A paddlewheel in the cylinder stirs the air at constant pressure. The final temperature of air is to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 3 There are no work interactions involved other than the boundary work. 4 The cylinder is wellinsulated and thus heat transfer is negligible. 5 The thermal energy stored in the cylinder itself and the paddlewheel is negligible. 6 The compression or expansion process is quasiequilibrium. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Also, Cp = 1.005 kJ/kg.K for air at room temperature (Table A2). The enthalpy of air at the initial temperature is h1 = [email protected] K = 298.18 kJ/kg Analysis We take the air in the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Wpw,in  Wb,out = U Wpw,in = m(h2  h1 ) since U + Wb = H during a constant pressure quasiequilibrium process. The mass of air is m= PV (400 kPa)(0.1 m 3 ) 1 = = 0.468 kg RT1 (0.287 kPa m 3 / kg K)(298 K)
Wpw AIR P = const. Substituting into the energy balance, 15 kJ = (0.468 kg)(h2  298.18 kJ/kg) h2 = 330.23 kJ/kg From Table A17, T2 = 329.9 K Alternative solution Using specific heats at room temperature, Cp = 1.005 kJ/kg.C, the final temperature is determined to be Wpw,in = m(h2  h1 ) mC p (T2  T1 ) 15 kJ = (0.468 kg)(1.005 kJ/kg.C)(T2  25)C which gives T2 = 56.9C 531 Chapter 5 The First Law of Thermodynamics 535E A cylinder is initially filled with nitrogen gas at a specified state. The gas is cooled by transferring heat from it. The amount of heat transfer is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. 5 Nitrogen is an ideal gas with constant specific heats. Properties The specific heat of nitrogen at the average temperature of Tave = (700+ 140)/2 = 420F is Cpave = 0.252 Btu/lbm.F (Table A2Eb). Analysis We take the nitrogen gas in the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout  Wb,out = U = m(u 2  u1 )  Qout = m(h2  h1 ) = mC p (T2  T1 ) since U + Wb = H during a constant pressure quasiequilibrium process. The mass of nitrogen is (50 psia )( 25 ft 3 ) PV = 2.814 lbm m= 1 = RT1 (0.3830 psia ft 3 / lbm R )(1160 R ) Substituting, Qout = (2.814 lbm)(0.252 Btu/lbm.F)(700  140)F = 397 Btu N2 50 psia 700F Q 532 Chapter 5 The First Law of Thermodynamics 536 A cylinder is initially filled with air at a specified state. Air is heated electrically at constant pressure, and some heat is lost in the process. The amount of electrical energy supplied is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 Air is an ideal gas with variable specific heats. 3 The thermal energy stored in the cylinder itself and the resistance wires is negligible. 4 The compression or expansion process is quasiequilibrium. Properties The initial and final enthalpies of air are (Table A17) h1 = h@ 298 K = 298.18 kJ / kg h2 = h@ 350 K = 350.49 kJ / kg Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in  Qout  Wb,out = U We,in = m(h2  h1 ) + Qout since U + Wb = H during a constant pressure quasiequilibrium process. Substituting, We,in = (15 kg)(350.49  298.18)kJ/kg + (60 kJ) = 845 kJ or, 1 kWh We,in = (845kJ) 3600 kJ = 0.235 kWh We AIR P = const. Q Alternative solution The specific heat of air at the average temperature of Tave = (25+ 77)/2 51C = 324 K is, from Table A2b, Cp,ave = 1.0065 kJ/kg.C. Substituting, We,in = mC p (T2  T1 ) + Qout = (15 kg)(1.0065 kJ/kg.C)( 77  25)C + 60 kJ = 845 kJ or, 1 kWh We,in = (845 kJ ) 3600 kJ = 0.235 kWh Discussion Note that for small temperature differences, both approaches give the same result. 533 Chapter 5 The First Law of Thermodynamics 537 An insulated cylinder initially contains CO2 at a specified state. The CO2 is heated electrically for 10 min at constant pressure until the volume doubles. The electric current is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The CO2 is an ideal gas with constant specific heats. 3 The thermal energy stored in the cylinder itself and the resistance wires is negligible. 4 The compression or expansion process is quasiequilibrium. Properties The gas constant and molar mass of CO2 are R = 0.1889 kPa.m3/kg.K and M = 44 kg/kmol (Table A1). The specific heat of CO2 at the average temperature of Tave = (300 + 600)/2 = 450 K is Cp,ave = 0.978 kJ/kg.C (Table A2b). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies W e,in  Wb,out = U W e,in = m(h2  h1 ) mC p (T2  T1 ) since U + Wb = H during a constant pressure quasiequilibrium process. The final temperature of CO2 is P1V1 P2V 2 P V = T2 = 2 2 T1 = 1 2 (300K) = 600K T1 T2 P1 V1 The mass of CO2 is m= Substituting, We,in = (1.059 kg)(0.978 kJ/kg.K)(600  300)K = 311 kJ Then, I= We,in Vt = 1000VA 311kJ = 4.71A (110V)(10 60s) 1kJ/s P V1 (200 kPa)(0.3 m 3 ) 1 = = 1.059 kg RT1 (0.1889 kPa m 3 / kg K)(300 K) We CO2 200 kPa 27C 534 Chapter 5 The First Law of Thermodynamics 538 A cylinder initially contains nitrogen gas at a specified state. The gas is compressed polytropically until the volume is reduced by onehalf. The work done and the heat transfer are to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The N2 is an ideal gas with constant specific heats. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Properties The gas constant of N2 are R = 0.2968 kPa.m3/kg.K (Table A1). The Cv value of N2 at the average temperature (369+300)/2 = 335 K is 0.744 kJ/kg.K (Table A2b). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Wb,in  Qout = U = m(u2  u1 ) Wb,in  Qout = mCv (T2  T1 ) The final pressure and temperature of nitrogen are
1 P2V2 .3 N2 100 kPa 27C PV1.3 = C Q = P1V11.3 P1V1 P2V2 = T1 T2 V P2 = 1 P1 = 21.3 (100kPa) = 246.2kPa V 2 P V 246.2kPa T2 = 2 2 T1 = 0.5 (300K) = 369.3K 100kPa P1 V1 1.3 Then the boundary work for this polytropic process can be determined from Wb ,in =  P dV = 
1 P2V2  P1V1 mR(T2  T1 ) = 1 n 1 n (0.8 kg)(0.2968 kJ/kg K)(369.3  300)K = = 54.8 kJ 1  1.3 2 Substituting into the energy balance gives Qout = Wb,in  mCv (T2  T1 ) = 54.8 kJ  (0.8 kg)(0.744 kJ / kg.K)(369.3  360)K = 13.6 kJ 535 Chapter 5 The First Law of Thermodynamics 539 Problem 538 is reconsidered. The process is to be plotted on a PV diagram, and the effect of the polytropic exponent n on the boundary work and heat transfer as the polytropic exponent varies from 1.1 to 1.6 is to be investigated. The boundary work and the heat transfer are to be plotted versus the polytropic exponent. Procedure Work(P[2],V[2],P[1],V[1],n:W12) If n=1 then W12=P[1]*V[1]*ln(V[2]/V[1])"[kJ]" Else W12=(P[2]*V[2]P[1]*V[1])/(1n)"[kJ]" endif End "Input Data" Vratio=0.5 "V[2]/V[1] = Vratio" n=1.3 "Polytropic exponent" P[1] = 100"[kPa]" T[1] = (27+273)"[K]" m=0.8"[kg]" MM=molarmass(nitrogen)"[kg/kmol]" R_u=8.314"[kJ/kmolK]" R=R_u/MM"[kJ/kgK]" V[1]=m*R*T[1]/P[1]"[m^3]" "Process equations" V[2]=Vratio*V[1] P[2]*V[2]/T[2]=P[1]*V[1]/T[1]"The combined ideal gas law for states 1 and 2 plus the polytropic process relation give P[2] and T[2]" P[2]*V[2]^n=P[1]*V[1]^n "Conservation of Energy for the closed system:" "E_in  E_out = DeltaE, we neglect Delta KE and Delta PE for the system, the nitrogen." Q12  W12 = m*(u[2]u[1]) u[1]=intenergy(N2, T=T[1])"[kJ/kg]" "internal energy for nitrogen as an ideal gas, kJ/kg" u[2]=intenergy(N2, T=T[2])"[kJ/kg]" Call Work(P[2],V[2],P[1],V[1],n:W12) "The following is required for the Pv plots" {P_plot*spv_plot/T_plot=P[1]*V[1]/m/T[1]"The combined ideal gas law for states 1 and 2 plus the polytropic process relation give P[2] and T[2]" P_plot*spv_plot^n=P[1]*(V[1]/m)^n} {spV_plot=R*T_plot/P_plot"[m^3]"} n 1 1.111 1.222 1.333 1.444 1.556 1.667 1.778 1.889 2 Q12 [kJ] 49.37 37 23.59 9.067 6.685 23.81 42.48 62.89 85.27 109.9 W12 [kJ] 49.37 51.32 53.38 55.54 57.82 60.23 62.76 65.43 68.25 71.23 536 Chapter 5 The First Law of Thermodynamics
Pressure vs. specific volume as function of polytropic exponent
1800 1600 1400 1200 1000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 0.2 0.4 0.6 0.8 1 n=1.0 n=1.3 n=2 P plot [kPa] 800 600 400 200 0 0 spv plot [m^3] 125 90 Q12 [kJ] 55 20 15 50 1 1.2 1.4 1.6 1.8 P plot 2 n
45 50 55 60 65 70 75 1 W 12 [kJ] 1.2 1.4 1.6 1.8 2 n
537 Chapter 5 The First Law of Thermodynamics 540 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the heater operates continuously when the heat losses from the room amount to 8,000 kJ/h. The power rating of the heater is to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 The temperature of the room is said to remain constant during this process. Analysis We take the room as the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this system reduces to Ein  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies ROOM
Tair=const We ,in  Qout = U = 0 We,in = Qout since U = mCvT = 0 for isothermal processes of ideal gases. Thus, Q We 1 kW & & We ,in = Qout = (6500 kJ/h) 3600 kJ/h = 1.81 kW 541E A cylinder initially contains air at a specified state. Heat is transferred to the air, and air expands isothermally. The boundary work done is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The air is an ideal gas with constant specific heats. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb,out = U = m(u2  u1 ) = mCv (T2  T1 ) = 0 since u = u(T) for ideal gases, and thus u2 = u1 when T1 = T2 . Therefore, Wb,out = Qin = 40 Btu
AIR T = const. 40 Btu 538 Chapter 5 The First Law of Thermodynamics 542 A cylinder initially contains argon gas at a specified state. The gas is stirred while being heated and expanding isothermally. The amount of heat transfer is to be determined. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The air is an ideal gas with constant specific heats. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 15 kJ Qin + Wpw,in  Wb,out = U = m(u2  u1 ) = mCv (T2  T1 ) = 0 since u = u(T) for ideal gases, and thus u2 = u1 when T1 = T2 . Therefore, Qin = Wb,out  Wpw,in = 15  3 = 12 kJ
3 kJ Ar T = const. Q 539 Chapter 5 The First Law of Thermodynamics 543 A cylinder equipped with a set of stops for the piston is initially filled with air at a specified state. Heat is transferred to the air until the volume doubled. The work done by the air and the amount of heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 Air is an ideal gas with variable specific heats. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Analysis We take the air in the cylinder as the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 400 kPa Qin  Wb,out = U = m(u 3  u1 ) Qin = m(u 3  u1 ) + Wb,out The initial and the final volumes and the final temperature of air are V1 = mRT1 (3 kg)(0.287 kPa m 3 / kg K)(300 K) = = 1.29 m 3 200 kPa P 1 AIR 200 kPa Q V3 = 2V1 = 2 1.29 = 2.58 m 3 P V1 P3 V3 1 = T1 T3 T3 = P
2 3 P3 V3 400 kPa 2 (300 K) = 1200 K T1 = 200 kPa P V1 1 No work is done during process 12 since V1 = V2. The pressure remains constant during process 23 and the work done during this process is 1 v Wb ,out = 2 1 P dV = P2 (V3  V 2 ) = (400 kPa)(2.58  1.29)m = 516 kJ
3 The initial and final internal energies of air are (Table A17) u1 = u@ 300 K = 214.07 kJ / kg u2 = u@ 1200 K = 933.33 kJ / kg Then from the energy balance, Qin = (3 kg)(933.33  214.07)kJ/kg + 516 kJ = 2674 kJ Alternative solution The specific heat of air at the average temperature of Tave = (300 + 1200)/2 = 750 K is, from Table A2b, Cvave = 0.800 kJ/kg.K. Substituting, Qin = m(u 3  u1 ) + Wb,out mCv (T3  T1 ) + Wb,out Qin = (3 kg)(0.800 kJ/kg.K)(1200  300) K + 516 kJ = 2676 kJ 540 Chapter 5 The First Law of Thermodynamics 544 [Also solved by EES on enclosed CD] A cylinder equipped with a set of stops on the top is initially filled with air at a specified state. Heat is transferred to the air until the piston hits the stops, and then the pressure doubles. The work done by the air and the amount of heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 Air is an ideal gas with variable specific heats. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 There are no work interactions involved. 3 The thermal energy stored in the cylinder itself is negligible. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Analysis We take the air in the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb,out = U = m(u 3  u1 ) Qin = m(u 3  u1 ) + Wb,out The initial and the final volumes and the final temperature of air are determined from V1 = mRT1 (3 kg)(0.287 kPa m 3 / kg K)(300 K) = = 1.29 m 3 200 kPa P 1 P
3 Q
AIR 200 kPa V3 = 2V1 = 2 1.29 = 2.58 m 3 P V1 P3 V3 1 = T1 T3 P V 400 kPa 2 (300 K) = 1200 K T3 = 3 3 T1 = 200 kPa P V1 1 No work is done during process 23 since V2 = V3. The pressure remains constant during process 12 and the work done during this process is 1kJ Wb = P dV = P2 (V3  V2 ) = (200kPa)(2.58  1.29)m 3 1kPa m 3 1 2 = 258kJ 1 2 v The initial and final internal energies of air are (Table A17) u1 = u@ 300 K = 214.07 kJ / kg u2 = u@ 1200 K = 933.33 kJ / kg Substituting, Qin = (3 kg)(933.33  214.07)kJ/kg + 258 kJ = 2416 kJ Alternative solution The specific heat of air at the average temperature of Tave = (300 + 1200)/2 = 750 K is, from Table A2b, Cvave = 0.800 kJ/kg.K. Substituting Qin = m(u 3  u1 ) + Wb,out mCv (T3  T1 ) + Wb,out = (3 kg)(0.800 kJ / kg.K)(1200  300) K + 258 kJ = 2418 kJ 541 Chapter 5 The First Law of Thermodynamics Closed System Energy Analysis: Solids and Liquids 545 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determin ed. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the brass balls are given to be = 8522 kg/m3 and Cp = 0.385 kJ/kg.C. Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Brass balls, 120C Water bath, 5C Qout = U ball = m(u2  u1 ) Qout = mC (T1  T2 ) The total amount of heat transfer from a ball is m = V = Qout D 3 (0.05 m) 3 = (8522 kg / m 3 ) = 0.558 kg 6 6 = mC (T1  T2 ) = (0.558 kg)(0.385 kJ / kg. C)(120  74 ) C = 9.88 kJ / ball Then the rate of heat transfer from the balls to the water becomes & & Qtotal = nball Qball = (100 balls / min) (9.88 kJ / ball ) = 988 kJ / min Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50C since energy input must be equal to energy output for a system whose energy level remains constant. That is, Ein = Eout when Esystem = 0 . 542 Chapter 5 The First Law of Thermodynamics 546 A number of aluminum balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of aluminum at the average temperature of (120+74)/2 = 97C = 370 K are = 2700 kg/m3 and Cp = 0.937 kJ/kg.C (Table A3). Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Aluminum balls, 120C Water bath, 50C Qout = U ball = m(u2  u1 ) Qout = mC (T1  T2 ) The total amount of heat transfer from a ball is m = V = Qout (0.05 m) 3 D 3 = (2700 kg/m 3 ) = 0.1767 kg 6 6 = mC (T1  T2 ) = (0.1767 kg )(0.937 kJ/kg.C)(120  74)C = 7.62 kJ/ball Then the rate of heat transfer from the balls to the water becomes & & Qtotal = n ball Qball = (100 balls/min) (7.62 kJ/ball) = 762 kJ/min
Therefore, heat must be removed from the water at a rate of 762 kJ/min in order to keep its temperature constant at 50C since energy input must be equal to energy output for a system whose energy level remains constant. That is, Ein = Eout when Esystem = 0 . 543 Chapter 5 The First Law of Thermodynamics 547E A person shakes a canned of drink in a iced water to cool it. The mass of the ice that will melt by the time the canned drink is cooled to a specified temperature is to be determined. Assumptions 1 The thermal properties of the drink are constant, and are taken to be the same as those of water. 2 The effect of agitation on the amount of ice melting is negligible. 3 The thermal energy capacity of the can itself is negligible, and thus it does not need to be considered in the analysis. Properties The density and specific heat of water at the average temperature of (75+45)/2 = 60F are = 62.3 lbm/ft3, and Cp = 1.0 Btu/lbm.F (Table A3E). The heat of fusion of water is 143.5 Btu/lbm. Analysis We take a canned drink as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qout = Ucanned drink = m(u2  u1 ) Qout = mC (T1  T2 ) Noting that 1 gal = 128 oz and 1 ft3 = 7.48 gal = 957.5 oz, the total amount of heat transfer from a ball is Cola 75F 1 ft 3 1 gal m = V = (62.3 lbm/ft 3 )(12 oz/can) = 0.781 lbm/can 7.48 gal 128 fluid oz Qout = mC (T1  T2 ) = (0.781 lbm/can )(1.0 Btu/lbm.F)(75  45)F = 23.4 Btu/can Noting that the heat of fusion of water is 14.5 Btu/lbm, the amount of ice that will melt to cool the drink is mice = Qout 23.4 Btu / can = = 0.163 lbm (per can of drink) hif 143.5 Btu / lbm since heat transfer to the ice must be equal to heat transfer from the can. Discussion The actual amount of ice melted will be greater since agitation will also cause some ice to melt. 544 Chapter 5 The First Law of Thermodynamics 548 An iron whose base plate is made of an aluminum alloy is turned on. The minimum time for the plate to reach a specified temperature is to be determined. Assumptions 1 It is given that 85 percent of the heat generated in the resistance wires is transferred to the plate. 2 The thermal properties of the plate are constant. 3 Heat loss from the plate during heating is disregarded since the minimum heating time is to be determined. 4 There are no changes in kinetic and potential energies. 5 The plate is at a uniform temperature at the end of the process. Properties The density and specific heat of the aluminum alloy plate are given to be = 2770 kg/m3 and Cp = 875 kJ/kg.C. Analysis The mass of the iron's base plate is m = V = LA = (2770 kg / m3 )( 0.005 m)( 0.03 m 2 ) = 0.4155 kg
Noting that only 85 percent of the heat generated is transferred to the plate, the rate of heat transfer to the iron's base plate is & Qin = 0.85 1000 W = 850 W We take plate to be the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass Air 22C = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = Uplate = m(u2  u1 ) & Qin t = mC (T2  T1 ) Solving for t and substituting, IRON 1000 W t = mCTplate (0.4155 kg )(875 J/kg.C)(140  22)C = = 50.5 s & 850 J/s Q
in which is the time required for the plate temperature to reach the specified temperature. 545 Chapter 5 The First Law of Thermodynamics 549 Stainless steel ball bearings leaving the oven at a specified uniform temperature at a specified rate are exposed to air and are cooled before they are dropped into the water for quenching. The rate of heat transfer from the ball bearing to the air is to be determined. Assumptions 1 The thermal properties of the bearing balls are constant. 2 The kinetic and potential energy changes of the balls are negligible. 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the ball bearings are given to be = 8085 kg/m3 and Cp = 0.480 kJ/kg.C. Analysis We take a single bearing ball as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Furnace Water, 30C Steel balls, 900C Qout = U ball = m(u2  u1 ) Qout = mC (T1  T2 ) The total amount of heat transfer from a ball is m = V = Qout D 3 (0.012 m) 3 = (8085 kg / m 3 ) = 0.007315 kg 6 6 . = mC (T1  T2 ) = (0.007315 kg)(0.480 kJ / kg. C)(900  850 ) C = 01756 kJ / ball Then the rate of heat transfer from the balls to the air becomes & & Qtotal = nball Qout (per ball) = (1400 balls / min) (01756 kJ / ball ) = 245.8 kJ / min = 4.10 kW . Therefore, heat is lost to the air at a rate of 4.10 kW. 546 Chapter 5 The First Law of Thermodynamics 550 Carbon steel balls are to be annealed at a rate of 2500/h by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The total rate of heat transfer from the balls to the ambient air is to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 There are no changes in kinetic and potential energies. 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the balls are given to be = 7833 kg/m3 and Cp = 0.465 kJ/kg.C. Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Furnace Air, 35C Steel balls, 900C Qout = U ball = m(u2  u1 ) Qout = mC (T1  T2 ) (b) The amount of heat transfer from a single ball is m = V = Qout (0.008 m) 3 D 3 = (7833 kg/m 3 ) = 0.00210 kg 6 6 = mC p (T1  T2 ) = (0.0021 kg )(0.465 kJ/kg.C)(900  100)C = 0.781 kJ (per ball) Then the total rate of heat transfer from the balls to the ambient air becomes & & Q = n Q = (2500 balls / h) (0.781 kJ / ball) = 1,953 kJ / h = 542 W
out ball out 547 Chapter 5 The First Law of Thermodynamics 551 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at the end of the 5min operating period is to be determined for the cases of operation with and without a heat sink. Assumptions 1 The device and the heat sink are isothermal. 2 The thermal properties of the device and of the sink are constant. 3 Heat loss from the device during on time is disregarded since the highest possible temperature is to be determined. Properties The specific heat of the device is given to be Cp = 850 J/kg.C. The specific heat of aluminum at room temperature of 300 K is 902 J/kg.C (Table A3). Analysis We take the device to be the system. Noting that electrical energy is supplied, the energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Electronic device 25C We,in = Udevice = m(u2  u1 ) & t = mC (T  T ) We,in 2 1 Substituting, the temperature of the device at the end of the process is determined to be (30 J / s)(5 60 s) = (0.020 kg)(850 J / kg. C)(T2  25) C We,in = Udevice + Uheat sink & t = mC (T  T ) We,in 2 1 device + mC (T2  T ) heat sink 1 Substituting, the temperature of the deviceheat sink combination is determined to be (30 J / s)(5 60 s) = (0.020 kg)(850 J / kg. C)(T2  25) C + (0.200 kg)(902 J / kg. C)(T2  25) C T2 = 70.6 C (with heat sink) Discussion These are the maximum temperatures. In reality, the temperatures will be lower because of the heat losses to the surroundings. T2 = 554 C (without the heat sink) Case 2 When a heat sink is attached, the energy balance can be expressed as 548 Chapter 5 The First Law of Thermodynamics 552 Problem 552 is reconsidered. The effect of the mass of the heat sink on the maximum device temperature as the mass of heat sink varies from 0 kg to 1 kg is to be investigated. The maximum temperature is to be plotted against the mass of heat sink. "Knowns:" "T_1 is the maximum temperature of the device" Q_dot_out = 30"[W]" m_device=20"[g]" Cp_device=850"[J/kgC]" A=5"[cm^2]" DELTAt=5"[min]" T_amb=25"[C]" {m_sink=0.2"[kg]"} "Cp_al taken from Table A3(b) at 300K" Cp_al=0.902"[kJ/kgC]" T_2=T_amb"[C]" "Solution:" "The device without the heat sink is considered to be a closed system." "Conservation of Energy for the closed system:" "E_dot_in  E_dot_out = DELTAE_dot, we neglect DELTA KE and DELTA PE for the system, the device." E_dot_in  E_dot_out = DELTAE_dot E_dot_in =0"[W]" E_dot_out = Q_dot_out"[W]" "Use the solid material approximation to find the energy change of the device." DELTAE_dot= m_device*convert(g,kg)*Cp_device*(T_2T_1_device)/(DELTAt*convert(min,s))"[W]" "The device with the heat sink is considered to be a closed system." "Conservation of Energy for the closed system:" "E_dot_in  E_dot_out = DELTAE_dot, we neglect DELTA KE and DELTA PE for the device with the heat sink." E_dot_in  E_dot_out = DELTAE_dot_combined "Use the solid material approximation to find the energy change of the device." DELTAE_dot_combined= (m_device*convert(g,kg)*Cp_device*(T_2T_1_device&sink)+m_sink*Cp_al*(T_2T_1_device&sink)*convert(kJ,J))/(DELTAt*convert(min,s))"[W]" msink [kg] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 T1,device&sink [C] 554.4 109 70.59 56.29 48.82 44.23 41.12 38.88 37.19 35.86 34.79 549 Chapter 5 The First Law of Thermodynamics
600 550 500 450 400 T 1,device&sink [C] 350 300 250 200 150 100 50 0 0 0.2 0.4 0.6 0.8 1 m sink [kg] 550 Chapter 5 The First Law of Thermodynamics 553 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked is to be determined. Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 The thermal properties of the egg are constant. 3 Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible. 4 There are no changes in kinetic and potential energies. Properties The density and specific heat of the egg are given to be = 1020 kg/m3 and Cp = 3.32 kJ/kg.C. Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = U egg = m(u2  u1 ) = mC (T2  T1 ) Then the mass of the egg and the amount of heat transfer become (0.055 m) 3 D 3 = (1020 kg/m 3 ) = 0.0889 kg m = V = 6 6 Qin = mC p (T2  T1 ) = (0.0889 kg )(3.32 kJ/kg.C)(70  8)C = 18.3 kJ Boiling Water Egg 8C 551 Chapter 5 The First Law of Thermodynamics 554E Large brass plates are heated in an oven at a rate of 300/min. The rate of heat transfer to the plates in the oven is to be determined. Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the brass are given to be = 532.5 lbm/ft3 and Cp = 0.091 Btu/lbm.F. Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Plates 75F Qin = Uplate = m(u2  u1 ) = mC (T2  T1 ) The mass of each plate and the amount of heat transfer to each plate is m = V = LA = (532.5 lbm / ft 3 )[(1.2 / 12 ft )(2 ft)(2 ft)] = 213 lbm Qin = mC (T2  T1 ) = (213 lbm / plate)(0.091 Btu / lbm. F)(1000  75) F = 17,930 Btu / plate Then the total rate of heat transfer to the plates becomes & & Qtotal = nplate Qin, per plate = (300 plates / min) (17,930 Btu / plate) = 5,379,000 Btu / min = 89,650 Btu / s 555 Long cylindrical steel rods are heattreated in an oven. The rate of heat transfer to the rods in the oven is to be determined. Assumptions 1 The thermal properties of the rods are constant. 2 The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the steel rods are given to be = 7833 kg/m3 and Cp = 0.465 kJ/kg.C. Analysis Noting that the rods enter the oven at a velocity of 3 m/min and exit at the same velocity, we can say that a 3m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is m = V = LA = L(D 2 / 4 ) = (7833 kg / m 3 )( 3 m)[ (0.1 m) 2 / 4 ] = 184.6 kg We take the 3m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Oven, 900C Qin = U rod = m(u2  u1 ) = mC (T2  T1 ) Substituting, Qin = mC (T2  T1 ) = (184 .6 kg)(0.465 kJ / kg. C)(700  30) C = 57,512 kJ Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes & Qin = Qin / t = (57,512 kJ) / (1 min) = 57,512 kJ / min = 958.5 kW Steady Flow Energy Balance: Nozzles and Diffusers
Steel rod, 30C 552 Chapter 5 The First Law of Thermodynamics 556C A steadyflow system involves no changes with time anywhere within the system or at the system boundaries 557C No. 558C It is mostly converted to internal energy as shown by a rise in the fluid temperature. 559C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature. 560C Heat transfer to the fluid as it flows through a nozzle is desirable since it will probably increase the kinetic energy of the fluid. Heat transfer from the fluid will decrease the exit velocity. 553 Chapter 5 The First Law of Thermodynamics 561 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The specific heat of air at the anticipated average temperature of 450 K is C p = 1.02 kJ/kg.C (Table A2). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Using the ideal gas relation, the specific volume and the mass flow rate of air are determined to be v1 = RT1 (0.287 kPa m 3 /kg K )(473 K ) = = 0.4525 m 3 /kg P1 300 kPa 1 1 & m = A1 V1 = (0.008m 2 )(30m/s) = 0.5304 kg/s v1 0.4525m 3 /kg (b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & & Ein  Eout = Esystem 0 (steady) =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
Substituting, V22  V12 V 2  V12 0 = C p , ave (T2  T1 ) + 2 2 2 0 = (1.02 kJ/kg K )(T2  200 o C) +
It yields (180 m/s) 2  (30 m/s) 2 2 1 kJ/kg 1000 m 2 /s 2 T2 = 184.6C (c) The specific volume of air at the nozzle exit is v2 =
& m= RT2 (0.287 kPa m 3 /kg K )(184.6 + 273 K ) = = 1.313 m 3 /kg 100 kPa P2
1 1 A2 V2 0.5304kg/s = A2 (180m/s ) v2 1.313m 3 /kg A2 = 0.00387 m 2 = 38.7 cm2 P1 = 300 kPa T1 = 200C V1 = 30 m/s A1 = 80 cm 2 AIR P2 = 100 kPa V2 = 180 m/s 554 Chapter 5 The First Law of Thermodynamics 562 Problem 561 is reconsidered. The effect of the inlet area on the mass flow rate, exit velocity, and the exit area as the inlet area varies from 50 cm^2 to 150 cm^2 is to be investigated, and the final results are to be plotted against the inlet area. Function HCal(WorkFluid$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'Air' = WorkFluid$ then HCal:=ENTHALPY('Air',T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ." endif end HCal "System: control volume for the nozzle, Property relation: Air is an ideal gas" "Process: Steady state, steady flow, adiabatic, no work" "Knowns  obtain from the input diagram" WorkFluid$ = 'Air' T[1] = 200 "[C]" P[1] = 300 "[kPa]" Vel[1] = 30 "[m/s]" P[2] = 100 "[kPa]" Vel[2] = 180 "[m/s]" A[1]=80 "[cm^2]" Am[1]=A[1]*convert(cm^2,m^2)"[m^2]" "Property Data  since the Enthalpy function has different parameters for ideal gas and real fluids, a function was used to determine h." h[1]=HCal(WorkFluid$,T[1],P[1]) h[2]=HCal(WorkFluid$,T[2],P[2]) "The Volume function has the same form for an ideal gas as for a real fluid." v[1]=volume(workFluid$,T=T[1],p=P[1]) v[2]=volume(WorkFluid$,T=T[2],p=P[2]) "Conservation of mass: " m_dot[1]= m_dot[2] "Mass flow rate" m_dot[1]=Am[1]*Vel[1]/v[1] m_dot[2]= Am[2]*Vel[2]/v[2] "Conservation of Energy  SSSF energy balance" h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000) "Definition" A_ratio=A[1]/A[2] A[2]=Am[2]*convert(m^2,cm^2) A1 [cm ] 50 60 70 80 90 100 110 120 130 140 150
2 A2 [cm ] 24.19 29.02 33.86 38.7 43.53 48.37 53.21 58.04 62.88 67.72 72.56 2 m1 0.3314 0.3976 0.4639 0.5302 0.5964 0.6627 0.729 0.7952 0.8615 0.9278 0.9941 T2 184.6 184.6 184.6 184.6 184.6 184.6 184.6 184.6 184.6 184.6 184.6 555 Chapter 5 The First Law of Thermodynamics
1 0.9 0.8 0.7 m [1] 0.6 0.5 0.4 0.3 50 70 90 110 130 150 A[1] [cm ^2]
80 70 A[2] [cm ^2] 60 50 40 30 20 50 70 90 110 130 150 A[1] [cm ^2] 556 Chapter 5 The First Law of Thermodynamics 563 Steam is accelerated in a nozzle from a velocity of 80 m/s. The mass flow rate, the exit velocity, and the exit area of the nozzle are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. Properties From the steam tables (Table A6) P1 = 5 MPa v1 = 0.06857 m 3 /kg T1 = 500 o C h1 = 3433.8 kJ/kg
and P2 = 2MPa v 2 = 0.15120m 3 /kg o T2 = 400 C h2 = 3247.6kJ/kg & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The mass flow rate of steam is 1 90 kJ/s Steam 2 & m= 1 1 V1 A1 = (80 m/s)(50 10  4 m 2 ) = 5.833 kg/s v1 0.06857 m 3 /kg (b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & & Ein  Eout = Esystem 0 (steady) =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & & m(h1 + V12 / 2) = Qout + m(h2 + V22 /2) (since W pe 0) V 2  V12 & &  Qout = m h2  h1 + 2 2 1kJ/kg 1000m 2 /s 2 Substituting, the exit velocity of the steam is determined to be V 2  (80m/s) 2  90kJ/s = (5.833kg/s ) 3247.6  3433.8 + 2 2 It yields V2 = 589.9 m/s (c) The exit area of the nozzle is determined from & m= & mv 2 (5.833 kg/s ) 0.1512 m 3 /kg 1 V2 A2 A2 = = = 15.0 10 4 m 2 v2 V2 589.9 m/s ( ) 557 Chapter 5 The First Law of Thermodynamics 564E Air is accelerated in a nozzle from 150 ft/s to 900 ft/s. The exit temperature of air and the exit area of the nozzle are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. Properties The enthalpy of air at the inlet is h1 = 143.47 Btu/lbm (Table A17E). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & & Esystem 0 (steady) Ein  Eout = =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout 6.5 Btu/s & m(h1 + V12 & & & / 2) = Qout + m(h2 + V22 /2) (since W pe 0) 1
AIR V 2  V12 & &  Qout = m h2  h1 + 2 2 or, 2 h2 = q out + h1  V22  V12 2 (900 ft/s) 2  (150 ft/s) 2 2 1 Btu/lbm 25,037 ft 2 /s 2 = 6.5 Btu/lbm + 143.47 Btu/lbm  = 121.2 Btu/lbm
Thus, from Table A17E, T2 = 507 R (b) The exit area is determined from the conservation of mass relation, RT / P v V 1 1 A2 V2 = A1 V1 A2 = 2 1 A1 = 2 2 RT / P v2 v1 v1 V2 1 1 A2 = V1 V A1 2 (508/14.7 )(150ft/s ) ( 2 ) 0.1ft = 0.048ft 2 (600/50 )(900ft/s ) 558 Chapter 5 The First Law of Thermodynamics 565 [Also solved by EES on enclosed CD] Steam is accelerated in a nozzle from a velocity of 40 m/s to 300 m/s. The exit temperature and the ratio of the inlettoexit area of the nozzle are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Table A6), P1 = 3MPa v1 = 0.09936m 3 /kg o T1 = 400 C h1 = 3230.9kJ/kg & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0
P1 = 3 MPa T1 = 400C V1 = 40 m/s Steam & & Ein = Eout P2 = 2.5 MPa V2 = 300 m/s & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
or, V22  V12 2 h2 = h1 
Thus, V22  V12 (300 m/s) 2  (40 m/s) 2 = 3230.9 kJ/kg  2 2 1 kJ/kg 1000 m 2 /s 2 = 3186.7 kJ/kg o T2 = 376.7 C h2 = 3186.7 kJ/kg v 2 = 0.1153m 3 /kg P2 = 2.5MPa (b) The ratio of the inlet to exit area is determined from the conservation of mass relation, A v V (0.09936 m 3 /kg )(300 m/s) 1 1 A2 V2 = A1 V1 1 = 1 2 = = 6.46 v2 v1 A2 v 2 V1 (0.1153 m 3 /kg )(40 m/s) 559 Chapter 5 The First Law of Thermodynamics 566 Air is accelerated in a nozzle from 120 m/s to 380 m/s. The exit temperature and pressure of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The enthalpy of air at the inlet temperature of 500 K is h1 = 503.02 kJ/kg (Table A17). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 1
AIR & & Ein = Eout 2 & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
or, h2 = h1  V22  V12 2 1kJ/kg 1000m 2 /s 2 = 438.02kJ/kg 2 V2  V12 (380m/s )2  (120m/s )2 = 503.02kJ/kg  2 2 Then from Table A17 we read T2 = 436.5 K (b) The exit pressure is determined from the conservation of mass relation, 1 1 1 1 A2 V2 = A1 V1 A2 V2 = A1 V1 v2 v1 RT2 / P2 RT1 / P1
Thus, P2 = A1T2 V1 2 (436.5 K )(120 m/s) (600 kPa ) = 330.8 kPa P1 = A2 T1 V2 1 (500 K )(380 m/s) 560 Chapter 5 The First Law of Thermodynamics 567 Air is decelerated in a diffuser from 230 m/s to 30 m/s. The exit temperature of air and the exit area of the diffuser are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The enthalpy of air at the inlet temperature of 400 K is h1 = 400.98 kJ/kg (Table A17). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 1
AIR & & Ein = Eout 2 & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
or, h2 = h1  From Table A17, V22  V12 2 , 2 V2  V12 (30m/s )2  (230m/s )2 = 400.98kJ/kg  2 2 1kJ/kg 1000m 2 /s 2 = 426.98kJ/kg T2 = 425.6 K (b) The specific volume of air at the diffuser exit is v2 = RT2 0.287kPa m 3 /kg K (425.6K ) = = 1.221m 3 /kg (100kPa ) P2 ( ) From conservation of mass, & m= & mv 2 (6000 3600 kg/s )(1.221 m 3 /kg ) 1 A2 V2 A2 = = = 0.0678 m 2 v2 V2 30 m/s 561 Chapter 5 The First Law of Thermodynamics 568E Air is decelerated in a diffuser from 600 ft/s to a low velocity. The exit temperature and the exit velocity of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The enthalpy of air at the inlet temperature of 20F is h1 = 114.69 Btu/lbm (Table A17E). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 1
AIR & & Ein = Eout 2 & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
or, h2 = h1  V22  V12 2 , 2 V2  V12 0  (600ft/s )2 = 114.69Btu/lbm  2 2 1Btu/lbm 25,037ft 2 /s 2 = 121.88Btu/lbm From Table A17E, T2 = 510.0 R (b) The exit velocity of air is determined from the conservation of mass relation, 1 1 1 1 A2 V2 = A1 V1 A2 V2 = A1 V1 v2 v1 RT2 / P2 RT1 / P1
Thus, V2 = A1T2 P1 1 (510 R )(13 psia ) V1 = (600 ft/s) = 114.3 ft/s A2 T1 P2 5 (480 R )(14.5 psia ) 562 Chapter 5 The First Law of Thermodynamics 569 CO2 gas is accelerated in a nozzle to 450 m/s. The inlet velocity and the exit temperature are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 CO2 is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of CO2 is 0.1889 kPa.m3/kg.K (Table A1). The enthalpy of CO2 at 500C is h1 = 30,797 kJ/kmol (Table A20). & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Using the ideal gas relation, the specific volume is determined to be v1 =
Thus, & m= RT1 0.1889 kPa m 3 /kg K (773 K ) = = 0.146 m 3 /kg P1 1000 kPa
& mv1 (6000/3600kg/s ) 0.146m 3 /kg 1 A1 V1 V1 = = = 60.8m/s v1 A1 40 10  4 m 2 ( ) 1 CO2 2 ( ) (b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & & Ein  Eout = Esystem 0 (steady) =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0) 0 = h2  h1 +
Substituting, h2 = h1 
2 V2  V12 M 2 V22  V12 2 = 30,797 kJ/kmol  = 26,423kJ/kmol (450m/s )2  (60.8m/s )2 2 1kJ/kg 1000m 2 /s 2 (44kg/kmol ) Then the exit temperature of CO2 from Table A20 is obtained to be T2 = 685.8 K 563 Chapter 5 The First Law of Thermodynamics 570 R134a is accelerated in a nozzle from a velocity of 20 m/s. The exit velocity of the refrigerant and the ratio of the inlettoexit area of the nozzle are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From the refrigerant tables (Table A13) P1 = 700kPa v1 = 0.04064m 3 /kg T1 = 100 o C h1 = 338.19kJ/kg and
R134a 1 2 P2 = 300 kPa v 2 = 0.07767 m 3 /kg T2 = 30 o C h2 = 274.70 kJ/kg
& & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & & Ein  Eout = Esystem 0 (steady) =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout
2 & & & & m(h1 + V12 / 2) = m(h2 + V21 /2) (since Q W pe 0) 0 = h2  h1 +
Substituting, 0 = (274.70  338.19 )kJ/kg + It yields V2 = 356.9 m/s V22  V12 2 1kJ/kg 1000m 2 /s 2 2 V2  (20m/s )2 2 (b) The ratio of the inlet to exit area is determined from the conservation of mass relation, A v V 0.04064m 3 /kg (356.9m/s ) 1 1 1 = 2 1 = = 9.34 A2 V2 = A1 V1 v2 v1 A2 v1 V2 0.07767 m 3 /kg (20m/s ) ( ( ) ) 564 Chapter 5 The First Law of Thermodynamics 571 Air is decelerated in a diffuser from 220 m/s. The exit velocity and the exit pressure of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The enthalpies are (Table A17) T1 = 27 C = 300 K T2 = 42 C = 315 K h1 = 300.19 kJ / kg h2 = 315.27 kJ / kg & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 18 kJ/s & & Ein = Eout 1 AIR 2 & m(h1 + V12 & & & / 2) = Qout + m(h2 + V22 /2) (since W pe 0) 1 kJ/kg 1000 m 2 /s 2 V 2  V12 & &  Qout = m h2  h1 + 2 2 Substituting, the exit velocity of the air is determined to be V 2  (220 m/s) 2 18 kJ/s = (2.5 kg/s ) 315.27  300.19 + 2  2 It yields V2 = 62.0 m/s (b) The exit pressure of air is determined from the conservation of mass and the ideal gas relations, & m= and P2 = P2 v 2 = RT2 A V 0.04m 2 (62m/s ) 1 v 2 = 2 2 = = 0.992m 3 /kg A2 V2 & 2.5kg/s m v2 RT2 0.287kPa m 3 /kg K (315K ) = = 91.1kPa v2 0.992m 3 /kg ( ) ( ) 565 Chapter 5 The First Law of Thermodynamics 572 Nitrogen is decelerated in a diffuser from 200 m/s to a lower velocity. The exit velocity of nitrogen and the ratio of the inlettoexit area are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Nitrogen is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The molar mass of nitrogen is M = 28 kg/kmol (Table A1). The enthalpies are (Table A18) T1 = 7C = 280 K h1 = 8141 kJ/kmol T2 = 22C = 295 K h2 = 8580 kJ/kmol
& & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q W pe 0)
1 V22  V12 h2  h1 V22  V12 , = + 0 = h2  h1 + 2 M 2 Substituting, N2 2 0=
It yields (8580  8141) kJ/kmol + V22  (200m/s )2 28 kJ/kmol 2
V2 = 93.0 m/s 1kJ/kg 1000m 2 /s 2 (b) The ratio of the inlet to exit area is determined from the conservation of mass relation, RT / P A v V 1 1 1 = 1 2 = 1 1 A2 V2 = A1 V1 v2 v1 A2 v 2 V1 RT2 / P2 or, V2 V 1 A1 T1 / P1 = A2 T2 / P2 V2 (280 K/60 kPa )(93.0 m/s ) V = (295 K/85 kPa )(200 m/s ) = 0.625 1 566 Chapter 5 The First Law of Thermodynamics 573 Problem 572 is reconsidered. The effect of the inlet velocity on the exit velocity and the ratio of the inlettoexit area as the inlet velocity varies from 180 m/s to 260 m/s is to be investigated. The final results are to be plotted against the inlet velocity. Function HCal(WorkFluid$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'N2' = WorkFluid$ then HCal:=ENTHALPY(WorkFluid$,T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ." endif end HCal "System: control volume for the nozzle" "Property relation: Nitrogen is an ideal gas" "Process: Steady state, steady flow, adiabatic, no work" "Knowns" WorkFluid$ = 'N2' T[1] = 7 "[C]" P[1] = 60 "[kPa]" {Vel[1] = 200 "[m/s]"} P[2] = 85 "[kPa]" T[2] = 22 "[C]" "Property Data  since the Enthalpy function has different parameters for ideal gas and real fluids, a function was used to determine h." h[1]=HCal(WorkFluid$,T[1],P[1])"[kJ/kg]" h[2]=HCal(WorkFluid$,T[2],P[2])"[kJ/kg]" "The Volume function has the same form for an ideal gas as for a real fluid." v[1]=volume(workFluid$,T=T[1],p=P[1])"[m^3/kg]" v[2]=volume(WorkFluid$,T=T[2],p=P[2])"[m^3/kg]" "From the definition of mass flow rate, m_dot = A*Vel/v and conservation of mass the area ratio A_Ratio = A_1/A_2 is:" A_Ratio*Vel[1]/v[1] =Vel[2]/v[2] "Conservation of Energy  SSSF energy balance" h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000) 567 Chapter 5 The First Law of Thermodynamics ARatio 0.2603 0.4961 0.6312 0.7276 0.8019 0.8615 0.9106 0.9518 0.9869
1 0.9 0.8 0.7 Vel1 [m/s] 180 190 200 210 220 230 240 250 260 Vel2 [m/s] 34.84 70.1 93.88 113.6 131.2 147.4 162.5 177 190.8 A Ratio 0.6 0.5 0.4 0.3 0.2 180 190 200 210 220 230 240 250 260 Vel[1] [m /s]
200 180 160 Vel[2] [m /s] 140 120 100 80 60 40 20 180 190 200 210 220 230 240 250 260 Vel[1] [m /s] 568 Chapter 5 The First Law of Thermodynamics 574 R134a is decelerated in a diffuser from a velocity of 140 m/s. The exit velocity of R134a and the mass flow rate of the R134a are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. Properties From the R134a tables (Tables A11 through A13) P1 = 700kPa v1 = 0.0292m 3 /kg sat.vapor h1 = 261.85kJ/kg and P2 = 800kPa v 2 = 0.02691m 3 /kg o T2 = 40 C h2 = 273.66kJ/kg & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit velocity of R134a is determined from the steadyflow mass balance to be v A 1 1 1 0.02691m 3 /kg A2 V2 = A1 V1 V2 = 2 1 V1 = (140m/s ) = 71.7m/s v1 A2 v2 v1 1.8 0.02920m 3 /kg (b) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass 3 kJ/s 1 R134a 2 = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & Qin + m(h1 + V12 / 2) = m(h2 + V22 /2) (since W pe 0) V 2  V12 & & Qin = m h2  h1 + 2 2 (71.7m/s )2  (140m/s) 2 & 3kJ/s = m 273.66  261.85 + 2 It yields & m = 0.655 kg/s 1kJ/kg 1000m 2 /s 2 Substituting, the mass flow rate of the refrigerant is determined to be 569 Chapter 5 The First Law of Thermodynamics Turbines and Compressors 575C Yes. 576C The volume flow rate at the compressor inlet will be greater than that at the compressor exit. 577C Yes. Because energy (in the form of shaft work) is being added to the air. 578C No. 579 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A4 through 6) P1 = 10MPa v1 = 0.02975m 3 /kg o T1 = 450 C h1 = 3240.9kJ/kg and P1 = 10 MPa T1 = 450C V1 = 80 m/s P2 = 10 kPa h2 = h f + x 2 h fg = 191.83 + 0.92 2392.8 = 2393.2kJ/kg x 2 = 0.92 Analysis (a) The change in kinetic energy is determined from V 2  V12 (50m/s )2  (80m/s) 2 ke = 2 = 2 2 1kJ/kg 1000m 2 /s 2 = 1.95kJ/kg P2 = 10 kPa x2 = 0.92 V2 = 50 m/s = 12 kg/s m
W STEAM & & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & m(h1 + V12 / 2) = Wout + m(h2 + V22 /2) (since Q pe 0)
2 2 & =  m h  h + V2  V1 & 2 Wout 1 2 Then the power output of the turbine is determined by substitution to be & Wout = (12 kg/s)(2393.2  3240.9  1.95)kJ/kg = 10.2 MW
(c) The inlet area of the turbine is determined from the mass flow rate relation, & m= & mv (12 kg/s)(0.02975 m 3 /kg ) 1 A1 V1 A1 = 1 = = 0.00446 m 2 v1 V1 80 m/s 580 Problem 579 is reconsidered. The effect of the turbine exit pressure on the power output of the turbine as the exit pressure varies from 10 kPa to 200 kPa is to be investigated. The power output is to be plotted against the exit pressure. 570 Chapter 5 The First Law of Thermodynamics "Knowns " T[1] = 450 "[C]" P[1] = 10000 "[kPa]" Vel[1] = 80 "[m/s]" P[2] = 10 "[kPa]" X_2=0.92 Vel[2] = 50 "[m/s]" m_dot[1]=12"[kg/s]" "Property Data" h[1]=enthalpy(Steam,T=T[1],P=P[1])"[kJ/kg]" h[2]=enthalpy(Steam,P=P[2],x=x_2)"[kJ/kg]" T[2]=temperature(Steam,P=P[2],x=x_2)"[C]" v[1]=volume(Steam,T=T[1],p=P[1])"[m^3/kg]" v[2]=volume(Steam,P=P[2],x=x_2)"[m^3/kg]" "Conservation of mass: " m_dot[1]= m_dot[2] "Mass flow rate" m_dot[1]=A[1]*Vel[1]/v[1] "[kg/s]" m_dot[2]= A[2]*Vel[2]/v[2] "[kg/s]" "Conservation of Energy  Steady Flow energy balance" m_dot[1]*(h[1]+Vel[1]^2/(2*1000)) = m_dot[2]*(h[2]+Vel[2]^2/(2*1000))+W_dot_turb*convert(MW,kJ/s) DELTAke=Vel[2]^2/(2*1000)Vel[1]^2/(2*1000)"[kJ/kg]" P2 [kPa] 10 31.11 52.22 73.33 94.44 115.6 136.7 157.8 178.9 200 Wturb [MW] 10.21 9.645 9.362 9.167 9.018 8.895 8.792 8.703 8.624 8.553 T2 [C] 45.79 69.92 82.4 91.16 98.02 103.7 108.6 112.9 116.7 120.2 571 Chapter 5 The First Law of Thermodynamics
10.25 9.9 W turb [Mw ] 9.55 9.2 8.85 8.5 0 40 80 120 160 200 P[2] [kPa]
130 120 110 T[2] [C] 100 90 80 70 60 50 40 0 40 80 120 160 200 P[2] [kPa] 572 Chapter 5 The First Law of Thermodynamics 581 Steam expands in a turbine. The mass flow rate of steam for a power output of 5 MW is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A4 through 6) P1 = 10MPa h1 = 3096.5kJ/kg o T1 = 400 C P2 = 20kPa h2 = h f + x 2 h fg = 251.40 + 0.90 2358.3 = 2373.9kJ/kg x 2 = 0.90 & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass 1 H2O 2 = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & mh = W + mh
1 out 2 & (since Q ke pe 0) & & Wout =  m(h2  h1 ) Substituting, the required mass flow rate of the steam is determined to be & 5000 kJ/s =  m(2373.9  3096.5) kJ/kg & m = 6.919 kg/s 573 Chapter 5 The First Law of Thermodynamics 582E Steam expands in a turbine. The rate of heat loss from the steam for a power output of 4 MW is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties From the steam tables (Tables A4 through 6) P1 = 1000psia h1 = 1448.1Btu/lbm o T1 = 900 F P2 = 5psia h2 = 1131.0Btu/lbm sat.vapor & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass 1 H2O 2 = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & (since ke pe 0) 1 out out + mh2 & =  m (h  h )  W & & 2 1 Qout out Substituting, & & Ein = Eout & & & mh = Q + W 1 Btu & Qout = (45000/3600 lbm/s)(45000/3600 lbm/s)Btu/lbm  4000 kJ/s 1.055 kJ = 172.3 Btu/s 574 Chapter 5 The First Law of Thermodynamics 583 Steam expands in a turbine. The exit temperature of the steam for a power output of 2 MW is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A4 through 6) P1 = 10MPa h1 = 3373.7 kJ/kg T1 = 500 o C & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 1 & & Ein = Eout & & & mh1 = Wout + mh2 & & Wout = m(h1  h2 ) Substituting, & (since Q ke pe 0) H2O 2000kJ/s = (3kg/s )(3373.7  h2 )kJ/kg h2 = 2707kJ/kg 2 Then the exit temperature becomes P2 = 20kPa o T2 = 110.8 C h2 = 2707kJ/kg 575 Chapter 5 The First Law of Thermodynamics 584 Argon gas expands in a turbine. The exit temperature of the argon for a power output of 250 kW is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats. Properties The gas constant of Ar is R = 0.2081 kPa.m3/kg.K. The constant pressure specific heat of Ar is Cp = 0.5203 kJ/kgC (Tables A2a) & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of argon and its mass flow rate are v1 = Thus, & m= RT1 0.2081kPa m 3 /kg K (723K ) = = 0.167 m 3 /kg P1 900kPa 1 1 A1V1 = 0.006m 2 (80m/s ) = 2.874kg/s 3 v1 0.167m /kg ( ) ( ) A1 = 60 cm2 P1 = 900 kPa T1 = 450C V1 = 80 m/s We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 ARGON 250 kW & & Ein = Eout 2 & & & & m(h1 + V12 / 2) = Wout + m(h2 + V2 /2) (since Q pe 0) & & Wout = m h 2  h1 + Substituting,
2 V2  V12 2 P2 = 150 kPa V2 = 150 m/s (150 m/s) 2  (80 m/s) 2 250 kJ/s = (2.874 kg/s ) (0.5203 kJ/kg o C)(T2  450 o C) + 2 It yields T2 = 267.3C 1 kJ/kg 1000 m 2 /s 2 576 Chapter 5 The First Law of Thermodynamics 585E Air expands in a turbine. The mass flow rate of air and the power output of the turbine are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.3704 psia.ft 3/lbm.R. The constant pressure specific heat of air at the average temperature of (900 + 300)/2 = 600F is Cp = 0.25 Btu/lbmF (Tables A2a) & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of air and its mass flow rate are v1 = RT1 0.3704psia ft 3 /lbm R (1360R ) = = 3.358ft 3 /lbm P1 150psia 1 1 & m = A1V1 = 0.1ft 2 (350ft/s ) = 10.42lbm/s 3 v1 3.358ft /lbm ( ) 1 ( ) (b) We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass AIR = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 2 & & Ein = Eout & & & & m(h1 + V12 / 2) = Wout + m(h2 + V22 /2) (since Q pe 0) V 2  V12 & & Wout =  m h2  h1 + 2 2 Substituting, V 2  V12 = m C p (T2  T1 ) + 2 & 2 2 2 & = (10.42 lbm/s) 0.250 Btu/lbmo F (300  900 )o F + (700 ft/s )  (350 ft/s ) Wout 2 = 1486.5 Btu/s = 1568 kW ( ) 1 Btu/lbm 25,037 ft 2 /s 2 577 Chapter 5 The First Law of Thermodynamics 586 Refrigerant134a is compressed steadily by a compressor. The power input to the compressor and the volume flow rate of the refrigerant at the compressor inlet are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the refrigerant tables (Tables A11 through 13) T1 = 20 o C v1 = 0.1464m 3 /kg sat.vapor h1 = 235.31kJ/kg P2 = 0.7MPa h2 = 307.01kJ/kg T2 = 70 o C & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass 2 R134a = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 1 & & Ein = Eout & & & & Win + mh1 = mh2 (since Q ke pe 0) & & Win = m(h2  h1 ) Substituting, & Win = (1.2kg/s )(307.01  235.31)kJ/kg = 86.04kJ/s (b) The volume flow rate of the refrigerant at the compressor inlet is & & V1 = mv1 = (1.2kg/s )(0.1464m 3 /kg ) = 0.176m 3 /s 578 Chapter 5 The First Law of Thermodynamics 587 Air is compressed by a compressor. The mass flow rate of air through the compressor is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The inlet and exit enthalpies of air are (Table A17) T1 = 25C = 298 K T2 = 347C = 620 K h1 = h@ 298 K = 298.2 kJ/kg h2 = h@ 620 K = 628.07 kJ/kg 2 Analysis We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass AIR = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0
1,500 kJ/min & & Ein = Eout 1 & & & & Win + m(h1 + V12 / 2) = Qout + m(h2 + V22 /2) (since pe 0) V 2  V12 & & & Win  Qout = m h2  h1 + 2 2 Substituting, the mass flow rate is determined to be (90m/s )2  0 1kJ/kg & 250 kJ/s  (1500/60 kJ/s) = m 628.07  298.2 + 1000m 2 /s 2 2 & = 0.674 kg/s m 579 Chapter 5 The First Law of Thermodynamics 588E Air is compressed by a compressor. The mass flow rate of air through the compressor and the exit temperature of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.3704 psia.ft 3/lbm.R (Table A1). The inlet enthalpy of air is (Table A17E) T1 = 60F = 520 R h1 = h@ 520 R = 124.27 Btu/lbm & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of air and its mass flow rate are RT1 0.3704psia ft 3 /lbm R (520R ) = = 13.1ft 3 /lbm P1 14.7 psia & V 5000 ft 3 / min & m= 1 = = 381.7 lbm / min = 6.36 lbm / s v1 13.1 ft 3 / lbm (b) We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as v1 = & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass ( ) 2 AIR = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 10 Btu/lbm & & Ein = Eout & & & & Win + mh1 = Qout + mh2 (since ke pe 0) & & & W  Q = m (h  h )
in out 2 1 1 Substituting, (700hp ) 0.7068Btu/s  (6.36 lbm/s) (10Btu/lbm) = (6.36lbm/s )(h2  124.27Btu/lbm ) 1hp h2 = 192.06Btu/lbm Then the exit temperature is determined from Table A17E to be T2 = 801 R = 341F 580 Chapter 5 The First Law of Thermodynamics 589E Problem 588E is reconsidered. The effect of the rate of cooling of the compressor on the exit temperature of air as the cooling rate varies from 0 to 100 Btu/lbm is to be investigated. The air exit temperature is to be plotted against the rate of cooling. "Knowns " T[1] = 60 "[F]" P[1] = 14.7 "[psia]" V_dot[1] = 5000 "[ft^3/min]" P[2] = 150 "[psia]" {q_out=10"[Btu/lbm]"} W_dot_in=700"[hp]" "Property Data" h[1]=enthalpy(Air,T=T[1])"[Btu/lbm]" h[2]=enthalpy(Air,T=T[2])"[Btu/lbm]" TR_2=T[2]+460"[R]" v[1]=volume(Air,T=T[1],p=P[1])"[ft^3/lbm]" v[2]=volume(Air,T=T[2],p=P[2])"[ft^3/lbm]" "Conservation of mass: " m_dot[1]= m_dot[2] "Mass flow rate" m_dot[1]=V_dot[1]/v[1] *convert(ft^3/min,ft^3/s)"[lbm/s]" m_dot[2]= V_dot[2]/v[2]*convert(ft^3/min,ft^3/s) "[lbm/s]" "Conservation of Energy  Steady Flow energy balance" W_dot_in*convert(hp,Btu/s)+m_dot[1]*(h[1]) = m_dot[1]*q_out+m_dot[1]*(h[2])
400 350 300 qout [Btu/lbm] 0 10 20 30 40 50 60 70 80 90 100 T2 [F] 382 340.9 299.7 258.3 216.9 175.4 133.8 92.26 50.67 9.053 32.63 T[2] [F] 250 200 150 100 50 0 50 0 20 40 60 80 100 q out [Btu/lbm ] 581 Chapter 5 The First Law of Thermodynamics 590 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. Properties The constant pressure specific heat of helium is Cp = 5.1926 kJ/kgK (Table A2a). & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 P2 = 700 kPa T2 = 430 K & & Ein = Eout & & & & Win + mh1 = Qout + mh2 (since ke pe 0) & & & & Win  Qout = m(h2  h1 ) = mC p (T2  T1 ) Thus, & & & Win = Qout + mC p (T2  T1 ) = (90/60kg/s)(20 kJ/kg) + (90/60kg/s)(5.1926kJ/kg K)(430  310)K = 965kW
P1 = 120 kPa T1 = 310 K
He m=90kg/mi W 582 Chapter 5 The First Law of Thermodynamics 591 CO2 is compressed by a compressor. The volume flow rate of CO 2 at the compressor inlet and the power input to the compressor are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with variable specific heats. 4 The device is adiabatic and thus heat transfer is negligible. Properties The gas constant of CO2 is R = 0.1889 kPa.m3/kg.K, and its molar mass is M = 44 kg/kmol (Table A1). The inlet and exit enthalpies of CO2 are (Table A20) T1 = 300 K h1 = 9,431 kJ / kmol h2 = 15,483 kJ / kmol T2 = 450 K & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of air and its volume flow rate are v1 = RT1 0.1889kPa m 3 /kg K (300K ) = = 0.5667 m 3 /kg P1 100kPa ( ) & & V = mv1 = (0.5 kg / s)(0.5667 m 3 / kg) = 0.283 m 3 / s (b) We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 2 & & Ein = Eout & & & & Win + mh1 = mh2 (since Q ke pe 0) & & & Win = m(h2  h1 ) = m(h2  h1 ) / M & Win CO2 (0.5kg/s )(15,483  9,431kJ/kmol ) = 68.8kW =
44kg/kmol 1 583 Chapter 5 The First Law of Thermodynamics Throttling Valves 592C Because usually there is a large temperature drop associated with the throttling process. 593C Yes. 594C No. Because air is an ideal gas and h = h(T) for ideal gases. Thus if h remains constant, so does the temperature. 595C If it remains in the liquid phase, no. But if some of the liquid vaporizes during throttling, then yes. 596 Refrigerant134a is throttled by a valve. The temperature drop of the refrigerant and specific volume after expansion are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of R134a is, from the refrigerant tables (Tables A11 through 13), P1 = 0.8MPa T1 = Tsat = 31.33o C sat.liquid h1 = h f = 93.42kJ / kg & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
0 & & & Ein  Eout = Esystem (steady) =0 & & Ein = Eout & & mh1 = mh2 h1 = h2 & & since Q W = ke pe 0 . Then, P2 = 0.14MPa h f = 25.77kJ/kg, Tsat = 18.8 o C (h2 = h1 ) h g = 236.04kJ/kg Obviously hf <h2 <hg, thus the refrigerant exists as a saturated mixture at the exit state and thus T2 = Tsat = 18.8C. Then the temperature drop becomes T = T2  T1 = 18.8  3133 = 50.13o C . The quality at this state is determined from x2 = Thus, v 2 = v f + x 2 v fg = 0.0007381 + 0.322 0.13876 = 0.0454m 3 /kg h2  h f h fg = 93.42  25.77 = 0.322 210.27  25.77
P2 = 140 kPa P1 = 800 kPa Sat. liquid R134a 584 Chapter 5 The First Law of Thermodynamics 597 [Also solved by EES on enclosed CD] Refrigerant134a is throttled by a valve. The pressure and internal energy after expansion are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of R134a is, from the refrigerant tables (Tables A11 through 13), P1 = 0.8MPa h1 h f @ 25o C = 84.33kJ/kg o T1 = 25 C & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
0 & & & Ein  Eout = Esystem (steady) P1 = 0.8 MPa T1 = 25C =0 & & Ein = Eout & & mh1 = mh2 h1 = h2 R134a & & since Q W = ke pe 0 . Then, T2 = 20 C h f = 24.26kJ/kg , u f = 24.17kJ/kg (h2 = h1 ) h g = 235.31kJ/kg u g = 215.84kJ/kg o T2 = 20C Obviously hf <h2 <hg, thus the refrigerant exists as a saturated mixture at the exit state, and thus P2 = Psat @ 20C = 0.13299 MPa Also, Thus, x2 = h2  h f h fg = 84.33  24.26 = 0.285 21105 . u 2 = u f + x 2 u fg = 24.17 + 0.285 (215.84  24.17 ) = 78.8kJ/kg 585 Chapter 5 The First Law of Thermodynamics 598 Steam is throttled by a wellinsulated valve. The temperature drop of the steam after the expansion is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of steam is (Tables A6), P1 = 8MPa h1 = 3398.3kJ/kg T1 = 500 o C & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
0 & & & Ein  Eout = Esystem (steady) P1 = 8 MPa T1 = 500C H2O =0 & & Ein = Eout & & mh1 = mh2 h1 = h2 P2 = 6 MPa & & since Q W = ke pe 0 . Then the exit temperature of steam becomes (h2 = h1 ) P2 = 6MPa o T2 = 490.1 C 586 Chapter 5 The First Law of Thermodynamics 599 Problems 598 is reconsidered. The effect of the exit pressure of steam on the exit temperature after throttling as the exit pressure varies from 6 MPa to 1 MPa is to be investigated. The exit temperature of steam is to be plotted against the exit pressure. "Input information from Diagram Window" {WorkingFluid$='Steam' "WorkingFluid: can be changed to ammonia or other fluids" P_in=8000 "[kPa]" T_in=500 "[C]" P_out=6000 "[C]"} $Warning off "Analysis" m_dot_in=m_dot_out "steadystate mass balance" m_dot_in=1 "mass flow rate is arbitrary" m_dot_in*h_in+Q_dotW_dotm_dot_out*h_out=0 "steadystate energy balance" Q_dot=0 "assume the throttle to operate adiabatically" W_dot=0 "throttles do not have any means of producing power" h_in=enthalpy(WorkingFluid$,T=T_in,P=P_in) "property table lookup" T_out=temperature(WorkingFluid$,P=P_out,h=h_out) "property table lookup" x_out=quality(WorkingFluid$,P=P_out,h=h_out) "x_out is the quality at the outlet" P[1]=P_in; P[2]=P_out; h[1]=h_in; h[2]=h_out "use arrays to place points on property plot" Pout [kPa] 1000 2000 3000 4000 5000 6000 Tout [C] 463 468.6 474.2 479.6 484.9 490 10 6 10 5 10 4 Steam 300C P [kPa] 10 3 10 2 10 1 10 0 0 100C 0.2 0.4 0.6 0.8 500 1000 1500 2000 2500 3000 3500 h [kJ/kg] 587 Chapter 5 The First Law of Thermodynamics Throttle exit T vs exit P for steam
495 490 485 T out [C] 480 475 470 465 460 1000 2000 3000 4000 5000 6000 P out [kPa] 588 Chapter 5 The First Law of Thermodynamics 5100E Highpressure air is throttled to atmospheric pressure. The temperature of air after the expansion is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. 5 Air is an ideal gas. & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
0 & & & Ein  Eout = Esystem (steady) P1 = 200 psia T1 = 90F =0 & & Ein = Eout & & mh1 = mh2 h1 = h2
Air & & since Q W = ke pe 0 . For an ideal gas, h = h(T). Therefore, T2 = T1 = 90F
P2 = 14.7 psia 589 Chapter 5 The First Law of Thermodynamics Mixing Chambers and Heat Exchangers 5101C Yes, if the mixing chamber is losing heat to the surrounding medium. 5102C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium. 5103C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium. 5104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is wellinsulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < Tsat @ 250 kPa = 127.44C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1 hf @ 80C = 334.91 kJ/kg h2 hf @ 20C = 83.96 kJ/kg h3 hf @ 42C = 175.92 kJ/kg Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance: Energy balance:
0 & & & min  mout = Esystem (steady) =0 & & & m1 + m2 = m 3
T1 = 80C m1 = 0.5 kg/s H2O (P = 250 kPa) T3 = 42C T2 = 20C m2 & & E in  E out 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & E system 144 44 2 3
0 (steady) Rate of change in internal, kinetic, potential, etc. energies =0 & & E in = E out & & & & & m1 h1 + m 2 h2 = m3 h3 (since Q = W = ke pe 0)
& Combining the two relations and solving for m2 gives & & & & m1h1 + m2h2 = (m1 + m2 )h3 h1  h3 & m1 h3  h2 Substituting, the mass flow rate of cold water stream is determined to be & m2 = & m2 = (334.91  175.92 )kJ/kg (0.5 kg/s ) = 0.864 (175.92  83.96 )kJ/kg kg/s 5105 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam is to be determined. 590 Chapter 5 The First Law of Thermodynamics Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties Noting that T < Tsat @ 300 kPa = 133.55C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1 hf @ 20C = 83.96 kJ/kg h3 hf @ 60C = 251.13 kJ/kg and P2 = 300kPa h2 = 3069.3kJ/kg o T2 = 300 C Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance: & & & min  mout = msystem 0 =
(steady) =0 =0 & & min = mout & & & m1 + m2 = m3 Energy balance: & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & m1h1 + m2 h2 = m3h3 Combining the two, & Solving for m2 : Substituting, & & (since Q W ke pe 0) T1 = 20C m1 = 1.8 kg/s H2O (P = 300 kPa) T3 = 60C T2 = 300C m2 & & & & m1h1 + m2h2 = (m1 + m2 )h3 & m2 = h1  h3 & m1 h3  h2 & m2 = 83.96  251.13 (1.8 kg/s ) = 0.107 kg/s 251.13  3069.3 591 Chapter 5 The First Law of Thermodynamics 5106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturated liquid, the ratio of the mass flow rates of the feedwater and the superheated vapor is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties Noting that T < Tsat @ 800 kPa = 170.43C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1 hf @ 50C = 209.33 kJ/kg h3 hf @ 800 kPa = 721.11 kJ/kg and P2 = 800kPa h2 = 2839.3kJ/kg o T2 = 200 C Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance: & & & min  mout = msystem 0 =
(steady) =0 =0 & & min = mout & & & m1 + m2 = m3
T1 = 50C m1 H2O (P = 800 kPa) Sat. liquid T2 = 200C m2 Energy balance: & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & m1h1 + m2 h2 = m3h3 Combining the two, & Dividing by m2 yields Solving for y: & & (since Q W ke pe 0) & & & & m1h1 + m2h2 = (m1 + m2 )h3 y h1 + h 2 = ( y + 1)h3 y= h3  h2 h1  h3 & & where y = m1 / m2 is the desired mass flow rate ratio. Substituting, y= 721.11  2839.3 = 4.14 209.33  721.11 592 Chapter 5 The First Law of Thermodynamics 5107E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From steam tables (Tables A5 through A6), h1 hf @ 50F = 18.06 Btu/lbm h2 = hg @ 50 psia = 1174.4 Btu/lbm Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steadyflow system can be expressed in the rate form as & & & Mass balance: min  mout = msystem 0 Energy balance: & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass (steady) & & & & & & & & & = 0 min = mout m1 + m2 = m3 = 2 m m1 = m2 = m =0 = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies T1 = 50F H2O (P = 50 psia) T3 , x3 Sat. vapor m2 = m1 & & Ein = Eout & & & m1h1 + m2 h2 = m3h3 Combining the two gives Substituting, h3 = (18.06 + 1174.4)/2 = 596.23 Btu/lbm & & (since Q W ke pe 0) & & & mh1 + mh2 = 2mh3 or h3 = (h1 + h2 ) / 2 At 50 psia, hf = 250.24 Btu/lbm and h g = 1174.4 Btu/lbm. Thus the exit stream is a saturated mixture since hf < h3 < hg. Therefore, T3 = Tsat @ 50 psia = 281.03F and x3 = h3  h f h fg = 596.23  250.24 = 0.374 924.2 593 Chapter 5 The First Law of Thermodynamics 5108 Two streams of refrigerant134a are mixed in a chamber. If the cold stream enters at twice the rate of the hot stream, the temperature and quality (if saturated) of the exit stream are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From R134a tables (Tables A11 through A13), h1 hf @ 12C = 66.18 kJ/kg h2 = h @ 1 MPa, 60C = 291.36 kJ/kg Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steadyflow system can be expressed in the rate form as & & & Mass balance: min  mout = msystem 0 Energy balance: & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass (steady) & & & & & & & & = 0 min = mout m1 + m2 = m3 = 3m2 since m1 = 2 m2 =0 = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies T1 = 12C m1 = 2m2 R134a (P = 1 MPa) T3 , x3 T2 = 60C & & Ein = Eout & & & m1h1 + m2 h2 = m3h3 Combining the two gives Substituting, h3 = (266.18 + 291.36)/3 = 141.24 kJ/kg At 1 MPa, hf = 105.29 kJ/kg and h g = 267.97 kJ/kg. Thus the exit stream is a saturated mixture since hf < h3 < hg. Therefore, T3 = Tsat @ 1 MPa = 39.39C and x3 = h3  h f h fg = 14124  105.29 . = 0.221 162.68 & & (since Q W ke pe 0) & & & 2m 2 h1 + m 2 h2 = 3m 2 h3 or h3 = (2h1 + h2 ) / 3 594 Chapter 5 The First Law of Thermodynamics 5109 Problem 5108 is reconsidered. The effect of the mass flow rate of the cold stream of R134a on the temperature and the quality of the exit stream as the ratio of the mass flow rate of the cold stream to that of the hot stream varies from 1 to 4 is to be investigated. The mixture temperature and quality are to be plotted against the coldtohot mass flow rate ratio. "Input Data" m_frac = 2 "m_frac =m_dot_cold/m_dot_hot= m_dot_1/m_dot_2" T[1]=12"[C]" P[1]=1000"[kPa]" T[2]=60"[C]" P[2]=1000"[kPa]" m_dot_1=m_frac*m_dot_2 P[3]=1000"[kPa]" m_dot_1=1 "Conservation of mass for the R134a: Sum of m_dot_in=m_dot_out" m_dot_1+ m_dot_2 =m_dot_3 "Conservation of Energy for steadyflow: neglect changes in KE and PE" "We assume no heat transfer and no work occur across the control surface." E_dot_in  E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "Steadyflow requirement" E_dot_in=m_dot_1*h[1] + m_dot_2*h[2] E_dot_out=m_dot_3*h[3] "Property data are given by:" h[1] =enthalpy(R134a,T=T[1],P=P[1]) "R134a data" h[2] =enthalpy(R134a,T=T[2],P=P[2]) T[3] =temperature(R134a,P=P[3],h=h[3]) x_3=QUALITY(R134a,h=h[3],P=P[3]) mfrac 1 1.333 1.667 2 2.333 2.667 3 3.333 3.667 4 T3 [C] 39.38 39.38 39.38 39.38 39.38 39.38 39.38 39.38 39.38 39.38 x3 0.4523 0.3538 0.28 0.2225 0.1766 0.139 0.1077 0.08117 0.05845 0.03876 595 Chapter 5 The First Law of Thermodynamics
0.5 0.4 0.3 x3 0.2 0.1 0 1 1.5 2 2.5 3 3.5 4 m frac 40 38 T[3] [C] 36 34 32 30 1 1.5 2 2.5 3 3.5 4 m frac 596 Chapter 5 The First Law of Thermodynamics 5110 Refrigerant134a is to be cooled by air in the condenser. For a specified volume flow rate of air, the mass flow rate of the refrigerant is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The constant pressure specific heat of air is Cp = 1.005 kJ/kgC (Table A2). The enthalpies of the R134a at the inlet and the exit states are (Tables A11 through A13) P3 = 1 MPa h3 = 313.20 kJ/kg T3 = 80 o C P4 = 1 MPa h4 h f @ 30oC = 91.49 kJ/kg o T4 = 30 C Analysis The inlet specific volume and the mass flow rate of air are RT 0.287 kPa m 3 /kg K (300K ) = 0.861m 3 /kg v1 = 1 = 100kPa P1 and & V 800m 3 /min & = 929.2kg/min m= 1 = v1 0.861m 3 /kg
AIR 1
R134a 3 4 ( ) 2 We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream):
0 & & & min  mout = msystem (steady) & & = 0 min = mout & & & & & & m1 = m2 = ma and m 3 = m4 = m R Energy balance (for the entire heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, & Solving for mR : Substituting, & & m a (h2  h1 ) = m R (h3  h 4 ) & mR = C p (T2  T1 ) h2  h1 & & ma ma h3  h4 h3  h4 & mR = (1.005 kJ/kg o C)(60  27) o C (929.2 kg/min ) = 139.0 kg/min (313.20  91.49) kJ/kg 597 Chapter 5 The First Law of Thermodynamics 5111E Refrigerant134a is vaporized by air in the evaporator of an airconditioner. For specified flow rates, the exit temperature of the air and the rate of heat transfer from the air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A1E). The constant pressure specific heat of air is Cp = 0.240 Btu/lbmF (Table A2E). The enthalpies of the R134a at the inlet and the exit states are (Tables A11E through A13E) P3 = 20 psia h3 = h f + x 3 h fg = 10.89 + 0.3 90.50 = 38.04 Btu/lbm x 3 = 0.3 P4 = 20 psia h 4 = h g @ 20 psia = 101.39 Btu/lbm sat.vapor Analysis The inlet specific volume and the mass flow rate of air are v1 = and 0.3704 psia ft 3 /lbm R (550R ) RT1 = = 13.86 ft 3 /lbm 14.7 psia P1 & 200 ft 3 /min V & m= 1 = = 14.43l bm/min v1 13.86 ft 3 /lbm
R134a AIR 1 3 4 ( ) 2 We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream):
0 & & & min  mout = msystem (steady) & & = 0 min = mout & & & & & & m1 = m2 = ma and m 3 = m4 = m R Energy balance (for the entire heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, Solving for T2 : Substituting, & & & mR (h3  h4 ) = ma (h2  h1 ) = maC p (T2  T1 ) T2 = T1 + & m R (h3  h4 ) & ma C p T2 = 90 o F + (4 lbm/min )(38.04  101.39 )Btu/lbm = 16.8 o F (14.43 Btu/min )(0.24 Btu/lbmo F) (b) The rate of heat transfer from the air to the refrigerant is determined from the steadyflow energy balance applied to the air only. It yields & & &  Q air ,out = m a (h 2  h1 ) = m a C p (T2  T1 ) & Q air ,out = (14.43lbm/min )(0.24 Btu/lbm o F )(16.8  90) o F = 253.5 Btu/min 598 Chapter 5 The First Law of Thermodynamics 5112 Refrigerant134a is condensed in a watercooled condenser. The mass flow rate of the cooling water required is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The enthalpies of R134a at the inlet and the exit states are (Tables A5 and A6) P3 = 800kPa h3 = 305.50kJ/kg T3 = 70 o C P4 = 800kPa h4 = h f @800kPa = 93.42kJ/kg sat.liquid Water exists as compressed liquid at both states, and thus h1 hf @ 15C = 62.99 kJ/kg h2 hf @ 30C = 125.79 kJ/kg 2
Water 1
R134a 3 4 Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance (for each fluid stream):
0 & & & min  mout = msystem (steady) & & & & & = 0 min = mout m1 = m2 = mw and & & & m3 = m4 = m R Energy balance (for the heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, & Solving for mw : Substituting, & mw = (305.50  93.42) kJ / kg (8 kg / min) = 27.0 kg / min (125.79  62.99) kJ / kg & & m w (h2  h1 ) = m R (h3  h4 ) & mw = h3  h4 & mR h2  h1 599 Chapter 5 The First Law of Thermodynamics 5113E [Also solved by EES on enclosed CD] Air is heated in a steam heating system. For specified flow rates, the volume flow rate of air at the inlet is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A1E). The constant pressure specific heat of air is Cp = 0.240 Btu/lbmF (Table A2E). The enthalpies of steam at the inlet and the exit states are (Tables A4E through A6E) AIR P3 = 30 psia h3 = 1237.8Btu/lbm T3 = 400 o F P4 = 25psia h 4 h f @ 212o F = 180.16 Btu/lbm o T4 = 212 F 1
Steam 3 4 2 Analysis We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream):
0 (steady) & & & & & & & & & & & min  mout = msystem = 0 min = mout m1 = m 2 = ma and m3 = m4 = m s Energy balance (for the entire heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, & Solving for ma : Substituting, & ma = Also, v1 = (1237.8  180.16 ) Btu / lbm (0.240 Btu / lbmo F )(130  80)o F (15 lbm / min ) = 1322 lbm / min = 22.03 lbm / s & & m a (h 2  h1 ) = m s (h3  h4 ) & ma = h3  h4 h3  h4 & & ms ms C p (T2  T1 ) h2  h1 RT1 (0.3704 psia ft 3 / lbm R )(540 R ) = = 13.61 ft 3 / lbm P 14.7 psia 1 Then the volume flow rate of air at the inlet becomes & & V1 = ma v1 = (22.03 lbm / s)(13.61 ft 3 / lbm) = 299.8 ft 3 / s 5100 Chapter 5 The First Law of Thermodynamics 5114 Steam is condensed by cooling water in the condenser of a power plant. If the temperature rise of the cooling water is not to exceed 10C, the minimum mass flow rate of the cooling water required is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Liquid water is an incompressible substance with constant specific heats at room temperature. Properties The cooling water exists as compressed liquid at both states, and its specific heat at room temperature is C = 4.18 kJ/kgC (Table A3). The enthalpies of the steam at the inlet and the exit states are (Tables A5 and A6) P3 = 20kPa h3 = h f + x3 h fg = 251.40 + 0.95 2358.3 = 2491.8kJ/kg x 3 = 0.95 P4 = 20kPa h4 h f @ 20kPa = 251.40kJ/kg sat.liquid Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance (for each fluid stream): & & & min  mout = msystem 0
(steady) & & & & & = 0 min = mout m1 = m2 = mw and & & & m3 = m4 = ms Energy balance (for the heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 Steam 20 kPa & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, & Solving for mw : Substituting, & mw = (2491.8  251.4 ) kJ/kg (4.18kJ/kg o C)(10 o C) (20,000/3600 kg/s ) = 298 kg/s = 17,866 kg/min & & m w (h2  h1 ) = m s (h3  h4 ) & mw = h3  h4 h3  h4 & & ms ms C p (T2  T1 ) h2  h1
Water 5101 Chapter 5 The First Law of Thermodynamics 5115 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The heat of vaporization of water at 50C is hfg = 2382.7 kJ/kg and specific heat of cold water is Cp = 4.18 kJ/kg.C (Tables A3 and A4). Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass Steam 50C 27C = 0 (steady) & Esystem 1442443 4 4 Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) Qin & p 2 1 Then the heat transfer rate to the cooling water in the condenser becomes
50C 18C Water & & Q = [mC p (Tout  Tin )] cooling water = (101 kg/s)(4.18 kJ/kg.C)(27C  18C) = 3800 kJ/s
The rate of condensation of steam is determined to be & Q 3800 kJ/s & & Q = (mh fg ) steam msteam = & = = 1.59 kg/s h fg 2382.7 kJ/kg 5102 Chapter 5 The First Law of Thermodynamics 5116 Problem 5115 is reconsidered. The effect of the inlet temperature of cooling water on the rate of condensation of steam as the inlet temperature varies from 10C to 20C at constant exit temperature is to be investigated. The rate of condensation of steam is to be plotted against the inlet temperature of the cooling water. "Input Data" T_s[1]=50"[C]" T_s[2]=50"[C]" m_dot_water=101 "[kg/s]" T_water[1]=18"[C]" T_water[2]=27"[C]" C_P_water = 4.20 "[kJ/kgC]" "Conservation of mass for the steam: m_dot_s_in=m_dot_s_out=m_dot_s" "Conservation of mass for the water: m_dot_water_in=lm_dot_water_out=m_dot_water" "Conservation of Energy for steadyflow: neglect changes in KE and PE" "We assume no heat transfer and no work occur across the control surface." E_dot_in  E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "[kW]" "Steadyflow requirement" E_dot_in=m_dot_s*h_s[1] + m_dot_water*h_water[1] "[kW]" E_dot_out=m_dot_s*h_s[2] + m_dot_water*h_water[2] "[kW]" "Property data are given by:" h_s[1] =enthalpy(steam,T=T_s[1],x=1)"[kJ/kg]" "steam data" h_s[2] =enthalpy(steam,T=T_s[2],x=0)"[kJ/kg]" h_water[1] =C_P_water*T_water[1]"[kJ/kg]" "water data" h_water[2] =C_P_water*T_water[2]"[kJ/kg]" h_fg_s=h_s[1]h_s[2] "kJ/kg]" "h_fg is found from the EES functions rather than using h_fg = 2305 kJ/kg" ms [kg/s] 3.028 2.671 2.315 1.959 1.603 1.247 Twater,1 [C] 10 12 14 16 18 20 5103 Chapter 5 The First Law of Thermodynamics
3.5 3 m s [kg/s] 2.5 2 1.5 1 10 12 14 16 18 20 T w ater[1] [C] 5104 Chapter 5 The First Law of Thermodynamics 5117 Water is heated in a heat exchanger by geothermal water. The rate of heat transfer to the water and the exit temperature of the geothermal water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and geothermal fluid are given to be 4.18 and 4.31 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Brine 140C 60C & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the rate of heat transfer to the cold water in the heat exchanger becomes & & Q = [ mC p (Tout  Tin )]water = (0.2 kg / s)(4.18 kJ / kg. C)(60 C  25 C) = 29.26 kW Water 25C Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water, the outlet temperature of the geothermal water is determined from & Q & & Q = [ mC p (Tin  Tout )]geot.water Tout = Tin  & mC p = 140 C  29.26 kW = 117.4 C (0.3 kg / s)(4.31 kJ / kg. C) 5105 Chapter 5 The First Law of Thermodynamics 5118 Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer in the heat exchanger and the mass flow rate of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 kJ/kg.C, respectively. Analysis (a) We take the ethylene glycol tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Hot Glycol 80C 2 kg/s Cold Water 20C & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mC (T  T ) Qout & p 1 2 Then the rate of heat transfer becomes 40C & & Q = [mC p (Tin  Tout )]glycol = (2 kg / s)(2.56 kJ / kg. C)(80 C  40 C) = 204.8 kW (b) The rate of heat transfer from water must be equal to the rate of heat transfer to the glycol. Then, & & & Q = [mC p (Tout  Tin )] water m water = = & Q C p (Tout  Tin ) 204.8 kJ/s = 1.4 kg/s (4.18 kJ/kg.C)(55C  20C) 5106 Chapter 5 The First Law of Thermodynamics 5119 Problem 5118 is reconsidered. The effect of the inlet temperature of cooling water on the mass flow rate of water as the inlet temperature varies from 10C to 40C at constant exit temperature) is to be investigated. The mass flow rate of water is to be plotted against the inlet temperature. "Input Data" {T_w[1]=20"[C]"} T_w[2]=55"[C]" "w: water" m_dot_eg=2"[kg/s]" "eg: ethylene glycol" T_eg[1]=80"[C]" T_eg[2]=40"[C]" C_p_w=4.18"[kJ/kgK]" C_p_eg=2.56"[kJ/kgK]" "Conservation of mass for the water: m_dot_w_in=m_dot_w_out=m_dot_w" "Conservation of mass for the ethylene glycol: m_dot_eg_in=m_dot_eg_out=m_dot_eg" "Conservation of Energy for steadyflow: neglect changes in KE and PE in each mass steam" "We assume no heat transfer and no work occur across the control surface." E_dot_in  E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 "[kW]" "Steadyflow requirement" E_dot_in=m_dot_w*h_w[1] + m_dot_eg*h_eg[1] "[kW]" E_dot_out=m_dot_w*h_w[2] + m_dot_eg*h_eg[2] "[kW]" Q_exchanged =m_dot_eg*h_eg[1]  m_dot_eg*h_eg[2] "[kW]" "Property data are given by:" h_w[1] =C_p_w*T_w[1] " liquid approximation applied for water and ethylene glycol" h_w[2] =C_p_w*T_w[2] h_eg[1] =C_p_eg*T_eg[1] h_eg[2] =C_p_eg*T_eg[2] mw [kg/s] 1.089 1.225 1.4 1.633 1.96 2.45 3.266 Tw,1 [C] 10 15 20 25 30 35 40 5107 Chapter 5 The First Law of Thermodynamics 3.5 3 m w [kgS] 2.5 2 1.5 1 10 15 20 25 30 35 40 T w [1] [C] 5108 Chapter 5 The First Law of Thermodynamics 5120 Oil is to be cooled by water in a thinwalled heat exchanger. The rate of heat transfer in the heat exchanger and the exit temperature of water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4.18 and 2.20 kJ/kg.C, respectively. Analysis We take the oil tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Hot oil 150C 2 kg/s Cold water 22C 1.5 kg/s 40C & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mC (T  T ) Qout & p 1 2 Then the rate of heat transfer from the oil becomes & & Q = [ mC p (Tin  Tout )]oil = (2 kg / s)(2.2 kJ / kg. C)(150 C  40 C) = 484 kW Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is determined from & & Q = [mC p (Tout  Tin )] water Tout = Tin + & Q & m water C p = 22C + 484 kJ/s = 99.2C (1.5 kg/s)(4.18 kJ/kg.C) 5109 Chapter 5 The First Law of Thermodynamics 5121 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the exit temperature of hot water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Hot water 100C 3 kg/s Cold Water 15C 0.60 kg/s & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) Qin & p 2 1 Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [mC p (Tout  Tin )]cold water = (0.60 kg/s)(4.18 kJ/kg.C)(45C  15C) = 75.24 kW
Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is determined to be & Q & & Q = [mC p (Tin  Tout )] hot water Tout = Tin  & mC p = 100C  75.24 kW = 94.0C (3 kg/s)(4.19 kJ/kg.C) 5110 Chapter 5 The First Law of Thermodynamics 5122 Air is preheated by hot exhaust gases in a crossflow heat exchanger. The rate of heat transfer and the outlet temperature of the air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.C, respectively. Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Air 95 kPa 20C 0.8 m3/s & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mC (T  T ) Qout & p 1 2 Then the rate of heat transfer from the exhaust gases becomes & & Q = [mC p (Tin  Tout )]gas. = (11 kg / s)(1.1 kJ / kg. C)(180 C  95 C) = 102.85 kW . The mass flow rate of air is & m= & (95 kPa)(0.8 m3 / s) PV = = 0.904 kg / s RT (0.287 kPa.m 3 / kg.K) 293 K Exhaust gases 1.1 kg/s, 95C Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes & Q & & Q = mC p (Tc,out  Tc,in ) Tc,out = Tc,in + & mC p = 20 C + 102.85 kW = 133.2 C (0.904 kg / s)(1.005 kJ / kg. C) 5111 Chapter 5 The First Law of Thermodynamics 5123 Water is heated by hot oil in a heat exchanger. The rate of heat transfer in the heat exchanger and the outlet temperature of oil are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in Oil the rate form as 170C & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
70C Water 20C 4.5 kg/s 10 kg/s & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) Qin & p 2 1 Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [ mC p (Tout  Tin )]water = (4.5 kg / s)(4.18 kJ / kg. C)(70 C  20 C) = 940.5 kW Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined from & Q 940.5 kW & & Q = [mC p (Tin  Tout )] oil Tout = Tin  = 170C  = 129.1C & mC p (10 kg/s)(2.3 kJ/kg.C) 5112 Chapter 5 The First Law of Thermodynamics 5124E Steam is condensed by cooling water in a condenser. The rate of heat transfer in the heat exchanger and the rate of condensation of steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heat of water is 1.0 Btu/lbm.F. The enthalpy of vaporization of water at 90F is 1042.7 Btu/lbm (Table A4E). Analysis We take the tubeside of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass Steam 90F 73F = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [ mC p (Tout  Tin )]water = (115.3 lbm / s)(1.0 Btu / lbm. F)(73 F  60 F) = 1499 Btu / s
90F 60F Water Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the steam in the heat exchanger is determined from & Q 1499 Btu/s & & & Q = (mh fg ) steam = m steam = = = 1.44 lbm/s h fg 1042.7 Btu/lbm 5113 Chapter 5 The First Law of Thermodynamics Pipe and duct Flow 5125 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The air flow rate of the fan and the diameter of the casing are to be determined. Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The specific heat of air at the average temperature of Tave = (45+60)/2 = 52.5C = 325.5 K is Cp = 1.0065 kJ/kg.C (Table A2b) Analysis The fan selected must be able to meet the cooling requirements of the computer at worst conditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45 C, and leave at 60C. We take the air space in the computer as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) Qin & p 2 1 Then the required mass flow rate of air to absorb heat at a rate of 60 W is determined to be & & & Q = mC p (Tout  Tin ) m = & Q C p (Tout  Tin ) = 60 W (1006.5 J/kg.C)(60  45)C = 0.00397 kg/s = 0.238 kg/min The density of air entering the fan at the exit and its volume flow rate are P 66.63 kPa = = 0.6972 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(60 + 273)K & & m 0.238 kg/min = 0.341 m 3 /min V= = 0.6972 kg/m 3 = For an average exit velocity of 110 m/min, the diameter of the casing of the fan is determined from & D 2 4V & V = Ac V = VD= = 4 V (4 )(0.341 m 3 /min) = 0.063 m = 6.3 cm (110 m/min) 5114 Chapter 5 The First Law of Thermodynamics 5126 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The air flow rate of the fan and the diameter of the casing are to be determined. Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The specific heat of air at the average temperature of Tave = (45+60)/2 = 52.5C is Cp = 1.0065 kJ/kg.C (Table A3). Analysis The fan selected must be able to meet the cooling requirements of the computer at worst conditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45 C, and leave at 60C. We take the air space in the computer as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the required mass flow rate of air to absorb heat at a rate of 100 W is determined to be & & & Q = mC p (Tout  Tin ) m = & Q C p (Tout  Tin ) = 100 W (1006.5 J/kg.C)(60  45)C = 0.006624 kg/s = 0.397 kg/min
The density of air entering the fan at the exit and its volume flow rate are P 66.63 kPa = = 0.6972 kg/m 3 3 RT (0.287 kPa.m /kg.K)(60 + 273)K & & m 0.397 kg/min = 0.57 m 3 /min V= = 0.6972 kg/m 3 =
For an average exit velocity of 110 m/min, the diameter of the casing of the fan is determined from & (4)(0.57 m 3 /min) D 2 4V & V = Ac V = V D= = = 0.081 m = 8.1 cm 4 V (110 m/min) 5115 Chapter 5 The First Law of Thermodynamics 5127E Electronic devices mounted on a cold plate are cooled by water. The amount of heat generated by the electronic devices is to be determined. Assumptions 1 Steady operating conditions exist. 2 About 15 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation. 3 Kinetic and potential energy changes are negligible. Properties The properties of water at room temperature are = 62.1 lbm/ft3 and Cp = 1.00 Btu/lbm.F (Table A3). Analysis We take the tubes of the cold plate to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Cold plate Water inlet 1 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then mass flow rate of water and the rate of heat removal by the water are determined to be
2 (0.25 / 12 ft) 2 D 2 V = (62.1 lbm/ft 3 ) (60 ft/min) = 1.270 lbm/min = 76.2 lbm/h 4 4 & & Q = mC p (Tout  Tin ) = (76.2 lbm/h)(1.00 Btu/lbm.F)(105  95)F = 762 Btu/h & m = AV = which is 85 percent of the heat generated by the electronic devices. Then the total amount of heat generated by the electronic devices becomes & 762 Btu / h = 896 Btu / h = 263 W Q= 0.85 5116 Chapter 5 The First Law of Thermodynamics 5128 A sealed electronic box is to be cooled by tap water flowing through channels on two of its sides. The mass flow rate of water and the amount of water used per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 Entire heat generated is dissipated by water. 3 Water is an incompressible substance with constant specific heats at room temperature. 4 Kinetic and potential energy changes are negligible. Properties The specific heat of water at room temperature is Cp = 4.18 kJ/kg.C (Table A3). Analysis We take the water channels on the sides to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Water 1nlet 1 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the mass flow rate of tap water flowing through the electronic box becomes & & Q = mC p T m = & & Q 2 kJ / s = = 0.1196 kg / s C p T (4.18 kJ / kg. C)(4 C)
Electronic box 2 kW Water exit Therefore, 0.11962 kg of water is needed per second to cool this electronic box. Then the2 amount of cooling water used per year becomes & m = mt = (0.1196 kg/s)(365 days/yr 24 h/day 3600 s/h) = 3,772,000 kg/yr = 3,772 tons/yr 5117 Chapter 5 The First Law of Thermodynamics 5129 A sealed electronic box is to be cooled by tap water flowing through channels on two of its sides. The mass flow rate of water and the amount of water used per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 Entire heat generated is dissipated by water. 3 Water is an incompressible substance with constant specific heats at room temperature. 4 Kinetic and potential energy changes are negligible Properties The specific heat of water at room temperature is Cp = 4.18 kJ/kg.C (Table A3). Analysis We take the water channels on the sides to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Water 1nlet 1 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the mass flow rate of tap water flowing through the electronic box becomes & & Q = mC p T m = & & Q 3 kJ / s = = 0.1794 kg / s C p T (4.18 kJ / kg. C)(4 C) Electronic box 3 kW Therefore, 0.1794 kg of water is needed per second to cool this electronic box. Then the amount of cooling water used per year becomes 2 & m = mt = (0.1794 kg/s)(365 days/yr 24 h/day 3600 s/h) = 5,658,400 kg/yr = 5,658 tons/yr Water exit 5118 Chapter 5 The First Law of Thermodynamics 5130 A long roll of large 1Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The rate at which heat needs to be removed from the oil to keep its temperature constant is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the roll are constant. 3 Kinetic and potential energy changes are negligible Properties The properties of the steel plate are given to be = 7854 kg/m3 and Cp = 0.454 kJ/kg.C. Analysis The mass flow rate of the sheet metal through the oil bath is & & m = V = wtV = (7854 kg/m 3 )(2 m)(0.005 m)(10 m/min) = 785.4 kg/min
We take the volume occupied by the sheet metal in the oil bath to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed I n the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Steel plate 10 m/min Oil bath 45C & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mC (T  T ) Qout & p 1 2 Then the rate of heat transfer from the sheet metal to the oil bath becomes & & Qout = mC p [Tin  Tout ] metal = (785.4 kg/min )(0.434 kJ/kg.C)(820  51.1)C = 262,090 kJ/min = 4368 kW
This is the rate of heat transfer from the metal sheet to the oil, which is equal to the rate of heat removal from the oil since the oil temperature is maintained constant. 5119 Chapter 5 The First Law of Thermodynamics 5131 Problem 5130 is reconsidered. The effect of the moving velocity of the steel plate on the rate of heat transfer from the oil bath as the velocity varies from 5 to 50 m/min is to be investigated. Tate of heat transfer is to be plotted against the plate velocity. "Knowns" Vel = 10 "[m/min]" T_bath = 45"[C]" T_1 = 820"[C]" T_2 = 51.1 "[C]" rho = 785 "[kg/m^3]" C_P = 0.454 "[kJ/kgC]" width = 2"[m]" thick = 0.5"[cm]" "Analysis: The mass flow rate of the sheet metal through the oil bath is:" Vol_dot = width*thick*convert(cm,m)*Vel/convert(min,s)"[m^3/s]" m_dot = rho*Vol_dot"[kg/s]" "We take the volume occupied by the sheet metal in the oil bath to be the system, which is a control volume. The energy balance for this steadyflow systemthe metal can be expressed in the rate form as:" E_dot_metal_in = E_dot_metal_out E_dot_metal_in=m_dot*h_1 E_dot_metal_out=m_dot*h_2+Q_dot_metal_out h_1 = C_P*T_1"[kJ/kg]" h_2 = C_P*T_2"[kJ/kg]" Q_dot_oil_out = Q_dot_metal_out"[KW]" Qoilout [kW] 228.4 456.7 685.1 913.4 1142 1370 1598 1827 2055 2284 Vel [m/min] 5 10 15 20 25 30 35 40 45 50
2500 2000 Q oil,out [KW ] 1500 1000 500 0 5 10 15 20 25 30 35 40 45 50 Vel [m/min] 5120 Chapter 5 The First Law of Thermodynamics 5132 [Also solved by EES on enclosed CD] The components of an electronic device located in a horizontal duct of rectangular cross section are cooled by forced air. The heat transfer from the outer surfaces of the duct is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible Properties The gas constant of air is R = 0.287 kJ/kg.C (Table A1). The specific heat of air at room temperature is Cp = 1.005 kJ/kg.C (Table A2). Analysis The density of air entering the duct and the mass flow rate are P 101.325 kPa = = 1165 kg / m 3 . RT (0.287 kPa.m 3 / kg.K)(30 + 273)K & & m = V = (1165 kg / m 3 )(0.6 m 3 / min) = 0. 700 kg / min . = We take the channel, excluding the electronic components, to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass 40C = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) Qin & p 2 1 Air 30C 0.6 m3/min 180 W 25C Then the rate of heat transfer to the air passing through the duct becomes & & Qair = [ mC p (Tout  Tin )]air = (0.700 / 60 kg / s)(1.005 kJ / kg. C)( 40  30) C = 0117 kW = 117 W . The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by natural convection and radiation, & & & Q =Q Q = 180  117 = 63 W
external total internal 5121 Chapter 5 The First Law of Thermodynamics 5133 The components of an electronic device located in a horizontal duct of circular cross section is cooled by forced air. The heat transfer from the outer surfaces of the duct is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible Properties The gas constant of air is R = 0.287 kJ/kg.C (Table A1). The specific heat of air at room temperature is Cp = 1.005 kJ/kg.C (Table A2). Analysis The density of air entering the duct and the mass flow rate are P 101.325 kPa = = 1165 kg / m 3 . RT (0.287 kPa.m 3 / kg.K)(30 + 273)K & & m = V = (1165 kg / m 3 )(0.6 m 3 / min) = 0. 700 kg / min . = We take the channel, excluding the electronic components, to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 1 Air 30C 0.6 m3/s 2 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the rate of heat transfer to the air passing through the duct becomes & & Qair = [ mC p (Tout  Tin )]air = (0.700 / 60 kg / s)(1.005 kJ / kg. C)( 40  30) C = 0117 kW = 117 W . The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by natural convection and radiation, & & & Q =Q Q = 180  117 = 63 W
external total internal 5122 Chapter 5 The First Law of Thermodynamics 5134E Water is heated in a parabolic solar collector. The required length of parabolic collector is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat loss from the tube is negligible so that the entire solar energy incident on the tube is transferred to the water. 3 Kinetic and potential energy changes are negligible Properties The specific heat of water at room temperature is Cp = 1.00 Btu/lbm.F (Table A2E). Analysis We take the thin aluminum tube to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
1 Water 55F 4 lbm/s 2 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mwater C p (T2  T1 ) 200F Then the total rate of heat transfer to the water flowing through the tube becomes & & Q total = mC p (Te  Ti ) = (4 lbm/s)(1.00 Btu/lbm.F)(200  55)F = 580 Btu/s = 2,088,000 Btu/h The length of the tube required is L= & Q total 2,088,000 Btu/h = = 5966 ft & 350 Btu/h.ft Q 5123 Chapter 5 The First Law of Thermodynamics 5135 Air enters a hollowcore printed circuit board. The exit temperature of the air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 The local atmospheric pressure is 1 atm. 4 Kinetic and potential energy changes are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.C (Table A1). The specific heat of air at room temperature is Cp = 1.005 kJ/kg.C (Table A2). Analysis The density of air entering the duct and the mass flow rate are P 101.325 kPa = = 116 kg / m 3 . RT (0.287 kPa.m 3 / kg.K)(32 + 273)K & & m = V = (116 kg / m 3 )(0.0008 m 3 / s) = 0.000928 kg / s . = We take the hollow core to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
1 Air 32C 0.8 L/s 2 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the exit temperature of air leaving the hollow core becomes & Q 20 J/s & & Qin = mC p (T2  T1 ) T2 = T1 + in = 32 C + = 53.4C & mC p (0.000928 kg/s)(1005 J/kg.C) 5124 Chapter 5 The First Law of Thermodynamics 5136 A computer is cooled by a fan blowing air through the case of the computer. The required flow rate of the air and the fraction of the temperature rise of air that is due to heat generated by the fan are to be determined. Assumptions 1 Steady flow conditions exist. 2 Air is an ideal gas with constant specific heats. 3 The pressure of air is 1 atm. 4 Kinetic and potential energy changes are negligible Properties The specific heat of air at room temperature is C p = 1.005 kJ/kg.C (Table A2). Analysis (a) We take the air space in the computer as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & & Qin + Win + mh1 = mh2 (since ke pe 0) & & & Qin + Win = mC p (T2  T1 )
Noting that the fan power is 25 W and the 8 PCBs transfer a total of 80 W of heat to air, the mass flow rate of air is determined to be & & Q + Win 25 + (8 10) W & & & & = = 0.0104 kg / s Qin + Win = mC p (Te  Ti ) m = in C p (Te  Ti ) (1005 J / kg. C)(10 C) (b) The fraction of temperature rise of air that is due to the heat generated by the fan and its motor can be determined from & Q 25 W & & = = 2.4C Q = mC p T T = & mC p (0.0104 kg/s)(1005 J/kg.C) f = 2.4C = 0.24 = 24% 10C 5125 Chapter 5 The First Law of Thermodynamics 5137 Hot water enters a pipe whose outer surface is exposed to cold air in a basement. The rate of heat loss from the water is to be determined. Assumptions 1 Steady flow conditions exist. 2 Water is an incompressible substance with constant specific heats. 3 The changes in kinetic and potential energies are negligible. Properties The properties of water at the average temperature of (90+88)/2 = 89C are = 965 kg/m 3 and Cp = 4.21 kJ/kg.C (Table A3). Analysis The mass flow rate of water is & m = AcV = (965 kg/m 3 ) (0.04 m) 2 (0.8 m/s) = 0.970 kg/s 4 We take the section of the pipe in the basement to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
1 Water 90C 0.8 m/s & Q
2 & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = mC (T  T ) Qout & p 1 2 88C Then the rate of heat transfer from the hot water to the surrounding air becomes & & Qout = mC p [Tin  Tout ]water = (0.970 kg / s)( 4.21 kJ / kg. C)(90  88) C = 8.17 kW 5126 Chapter 5 The First Law of Thermodynamics 5138 Problem 5137 is reconsidered. The effect of the inner pipe diameter on the rate of heat loss as the pipe diameter varies from 1.5 cm to 7.5 cm is to be investigated. The rate of heat loss is to be plotted against the diameter. "Knowns:" {D = 0.04 "[m]"} rho = 965 "[kg/m^3]" Vel = 0.8"[m/s]" T_1 = 90"[C]" T_2 = 88"[C]" C_P = 4.21"[kJ/kgC]" "Analysis:" "The mass flow rate of water is:" Area = pi*D^2/4"[m^2]" m_dot = rho*Area*Vel"[kg/s]" "We take the section of the pipe in the basement to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as" E_dot_in  E_dot_out = DELTAE_dot_sys DELTAE_dot_sys = 0"[kW]" "Steadyflow assumption" E_dot_in = m_dot*h_in"[kW]" E_dot_out =Q_dot_out+m_dot*h_out"[kW]" h_in = C_P * T_1"[kJ/kg]" h_out = C_P * T_2"[kJ/kg]" D [m] 0.015 0.025 0.035 0.045 0.055 0.065 0.075
30 25 20 Qout [kW] 1.149 3.191 6.254 10.34 15.44 21.57 28.72 Q out [kW ] 15 10 5 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 D [m ] 5127 Chapter 5 The First Law of Thermodynamics 5139 A room is to be heated by an electric resistance heater placed in a duct in the room. The power rating of the electric heater and the temperature rise of air as it passes through the heater are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is negligible. 5 No air leaks in and out of the room. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The specific heats of air at room temperature are Cp = 1.005 and Cv = 0.718 kJ/kgK (Table A2). Analysis (a) The total mass of air in the room is V = 5 6 8m 3 = 240m 3
. PV (98 kPa )(240 m 3 ) = 284.6 kg m= 1 = 3 RT1 (0.287 kPa m /kg K )(288 K ) We first take the entire room as our system, which is a closed system since no mass leaks in or out. The power rating of the electric heater is determined by applying the conservation of energy relation to this constant volume closed system: E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass 200 kJ/min 568 m3 We
200 W = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies & & & t We,in + W fan ,in  Qout = mC v, ave (T2  T1 ) Solving for the electrical work input gives ( We,in + W fan,in  Qout = U ) (since KE = PE = 0) & & & We ,in = Qout  W fan ,in + mC v (T2  T1 ) / t = (200/60 kJ/s)  (0.2 kJ/s) + (284.6 kg )(0.718kJ/kg o C)(25  15) o C/(15 60s) = 5.40 kW
(b) We now take the heating duct as the system, which is a control volume since mass crosses the boundary. & & & There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this adiabatic steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & & & We,in + W fan,in + mh1 = mh2 (since Q = ke pe 0) & & & & We,in + W fan,in = m(h2  h1 ) = mC p (T2  T1 ) Thus, T = T2  T1 = & & We,in + W fan ,in & mC p = (5.40 + 0.2)kJ/s = 6.7 o C (50/60kg/s )(1.005kJ/kg K ) 5128 Chapter 5 The First Law of Thermodynamics 5140 A house is heated by an electric resistance heater placed in a duct. The power rating of the electric heater is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kgK (Table A2) Analysis We take the heating duct as the system, which is a control volume since mass crosses the & & & boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
400 W & & We,in + W fan,in & & Ein = Eout & & & + mh1 = Qout + mh2 (since ke pe 0) & & = Qout + mC p (T2  T1 ) We
300 W & & & & We,in + W fan,in = Qout + m(h2  h1 ) Substituting, the power rating of the heating element is determined to be & & & & We ,in = Qout + mC p T  W fan ,in = (0.4 kJ/s) + (0.6 kg/s) (1.005 kJ/kg o C)(5 o C)  0.3 kW = 3.12 kW 5129 Chapter 5 The First Law of Thermodynamics 5141 A hair dryer consumes 1200 W of electric power when running. The inlet volume flow rate and the exit velocity of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The power consumed by the fan and the heat losses are negligible. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kgK (Table A2) Analysis We take the hair dryer as the system, which is a control volume since mass crosses the boundary. & & & There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
T2 = 47C A2 = 60 cm2 P1 = 100 kPa T1 = 22C & & Ein = Eout & & & & We,in + mh1 = mh2 (since Qout ke pe 0) & & & We,in = m(h2  h1 ) = mC p (T2  T1 ) Substituting, the mass and volume flow rates of air are determined to be & m= v1 = C p (T2  T1 ) RT1 P1 & We,in
We = 1200 W (100 kPa ) & & V1 = mv1 = (0.04776 kg/s )(0.8467 m 3 /kg ) = 0.0404 m 3 /s (1.005kJ/kg C)(47  22) C = 0.04776kg/s (0.287 kPa m /kg K )(295 K ) = 0.8467 m /kg =
o o 3 3 = 1.2 kJ/s & & & (b) The exit velocity of air is determined from the mass balance m1 = m2 = m to be v2 = & m= RT2 (0.287 kPa m 3 /kg K )(320K ) = = 0.9184 m 3 /kg (100 kPa) P2 & mv 2 (0.04776 kg/s)(0.9184 m 3 /kg ) 1 V2 = = = 7.31 m/s A2 V2 v2 A2 60 10  4 m 2 5130 Chapter 5 The First Law of Thermodynamics 5142 Problem 5141 is reconsidered. The effect of the exit crosssectional area of the hair drier on the exit velocity as the exit area varies from 25 cm^2 to 75 cm^2 is to be investigated. The exit velocity is to be plotted against the exit crosssectional area, "Knowns:" R=0.287 "[kPam^3/kgK]" P= 100"[kPa]" T_1 = 22"[C]" T_2= 47"[C]" {A_2 = 60"[cm^2]"} A_1 = 53.35"[cm^2]" W_dot_ele=1200"[W]" "Analysis: We take the hair dryer as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit. Thus, the energy balance for this steadyflow system can be expressed in the rate form as:" E_dot_in = E_dot_out E_dot_in = W_dot_ele*convert(W,kW) + m_dot_1*h_1+Vel_1^2/2*convert(m^2/s^2,kJ/kg)"[kW]" E_dot_out = m_dot_2*h_2+Vel_2^2/2*convert(m^2/s^2,kJ/kg)"[kW]" h_2 = enthalpy(air, T = T_2)"[kJ/kg]" h_1= enthalpy(air, T = T_1)"[kJ/kg]" "The volume flow rates of air are determined to be:" V_dot_1 = m_dot_1*v_1"[m^3/s]" P*v_1=R*(T_1+273) V_dot_2 = m_dot_2*v_2"[m^3/s]" P*v_2=R*(T_2+273)"[m^3/s]" m_dot_1 = m_dot_2 Vel_1=V_dot_1/(A_1*convert(cm^2,m^2))"[m/s]" "(b) The exit velocity of air is determined from the mass balance to be" Vel_2=V_dot_2/(A_2*convert(cm^2,m^2))"[m/s]" A2 [cm ] 25 30 35 40 45 50 55 60 65 70 75 2 18 Vel2 [m/s] 16 13.75 12.03 10.68 9.583 8.688 7.941 7.31 6.77 6.303 5.896 16 14 Neglect KE Changes Set A = 53.35cm ^2
1 Include KE Changes Vel 2 [m /s] 12 10 8 6 4 20 30 40 50 60 70 80 A 2 [cm ^2] 5131 Chapter 5 The First Law of Thermodynamics 5143 The ducts of a heating system pass through an unheated area. The rate of heat loss from the air in the ducts is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved. Properties The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kgK (Table A2) Analysis We take the heating duct as the system, which is a control volume since mass crosses the & & & boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & & mh1 = Qout + mh2 (since W ke pe 0) & & & Qout = m(h1  h2 ) = mC p (T1  T2 ) Substituting, & Qout = (120 kg / min)(1.005 kJ / kgo C)(4 o C) = 482 kJ / min 120 kg/min AIR Q 5132 Chapter 5 The First Law of Thermodynamics 5144E The ducts of an airconditioning system pass through an unconditioned area. The inlet velocity and the exit temperature of air are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved. Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A1E). The constant pressure specific heat of air at room temperature is Cp = 0.240 Btu/lbm.R (Table A2E) Analysis (a) The inlet velocity of air through the duct is V1 = & & V1 V 450 ft 3 /min = 12 = = 825 ft/min A1 r (5/12 ft ) 2 Then the mass flow rate of air becomes RT 0.3704psia ft 3 /lbm R (510R ) = 12.6ft 3 /lbm v1 = 1 = (15psia ) P1 & V 450ft 3 /min & = 35.7lbm/min = 0.595lbm/s m= 1 = v1 12.6ft 3 /lbm ( ) 450 ft3/min AIR 2 Btu/s D = 10 in (b) We take the airconditioning duct as the system, which is a control volume since mass crosses the & & & boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & & Qin + mh1 = mh2 (since W ke pe 0) & & & Qin = m(h2  h1 ) = mC p (T2  T1 ) Then the exit temperature of air becomes T2 = T1 + & Qin 2 Btu/s = 50 o F + = 64.0 o F & mC p (0.595 lbm/s)(0.24 Btu/lbmo F) 5133 Chapter 5 The First Law of Thermodynamics 5145 Water is heated by a 7kW resistance heater as it flows through an insulated tube. The mass flow rate of water is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Water is an incompressible substance with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The tube is adiabatic and thus heat losses are negligible. Properties The specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). Analysis We take the water pipe as the system, which is a control volume since mass crosses the boundary. & & & There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & E in  E out 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & E system 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & E in = E out & & & & We,in + mh1 = mh2 (since Qout ke pe 0) & & & & We,in = m(h2  h1 ) = m[C (T2  T1 ) + vP 0 ] = mC (T2  T1 ) WATER 20C 75C 7 kW Substituting, the mass flow rates of water is determined to be & m= & We ,in C (T2  T1 ) = 7 kJ/s (4.184 kJ/kgo C)(75  20) o C = 0.0304 kg/s 5134 Chapter 5 The First Law of Thermodynamics 5146 Steam pipes pass through an unheated area, and the temperature of steam drops as a result of heat losses. The mass flow rate of steam and the rate of heat loss from are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved. Properties From the steam tables (Table A6), P1 = 1MPa v1 = 0.2327 m 3 /kg o T1 = 250 C h1 = 2942.6kJ/kg and P2 = 800kPa h2 = 2839.3kJ/kg o T2 = 200 C 1 MPa 250C STEAM 800 kPa 200C Q Analysis (a) The mass flow rate of steam is determined directly from & m= 1 1 A1 V1 = (0.06m )2 (2m/s ) = 0.0972 kg/s 3 v1 0.2327 m /kg [ (b) We take the steam pipe as the system, which is a control volume since mass crosses the boundary. There & & & is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & mh1 = Qout + mh2 & & Q = m (h  h )
out 1 2 & (since W ke pe 0) Substituting, the rate of heat loss is determined to be & Qloss = (0.0972 kg/s )(2942.6  2839.3) kJ/kg = 10.04 kJ/s 5135 Chapter 5 The First Law of Thermodynamics Energy Balance for Charging and Discharging Processes 5147 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem mi = m2 (since mout = minitial = 0) Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2 (since W Eout = Einitial = ke pe 0) Combining the two balances: Qin = m 2 (u 2  hi ) where
100 kPa 17C P2V (100 kPa )(0.008 m 3 ) = = 0.0096 kg RT2 (0.287 kPa m 3 /kg K )(290 K ) h = 290.16 kJ/kg 17 Ti = T2 = 290 K Table A i u 2 = 206.91 kJ/kg m2 =
Substituting, Qin = (0.0096 kg)(206.91  290.16) kJ/kg =  0.8 kJ 8L Evacuated Qout = 0.8 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction. 5136 Chapter 5 The First Law of Thermodynamics 5148 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fill the tank until mechanical equilibrium is established. The final temperature in the tank is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The device is adiabatic and thus heat transfer is negligible. Properties The specific heat ratio air at room temperature is k = 1.4 (Table A2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2 Esystem 1 24 4 3 (since mout = minitial = 0) = Change in internal, kinetic, potential, etc. energies mi hi = m2 u2 (since Q W Eout = Einitial = ke pe 0) Combining the two balances: u2 = hi Substituting, T2 = 1.4 290K = 406 K = 133 o C
initially evacuated Cv T2 = C p Ti T2 = (C p / Cv )Ti = kTi Air 5137 Chapter 5 The First Law of Thermodynamics 5149 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until mechanical equilibrium is established. The mass of air that entered and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the tank (will be verified). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The properties of air are (Table A17) hi = 295.17 kJ/kg Ti = 295 K u1 = 210.49 kJ/kg T1 = 295 K u 2 = 250.02 kJ/kg T2 = 350 K Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem mi = m2  m1 Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2  m1u1 (since W ke pe 0) The initial and the final masses in the tank are m1 = m2 = (100 kPa )(2 m 3 ) P1V = = 2.362 kg RT1 (0.287 kPa m 3 /kg K )(295 K )
Pi = 600 kPa Ti = 22C (600 kPa )(2m 3 ) P2V = = 11.946 kg RT2 (0.287 kPa m 3 /kg K )(350K ) Then from the mass balance, mi = m2  m1 = 11.946  2.362 = 9.584 kg (b) The heat transfer during this process is determined from V1 = 2 m3 P1 = 100 kPa T1 = 22C Q Qin =  mi hi + m 2 u 2  m1u1 = (9.584 kg )(295.17 kJ/kg ) + (11.946 kg )(250.02 kJ/kg )  (2.362 kg )(210.49 kJ/kg ) = 339 kJ Qout = 339 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction. 5138 Chapter 5 The First Law of Thermodynamics 5150 A rigid tank initially contains saturated R134a vapor. The tank is connected to a supply line, and R134a is allowed to enter the tank. The final temperature in the tank, the mass of R134a that entered, and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A11 through A13)
3 T1 =8 o C v1 = v f + x1v fg = 0.0007884 + 0.6 (0.0525  0.0007884 ) = 0.03182m /kg u1 = u f + x1u fg = 60.43 + 0.6 (231.46  60.43) = 163.05kJ/kg x1 = 0.6 P2 = 800kPa v 2 = v g @800kPa = 0.0255m 3 /kg sat.vapor u 2 = u g @800 kPa = 243.78kJ/kg Pi = 1.0MPa hi = 356.52kJ/kg Ti = 120 o C Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2  m1 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2  m1u1 (since W ke pe 0) (a) The tank contains saturated vapor at the final state at 800 kPa, and thus the final temperature is the saturation temperature at this pressure, T2 = Tsat @ 800 kPa = 31.33o C (b) The initial and the final masses in the tank are m1 = m2 = V 0.2 m3 = = 6.29 kg v1 0.03182 m3 / kg
3 R134a 1 MPa 120C V 0.2 m = = 7.84 kg v2 0.0255 m3 / kg Then from the mass balance mi = m2  m1 = 7.84  6.29 = 1.55 kg (c) The heat transfer during this process is determined from the energy balance to be 0.2 m3 R134a Qin =  mi hi + m 2 u 2  m1u1 = 333 kJ = (1.55 kg )(356.52 kJ/kg ) + (7.84 kg )(243.78 kJ/kg ) (6.29 kg )(163.05 kJ/kg ) 5139 Chapter 5 The First Law of Thermodynamics 5151E A rigid tank initially contains saturated water vapor. The tank is connected to a supply line, and water vapor is allowed to enter the tank until onehalf of the tank is filled with liquid water. The final pressure in the tank, the mass of steam that entered, and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A4E through A6E)
3 T1 = 250 o F v1 = v g @ 250o F = 13.826ft /lbm sat.vapor u1 = u g @ 250o F = 1087.9Btu/lbm 3 T2 = 250 o F v f = 0.017001, v g = 13.826ft /lbm sat.mixture u f = 218.49, u g = 1087.9Btu/lbm Steam 160 psia 400F Water
4 ft3 250F Sat. vapor Pi = 160psia hi = 1217.8Btu/lbm o Ti = 400 F Q Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem mi = m2  m1 Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2  m1u1 (since W ke pe 0) (a) The tank contains saturated mixture at the final state at 250F, and thus the exit pressure is the saturation pressure at this temperature, P2 = Psat @ 250oF = 29.82 psia
(b) The initial and the final masses in the tank are 4 ft 3 V m1 = = = 0.289 lbm v1 13.826 ft 3 /lbm m2 = m f + m g = Vf vf + Vg vg = 2 ft 3 0.017001 ft 3 /lbm + 2 ft 3 13.826 ft 3 /lbm = 117.64 + 0.14 = 117.78 lbm Then from the mass balance: mi = m2  m1 = 117.78  0.289 = 117.49 lbm (c) The heat transfer during this process is determined from the energy balance to be Qin = mi hi + m 2 u 2  m1u 1 = (117.49 lbm )(1217.8 Btu/lbm ) + 25,855 Btu  (0.289 lbm )(1087.9 Btu/lbm ) = 117,539 Btu Qout = 117,539 Btu since U 2 = m2 u2 = m f u f + mg u g = 117.64 218.49 + 0.14 1087.9 = 25,855 Btu Discussion A negative result for heat transfer indicates that the assumed direction is wrong, and should be reversed. 5152 A cylinder initially contains superheated steam. The cylinder is connected to a supply line, and is superheated steam is allowed to enter the cylinder until the volume doubles at constant pressure. The final temperature in the cylinder and the mass of the steam that entered are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains 5140 Chapter 5 The First Law of Thermodynamics constant. 2 The expansion process is quasiequilibrium. 3 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A4 through A6) P1 = 500 kPa v1 = 0.4249m 3 /kg o T1 = 200 C u 1 = 2642.9 kJ/kg Pi = 1MPa h = 3157.7 kJ/kg o i Ti = 350 C Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem mi = m2  m1 Energy balance: E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 123 4 4
Change in internal, kinetic, potential, etc. energies mi hi = Wb ,out + m 2 u 2  m1u1 (since Q ke pe 0)
Combining the two relations gives 0 = Wb,out  (m 2  m1 )hi + m 2 u 2  m1u1 The boundary work done during this process is 1 kJ = 5 kJ PdV = P(V2  V1 ) = (500 kPa )(0.02  0.01)m 3 1 kPa m 3 1 The initial and the final masses in the cylinder are 0.01 m 3 V m1 = 1 = = 0.0235 kg v1 0.4249 m 3 /kg Wb,out = 2 m2 = Substituting, 3 V2 0.02 m = v2 v2 0.02 0.02 = 5 v  0.0235 (3157.7 ) + v u 2  (0.0235)(2642.9 ) 2 2 Then by trial and error, T2 = 262.6C and v2 = 0.4865 m3/kg (b) The final mass in the cylinder is V 0.02 m 3 m2 = 2 = = 0.0411 kg v2 0.4865 m 3 / kg Then, mi = m2  m1 = 0.0411  0.0235 = 0.0176 kg 5141 Chapter 5 The First Law of Thermodynamics 5153 A cylinder initially contains saturated liquidvapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The final temperature in the cylinder and the mass of the steam that entered are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasiequilibrium. 3 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A4 through A6) P1 = 300kPa h1 = h f + x1 h fg = 561.47 + 0.8 2163.8 = 2292.51kJ/kg x1 = 0.8 P2 = 300kPa h2 = h g @300 kPa = 2725.3kJ/kg Pi = 0.5MPa hi = 3167.7 kJ/kg o Ti = 350 C sat.vapor (P = 300 kPa) m1 = 10 kg H2O Pi = 0.5 MPa Ti = 350C Analysis (a) The cylinder contains saturated vapor at the final state at a pressure of 300 kPa, thus the final temperature in the cylinder must be T2 = Tsat @ 300 kPa = 133.6C (b) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2  m1 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies mi hi = Wb,out + m2 u2  m1u1 (since Q ke pe 0) Combining the two relations gives 0 = Wb,out  (m 2  m1 )hi + m 2 u 2  m1u1 or, 0 = (m2  m1 )hi + m2h2  m1h1 since the boundary work and U combine into H for constant pressure expansion and compression processes. Solving for m2 and substituting, m2 = Thus, mi = m2  m1 = 19.78  10 = 9.78 kg hi  h1 (3167.7  2292.51)kJ/kg (10kg ) = 19.78kg m1 = hi  h2 (3167.7  2725.3)kJ/kg 5142 Chapter 5 The First Law of Thermodynamics 5154 A rigid tank initially contains saturated R134a vapor. The tank is connected to a supply line, and R134a is allowed to enter the tank. The mass of the R134a that entered and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A11 through A13) P1 = 1 MPa v1 = v g @1 MPa = 0.0202 m 3 /kg sat .vapor u1 = u g @1 MPa = 247.77 kJ/kg P2 = 1.2 MPa v 2 = v f @1.2 MPa = 0.0008928 m 3 /kg sat .liquid u 2 = u f @1.2 MPa = 114.69 kJ/kg Pi = 1.2 MPa hi = h f @ 30o C = 91.49 kJ/kg Ti = 30 o C R134a 1.2 MPa 30C R134a
0.1 m3 1 MPa Sat. vapor Q Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2  m1 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2  m1u1 (since W ke pe 0) (a) The initial and the final masses in the tank are V 0.1 m 3 = 4.95 kg m1 = 1 = v1 0.0202 m 3 / kg V2 0.1 m 3 = = 112.01 kg v2 0.0008928 m 3 / kg Then from the mass balance m2 = mi = m2  m1 = 112.01  4.95 = 107.06 kg (c) The heat transfer during this process is determined from the energy balance to be Qin =  mi hi + m 2 u 2  m1u1 = 1825 kJ = (107.06 kg )(91.49 kJ/kg ) + (112.01 kg )(114.69 kJ/kg )  (4.95 kg )(247.77 kJ/kg ) 5143 Chapter 5 The First Law of Thermodynamics 5155 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of the mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A4 through A6)
3 T1 = 200 o C v1 = v f @ 200o C = 0.001157 m /kg sat .liquid u1 = u f @ 200o C = 850.65 kJ/kg H2O Sat. liquid T = 200C V = 0.3 m3 Q Te = 200 o C he = h f @ 200o C = 852.45 kJ/kg sat .liquid Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = me he + m2 u2  m1u1 (since W ke pe 0) The initial and the final masses in the tank are m1 = V1 0.3m 3 = = 259.3kg v1 0.001157 m 3 /kg
1 2 m 2 = 1 m1 = 2 (259.3kg ) = 129.65kg Then from the mass balance, me = m1  m2 = 259.3  129.65 = 129.65 kg Now we determine the final internal energy, v2 = x2 = 0.3m 3 V = = 0.002314 m 3 /kg m 2 129.65kg v2  v f v fg = 0.002314  0.001157 = 0.00849 0.13736  0.001157 T2 = 200 o C u 2 = u f + x 2 u fg = 850.65 + (0.00849 )(1744.7 ) = 865.46 kJ/kg x 2 = 0.00849 Then the heat transfer during this process is determined from the energy balance by substitution to be Q = (129.65 kg )(852.45 kJ/kg ) + (129.65 kg )(865.46 kJ/kg )  (259.3 kg )(850.65 kJ/kg ) = 2153 kJ 5144 Chapter 5 The First Law of Thermodynamics 5156 A rigid tank initially contains saturated liquidvapor mixture of refrigerant134a. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquid remains inside. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of R134a are (Tables A11 through A13) P1 = 800kPa v f =0.0008454m 3 /kg, v g = 0.0255m 3 /kg u f =92.75kJ/kg, u g = 243.78kJ/kg P2 = 800kPa v 2 = v g @800kPa = 0.0255m 3 /kg sat.vapor u 2 = u g @800 kPa = 243.78kJ/kg Pe = 800kPa he = h f @800 kPa = 93.42kJ/kg sat.liquid Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass R134a Sat. vapor P = 800 kPa V = 0.1 m3 Q me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin = me he + m2 u2  m1u1 (since W ke pe 0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + m g = Vf vf + Vg vg = 0.1 0.4m 3 0.0008454m 3 /kg + 0.1 0.6m 3 0.0255m 3 /kg = 47.32 + 2.35 = 49.67 kg U 1 = m1u1 = m f u f + m g u g = (47.32 )(92.75) + (2.35)(243.78 ) = 4962kJ m2 = 0.1m 3 V = = 3.92kg v 2 0.0255m 3 /kg Then from the mass and energy balances, me = m1  m2 = 49.67  3.92 = 45.75 kg Qin = (45.75 kg )(93.42 kJ/kg ) + (3.92 kg )(243.78 kJ/kg )  4962 kJ = 267.6 kJ 5145 Chapter 5 The First Law of Thermodynamics 5157E A rigid tank initially contains saturated liquidvapor mixture of R134a. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until all the liquid in the tank disappears. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of R134a are (Tables A11E through A13E) P1 = 100psia, v f = 0.01332ft 3 /lbm, v g = 0.4747ft 3 /lbm u f = 36.75Btu/lbm, u g = 103.68Btu/lbm P2 = 100psia v 2 = v g @100 psia = 0.4747ft /lbm sat.vapor u 2 = u g @100 psia = 103.68Btu/lbm
3 R134a Sat. vapor P = 100 psia V = 4 ft3 Q Pe = 100psia he = h g @100 psia = 112.46Btu/lbm sat.vapor Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  me he = m2 u2  m1u1 (since W ke pe 0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + m g = Vf vf + Vg vg = 4 0.2ft 3 0.01332ft 3 /lbm + 4 0.8ft 3 0.4747 ft 3 /lbm = 60.06 + 6.74 = 66.8lbm U 1 = m1u1 = m f u f + m g u g = (60.06 )(36.75) + (6.74 )(103.68) = 2906Btu m2 = 4ft 3 V = = 8.426lbm v 2 0.4747ft 3 /lbm Then from the mass and energy balances, me = m1  m2 = 66.8  8.426 = 58.374 lbm Qin = m e he + m 2 u 2  m1u1 = 4532 Btu = (58.374 lbm )(112.46 Btu/lbm ) + (8.426 lbm )(103.68 Btu/lbm )  2906 Btu 5146 Chapter 5 The First Law of Thermodynamics 5158 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until the temperature rises to 500C. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process by using constant average properties for the steam leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A4 through A6) P1 = 2MPa v1 = 0.12547m 3 /kg o T1 = 300 C u1 = 2772.6kJ/kg, h1 = 3023.5kJ/kg P2 = 2MPa v 2 = 0.17568m 3 /kg T2 = 500 o C u 2 = 3116.2kJ/kg, h2 = 3467.6kJ/kg STEAM 2 MPa Q Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  me he = m2 u2  m1u1 (since W ke pe 0) The state and thus the enthalpy of the steam leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Thus, he h1 + h2 3023.5 + 3467.6 kJ / kg = = 3245.55 kJ / kg 2 2 The initial and the final masses in the tank are m1 = m2 = V1 0.2 m 3 = = 1.594 kg v1 0.12547 m 3 / kg V2 0.2 m 3 = = 1.138 kg v2 0.17568 m 3 / kg Then from the mass and energy balance relations, me = m1  m2 = 1.594  1138 = 0.456 kg . Qin = me he + m 2 u 2  m1u1 = 606.7 kJ = (0.456 kg )(3245.55 kJ/kg ) + (1.138 kg )(3116.2 kJ/kg )  (1.594 kg )(2772.6 kJ/kg ) 5147 Chapter 5 The First Law of Thermodynamics 5159 A pressure cooker is initially halffilled with liquid water. If the pressure cooker is not to run out of liquid water for 1 h, the highest rate of heat transfer allowed is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of water are (Tables A4 through A6) P1 = 175kPa v f = 0.001057 m 3 /kg, v g = 1.0036m 3 /kg u f = 486.8kJ/kg, u g = 2524.9 kJ/kg P2 = 175kPa v 2 = v g @175kPa = 1.0036m 3 /kg sat.vapor u 2 = u g @175kPa = 2524.9kJ/kg Pe = 175kPa he = h g @175kPa = 2700.6kJ/kg sat.vapor Pressure Cooker 4L 175 kPa & Q Analysis We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  me he = m2 u2  m1u1 (since W ke pe 0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + m g = Vf vf + Vg vg = 0.002m 3 0.001057 m 3 /kg + 0.002m 3 1.0036m 3 /kg = 1.892 + 0.002 = 1.894kg U 1 = m1u1 = m f u f + m g u g = (1.892 )(486.8) + (0.002 )(2524.9 ) = 926.1kJ m2 = 0.004m 3 V = = 0.004kg v 2 1.0036m 3 /kg Then from the mass and energy balances, me = m1  m2 = 1.894  0.004 = 1.890 kg Qin = me he + m 2 u 2  m1u1 Thus, & Q = 4188 kJ = 1.163 kW Q= t 3600 s = (1.890kg )(2700.6kJ/kg ) + (0.004kg )(2524.9kJ/kg )  926.1kJ = 4188kJ 5148 Chapter 5 The First Law of Thermodynamics 5160 An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half of the mass of helium is allowed to escape. The final temperature and pressure in the tank are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k =1.667 (Table A2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem m2 = 1 m1 (given) 2 Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 me = m2 = 1 m1 2
He 0.08 m3 2 MPa 80C = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies  me he = m2 u2  m1u1 (since W Q ke pe 0) Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Combining the mass and energy balances: Dividing by 2m1: Dividing by Cv: Solving for T2:
1 1 0 = 2 m1he + 2 m1u2  m1u1 0 = he + u 2  2u1 or 0 = C p 0 = k (T1 + T2 ) + 2T2  4T1 T2 = T1 + T2 + C vT2  2C v T1 2 since k = C p / C v (4  k ) (4  1.667 ) T1 = (353 K ) = 225 K (2 + k ) (2 + 1.667 ) The final pressure in the tank is P1V m RT m T 1 225 (2000 kPa ) = 637 kPa = 1 1 P2 = 2 2 P1 = 2 353 P2V m 2 RT2 m1T2 5149 Chapter 5 The First Law of Thermodynamics 5161E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia. The amount of electrical work transferred is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R =0.3704 psia.ft 3/lbm.R (Table A1E). The properties of air are (Table A17E) Ti = 580 R T1 = 580 R T2 = 580 R hi = 138.66 Btu / lbm u1 = 98.90 Btu / lbm u2 = 98.90 Btu / lbm Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2
AIR 60 ft3 75 psia 120F = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in  me he = m2 u2  m1u1 (since Q ke pe 0) The initial and the final masses of air in the tank are m1 = m2 P1V (75psia ) 60ft 3 = = 20.95lbm RT1 0.3704psia ft 3 /lbm R (580R ) We ) PV (30psia )(60ft ) = = RT (0.3704psia ft /lbm R )(580R ) = 8.38lbm
3 2 2 3 ( ( ) Then from the mass and energy balances, me = m1  m2 = 20.95  8.38 = 12.57 lbm We ,in = me he + m2 u 2  m1u1 = (12.57 lbm )(138.66 Btu/lbm) + (8.38 lbm )(98.90 Btu/lbm)  (20.95 lbm )(98.90 Btu/lbm) = 500 Btu 5150 Chapter 5 The First Law of Thermodynamics 5162 A vertical cylinder initially contains air at room temperature. Now a valve is opened, and air is allowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half. The amount air that left the cylinder and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specific heats. 5 The direction of heat transfer is to the cylinder (will be verified). Properties The gas constant of air is R =0.287 kPa.m3/kg.K. Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + Wb,in  me he = m2 u2  m1u1 (since ke pe 0) The initial and the final masses of air in the cylinder are m1 = m2 = P1V1 (300kPa ) 0.2m 3 = = 0.714kg RT1 0.287 kPa m 3 /kg K (293K ) P2V2 (300kPa ) 0.1m 3 = = 0.357 kg = 1 m1 2 RT2 0.287kPa m 3 /kg K (293K ) Then from the mass balance, ( ( ( ( ) ) AIR 300 kPa 0.2 m3 20C ) ) me = m1  m2 = 0.714  0.357 = 0.357 kg (b) This is a constant pressure process, and thus the Wb and the U terms can be combined into h to yield Q = me he + m2 h2  m1h1 Noting that the temperature of the air remains constant during this process, we have hi = h1 = h2 = h. Also, me = m2 = 1 m1 . Thus, 2 Q= (1 m1 + 1 m1  m1 )h = 0 2 2 5151 Chapter 5 The First Law of Thermodynamics 5163 A balloon is initially filled with helium gas at atmospheric conditions. The tank is connected to a supply line, and helium is allowed to enter the balloon until the pressure rises from 100 to 150 kPa. The final temperature in the balloon is to be determined. Assumptions 1 This is an unsteady process since the conditions within process, but it can be analyzed as a uniformflow process since the constant. 2 Helium is an ideal gas with constant specific heats. 3 equilibrium. 4 Kinetic and potential energies are negligible. 5 There other than boundary work. 6 Heat transfer is negligible. the device are changing during the state of fluid at the inlet remains The expansion process is quasiare no work interactions involved Properties The gas constant of helium is R = 2.0769 kJ/kgK (Table A1). The specific heats of helium are Cp = 5.1926 and Cv = 3.1156 kJ/kgK (Table A2a). Analysis We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2  m1
He 25C 150 kPa = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies mi hi = Wb,out + m2 u2  m1u1 (since Q ke pe 0) m1 = P1V1 (100kPa ) 65m 3 = = 10.61kg RT1 2.0769kPa m 3 /kg K (295K ) P P1 V1 150kPa 65m 3 = 97.5m 3 = V2 = 2 V1 = 100kPa P1 P2 V2 m2 = P2V2 RT2
3 3 ( ) (150kPa )(97.5m ) 7041.74 = (2.0769kPa m /kg K )(T K ) = T kg
2 2 ( ( ) ) He 22C 100 kPa Then from the mass balance, mi = m2  m1 = 7041.74  10.61 kg T2 Noting that P varies linearly with V, the boundary work done during this process is P1 + P2 (V2  V1 ) = (100 + 150 )kPa (97.5  65)m 3 = 4062.5kJ 2 2 Using specific heats, the energy balance relation reduces to Wb = Wb,out = mi C p Ti  m2 Cv T2 + m1Cv T1 Substituting, 7041.74 7041.74 4062.5 = (3.1156 )T2 + (10.61)(3.1156)(295)  10.61(5.1926 )(298)  T T2 2 It yields T2 = 333.6 K 5152 Chapter 5 The First Law of Thermodynamics 5164 A balloon is initially filled with pressurized helium gas. Now a valve is opened, and helium is allowed to escape until the pressure inside drops to atmospheric pressure. The final temperature of helium in the balloon and the amount helium that has escaped are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process by assuming exit properties to be constant at average conditions. 2 Kinetic and potential energies are negligible. 3 There are no work interactions other than boundary work. 4 Helium is an ideal gas with constant specific heats. 5 Heat transfer is negligible. Properties The gas constant of helium is R =2.0769 kPa.m3/kg.K (Table A1). The specific heats of helium are Cp = 5.1926 and Cv = 3.1156 kJ/kgK (Table A2). Analysis The properties of helium leaving the balloon are changing during this process. But we will treat them as a constant at the average temperature. Thus Te (T1 + T2)/2. Also h = CpT and u = CvT. We take the balloon as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem me = m1  m2 Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Wb,in  me he = m2 u2  m1u1 (since Q ke pe 0) or T1 + T2 + m2 Cv T2  m1Cv T1 2 The initial and the final masses in the balloon are Wb,in = me C p m1 = m2 = P1V1 (150kPa ) 10m 3 = = 2.4074kg RT1 2.0769kPa m 3 /kg K (300K ) P2V2 (100kPa ) 8.5m 3 409.264 = = 3 RT2 T2 2.0769kPa m /kg K T2 Then from the mass balance, 409.264 me = m1  m2 = 2.4074  T2 Noting that the pressure changes linearly with volume, the boundary work done during this process is ( ( ( ( ) ) ) He 27C 150 kPa ) Wb = P1 + P2 (V2  V1 ) = (150 + 100 )kPa (8.5  10)m 3 = 187.5kJ 2 2 Combining mass and energy balances and substituting, 409.264 300 + T2 409.264 (5.1926 )  ( 187.5) = 2.4074  + (3.1156)T2  (2.4074)(3.1156)(300) T2 2 T2 It yields T22  56T2  51,003.5 = 0 T2 = 256 K (b) The amount of helium that has escaped is m e = m1  m 2 = 2.4074  409.264 409.264 = 2.1074  = 0.509 kg T2 256 5153 Chapter 5 The First Law of Thermodynamics 5165 Problem 5164 is reconsidered. The effect of the percent change of the volume of the balloon (in the range of 0 to 15%) on the final temperature in the balloon and the amount of mass that has escaped is to be investigated. The final temperature and the amount of discharged helium are to be plotted against the percent change in volume. "Knowns:" C_P = 5.1926"[kJ/kgK ]" C_V = 3.1156 "[kJ/kgK ]" R=2.0769 "[kPam^3/kgK]" P_1= 150"[kPa]" P_2= 100"[kPa]" T_1 = 300"[K]" V_1 = 10"[m^3]" {PCVolChange =15 "[%]"} "Percent Volume Change" {V_2 = 8.5"[m^3]"} T_out = (T_1 + T_2)/2"[K]" V_1*PCVolChange=(V_1V_2)*100 "Analysis: "Mass balance:" m_in = 0"[kg]" m_in  m_out = m_2  m_1 "Energy balance:" E_in  E_out = DELTAE_sys E_in = W_b_in"[kJ]" E_out = m_out*h_out"[kJ]" h_out = C_P*T_out"[kJ/kg]" DELTAE_sys = m_2*u_2m_1*u_1 u_1=C_V*T_1"[kJ/kg]" u_2= C_V*T_2"[kJ/kg]" "The volume flow rates of air are determined to be:" P_1*V_1=m_1*R*T_1 P_2*V_2=m_2*R*T_2 "Boundary Work: Due to pressure changing linearly with volume. Note the minus sign for work in" W_b_in = (P_1+P_2)/2*(V_2V_1) mout [kg] 0.5204 0.5775 0.6347 0.6918 0.7489 0.806 PCVolChange [%] 0 3 6 9 12 15 T2 [K] 255.2 255.2 255.3 255.4 255.5 255.6 5154 Chapter 5 The First Law of Thermodynamics
1 0.95 0.9 0.85 0.8 m out [kg] 0.75 0.7 0.65 0.6 0.55 0.5 0 2 4 6 8 10 12 14 16 Percent Volum e Change [%] 256 255.9 255.8 255.7 255.6 T 2 [K] 255.5 255.4 255.3 255.2 255.1 255 0 2 4 6 8 10 12 14 16 Percent Volum e Change [%] 5155 Chapter 5 The First Law of Thermodynamics 5166 A vertical pistoncylinder device equipped with an external spring initially contains superheated steam. Now a valve is opened, and steam is allowed to escape until the volume of the cylinder goes down by half. The initial and final masses of steam in the cylinder and the amount of heat transferred are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process by assuming the properties of steam that escape to be constant at average conditions. 2 Kinetic and potential energies are negligible. 3 The spring is a linear spring. 4 The direction of heat transfer is to the cylinder (will be verified). Properties From the steam tables (Tables A4 through A6), P1 = 1 MPa v1 = 0.2327m 3 /kg T1 = 250 o C u 1 = 2709.9 kJ/kg, h1 = 2942.6 kJ/kg P2 = 800 kPa v 2 = 0.2404 m 3 /kg sat .vapor u 2 = 2576.8 kJ/kg, h 2 = 2769.1 kJ/kg Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem me = m1  m2 Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + Wb,in  me he = m2 u2  m1u1 (since ke pe 0) The state and thus the enthalpy of the steam leaving the cylinder is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Thus, h +h 2942.6 + 2769.1 kJ / kg he 1 2 = = 2855.9 kJ / kg 2 2 The initial and the final masses in the tank are m1 = m2 = V1 0.2 m 3 = = 0.859 kg v1 0.2327 m 3 / kg V2 0.1 m 3 = = 0.416 kg v2 0.2404 m 3 / kg
Steam 1 MPa 0.2 m3 250C Then from the mass balance, me = m1  m2 = 0.859  0.416 = 0.443 kg (b) The boundary work done during this process is (1000 + 800)kPa P + P2 (V1  V2 ) = (0.2  0.1)m 3 = 90 kJ Wb,in = 1 2 2 Then the heat transfer during this process becomes Qin = Wb,in + m e he + m 2 u 2  m1u1 = 80.7 kJ Qout = 80.7kJ = 90 kJ + (0.443kg )(2855.9 kJ/kg ) + (0.416 kg )(2576.8kJ/kg ) (0.859 kg )(2709.9 kJ/kg ) 5156 Chapter 5 The First Law of Thermodynamics 5167 A vertical pistoncylinder device initially contains steam at a constant pressure of 300 kPa. Now a valve is opened, and steam is allowed to escape at constant temperature and pressure until the volume reduces to onethird. The mass of steam that escaped and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the properties of steam that escape remain constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions other than boundary work. 4 The direction of heat transfer is to the cylinder (will be verified). Properties From the steam tables (Tables A4 through A6), he = 2967.6kJ/kg Pe = P1 = P2 = 300kPa 3 v1 = v 2 = 0.7964m /kg Te = T1 = T2 = 250 o C u = u = 2728.7 kJ/kg 1 2 Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + Wb,in  me he = m2 u2  m1u1 (since ke pe 0) or Qin = me he + m2 h2  m1 h1 since for a constant pressure process, the Wb and the U terms can be combined into H. The initial and final masses of steam in the cylinder are m1 = m2 = V1 0.3 m 3 = = 0.377 kg v1 0.7964 m 3 / kg V2 0.1 m 3 = = 0.126 kg v2 0.7964 m 3 / kg
STEAM 300 kPa 0.3 m3 250C Q Then from the mass balance, me = m1  m2 = 0.377  0126 = 0.251 kg . (b) Noting that he = h1 = h2 = h and me = m1  m2, the energy balance relation reduces to Qin = m e he + m 2 h2  m1 h1 = (m e + m 2  m1 )h =0 Therefore, there will be no heat transfer during this process. 5157 Chapter 5 The First Law of Thermodynamics Review Problems 5168 A cylinder is initially filled with saturated R134a vapor at a specified pressure. The refrigerant is heated both electrically and by heat transfer at constant pressure for 6 min. The electric current is to be determined, and the process is to be shown on a Tv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2 The thermal energy stored in the cylinder itself and the wires is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + We,in  Wb,out = U R134a P= const. (since Q = KE = PE = 0) W Qin + We,in = m(h2  h1 ) Qin + (VIt ) = m(h2  h1 ) since U + Wb = H during a constant pressure quasiequilibrium process. The properties of R134a are (Tables A11 through A13) P1 = 200kPa h1 = h g @ 200 kPa = 241.30kJ/kg sat.vapor P2 = 200kPa h 2 = 314.02kJ/kg o T1 = 70 C Substituting, T
2 1 v 1000VA 250,000VA + (110V )(I )(6 60s ) = (12kg )(314.02  241.30 )kJ/kg 1kJ/s I = 15.72A 5158 Chapter 5 The First Law of Thermodynamics 5169 A cylinder is initially filled with saturated liquidvapor mixture of R134a at a specified pressure. Heat is transferred to the cylinder until the refrigerant vaporizes completely at constant pressure. The initial volume, the work done, and the total heat transfer are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis (a) Using property data from R134a tables (Tables A11 through A13), the initial volume of the refrigerant is determined to be P1 = 200kPa v f = 0.0007532, v g = 0.0993m 3 /kg x1 = 0.25 u g = 221.43kJ/kg u f = 36.69, v1 = v f + x1 v fg = 0.0007532 + 0.25 (0.0993  0.0007532) = 0.02539m 3 /kg R134a
200 kPa u1 = u f + x1u fg = 36.69 + 0.25 (221.43  36.69 ) = 82.88kJ/kg V1 = mv1 = (0.2kg ) 0.02539m 3 /kg = 0.005078m 3 (b) The work done during this constant pressure process is Q ( ) P P2 = 200kPa v 2 = v g @ 200kPa = 0.0993m 3 /kg sat.vapor u 2 = u g @ 200 kPa = 221.43kJ/kg Wb ,out = PdV = P(V
2 1 2  V1 ) = mP(v 2  v1 ) = 2.96 kJ 1 2 1 kJ = (0.2 kg )(200 kPa )(0.0993  0.02539)m 3 /kg 1 kPa m 3 v (c) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb,out = U Qin = m(u2  u1 ) + Wb,out Substituting, Qin = (0.2 kg)(221.43  82.88)kJ/kg + 2.96 = 30.67 kJ 5159 Chapter 5 The First Law of Thermodynamics 5170 A cylinder is initially filled with helium gas at a specified state. Helium is compressed polytropically to a specified temperature and pressure. The heat transfer during the process is to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasiequilibrium. Properties The gas constant of helium is R = 2.0769 kPa.m3/kg.K (Table A1). Also, Cv = 3.1156 kJ/kg.K (Table A2). Analysis The mass of helium and the exponent n are determined to be m= P1V1 (150kPa ) 0.5m 3 = = 0.123kg RT1 2.0769kPa m 3 /kg K (293K ) ( ( ) ) P1V1 P2V2 T P 413K 150kPa = V2 = 2 1 V1 = 0.5m 3 = 0.264m 3 293K 400kPa RT1 RT2 T1 P2 P P2V2n = P1V1n 2 P1 V1 = V 2 400 0.5 n = 1.536 150 = 0.264 n n Then the boundary work for this polytropic process can be determined from Wb ,in =  P dV =  P2V 2  P1V1 mR(T2  T1 ) = 1 1 n 1 n (0.123kg )(2.0769kJ/kg K )(413  293)K = = 57.2 kJ 1  1.536 2 We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Taking the direction of heat transfer to be to the cylinder, the energy balance for this stationary closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin + Wb,in = U = m(u2  u1 ) Qin = m(u2  u1 )  Wb,in = mCv (T2  T1 )  Wb,in Substituting, Qin = (0.123 kg)(3.1156 kJ/kgK)(413  293)K  (57.2 kJ) = 11.2 kJ The negative sign indicates that heat is lost from the system.
He PVn = C Q 5160 Chapter 5 The First Law of Thermodynamics 5171 A cylinder and a rigid tank initially contain the same amount of an ideal gas at the same state. The temperature of both systems is to be raised by the same amount. The amount of extra heat that must be transferred to the cylinder is to be determined. Analysis In the absence of any work interactions, other than the boundary work, the H and U represent the heat transfer for ideal gases for constant pressure and constant volume processes, respectively. Thus the extra heat that must be supplied to the air maintained at constant pressure is Qin, extra = H  U = mC p T  mCv T = m(C p  Cv )T = mRT where R 8.314 kJ / kmol K R= u = = 0.3326 kJ / kg K M 25 kg / kmol Substituting, Qin, extra = (12 kg)(0.3326 kJ/kgK)(15 K) = 59.9 kJ
IDEAL GAS P = const. IDEAL GAS V = const. Q Q 5161 Chapter 5 The First Law of Thermodynamics 5172 The heating of a passive solar house at night is to be assisted by solar heated water. The length of time that the electric heating system would run that night with or without solar heating are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house is maintained at 22C at all times. Properties The density and specific heat of water at room temperature are = 1 kg/L and C = 4.18 kJ/kgC (Table A3). Analysis The total mass of water is m w = V = (1kg/L )(50 20L ) = 1000kg Taking the contents of the house, including the water as our system, the energy balance relation can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass 50,000 kJ/h 22C = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies water 80C We,in  Qout = U = (U ) water + (U ) air = (U ) water = mC (T2  T1 ) water or, & We,in t  Qout = [ mC (T2  T1 )]water Substituting, (15 kJ/s)t  (50,000 kJ/h)(10 h) = (1000 kg)(4.18 kJ/kg C)(22  80)C It gives t = 17,170 s = 4.77 h (b) If the house incorporated no solar heating, the energy balance relation above would simplify further to & We,in t  Qout = 0 Substituting, (15 kJ/s)t  (50,000 kJ/h)(10 h) = 0 It gives t = 33,333 s = 9.26 h 5162 Chapter 5 The First Law of Thermodynamics 5173 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise the water temperature to a specified temperature is to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the container itself and the heater is negligible. 3 Heat loss from the container is negligible. Properties The density and specific heat of water at room temperature are = 1 kg/L and C = 4.18 kJ/kgC (Table A3). Analysis Taking the water in the container as the system, the energy balance can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies We,in = (U ) water & We,in t = mC (T2  T1 ) water Substituting, (800 J/s)t = (40 kg)(4180 J/kgC)(80  20)C Solving for t gives t = 12,540 s = 209.0 min = 3.483 h Resistance Heater
Water 5163 Chapter 5 The First Law of Thermodynamics 5174 One ton of liquid water at 80C is brought into a room. The final equilibrium temperature in the room is to be determined. Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). The specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). Analysis The volume and the mass of the air in the room are V = 4x5x6 = 120 m PV (100 kPa )(120P m 3 ) m air = 1 1 = = 141.7 kg RT1 (0.2870 kPa m 3 /kg K )( 295 K ) Taking the contents of the room, including the water, as our system, the energy balance can be written as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass 4m 5m 6m ROOM
22C 100 kPa Heat Water 80C = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies 0 = U = (U )water + (U )air or [mC (T2  T1 )]water + [mC v (T2  T1 )]air =0 Substituting, (1000 kg )(4.180 kJ/kgo C)(T f  80) o C + (147.7 kg )(0.718 kJ/kg o C)(T f  22) o C = 0
It gives Tf = 78.6C where Tf is the final equilibrium temperature in the room. 5164 Chapter 5 The First Law of Thermodynamics 5175 A room is to be heated by 1 ton of hot water contained in a tank placed in the room. The minimum initial temperature of the water is to be determined if it to meet the heating requirements of this room for a 25h period. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constant specific heats. 3 The energy stored in the container itself is negligible relative to the energy stored in water. 4 The room is maintained at 20C at all times. 5 The hot water is to meet the heating requirements of this room for a 25h period. Properties The specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). Analysis Heat loss from the room during a 25h period is Qloss = (10,000 kJ/h)(24 h) = 240,000 kJ Taking the contents of the room, including the water, as our system, the energy balance can be written as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout = U = (U ) water + (U )air 0
10,000 kJ/h or Qout = [mC(T2  T1)]water Substituting, 240,000 kJ = (1000 kg)(4.18 kJ/kgC)(20  T1) It gives T1 = 77.4C
water 20C where T1 is the temperature of the water when it is first brought into the room. 5165 Chapter 5 The First Law of Thermodynamics 5176 A sample of a food is burned in a bomb calorimeter, and the water temperature rises by 3.2C when equilibrium is established. The energy content of the food is to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constant specific heats. 3 The energy stored in the reaction chamber is negligible relative to the energy stored in water. 4 The energy supplied by the mixer is negligible. Properties The specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). The constant volume specific heat of air at room temperature is Cv = 0.718 kJ/kgC (Table A2). Analysis The chemical energy released during the combustion of the sample is transferred to the water as heat. Therefore, disregarding the change in the sensible energy of the reaction chamber, the energy content of the food is simply the heat transferred to the water. Taking the water as our system, the energy balance can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 12 3 4 4
Change in internal, kinetic, potential, etc. energies Qin = U or Substituting , Qin = (U )water = [mC (T2  T1 )]water Qin = (3 kg)(4.18 kJ/kgC)(3.2C) = 40.13 kJ Water Reaction chamber Food T = 3.2C for a 2g sample. Then the energy content of the food per unit mass is 40.13 kJ 1000 g = 20,060 kJ/kg 2 g 1 kg To make a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber, we treat the entire mass within the chamber as air and determine the change in sensible internal energy: (U )chamber = [mC v (T2  T1 )]chamber = (0.102 kg ) 0.718 kJ/kgo C 3.2 o C = 0.23 kJ ( )( ) which is less than 1% of the internal energy change of water. Therefore, it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis. 5166 Chapter 5 The First Law of Thermodynamics 5177 A man drinks one liter of cold water at 3C in an effort to cool down. The drop in the average body temperature of the person under the influence of this cold water is to be determined. Assumptions 1 Thermal properties of the body and water are constant. 2 The effect of metabolic heat generation and the heat loss from the body during that time period are negligible. Properties The density of water is very nearly 1 kg/L, and the specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). The average specific heat of human body is given to be 3.6 kJ/kg.C. Analysis. The mass of the water is mw = V = (1 kg/L )(1 L ) = 1 kg We take the man and the water as our system, and disregard any heat and mass transfer and chemical reactions. Of course these assumptions may be acceptable only for very short time periods, such as the time it takes to drink the water. Then the energy balance can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = 0 = U Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 0 = U body + U water [mCv (T2  T1 )]body + [mCv (T2  T1 )]water = 0
Substituting, (68 kg )(3.6 kJ/kgo C)(T f  39) o C + (1 kg )(4.18 kJ/kg o C)(T f  3) o C = 0
It gives Tf = 38.4C Therefore, the average body temperature of this person should drop about half a degree Celsius. 5167 Chapter 5 The First Law of Thermodynamics 5178 A 0.2L glass of water at 20C is to be cooled with ice to 5C. The amount of ice or cold water that needs to be added to the water is to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the glass is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy). Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is C = 4.18 kJ/kgC (Table A3). The specific heat of ice at about 0C is C = 2.11 kJ/kgC (Table A3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ/kg,. Analysis (a) The mass of the water is m w = V = (1 kg/L)(0.2 L) = 0.2kg We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = 0 = U Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Ice cubes 0C 0 = U ice + Uwater [mC (0 o C  T1 ) solid + mhif + mC (T2 0 o C) liquid ]ice + [mC (T2  T1 )] water = 0 Noting that T1, ice = 0C and T2 = 5C and substituting gives
Water 20C 0.2 L m[0 + 333.7 kJ/kg + (4.18 kJ/kgC)(50)C] + (0.2 kg)(4.18 kJ/kgC)(520)C = 0 m = 0.0354 kg = 35.4 g (b) When T1, ice = 8C instead of 0C, substituting gives m[(2.11 kJ/kgC)[0(8)]C + 333.7 kJ/kg + (4.18 kJ/kgC)(50)C] + (0.2 kg)(4.18 kJ/kgC)(520)C = 0 m = 0.0338 kg = 33.8 g Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by a term for cold water at 0C: (U )cold water + (U )water = 0 [mC (T2  T1 )]cold water + [mC (T2  T1 )]water =0 Substituting, [mcold water (4.18 kJ/kgC)(5  0)C] + (0.2 kg)(4.18 kJ/kgC)(520)C = 0 It gives m = 0.6 kg = 600 g Discussion Note that this is 17 times the amount of ice needed, and it explains why we use ice instead of water to cool drinks. Also, the temperature of ice does not seem to make a significant difference. 5168 Chapter 5 The First Law of Thermodynamics 5179 Problem 5178 is reconsidered. The effect of the initial temperature of the ice on the final mass of ice required as the ice temperature varies from 20C to 0C is to be investigated. The mass of ice is to be plotted against the initial temperature of ice. "Knowns" rho_water = 1"[kg/L]" V = 0.2 "[L]" T_1_ice = 0"[C]" T_1 = 20"[C]" T_2 = 5"[C]" C_ice = 2.11 "[kJ/kgC]" C_water = 4.18 "[kJ/kgC]" h_if = 333.7 "[kJ/kg]" T_1_ColdWater = 0"[C]" "The mass of the water is:" m_water = rho_water*V "[kg]" "The system is the water plus the ice. Assume a short time period and neglect any heat and mass transfer. The energy balance becomes:" E_in  E_out = DELTAE_sys "[kJ]" E_in = 0 "[kJ]" E_out = 0"[kJ]" DELTAE_sys = DELTAU_water+DELTAU_ice"[kJ]" DELTAU_water = m_water*C_water*(T_2  T_1)"[kJ]" DELTAU_ice = DELTAU_solid_ice+DELTAU_melted_ice"[kJ]" DELTAU_solid_ice =m_ice*C_ice*(0T_1_ice) + m_ice*h_if"[kJ]" DELTAU_melted_ice=m_ice*C_water*(T_2  0)"[kJ]" m_ice_grams=m_ice*convert(kg,g)"[g]" "Cooling with Cold Water:" DELTAE_sys = DELTAU_water+DELTAU_ColdWater"[kJ]" DELTAU_water = m_water*C_water*(T_2_ColdWater  T_1)"[kJ]" DELTAU_ColdWater = m_ColdWater*C_water*(T_2_ColdWater  T_1_ColdWater)"[kJ]" m_ColdWater_grams=m_ColdWater*convert(kg,g)"[g]" mice,grams [g] 31.6 32.47 33.38 34.34 35.36 T1,ice [C] 20 15 10 5 0 36 35 34 m ice,grams [g] 33 32 31 30 20 16 12 8 4 0 T 1,ice [C] 5169 Chapter 5 The First Law of Thermodynamics 5180 A 1 ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium temperature in the tank is to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy). Properties The specific heat of water at room temperature is C = 4.18 kJ/kgC, and the specific heat of ice at about 0C is C = 2.11 kJ/kgC (Table A3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ/kg. Analysis We take the ice and the water as our system, and disregard any heat transfer between the system and the surroundings. Then the energy balance for this process can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = 0 = U Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 0 = U ice + Uwater [mC (0 o C  T1 ) solid + mhif + mC (T2 0 o C) liquid ]ice + [mC (T2  T1 )] water = 0 Substituting, (80 kg){(2.11 kJ / kgo C)[0  (5)]o C + 333.7 kJ / kg + (4.18 kJ / kgo C)(T2  0) o C} + (1000 kg)(4.18 kJ / kgo C)(T2  20)o C = 0 It gives T2 = 12.4C
ice 5C 80 kg which is the final equilibrium temperature in the tank. WATER
1 ton 5170 Chapter 5 The First Law of Thermodynamics 5181 An insulated cylinder initially contains a saturated liquidvapor mixture of water at a specified temperature. The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder. The amount of ice that needs to be added is to be determined. Assumptions 1 Thermal properties of the ice are constant. 2 The cylinder is wellinsulated and thus heat transfer is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy). Properties The specific heat of ice at about 0C is C = 2.11 kJ/kgC (Table A3). The melting temperature and the heat of fusion of ice at 1 atm are given to be 0C and 333.7 kJ/kg. Analysis (a) We take the contents of the cylinder (ice and saturated water) as our system, which is a closed system. Noting that the temperature and thus the pressure remains constant during this phase change process and thus Wb + U = H, the energy balance for this system can be written as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Wb,in = U Hice + Hwater = 0 H = 0 [mC (0 o C  T1 ) solid + mhif + mC (T2 0 o C) liquid ] ice + [m(h 2  h1 )] water = 0 The properties of water at 100C are (Table A4) v f = 0.001044 , vg = 1.6729 m 3 / kg h f = 419.04, Then, h fg = 2257.0 kJ.kg
ice 0C v1 = v f + x1 v fg = 0.001044 + 0.2 (1.6729  0.001044 ) = 0.3354m 3 / kg h1 = h f + x1 h fg = 419.04 + 0.2 2257.0 = 870.4kJ / kg h2 = h f @100C = 419.04 kJ / kg msteam = V1 0.01 m 3 = = 0.0298 kg v1 0.3354 m 3 / kg WATER 0.01 m3 100C Noting that T1, ice = 0C and T2 = 100C and substituting gives m[0 + 333.7 kJ/kg + (4.18 kJ/kgC)(1000)C] + (0.0298 kg)(419.04 870.4) kJ/kg = 0 m = 0.0179 kg = 17.9 g ice 5171 Chapter 5 The First Law of Thermodynamics 5182 The cylinder of a steam engine initially contains saturated vapor of water at 100 kPa. The cylinder is cooled by pouring cold water outside of it, and some of the steam inside condenses. If the piston is stuck at its initial position, the friction force acting on the piston and the amount of heat transfer are to be determined. Assumptions The device is airtight so that no air leaks into the cylinder as the pressure drops. Analysis We take the contents of the cylinder (the saturated liquidvapor mixture) as the system, which is a closed system. Noting that the volume remains constant during this phase change process, the energy balance for this system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qout = U = m(u2  u1 ) The saturation properties of water at 100 kPa and at 30C are (Tables A4 and A5) P = 100 kPa 1 T2 = 30 o C v f = 0.001043 m 3 / kg, v g = 1.6940 m 3 / kg u f = 417.36 kJ / kg, u g = 2506.1 kJ / kg v f = 0.001004 m 3 /kg, v g = 32.89 m 3 /kg u f = 125.78 kJ/kg, u fg = 2290.8 kJ/kg Psat = 4.246 kPa
P2 = Psat @ 30 o C = 4.246 kPa v1 = vg @100 kPa = 1.694 m 3 / kg u1 = u g @ 100 kPa = 2506.1 kJ / kg
Cold water Then, and V1 0.05m = = 0.0295kg v1 1.6940m 3 /kg v 2  v f 1.694  0.001 v 2 = v1 x 2 = = = 0.05148 v fg 32.89  0.001 m= u 2 = u f + x 2 u fg = 125.78 + 0.05148 2290.8 = 243.7kJ/kg 3 Steam 0.05 m3 100 kPa The friction force that develops at the pistoncylinder interface balances the force acting on the piston, and is equal to 1000 N/m 2 F = A( P1  P2 ) = (0.1 m 3 )(100  4.246)kPa 1 kPa The heat transfer is determined from the energy balance to be = 9575 N Qout = m(u1  u 2 ) = (0.0295 kg )(2506.1  243.7 )kJ/kg = 66.7 kJ 5172 Chapter 5 The First Law of Thermodynamics 5183 Water is boiled at sea level (1 atm pressure) in a coffee maker, and half of the water evaporates in 25 min. The power rating of the electric heating element and the time it takes to heat the cold water to the boiling temperature are to be determined. Assumptions 1 The electric power consumption by the heater is constant. 2 Heat losses from the coffee maker are negligible. Properties The enthalpy of vaporization of water at the saturation temperature of 100C is hfg = 2257 kJ/kg (Table A4). At an average temperature of (100+18)/2 = 59C, the specific heat of water is C = 4.18 kJ/kg.C, and the density is about 1 kg/L (Table A3). Analysis The density of water at room temperature is very nearly 1 kg/L, and thus the mass of 1 L water at 18C is nearly 1 kg. Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a liquid at a specified temperature, the amount of electrical energy needed to vaporize 0.5 kg of water in 25 min is mh fg (0.5 kg)(2257 kJ/kg) & & We = We t = mh fg We = = = 0.752 kW t (25 60 s) Therefore, the electric heater consumes (and transfers to water) 0.752 kW of electric power. Noting that the specific heat of water at the average temperature of (18+100)/2 = 59C is C = 4.18 kJ/kgC, the time it takes for the entire water to be heated from 18C to 100C is determined to be mCT (1 kg)(4.18 kJ / kg C)(100  18) C & We = We t = mCT t = = = 456 s = 7.60 min & 0.752 kJ / s We Discussion We can also solve this problem using vf data (instead of density), and hf data instead of specific heat. At 100 C, we have vf = 0.001044 m 3/kg and hf = 419.04 kJ/kg. At 18C, we have hf = 75.57 kJ/kg (Table A4). The two results will be practically the same. Water
100C Heater We 5184 Water is boiled at a specified temperature by hot gases flowing through a stainless steel pipe submerged in water. The rate of evaporation of is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the outer surfaces of the boiler are negligible. Properties The enthalpy of vaporization of water at 150C is hfg = 2114.3 kJ/kg (Table A4). Analysis The rate of heat transfer to water is given to be 74 kJ/s. Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a unit mass of a liquid at a specified temperature, the rate of evaporation of water is determined to be & m evaporation = & Q boiling h fg = 74 kJ/s = 0.0350 kg/s 2114.3 kJ/kg Steam Water 150C Heater Cold water 5173 Chapter 5 The First Law of Thermodynamics 5185 Cold water enters a steam generator at 20C, and leaves as saturated vapor at Tsat = 100C. The fraction of heat used to preheat the liquid water from 20C to saturation temperature of 100C is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible. 3 The specific heat of water is constant at the average temperature. Properties The heat of vaporization of water at 100C is hfg = 2257 kJ/kg (Table A4), and the specific heat of liquid water at the average temperature of (20+100)/2 = 60 C is C = 4.18 kJ/kg.C (Table A3). Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature. Using the average specific heat, the amount of heat transfer needed to preheat a unit mass of water from 20C to 100C is qpreheating = CT = (4.18 kJ / kg C)(100  20) C = 334.4 kJ / kg and q total = qboiling + qpreheating = 2257 + 334.4 = 2591.4 kJ / kg Therefore, the fraction of heat used to preheat the water is Fraction to preheat = qpreheating q total = 334.4 = 0.129 (or 12.9%) 2591.4
Cold water 20C
Water 100C Heater Steam 5186 Cold water enters a steam generator at 20C and is boiled, and leaves as saturated vapor at boiler pressure. The boiler pressure at which the amount of heat needed to preheat the water to saturation temperature that is equal to the heat of vaporization is to be determined. Assumptions Heat losses from the steam generator are negligible. Properties The enthalpy of liquid water at 20C is 83.96 kJ/kg. Other properties needed to solve this problem are the heat of vaporization hfg and the enthalpy of saturated liquid at the specified temperatures, and they can be obtained from Table A4. Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature, and h represents the amount of heat needed to preheat a unit mass of water from 20C to the saturation temperature. Therefore, qpreheating = qboiling (h f @Tsat  h f @20C ) = h [email protected] h f @Tsat  83.96 kJ / kg = h [email protected] h f @Tsat  h [email protected] = 83.96 kJ / kg The solution of this problem requires choosing a boiling temperature, reading hf and hfg at that temperature, and substituting the values into the relation above to see if it is satisfied. By trial and error, (Table A4) At 310C: At 315C: h f @Tsat  h [email protected] = 1401.3 1326 = 75.3 kJ/kg h f @Tsat  h [email protected] = 1431.0 1283.5 = 147.5 kJ/kg
Water Steam The temperature that satisfies this condition is determined from the two values above by interpolation to be 310.6C. The saturation pressure corresponding to this temperature is 9.94 MPa.
Cold water 20C Heater 5174 Chapter 5 The First Law of Thermodynamics 5187 Saturated steam at 1 atm pressure and thus at a saturation temperature of Tsat = 100C condenses on a vertical plate maintained at 90C by circulating cooling water through the other side. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 The steam condenses and the condensate drips off at 100C. (In reality, the condensate temperature will be between 90 and 100, and the cooling of the condensate a few C should be considered if better accuracy is desired). Properties The enthalpy of vaporization of water at 100C is hfg = 2257 kJ/kg (Table A4). Analysis The rate of heat transfer during this condensation process is given to be 180 kJ/s. Noting that the heat of vaporization of water represents the amount of heat released as a unit mass of vapor at a specified temperature condenses, the rate of condensation of steam is determined from & m condensation = & Q h fg = 180 kJ/s = 0.0798 kg/s 2257 kJ/kg
Plate Steam & Q
90C 100C Condensate 5188 Water is boiled at Tsat = 100C by an electric heater. The rate of evaporation of water is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the outer surfaces of the water tank are negligible. Properties The enthalpy of vaporization of water at 100C is hfg = 2257 kJ/kg (Table A4). Analysis Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a unit mass of a liquid at a specified temperature, the rate of evaporation of water is determined to be & mevaporation = & We,boiling h fg
Steam 5 kJ / s = = 0.00222 kg / s = 7.98 kg / h 2257 kJ / kg Water 100C Heater 5175 Chapter 5 The First Law of Thermodynamics 5189 An insulated cylinder initially contains saturated liquidvapor mixture of water at a specified temperature. A cold copper block is dropped into the cylinder. The final equilibrium temperature inside the cylinder and the final mass of water vapor are to be determined. Assumptions 1 The copper block has a constant specific heat at the average temperature. 2 The cylinder is wellinsulated and thus heat transfer is negligible. 3 Only part of the vapor in the cylinder condenses as a result of copper block being dropped into the cylinder, so that the temperature inside the cylinder remains constant (will be verified). Properties The specific heat of copper at the average temperature of (30+120)/2 = 75C is C = 0.391 kJ/kgC (Table A3). The enthalpy of vaporization of water at 120C is hfg = 2202.6 kJ/kg (Table A4). Analysis We take the copper block as the system, which is closed system. Assuming the final temperature to be 120C, the energy balance for copper can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Cu 30C Qin = U = [ mC (T2  T1 )]copper Substituting, Qin = (5 kg)(0.391 kJ/kg.C)(120  30)C = 176.0 kJ WATER 5 kg 120C This energy will come from the water vapor inside. Noting that the enthalpy of vaporization represents the amount of energy released when a unit mass of vapor condenses at a specified temperature, the amount of vapor that condenses to release 176 kJ of heat is mcondensed = Q 176 kJ = = 0.080 kg h fg 2202.6 kJ / kg Initially, the cylinder contained 1 kg of vapor. Then the amount of water vapor in the cylinder at the end of the process becomes mfinal = minitial  mcondensed = 1.0  0.080 = 0.920 kg This verifies our assumption that the cylinder still contains some vapor at the end of the process. Then the final temperature in the cylinder must be the same as the initial temperature since saturation conditions still exist inside, T2 = T1 =120C 5190 The air pressure in a tire is measured before and after a trip. The temperature rise of air inside the tire is to be estimated. Assumptions 1 Air is an ideal gas. 2 The volume of the tire remains constant. 3 No air leaks out of the tire during the trip. Analysis Using the ideal gas relation between the two states, the final temperature in the tire is determined to be P V P2 V 1 = T1 T2 T2 = P2 (220 + 90) kPa T1 = (25 + 273 K) = 318.6 K = 45.6 C P (200 + 90) kPa 1
Tire 25C 200 kPa Therefore, the temperature rise of air in the tire during the trip is T = T2  T1 = 45.6  25 = 20.6 C 5176 Chapter 5 The First Law of Thermodynamics 5191 Two identical buildings in Los Angeles and Denver have the same infiltration rate. The ratio of the heat losses by infiltration at the two cities under identical conditions is to be determined. Assumptions 1 Both buildings are identical and both are subjected to the same conditions except the atmospheric conditions. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady flow conditions exist. Analysis We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the building. The energy balance for this imaginary steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Los Angeles: 101 kPa Denver: 83 kPa & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & & Qin = mC p (T2  T1 ) = VC p (T2  T1 ) Then the sensible infiltration heat loss (heat gain for the infiltrating air) can be expressed & & Qinfiltrati on = m air C p (Ti  To ) = o, air ( ACH )(V building )C p (Ti  To ) where ACH is the infiltration volume rate in air changes per hour. Therefore, the infiltration heat loss is proportional to the density of air, and thus the ratio of infiltration heat losses at the two cities is simply the densities of outdoor air at those cities, Infiltration heat loss ratio = = = & Qinfiltrati on, Los Angeles o, air, Los Angeles = & o, air, Denver Qinfiltrati on, Denver ( P0 / RT0 ) Los Angeles ( P0 / RT0 ) Denver 101 kPa = 1.22 83 kPa = Po, Los Angeles P0, Denver Therefore, the infiltration heat loss in Los Angeles will be 22% higher than that in Denver under identical conditions. 5177 Chapter 5 The First Law of Thermodynamics 5192 The ventilating fan of the bathroom of an electrically heated building in San Francisco runs continuously. The amount and cost of the heat "vented out" per month in winter are to be determined. Assumptions 1 We take the atmospheric pressure to be 1 atm = 101.3 kPa since San Francisco is at sea level. 2 The building is maintained at 22C at all times. 3 The infiltrating air is heated to 22C before it exfiltrates. 4 Air is an ideal gas with constant specific heats at room temperature. 5 Steady flow conditions exist. Properties The gas constant of air is R = 0.287 kPa.m3/kgK (Table A1). The specific heat of air at room temperature is Cp = 1.005 kJ/kgC (Table A2). Analysis The density of air at the indoor conditions of 1 atm and 22C is o = Po (101.3 kPa) = = 1.20 kg/m 3 RTo (0.287 kPa.m 3 /kg.K)(22 + 273 K)
30 L/s 12.2C Then the mass flow rate of air vented out becomes & & mair = Vair = (1.20 kg / m3 )(0.030 m 3 / s) = 0.036 kg / s
22C We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the house. The energy balance for this imaginary steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Noting that the indoor air vented out at 22C is replaced by infiltrating outdoor air at 12.2C, the sensible infiltration heat loss (heat gain for the infiltrating air) due to venting by fans can be expressed & & Qloss by fan = m air C p (Tindoors  Toutdoors ) = (0.036 kg/s)(1.005 kJ/kg.C)(22  12.2)C = 0.355 kJ/s = 0.355 kW Then the amount and cost of the heat "vented out" per month ( 1 month = 3024 = 720 h) becomes & Energy loss = Qloss by fan t = (0.355 kW)(720 h/month) = 256 kWh/month Money loss = (Energy loss)(Unit cost of energy) = (256 kWh/month )($0.09 /kWh ) = $23.0/mont h Discussion Note that the energy and money loss associated with ventilating fans can be very significant. Therefore, ventilating fans should be used with care. 5178 Chapter 5 The First Law of Thermodynamics 5193 Chilled air is to cool a room by removing the heat generated in a large insulated classroom by lights and students. The required flow rate of air that needs to be supplied to the room is to be determined. Assumptions 1 The moisture produced by the bodies leave the room as vapor without any condensing, and thus the classroom has no latent heat load. 2 Heat gain through the walls and the roof is negligible. 4 Air is an ideal gas with constant specific heats at room temperature. 5 Steady operating conditions exist. Properties The specific heat of air at room temperature is 1.005 kJ/kgC (Table A2). The average rate of sensible heat generation by a person is given to be 60 W. Analysis The rate of sensible heat generation by the people in the room and the total rate of sensible internal heat generation are & & Qgen, sensible = qgen, sensible (No. of people) = (60 W / person)(150 persons) = 9000 W & & & Qtotal, sensible = Qgen, sensible + Qlighting = 9000 + 4000 = 13,000 W
Both of these effects can be viewed as heat gain for the chilled air stream, which can be viewed as a steady stream of cool air that is heated as it flows in an imaginary duct passing through the room. The energy balance for this imaginary steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = 0 (steady) & Esystem 1442443 4 4 Rate of change in internal, kinetic, potential, etc. energies =0 Chilled air 15C Return air 25C & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & =Q & Qin & total, sensible = mC p (T2  T1 ) Then the required mass flow rate of chilled air becomes & mair = & Qtotal, sensible C p T = 13 kJ / s = 1.29 kg / s (1.005 kJ / kg C)(25  15) C Discussion The latent heat will be removed by the airconditioning system as the moisture condenses outside the cooling coils. 5179 Chapter 5 The First Law of Thermodynamics 5194 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the mass flow rate of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water are constant. Properties The specific heat of chicken are given to be 3.54 kJ/kg.C. The specific heat of water is 4.18 kJ/kg.C (Table A3). Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of & mchicken = (500 chicken / h)(2.2 kg / chicken) = 1100 kg / h = 0.3056 kg / s
Taking the chicken flow stream in the chiller as the system, the energy balance for steadily flowing chickens can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & =Q & & Q =m C (T  T )
out chicken chicken p 1 2 Then the rate of heat removal from the chickens as they are cooled from 15C to 3C becomes & & Qchicken =(mC p T ) chicken = (0.3056 kg/s)(3.54 kJ/kg. C)(15  3) C = 13.0 kW The chiller gains heat from the surroundings at a rate of 200 kJ/h = 0.0556 kJ/s. Then the total rate of heat gain by the water is & & & Qwater = Qchicken + Qheat gain = 13.0 + 0.056 = 13.056 kW Noting that the temperature rise of water is not to exceed 2C as it flows through the chiller, the mass flow rate of water must be at least & m water = & Q water 13.056 kW = = 1.56 kg/s (C p T ) water (4.18 kJ/kg. C)(2 C)
Chicken 15C Immersion chilling, 0.5C If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2C. 5180 Chapter 5 The First Law of Thermodynamics 5195 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the mass flow rate of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water are constant. 3 Heat gain of the chiller is negligible. Properties The specific heat of chicken are given to be 3.54 kJ/kg.C. The specific heat of water is 4.18 kJ/kg.C (Table A3). Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of & mchicken = (500 chicken / h)(2.2 kg / chicken) = 1100 kg / h = 0.3056 kg / s
Taking the chicken flow stream in the chiller as the system, the energy balance for steadily flowing chickens can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & =Q & & Q =m C (T  T )
out chicken chicken p 1 2 Then the rate of heat removal from the chickens as they are cooled from 15C to 3C becomes & & Qchicken =(mC p T ) chicken = (0.3056 kg/s)(3.54 kJ/kg. C)(15  3) C = 13.0 kW Heat gain of the chiller from the surroundings is negligible. Then the total rate of heat gain by the water is & & Qwater = Qchicken = 13.0 kW
Noting that the temperature rise of water is not to exceed 2C as it flows through the chiller, the mass flow rate of water must be at least Immersion chilling, 0.5C & m water = & Qwater 13.0 kW = = 1.56 kg/s (C p T ) water (4.18 kJ/kg. C)(2 C) Chicken 15C If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2C. 5181 Chapter 5 The First Law of Thermodynamics 5196 A regenerator is considered to save heat during the cooling of milk in a dairy plant. The amounts of fuel and money such a generator will save per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The properties of the milk are constant. Properties The average density and specific heat of milk can be taken to be milk water = 1 kg/L and Cp, milk= 3.79 kJ/kg.C (Table A3). Analysis The mass flow rate of the milk is & & mmilk = Vmilk = (1 kg / L)(12 L / s) = 12 kg / s = 43,200 kg / h
Taking the pasteurizing section as the system, the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Hot milk 72C & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = m C (T  T ) Qin & milk p 2 1 & Q
4C Cold milk Regenerator Therefore, to heat the milk from 4 to 72C as being done currently, heat must be transferred to the milk at a rate of & & Qcurrent = [ mCp (Tpasturization  Trefrigeration )]milk = (12 kg / s)(3.79 kJ / kg. C)(72  4) C = 3093kJ / s The proposed regenerator has an effectiveness of = 0.82, and thus it will save 82 percent of this energy. Therefore, & & Qsaved = Qcurrent = (0.82)( 3093 kJ / s) = 2536 kJ / s Noting that the boiler has an efficiency of boiler = 0.82, the energy savings above correspond to fuel savings of & Q (2536 kJ / s) (1therm) = 0.02931therm / s Fuel Saved = saved = boiler (0.82) (105,500 kJ) Noting that 1 year = 36524=8760 h and unit cost of natural gas is $0.52/therm, the annual fuel and money savings will be Fuel Saved = (0.02931 therms/s)(87603600 s) = 924,450 therms/yr Money saved = (Fuel saved)(Unit cost of fuel) = (924,450 therm / yr)($0.52 / therm) = $480,700 / yr 5182 Chapter 5 The First Law of Thermodynamics 5197E A refrigeration system is to cool eggs by chilled air at a rate of 10,000 eggs per hour. The rate of heat removal from the eggs, the required volume flow rate of air, and the size of the compressor of the refrigeration system are to be determined. Assumptions 1 Steady operating conditions exist. 2 The eggs are at uniform temperatures before and after cooling. 3 The cooling section is wellinsulated. 4 The properties of eggs are constant. 5 The local atmospheric pressure is 1 atm. Properties The properties of the eggs are given to = 67.4 lbm/ft3 and Cp = 0.80 Btu/lbm.F. The specific heat of air at room temperature Cp = 0.24 Btu/lbm. F (Table A2E). The gas constant of air is R = 0.3704 psia.ft3/lbm.R (Table A1E). Analysis (a) Noting that eggs are cooled at a rate of 10,000 eggs per hour, eggs can be considered to flow steadily through the cooling section at a mass flow rate of & m egg = (10,000 eggs/h)(0.14 lbm/egg) = 1400 lbm/h = 0.3889 lbm/s
Taking the egg flow stream in the cooler as the system, the energy balance for steadily flowing eggs can be expressed in the rate form as
Egg 0.14 lbm & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Air 34F & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = Q = m C (T  T ) & & egg p 1 2 Qout egg Then the rate of heat removal from the eggs as they are cooled from 90F to 50F at this rate becomes & & Qegg = (mC p T ) egg = (1400 lbm / h)(0.80 Btu / lbm. F)(90  50) F = 44,800 Btu / h (b) All the heat released by the eggs is absorbed by the refrigerated air since heat transfer through he walls of cooler is negligible, and the temperature rise of air is not to exceed 10F. The minimum mass flow and volume flow rates of air are determined to be & mair = air = & V air & Qair 44,800 Btu / h = = 18,667 lbm / h (C p T ) air (0.24 Btu / lbm. F)(10 F) 14.7 psia P = = 0.0803 lbm/ft 3 RT (0.3704 psia.ft 3 /lbm.R)(34 + 460)R & m 18,667 lbm/h = air = = 232,500 ft/h air 0.0803 lbm/ft 5183 Chapter 5 The First Law of Thermodynamics 5198 Dough is made with refrigerated water in order to absorb the heat of hydration and thus to control the temperature rise during kneading. The temperature to which the city water must be cooled before mixing with flour is to be determined to avoid temperature rise during kneading. Assumptions 1 Steady operating conditions exist. 2 The dough is at uniform temperatures before and after cooling. 3 The kneading section is wellinsulated. 4 The properties of water and dough are constant. Properties The specific heats of the flour and the water are given to be 1.76 and 4.18 kJ/kg.C, respectively. The heat of hydration of dough is given to be 15 kJ/kg. Analysis It is stated that 2 kg of flour is mixed with 1 kg of water, and thus 3 kg of dough is obtained from each kg of water. Also, 15 kJ of heat is released for each kg of dough kneaded, and thus 315 = 45 kJ of heat is released from the dough made using 1 kg of water.
Taking the cooling section of water as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as
Flour & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
Water 15C & Q
15 kJ/kg & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & =Q & & Qout water = mwater C p (T1  T2 ) Dough In order for water to absorb all the heat of hydration and end up at a temperatu re of 15C, its temperature before entering the mixing section must be reduced to Qin = Qdough = mC p (T2  T1 ) T1 = T2  Q 45 kJ = 15 C  = 4.2 C mC p (1 kg)(4.18 kJ / kg. C) That is, the water must be precooled to 4.2C before mixing with the flour in order to absorb the entire heat of hydration. 5184 Chapter 5 The First Law of Thermodynamics 5199 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat and water mass that need to be supplied to the water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The entire water body is maintained at a uniform temperature of 55C. 3 Heat losses from the outer surfaces of the bath are negligible. 4 Water is an incompressible substance with constant properties. Properties The specific heat of water at room temperature is Cp = 4.18 kJ/kg.C. Also, the specific heat of glass is 0.80 kJ/kg.C (Table A3). Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is & mbottle = mbottle Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s
Taking the bottle flow section as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Water bath 55C & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & & Qin = Qbottle = mwater C p (T2  T1 ) Then the rate of heat removal by the bottles as they are heated from 20 to 55C is & & Q bottle = m bottle C p T = (2 kg/s )(0.8kJ/kg. C )(55  20 ) C = 56,000 W The amount of water removed by the bottles is & m water,out = (Flow rateof bottles )(Water removed per bottle ) = (800 bottles / min )(0.2 g/bottle ) = 160 g/min = 2.6710  3 kg/s Noting that the water removed by the bottles is made up by fresh water entering at 15C, the rate of heat removal by the water that sticks to the bottles is & & Q water removed = m water removed C p T = (2.67 10 3 kg/s )(4180 J/kg C)(55  15) C= 446 W Therefore, the total amount of heat removed by the wet bottles is & & & Q total, removed = Qglass removed + Q water removed = 56,000 + 446 = 56,446 W Discussion In practice, the rates of heat and water removal will be much larger since the heat losses from the tank and the moisture loss from the open surface are not considered. & Q 5185 Chapter 5 The First Law of Thermodynamics 5200 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat and water mass that need to be supplied to the water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The entire water body is maintained at a uniform temperature of 50C. 3 Heat losses from the outer surfaces of the bath are negligible. 4 Water is an incompressible substance with constant properties. Properties The specific heat of water at room temperature is Cp = 4.18 kJ/kg.C. Also, the specific heat of glass is 0.80 kJ/kg.C (Table A3). Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is & mbottle = mbottle Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s
Taking the bottle flow section as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 Water bath 50C & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & & Qin = Qbottle = mwater C p (T2  T1 ) Then the rate of heat removal by the bottles as they are heated from 20 to 50C is & & Q bottle = m bottle C p T = (2 kg/s )(0.8 kJ/kg. C )(50  20 ) C= 48,000 W The amount of water removed by the bottles is & m water,out = (Flow rate of bottles )(Water removed per bottle ) = (800 bottles / min )(0.2 g/bottle )= 160 g/min = 2.6710  3 kg/s & Q Noting that the water removed by the bottles is made up by fresh water entering at 15C, the rate of heat removal by the water that sticks to the bottles is & & Q water removed = m water removed C p T = (82.67 10 3 kg/s)( 4180 J/kg C)(50  15) C= 391 W Therefore, the total amount of heat removed by the wet bottles is & & & Q total, removed = Qglass removed + Q water removed = 48,000 + 391 = 48,391 W Discussion In practice, the rates of heat and water removal will be much larger since the heat losses from the tank and the moisture loss from the open surface are not considered. 5186 Chapter 5 The First Law of Thermodynamics 5201 Long aluminum wires are extruded at a velocity of 10 m/min, and are exposed to atmospheric air. The rate of heat transfer from the wire is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant. Properties The properties of aluminum are given to be = 2702 kg/m3 and Cp = 0.896 kJ/kg.C. Analysis The mass flow rate of the extruded wire through the air is & & m = V = (r02 )V = (2702 kg/m 3 ) (0.0015 m) 2 (10 m/min) = 0.191 kg/min
Taking the volume occupied by the extruded wire as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
350C 10 m/min & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & =Q & & Qout 1 wire = mwire C p (T  T2 ) Then the rate of heat transfer from the wire to the air becomes Aluminum wire & & Q = mC p [T (t )  T ] = (0.191 kg / min)(0.896 kJ / kg. C)( 350  50) C = 51.3 kJ / min = 0.856 kW 5202 Long copper wires are extruded at a velocity of 10 m/min, and are exposed to atmospheric air. The rate of heat transfer from the wire is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant. Properties The properties of copper are given to be = 8950 kg/m3 and Cp = 0.383 kJ/kg.C. Analysis The mass flow rate of the extruded wire through the air is & & m = V = (r02 )V = (8950 kg/m 3 ) (0.0015 m) 2 (10 m/min) = 0.633 kg/min
Taking the volume occupied by the extruded wire as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
350C 10 m/min & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & =Q & & Qout wire = mwire C p (T  T2 ) 1 Then the rate of heat transfer from the wire to the air becomes Copper wire & & Q = mC p [T (t )  T ] = (0.633 kg / min)( 0.383 kJ / kg. C)( 350  50) C = 72.7 kJ / min = 1.21 kW 5187 Chapter 5 The First Law of Thermodynamics 5203 Steam at a saturation temperature of Tsat = 40C condenses on the outside of a thin horizontal tube. Heat is transferred to the cooling water that enters the tube at 25C and exits at 35C. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 Water is an incompressible substance with constant properties at room temperature. 3 The changes in kinetic and potential energies are negligible. Properties The properties of water at room temperature are = 997 kg/m3 and Cp = 4.18 kJ/kg.C. The enthalpy of vaporization of water at a saturation temperature of 40C is hfg = 2406.7 kJ/kg (Table A4). Analysis The mass flow rate of water through the tube is & mwater = VAc = (997 kg / m 3 )(2 m / s)[ (0.03 m) 2 / 4 ] = 1.409 kg / s Taking the volume occupied by the cold water in the tube as the system, which is a steadyflow control volume, the energy balance for this steadyflow system can be expressed in the rate form as
Steam 40C 35C Cold Water 25C & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = 0 (steady) & Esystem 1442443 4 4 Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & & Qin = Qwater = mwater C p (T2  T1 ) Then the rate of heat transfer to the water and the rate of condensation become & & Q = mC p (Tout  Tin ) = (1.409 kg/s)(4.18 kJ/kg C)(35  25)C = 58.9 kW & Q 58.9 kJ/s & & & Q = m evap h fg m evap = = = 0.0245 kg/s h fg 2406.7 kJ/kg 5188 Chapter 5 The First Law of Thermodynamics 5204E Saturated steam at a saturation pressure of 0.95 psia and thus at a saturation temperature of Tsat = 100F (Table A4E) condenses on the outer surfaces of 144 horizontal tubes by circulating cooling water arranged in a 12 12 square array. The rate of heat transfer to the cooling water and the average velocity of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The tubes are isothermal. 3 Water is an incompressible substance with constant properties at room temperature. 4 The changes in kinetic and potential energies are negligible. Properties The properties of water at room temperature are = 62.1 lbm/ft3 and Cp = 1.00 Btu/lbm.F. The enthalpy of vaporization of water at a saturation temperature of 100F is hfg = 1037 Btu/lbm (Table A4E). Analysis The rate of heat transfer from the steam to the cooling water is equal to the heat of vaporization released as the vapor condenses at the specified temperature, & & Q = mh fg = (6800 lbm / h)(1037 Btu / lbm) = 7,051,600 Btu / h = 1959 Btu / s All of this energy is transferred to the cold water. Therefore, the mass flow rate of cold water must be & & Q = mwater C p T & mwater = & Q 1959 Btu / s = = 244.8 lbm / s C p T (1.00 Btu / lbm. F)(8 F) Then the average velocity of the cooling water through the 144 tubes becomes & m = AV V= & & m m 244.8 lbm / s = = = 1.26 ft / s A (nD 2 / 4 ) (62.1 lbm / ft 3 )[144 (2 /12 ft) 2 / 4 ]
Saturated steam 0.95 psia Cooling water 5189 Chapter 5 The First Law of Thermodynamics 5205 Saturated refrigerant134a vapor at a saturation temperature of Tsat = 30C condenses inside a tube. The rate of heat transfer from the refrigerant for the condensate exit temperatures of 16C and 20C are to be determined. Assumptions 1 Steady flow conditions exist. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions involved. Properties The properties of saturated refrigerant134a at 30C are hf = 91.49 kJ/kg, are hg = 263.50 kJ/kg, and are hfg = 172.00 kJ/kg. The enthalpy of saturated liquid refrigerant at 16C is hf = 71.69 kJ/kg, (Table A11). Analysis We take the tube and the refrigerant in it as the system. This is a control volume since mass crosses the system boundary during the process. We note that there is only one inlet and one exit, and thus & & & m1 = m2 = m . Noting that heat is lost from the system, the energy balance for this steadyflow system can be expressed in the rate form as & & & Esystem 0 (steady) Ein  Eout = =0 1 24 4 3 1442443 4 4 Qout
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & mh1 = Qout + mh2 (since ke pe 0) & = m (h  h ) & Q
out 1 2 R134a 30C where at the inlet state h1 = hg = 263.50 kJ/kg. Then the rates of heat transfer during this condensation process for both cases become Case 1: T2 = 30C: h2 = [email protected] = 91.49 kJ/kg. & Qout = (0.1 kg / min)(263.5  91.49) kJ / kg = 17.2 kW Case 2: T2 = 16C: h2 [email protected] =71.69 kJ/kg. & Qout = (0.1 kg/min)(263.5  71.69) kJ/kg = 19.2 kW
Discussion Note that the rate of heat removal is greater in the second case since the liquid is subcooled in that case. 5190 Chapter 5 The First Law of Thermodynamics 5206E A winterizing project is to reduce the infiltration rate of a house from 2.2 ACH to 1.1 ACH. The resulting cost savings are to be determined. Assumptions 1 The house is maintained at 72F at all times. 2 The latent heat load during the heating season is negligible. 3 The infiltrating air is heated to 72F before it exfiltrates. 4 Air is an ideal gas with constant specific heats at room temperature. 5 The changes in kinetic and potential energies are negligible. 6 Steady flow conditions exist. Properties The gas constant of air is 0.3704 psia.ft3/lbmR (Table A1E). The specific heat of air at room temperature is 0.24 Btu/lbmF (Table A2E). Analysis The density of air at the outdoor conditions is o = Po 13.5 psia = = 0.0734 lbm/ft 3 RTo (0.3704 psia.ft 3 /lbm.R)(496.5 R) The volume of the house is V building = (Floor area)(Height) = (3000 ft 2 )(9 ft) = 27,000 ft 3 We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the house. The energy balance for this imaginary steadyflow system can be expressed in the rate form as 0 (steady) & & & Ein  Eout = Esystem =0 1 24 4 3 1442443 4 4
Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & = mC (T  T ) = VC (T  T ) & Qin & p 2 1 p 2 1 Cold air 36.5F Warm air
72F Warm air 72F The reduction in the infiltration rate is 2.2 1.1 = 1.1 ACH. The reduction in the sensible infiltration heat load corresponding to it is & Qinfiltrati on, saved = o C p ( ACH saved )(V building )(Ti  To ) = (0.0734 lbm/ft 3 )(0.24 Btu/lbm.F)(1.1/h)(27,000 ft 3 )(72  36.5)F = 18,573 Btu/h = 0.18573 therm/h since 1 therm = 100,000 Btu. The number of hours during a six month period is 63024 = 4320 h. Noting that the furnace efficiency is 0.65 and the unit cost of natural gas is $0.62/therm, the energy and money saved during the 6month period are & Energy savings = (Qinfiltrati on, saved )( No. of hours per year)/Efficiency = (0.18573 therm/h)(4320 h/year)/0.65 = 1234 therms/year Cost savings = (Energy savings)( Unit cost of energy) = (1234 therms/year)($0.62/therm) = $765/year Therefore, reducing the infiltration rate by onehalf will reduce the heating costs of this homeowner by $765 per year. 5191 Chapter 5 The First Law of Thermodynamics 5207 Outdoors air at 10C and 90 kPa enters the building at a rate of 35 L/s while the indoors is maintained at 20C. The rate of sensible heat loss from the building due to infiltration is to be determined. Assumptions 1 The house is maintained at 20C at all times. 2 The latent heat load is negligible. 3 The infiltrating air is heated to 20C before it exfiltrates. 4 Air is an ideal gas with constant specific heats at room temperature. 5 The changes in kinetic and potential energies are negligible. 6 Steady flow conditions exist. Properties The gas constant of air is R = 0.287 kPa.m3/kgK (Table A1). The specific heat of air at room temperature is Cp = 1.0 kJ/kgC (Table A2). Analysis The density of air at the outdoor conditions is o = Po 90 kPa = = 1.19 kg/m 3 3 RTo (0.287 kPa.m /kg.K)(10 + 273 K) We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the building. The energy balance for this imaginary steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout & & & Qin + mh1 = mh2 (since ke pe 0) & & Qin = mC p (T2  T1 ) Then the sensible infiltration heat load corresponding to an infiltration rate of 35 L/s becomes Cold air 10C 90 kPa 35 L/s Warm air
20C Warm air 20C & & Qinfiltrati on = oV air C p (Ti  To ) = (1.19 kg/m 3 )(0.035 m 3 /s)(1.005 kJ/kg.C)[20  (10)]C = 1.256 kW
Therefore, sensible heat will be lost at a rate of 1.335 kJ/s due to infiltration. 5192 Chapter 5 The First Law of Thermodynamics 5208 The maximum flow rate of a standard shower head can be reduced from 13.3 to 10.5 L/min by switching to lowflow shower heads. The ratio of the hottocold water flow rates and the amount of electricity saved by a family of four per year by replacing the standard shower heads by the lowflow ones are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus mCV = 0 and E CV = 0 . 2 The kinetic and potential energies are negligible, ke pe 0 . 3 Heat losses & from the system are negligible and thus Q 0. 4 There are no work interactions involved. 5 .Showers operate at maximum flow conditions during the entire shower. 6 Each member of the household takes a 5min shower every day. 7 Water is an incompressible substance with constant properties. 8 The efficiency of the electric water heater is 100%. Properties The density and specific heat of water at room temperature are = 1 kg/L and C = 4.18 kJ/kg.C (Table A3). Analysis (a) We take the mixing chamber as the system. This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit. The mass and energy balances for this steadyflow system can be expressed in the rate form as follows: Mass balance: & & & min  mout = msystem 0
(steady) =0 & & & & & min = mout m1 + m2 = m3 Energy balance: & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass Mixture 3 Cold water 1 Hot water 2 = & Esystem0 (steady) 1442443
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & m1h1 + m2h2 = m3h3 (since Q 0, W = 0, ke pe 0) Combining the mass and energy balances and rearranging, & & & & m1 h1 + m2 h2 = (m1 + m2 )h3 & & m2 (h2  h3 ) = m1 (h3  h1 ) Then the ratio of the mass flow rates of the hot water to cold water becomes & m2 h3  h1 C (T3  T1 ) T3  T1 (42  15) C = 2.08 = = = = & m1 h2  h3 C (T2  T3 ) T2  T3 (55  42) C (b) The lowflow heads will save water at a rate of & Vsaved = [(13.3 10.5) L / min](5 min / person.day)(4 persons)(365 days / yr) = 20,440 L / year & & m = V = (1 kg / L)(20,440 L / year) = 20,440 kg / year
saved saved Then the energy saved per year becomes & Energy saved = msaved CT = (20,440 kg / year)(4.18 kJ / kg. C)(42 15) C = 2,307,000 kJ / year = 641 kWh (since 1 kWh = 3600 kJ) Therefore, switching to lowflow shower heads will save about 641 kWh of electricity per year. 5193 Chapter 5 The First Law of Thermodynamics 5209 Problem 5208 is reconsidered. The effect of the inlet temperature of cold water on the energy saved by using the lowflow showerhead as the inlet temperature varies from 10C to 20C is to be investigated. The electric energy savings is to be plotted against the water inlet temperature. "Knowns:" C_P = 4.18"[kJ/kgK ]" density=1"[kg/L]" {T_1 = 15"[C]"} T_2 = 55"[C]" T_3 = 42"[C]" V_dot_old = 13.3"[L/min]" V_dot_new = 10.5"[L/min]" m_dot_1=1"[kg/s]" "We can set m_dot_1 = 1 without loss of generality." "Analysis:" "(a) We take the mixing chamber as the system. This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit. The mass and energy balances for this steadyflow system can be expressed in the rate form as follows:" "Mass balance:" m_dot_in  m_dot_out = DELTAm_dot_sys DELTAm_dot_sys=0"[kg/s]" m_dot_in =m_dot_1 + m_dot_2"[kg/s]" m_dot_out = m_dot_3"[kg/s]" "The ratio of the mass flow rates of the hot water to cold water is obtained by setting m_dot_1=1[kg/s]. Then m_dot_2 represents the ratio of m_dot_2/m_dot_1" "Energy balance:" E_dot_in  E_dot_out = DELTAE_dot_sys DELTAE_dot_sys=0"[kW]" E_dot_in = m_dot_1*h_1 + m_dot_2*h_2"[kW]" E_dot_out = m_dot_3*h_3"[kW]" h_1 = C_P*T_1"[kJ/kg]" h_2 = C_P*T_2"[kJ/kg]" h_3 = C_P*T_3"[kJ/kg]" "(b) The lowflow heads will save water at a rate of " V_dot_saved = (V_dot_old  V_dot_new)"L/min"*(5"min/personday")*(4"persons")*(365"days/year") "[L/year]" m_dot_saved=density*V_dot_saved"[kg/year]" "Then the energy saved per year becomes" E_dot_saved=m_dot_saved*C_P*(T_3  T_1)"kJ/year"*convert(kJ,kWh) "[kWh/year]" "Therefore, switching to lowflow shower heads will save about 641 kWh of electricity per year. " "Ratio of hottocold water flow rates:" m_ratio = m_dot_2/m_dot_1 Esaved [kWh/year] 759.5 712 664.5 617.1 569.6 522.1 T1 [C] 10 12 14 16 18 20 5194 Chapter 5 The First Law of Thermodynamics
800 750 E saved [kW h/year] 700 650 600 550 500 10 12 14 16 18 20 T 1 [C] 5195 Chapter 5 The First Law of Thermodynamics 5210 A fan is powered by a 0.5 hp motor, and delivers air at a rate of 85 m 3/min. The highest possible air velocity at the fan exit is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus mCV = 0 and E CV = 0 . 2 The inlet velocity and the change in potential energy are negligible, V1 0 and pe 0 . 3 There are no heat and work interactions other than the electrical power consumed by the fan motor. 4 The efficiencies of the motor and the fan are 100% since best possible operation is assumed. 5 Air is an ideal gas with constant specific heats at room temperature. Properties The density of air is given to be = 1.18 kg/m3. The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kg.C (Table A2). Analysis We take the fanmotor assembly as the system. This is a control volume since mass crosses the system boundary during the process. We note that there is only one inlet and one exit, and thus & & & m1 = m2 = m . The velocity of air leaving the fan will be highest when all of the entire electrical energy drawn by the motor is converted to kinetic energy, and the friction between the air layers is zero. In this best possible case, no energy will be converted to thermal energy, and thus the temperature change of air will be zero, T2 = T1 . Then the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0 & & Ein = Eout
2 & & & We,in + mh1 = m(h2 + V2 / 2) (since V1 0 and pe 0) Noting that the temperature and thus enthalpy remains constant, the relation above simplifies further to & & 2 We,in = mV2 / 2 where
0.5 hp 180 m3/min & & m = V = (1.18 kg/m 3 )(85 m 3 /min) = 100.3 kg/min = 1.67 kg/s
Solving for V2 and substituting gives V2 = & 2We,in 2(0.5 hp) 745.7 W 1 m 2 / s 2 = & m 1.67 kg/s 1 hp 1 W = 21.1 m/s Discussion In reality, the velocity will be less because of the inefficiencies of the motor and the fan. 5196 Chapter 5 The First Law of Thermodynamics 5211 The average air velocity in the circular duct of an airconditioning system is not to exceed 10 m/s. If the fan converts 70 percent of the electrical energy into kinetic energy, the size of the fan motor needed and the diameter of the main duct are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus mCV = 0 and E CV = 0 . 2 The inlet velocity is negligible, V1 0. 3 There are no heat and work interactions other than the electrical power consumed by the fan motor. 4 Air is an ideal gas with constant specific heats at room temperature. Properties The density of air is given to be = 1.20 kg/m3. The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kg.C (Table A2). Analysis We take the fanmotor assembly as the system. This is a control volume since mass crosses the system boundary during the process. We note that there is only one inlet and one exit, and thus & & & m1 = m2 = m . The change in the kinetic energy of air as it is accelerated from zero to 10 m/s at a rate of 3 180 m /s is & & m = V = (1.20 kg/m 3 )(180 m 3 /min) = 216 kg/min = 3.6 kg/s & & KE = m V 22  V12 (10 m/s) 2  0 1 kJ/kg = (3.6 kg/s ) = 0.18 kW 2 2 1000 m 2 / s 2 It is stated that this represents 70% of the electrical energy consumed by the motor. Then the total electrical power consumed by the motor is determined to be & & 0.7Wmotor = KE & KE 0.18 kW & Wmotor = = = 0.257 kW 0.7 0.7 10 m/s 180 m3/min The diameter of the main duct is & V = VA = V (D 2 / 4) D = & 4V = V 4(180 m 3 / min) 1 min = 0.618 m (10 m/s) 60 s Therefore, the motor should have a rated power of at least 0.257 kW, and the diameter of the duct should be at least 61.8 cm 5197 Chapter 5 The First Law of Thermodynamics 5212 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: min  mout = msystem Energy balance: Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = mi = m2 Esystem 1 24 4 3 (since mout = minitial = 0) Change in internal, kinetic, potential, etc. energies Qin + mi hi = m2 u2 (since W Eout = Einitial = ke pe 0) Combining the two balances: Qin = m2 (u2  hi ) = m2 (Cv T2  C p Ti ) But Ti = T2 = T0 and Cp  Cv = R. Substituting, P0 T0 Qin = m 2 C v  C p T0 = m 2 RT0 =  Therefore, Qout = P0V ( ) P0V RT0 =  P0V RT0 V Evacuated (Heat is lost from the tank) 5198 Chapter 5 The First Law of Thermodynamics 5213 An adiabatic air compressor is powered by a directcoupled steam turbine, which is also driving a generator. The net power delivered to the generator is to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The devices are adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. Properties From the steam tables (Tables A4 through 6) P3 = 12.5MPa h3 = 3341.8kJ/kg T3 = 500 o C and P4 = 10kPa h4 = h f + x 4 h fg = 191.83 + (0.92 )(2392.8 ) = 2393.2kJ/kg x 4 = 0.92 From the air table (Table A17), T1 = 295 K h1 = 295.17 kJ/kg T2 = 620 K h 2 = 628.07 kJ/kg & & & Analysis There is only one inlet and one exit for either device, and thus min = mout = m . We take either the turbine or the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for either steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0
Air comp 1 MPa 620 K 12.5 MPa 500C & & Ein = Eout For the turbine and the compressor it becomes Compressor: Turbine: Substituting, & Wcomp, in = (10kg/s )(628.07  295.17 )kJ/kg = 3329 kW & W turb, out = (25kg/s )(3341.8  2393.2 )kJ/kg = 23,715 kW Therefore, & & & W net, out = W turb, out  W comp,in = 23,715  3329 = 20,386 kW & & & Wcomp, in + mair h1 = mair h2 & & Wcomp, in = mair (h2  h1 ) & & Wturb, out = msteam (h3  h4 ) Steam turbine & & & msteam h3 = Wturb, out + msteam h4 98 kPa 295 K 10 kPa 5199 Chapter 5 The First Law of Thermodynamics 5214 Water is heated from 16C to 43C by an electric resistance heater placed in the water pipe as it flows through a showerhead steadily at a rate of 10 L/min. The electric power input to the heater, and the money that will be saved during a 10min shower by installing a heat exchanger with an effectiveness of 0.50 are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time at any point within the system and thus mCV = 0 and E CV = 0 ,. 2 Water is an incompressible substance with constant specific heats. 3 The kinetic and potential energy changes are negligible, ke pe 0 . 4 Heat losses from the pipe are negligible. Properties The density and specific heat of water at room temperature are = 1 kg/L and C = 4.18 kJ/kgC (Table A3). Analysis We take the pipe as the system. This is a control volume since mass crosses the system boundary & & & during the process. We observe that there is only one inlet and one exit and thus m1 = m2 = m . Then the energy balance for this steadyflow system can be expressed in the rate form as & & Ein  E out 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & E system 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = E out & & & We,in + mh1 = mh2 (since ke pe 0) & & & & We,in = m(h2  h1 ) = m[C (T2  T1 ) + v ( P2  P1 ) 0 ] = mC (T2  T1 ) where & & m = V = (1 kg/L )(10 L/min ) = 10 kg/min Substituting, & We,in = (10/60kg/s ) 4.18 kJ/kg o C (43  16 )o C = 18.8 kW
16C WATER 43C ( ) 7 kW The energy recovered by the heat exchanger is & & & Qsaved = Q max = mC (Tmax  Tmin ) = 0.5(10/60 kg/s ) 4.18 kJ/kg. o C (39  16 )o C = 8.0kJ/s = 8.0kW ( ) Therefore, 8.0 kW less energy is needed in this case, and the required electric power in this case reduces to & & & Win, new = Win, old  Qsaved = 18.8  8.0 = 10.8 kW The money saved during a 10min shower as a result of installing this heat exchanger is (8.0 kW )(10/60 h )(8.5 cents/kWh ) = 11.3 cents 5200 Chapter 5 The First Law of Thermodynamics 5215 Problem 5214 is reconsidered. The effect of the heat exchanger effectiveness on the money saved as the effectiveness ranges from 20 percent to 90 percent is to be investigated, and the money saved is to be plotted against the effectiveness, "Knowns:" density = 1 "[kg/L]" V_dot = 10 "[L/min]" C = 4.18 "[kJ/kgC]" T_1 = 16"[C]" T_2 = 43"[C]" T_max = 39"[C]" T_min = T_1"[C]" epsilon = 0.5 "heat exchanger effectiveness " EleRate = 8.5 "[cents/kWh]" "For entrance, one exit, steady flow m_dot_in = m_dot_out = m_dot_wat er:" m_dot_water= density*V_dot /convert(min,s)"[kg/s]" "Energy balance for the pipe:" W_dot_ele_in+ m_dot_water*h_1=m_dot_water*h_2 "Neglect ke and pe" "For incompressible fluid in a constant pressure process, the enthalpy is:" h_1 = C*T_1"[kJ/kg]" h_2 = C*T_2"[kJ/kg]" "The energy recovered by the heat exchanger is" Q_dot_saved=epsilon*Q_dot_max "[kW]" Q_dot_max = m_dot_water*C*(T_max  T_min)"[kW]" "Therefore, 8.0 kW less energy is needed in this case, and the required electric power in this case reduces to" W_dot_ele_new = W_dot_ele_in  Q_dot_saved "[kW]" "The money saved during a 10min shower as a result of installing this heat exchanger is" Costs_saved = Q_dot_saved*10"min"*convert(min,h)*EleRate"[cents]" 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Costs saved [cents] Costssaved [cents] 4.54 6.81 9.08 11.35 13.62 15.89 18.16 20.43 24 20.33 16.67 13 9.333 5.667 2 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Heat exchanger effectivensss 5201 Chapter 5 The First Law of Thermodynamics 5216 [Also solved by EES on enclosed CD] Steam expands in a turbine steadily. The mass flow rate of the steam, the exit velocity, and the power output are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. Properties From the steam tables (Tables A4 through 6) P1 = 10 MPa v1 = 0.03564 m 3 /kg T1 = 550 o C h1 = 3500.9 kJ/kg
and 1 30 kJ/kg P2 = 25 kPa v 2 = v f + x 2 v fg = 0.00102 + (0.95)(6.203) = 5.894 m 3 /kg x 2 = 0.95 h2 = h f + x 2 h fg = 271.93 + (0.95)(2346.3) = 2500.9 kJ/kg
Analysis (a) The mass flow rate of the steam is & m= 1 1 V1 A1 = (60m/s ) 0.015m 2 = 25.3kg/s v1 0.03564m 3 /kg H2O ( ) 2 & & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit velocity is determined from & m= 1 V2 A2 v2 V2 = & mv2 (25.3 kg / s)(5.894 m 3 / kg) = = 1065 m / s A2 0.14 m 2 (c) We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & m(h1 + V12 / 2) = Wout + Qout + m(h2 + V22 /2) (since pe 0) V 2  V12 & & & Wout = Qout  m h2  h1 + 2 2 1kJ/kg 1000m 2 /s 2 Then the power output of the turbine is determined by substituting to be (1065m/s )2  (60m/s )2 & Wout = (25.3 30)kJ/s  (25.3kg/s ) 2500.9  3500.9 + 2 = 10,240kW 5202 Chapter 5 The First Law of Thermodynamics 5217 Problem 5216 is reconsidered. The effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine as the exit pressure varies from 10 kPa to 50 kPa (with the same quality), and the exit area to varies from 1000 cm2 to 3000 cm2 is to be investigated. The exit velocity and the power output are to be plotted against the exit pressure for the exit areas of 1000, 2000, and 3000 cm2. A[1]=150 "[cm^2]" T[1]=550"[c]" P[1]=10000"[kPa]" Vel[1]= 60 "[m/s]" A[2]=1400"[cm^2]" P[2]=25"[kPa]" q_out = 30"[kJ/kg" m_dot = A[1]*Vel[1]/v[1]*convert(cm^2,m^2) "[kg/s]" v[1]=volume(steam, T=T[1], P=P[1]) "[m^3/kg]" "specific volume of steam at state 1" Vel[2]=m_dot*v[2]/(A[2]*convert(cm^2,m^2)) "[m/s]" v[2]=volume(steam, x=0.95, P=P[2]) "[m^3/kg]" "specific volume of steam at state 2" T[2]=temperature(steam, P=P[2], v=v[2]) "[C]" "not required, but good to know" "[conservation of Energy for steadyflow:" "Ein_dot  Eout_dot = DeltaE_dot" "For steadyflow, DeltaE_dot = 0" DELTAE_dot=0 "[kW]" "For the turbine as the control volume, neglecting the PE of each flow steam:" E_dot_in=E_dot_out "[kW]" h[1]=enthalpy(steam,T=T[1], P=P[1]) "[kJ/kg]" E_dot_in=m_dot*(h[1]+ Vel[1]^2/(2*1000) )"[kJ/kg]" h[2]=enthalpy(steam,x=0.95, P=P[2]) "[kJ/kg]" E_dot_out=m_dot*(h[2]+ Vel[2]^2/(2*1000) )+ m_dot *q_out+ W_dot_out"[kW]" Power=W_dot_out*convert(MW,kW) "[kW]" Q_dot_out=m_dot*q_out"[kW]" Power [kW] 1.377E+07 5.469E+06 1.298E+07 1.665E+07 1.870E+07 1.994E+07 2.073E+07 2.127E+07 2.164E+07 2.190E+07 P2 [kPa] 10 14.44 18.89 23.33 27.78 32.22 36.67 41.11 45.56 50 Vel2 [m/s] 1763 1247 969.6 795.1 675.2 587.5 520.5 467.6 424.7 389.2 5203 Chapter 5 The First Law of Thermodynamics
2.500 x 10 7 1.000 x 10 7 Pow er [kW ] 4.500 x 10 7 8.000 x 10 7 1.150 x 10 8 1.500 x 10 8 10 A =1000 cm^2 2 2000 3000 15 20 25 30 35 40 45 50 P[2] [kPa] 4000 3500 3000 A = 1000 cm ^2 2 2000 3000 Vel[2] [m /s] 2500 2000 1500 1000 500 0 10 15 20 25 30 35 40 45 50 P[2] [kPa] 5204 Chapter 5 The First Law of Thermodynamics 5218E Refrigerant134a is compressed steadily by a compressor. The mass flow rate of the refrigerant and the exit temperature are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 2 Properties From the refrigerant tables (Tables A11E through A13E) P1 = 15 psia v1 = 3.2468 ft 3 /lbm o T1 = 20 F h1 = 106.34 Btu/lbm Analysis (a) The mass flow rate of refrigerant is & m= & V1 10 ft 3 / s = = 3.08 lbm / s v1 3.2468 ft 3 / lbm R134a 1 & & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & Win + mh1 = mh2 (since Q ke pe 0) & & Win = m(h2  h1 ) Substituting, 0.7068Btu/s = (3.08lbm/s )(h2  106.34 )Btu/lbm (60 hp) 1hp h2 = 120.11Btu/lbm Then the exit temperature becomes o T2 = 114.6 F h2 = 120.11Btu/lbm P2 = 120psia 5205 Chapter 5 The First Law of Thermodynamics 5219 Air is preheated by the exhaust gases of a gas turbine in a regenerator. For a specified heat transfer rate, the exit temperature of air and the mass flow rate of exhaust gases are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the regenerator to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Exhaust gases can be treated as air. 6 Air is an ideal gas with variable specific heats. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The enthalpies of air are T1 = 550 K h1 = 554.71 kJ/kg T3 = 800 K h3 = 821.95 kJ/kg T4 = 600 K h4 = 607.02 kJ/kg
Analysis (a) We take the air side of the heat exchanger as the system, which is a control volume since mass & & & crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0
Exhaust Gases AIR 1 & & Ein = Eout & & & & Qin + mair h1 = mair h2 (since W = ke pe 0) & & Qin = mair (h2  h1 ) Substituting, 3 4 2 3200kJ/s = (800/60 kg/s )(h2  554.71kJ/kg ) h2 = 794.71 kJ/kg
Then from Table A17 we read T2 = 775.1 K (b) Treating the exhaust gases as an ideal gas, the mass flow rate of the exhaust gases is determined from the steadyflow energy relation applied only to the exhaust gases, & & Ein = Eout & & & & m h =Q +m h (since W = ke pe 0)
exhaust 3 out exhaust 4 & & Qout = mexhaust (h3  h4 ) & 3200kJ/s = mexhaust (821.95  607.02 )kJ/kg It yields & mexhaust = 14.9 kg / s 5206 Chapter 5 The First Law of Thermodynamics 5220 Water is to be heated steadily from 20C to 55C by an electrical resistor inside an insulated pipe. The power rating of the resistance heater and the average velocity of the water are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time at any point within the system and thus mCV = 0 and E CV = 0 . 2 Water is an incompressible substance with constant specific heats. 3 The kinetic and potential energy changes are negligible, ke pe 0 . 4 The pipe is insulated and thus the heat losses are negligible. Properties The density and specific heat of water at room temperature are = 1000 kg/m3 and C = 4.18 kJ/kgC (Table A3). Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the system & & & boundary during the process. Also, there is only one inlet and one exit and thus m1 = m2 = m . The energy balance for this steadyflow system can be expressed in the rate form as & & E in  E out 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & E system 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0
WATER 30 L/min D = 5 cm & & E in = E out & & & & We,in + mh1 = mh2 (since Qout ke pe 0) & & & & We,in = m(h2  h1 ) = m[C (T2  T1 ) + vP 0 ] = mC (T2  T1 ) The mass flow rate of water through the pipe is
We & & m = V1 = (1000 kg/m 3 )(0.030 m 3 /min ) = 30 kg/min
Therefore, & & We,in = mC (T2  T1 ) = (30/60 kg/s )(4.18 kJ/kg o C)(55  20) o C = 73.2 kW
(b) The average velocity of water through the pipe is determined from V1 = & & V1 0.030 m 3 /min V = 2 = = 15.3 m/min A1 r (0.025 m) 2 5207 Chapter 5 The First Law of Thermodynamics 5221 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the amount of heat transfer are to be determined. Assumptions 1 The tanks are stationary and thus the kinetic and potential energy changes are zero. 2 The tank is insulated and thus heat transfer is negligible. 3 There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  Qout = U = (U ) A + (U ) B Qout = [U 2, A+ B  U 1, A  U 1, B ] (since W = KE = PE = 0)
H2O 400 kPa = [m 2, total u 2  (m1u1 ) A  (m1u1 ) B ] A B H2O 200 kPa Q
The properties of water in each tank are (Tables A4 through A6) Tank A: P1 = 400kPa v f = 0.001084, v g = 0.4625m 3 /kg x1 = 0.80 u f = 604.31, u fg = 1949.3kJ/kg u1, A = u f + x1u fg = 604.31 + (0.8 1949.3) = 2163.75kJ/kg Tank B: v1, A = v f + x1v fg = 0.001084 + [0.8 (0.4625  0.001084 )] = 0.3702m 3 /kg P1 = 200 kPa v1, B = 1.1988 m 3 /kg T1 = 250 o C u1, B = 2731.2 kJ/kg V 0.2 m 3 = 0.540 kg m1, A = A = v1, A 0.3702 m 3 /kg m1, B = VB 0.5 m 3 = = 0.417 kg v1, B 1.1988 m 3 /kg mt = m1, A + m1, B = 0.540 + 0.417 = 0.957 kg v2 = Vt 0.7 m 3 = = 0.731 m 3 /kg mt 0.957 kg T2 = 25 o C v f = 0.001003, v g = 43.36 m 3 /kg 3 v 2 = 0.731 m /kg u f = 104.88, u fg = 2304.9 kJ/kg Thus at the final state the system will be a saturated liquidvapor mixture since vf < v2 < vg . Then the final pressure must be P2 = Psat @ 25 C = 3.169 kPa Also, v2  v f 0.731  0.001 x2 = = = 0.0168 v fg 43.36  0.001 u 2 = u f + x 2 u fg = 104.88 + (0.0168 2304.9 ) = 143.60kJ/kg Substituting, Qout = [(0.957)(143.6)  (0.540)(2163.75)  (0.417)(2731.2)] = 2170 kJ 5208 Chapter 5 The First Law of Thermodynamics 5222 Problem 5221 is reconsidered. The effect of the environment temperature on the final pressure and the heat transfer as the environment temperature varies from 0C to 50C is to be investigated. The final results are to be plotted against the environment temperature. "Knowns" Vol_A=0.2"[m^3]" P_A[1]=400"[kPa]" x_A[1]=0.8 T_B[1]=250"[C]" P_B[1]=200"[kPa]" Vol_B=0.5"[m^3]" T_final=25"[C]" "T_final = T_surroundings. To do the parametric study or to solve the problem when Q_out = 0, place this statement in {}." {Q_out=0"[kJ]"} "To determine the surroundings temperature that makes Q_out = 0, remove the {} and resolve the problem." "Solution" "Conservation of Energy for the combined tanks:" E_inE_out=DELTAE E_in=0"[kJ]" E_out=Q_out"[kJ]" DELTAE=m_A*(u_A[2]u_A[1])+m_B*(u_B[2]u_B[1])"[kJ]" m_A=Vol_A/v_A[1]"[kg]" m_B=Vol_B/v_B[1]"[kg]" u_A[1]=INTENERGY(Steam,P=P_A[1], x=x_A[1])"[kJ/kg]" v_A[1]=volume(Steam,P=P_A[1], x=x_A[1])"[m^3/kg]" T_A[1]=temperature(Steam,P=P_A[1], x=x_A[1])"[C]" u_B[1]=INTENERGY(Steam,P=P_B[1],T=T_B[1])"[kJ/kg]" v_B[1]=volume(Steam,P=P_B[1],T=T_B[1])"[m^3/kg]" "At the final state the steam has uniform properties through out the entire system." u_B[2]=u_final"[kJ/kg]" u_A[2]=u_final"[kJ/kg]" m_final=m_A+m_B"[kg]" Vol_final=Vol_A+Vol_B"[m^3]" v_final=Vol_final/m_final"[m^3/kg]" u_final=INTENERGY(Steam,T=T_final, v=v_final)"[kJ/kg]" P_final=pressure(Steam,T=T_final, v=v_final)"[kPa]" Pfinal [kPa] 3.169 12.34 38.56 101.3 232 261 277.5 293.9 310.1 326.3 Qout [kJ] 2170 1977 1641 1034 19.89 155.3 192.5 229.4 266.2 303.1 Tfinal [C] 25 50 75 100 125 150 175 200 225 250 5209 Chapter 5 The First Law of Thermodynamics
2500 2000 1500 Q out [kJ] 1000 500 0 500 25 70 115 160 205 250 T final [C]
350 300 250 P final [kPa] 200 150 100 50 0 25 70 115 160 205 250 T final [C] 5210 Chapter 5 The First Law of Thermodynamics 5223 A rigid tank filled with air is connected to a cylinder with zero clearance. The valve is opened, and air is allowed to flow into the cylinder. The temperature is maintained at 30C at all times. The amount of heat transfer with the surroundings is to be determined. Assumptions 1 Air is an ideal gas. 2 The kinetic and potential energy changes are negligible, ke pe 0 . 3 There are no work interactions involved other than the boundary work. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Analysis We take the entire air in the tank and the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  Wb,out = U = m(u2  u1 ) = 0 Qin = Wb,out since u = u(T) for ideal gases, and thus u2 = u1 when T1 = T2 . The initial volume of air is P V1 P2 V2 1 = T1 T2 V2 = Q Air T= 30C P2 T2 400 kPa 1 (0.4 m 3 ) = 0.80 m 3 V1 = 200 kPa P T1 1 The pressure at the piston face always remains constant at 200 kPa. Thus the boundary work done during this process is W b,out = PdV = P (V
1 2 2 2 1 kJ  V1 ) = (200kPa)(0.8  0.4)m 3 1 kPa m 3 = 80 kJ Therefore, the heat transfer is determined from the energy balance to be Wb,out = Qin = 80 kJ 5211 Chapter 5 The First Law of Thermodynamics 5224 A wellinsulated room is heated by a steam radiator, and the warm air is distributed by a fan. The average temperature in the room after 30 min is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The kinetic and potential energy changes are negligible. 3 The air pressure in the room remains constant and thus the air expands as it is heated, and some warm air escapes. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A1). Also, Cp = 1.005 kJ/kg.K for air at room temperature (Table A2). Analysis We first take the radiator as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies 10C 4m 4m 5m Qout = U = m(u2  u1 ) Qout = m(u1  u2 ) (since W = KE = PE = 0)
Steam radiator Using data from the steam tables (Tables A4 through A6), some properties are determined to be P1 = 200kPa v1 = 1.0803m 3 /kg T1 = 200 o C u1 = 2654.4kJ/kg P2 = 100kPa v f = 0.001043, v g = 1.6940m 3 /kg (v 2 = v1 ) u f = 417.36, u fg = 2088.7kJ/kg x2 = v2  v f v fg = 1.0803  0.001043 = 0.6375 1.6940  0.001043 u 2 = u f + x 2 u fg = 417.36 + 0.6375 2088.7 = 1749 kJ/kg m= V1 0.015 m 3 = = 0.0139 kg v1 1.0803 m 3 /kg Substituting, Qout = (0.0139 kg)( 2654.4  1749)kJ/kg = 12.6 kJ The volume and the mass of the air in the room are V = 445 = 80 m and m air = P1V1 (100kPa ) 80m 3 = = 98.5kg RT1 0.2870kPa m 3 /kg K (283K ) ( ( ) ) The amount of fan work done in 30 min is & W fan,in = W fan,in t = (0120 kJ / s)(30 60 s) = 216 kJ . We now take the air in the room as the system. The energy balance for this closed system is expressed as Ein  Eout = Esystem Qin + W fan,in  Wb,out = U Qin + W fan,in = H mC p (T2  T1 ) since the boundary work and U combine into H for a constant pressure expansion or compression process. It can also be expressed as & & (Qin + W fan,in )t = mC p, ave (T2  T1 ) Substituting, (12.6 kJ) + (216 kJ) = (98.5 kg)(1.005 kJ/kg C)(T2  10)C 5212 Chapter 5 The First Law of Thermodynamics which yields T2 = 12.3C Therefore, the air temperature in the room rises from 10C to 12.3C in 30 min. 5213 Chapter 5 The First Law of Thermodynamics 5225 A cylinder equipped with a set of stops for the piston is initially filled with saturated liquidvapor mixture of water a specified pressure. Heat is transferred to the water until the volume increases by 20%. The initial and final temperature, the mass of the liquid when the piston starts moving, and the work done during the process are to be determined, and the process is to be shown on a Pv diagram. Assumptions 1 The kinetic and potential energy changes are negligible, ke pe 0 . 2 The thermal energy stored in the cylinder itself is negligible. 3 The compression or expansion process is quasiequilibrium. Analysis (a) Initially the system is a saturated mixture at 100 kPa pressure, and thus the initial temperature is T1 = Tsat @100 kPa = 99.63 o C The total initial volume is V1 = m f v f + m g v g = 2 0.001043 + 3 1.6940 = 5.08 m 3
Then the total and specific volumes at the final state are
H2O 5 kg V3 = 1.2V1 = 1.2 5.08 = 6.10 m 3 v3 =
Thus, V3 6.10 m 3 = = 1.22 m 3 /kg 5 kg m o T3 = 259.0 C v3 = 1.22 m /kg 3 P3 = 200 kPa P
2 3 (b) When the piston first starts moving, P2 = 200 kPa and V2 = V1 = 5.08 m3. The specific volume at this state is v2 = V2 5.08 m 3 = = 1.016 m 3 /kg m 5kg
3 1 v which is greater than vg = 0.8857 m /kg at 200 kPa. Thus no liquid is left in the cylinder when the piston starts moving. (c) No work is done during process 12 since V1 = V2. The pressure remains constant during process 23 and the work done during this process is
3 1 kJ Wb = P dV = P2 (V3  V 2 ) = (200 kPa )(6.10  5.08)m 3 1 kPa m 3 2 = 204kJ 5214 Chapter 5 The First Law of Thermodynamics 5226 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely. Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is wellinsulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is Cv = 0.743 kJ/kgC for N2, and R = 2.0769 kPa.m3/kg.K is Cv = 3.1156 kJ/kgC for He (Tables A1 and A2) Analysis The mass of each gas in the cylinder is PV (500 kPa ) 1 m 3 m N2 = 1 1 = = 4.77 kg RT 0.2968kPa m 3 /kg K (353 K ) 1 N2 m He PV = 1 1 RT 1 He
3 3 ) ( ) (500 kPa )(1 m ) = (2.0769 kPa m /kg K )(298 K ) = 0.808 kg
Change in internal, kinetic, potential, etc. energies ( N2 1 m3 500 kPa 80C He 1 m3 500 kPa 25C Taking the entire contents of the cylinder as our system, the 1st law relation can be written as E in  E out = E system 1 24 4 3 1 24 4 3
Net energy transfer by heat, work, and mass 0 = U = (U ) N 2 + (U )He Substituting, 0 = [mC v (T2  T1 )] N 2 + [mC v (T2  T1 )] He (4.77kg )(0.743kJ/kgo C)(T f
Tf = 57.2C  80 o C + (0.808kg ) 3.1156kJ/kg o C T f  25 o C = 0 ) ( )( ) It gives where Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats. Discussion Using the relation PV = NRuT, it can be shown that the total number of moles in the cylinder is 0.170 + 0.202 = 0.372 kmol, and the final pressure is 510.6 kPa. 5215 Chapter 5 The First Law of Thermodynamics 5227 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely. Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself, except the piston, is negligible. 3 The cylinder is wellinsulated and thus heat transfer is negligible. 4 Initially, the piston is at the average temperature of the two gases. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is Cv = 0.743 kJ/kgC for N2, and R = 2.0769 kPa.m3/kg.K is Cv = 3.1156 kJ/kgC for He (Tables A1 and A2). The specific heat of copper piston is C = 0.386 kJ/kgC (Table A3). Analysis The mass of each gas in the cylinder is PV (500kPa ) 1m 3 = 4.77 kg m N2 = 1 1 = RT 0.2968kPa m 3 /kg K (353K ) 1 N2 ( ( ) PV (500kPa ) 1m 3 = 0.808kg m He = 1 1 = RT 2.0769kPa m 3 /kg K (353K ) 1 He ( ( ) ) N2 1 m3 500 kPa 80C He 1 m3 500 kPa 25C ) Taking the entire contents of the cylinder as our system, the 1st law relation can be written as E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass Copper = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies 0 = U = (U ) N 2 + (U )He + (U )Cu 0 = [mC v (T2  T1 )] N 2 + [mC v (T2  T1 )] He + [mC (T2  T1 )] Cu where T1, Cu = (80 + 25) / 2 = 52.5C Substituting, (4.77kg )(0.743kJ/kgo C)(T f  80)o C + (0.808kg )(3.1156kJ/kgo C)(T f + (5.0kg )(0.386kJ/kgo C )(T f  52.5)o C = 0  25 o C ) It gives Tf = 56.0C where Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats. 5216 Chapter 5 The First Law of Thermodynamics 5228 Problem 5227 is reconsidered. The effect of the mass of the copper piston on the final equilibrium temperature as the mass of piston varies from 1 kg to 10 kg is to be investigated. The final temperature is to be plotted against the mass of piston. "Knowns:" R_u=8.314"[kJ/kmolK]" V_N2[1]=1"[m^3]" Cv_N2=0.743"[kJ/kgK]" "From Table A2(a) at 27C" R_N2=0.2968"[kJ/kgK]" "From Table A2(a)" T_N2[1]=80"[C]" P_N2[1]=500"[kPa]" V_He[1]=1"[m^3]" Cv_He=3.1156"[kJ/kgK]" "From Table A2(a) at 27C" T_He[1]=25"[C]" P_He[1]=500"[kPa]" R_He=2.0769"[kJ/kgK]" "From Table A2(a)" m_Pist=5"[kg]" Cv_Pist=0.386"[kJ/kgK]" "Use Cp for Copper from Table A3(b) at 27C" "Solution:" "mass calculations:" P_N2[1]*V_N2[1]=m_N2*R_N2*(T_N2[1]+273) P_He[1]*V_He[1]=m_He*R_He*(T_He[1]+273) "The entire cylinder is considered to be a closed system, neglecting the piston." "Conservation of Energy for the closed system:" "E_in  E_out = DELTAE_negPist, we neglect DELTA KE and DELTA PE for the cylinder." E_in  E_out = DELTAE_neglPist E_in =0"[kJ]" E_out = 0"[kJ]" "At the final equilibrium state, N2 and He will have a common temperature." DELTAE_neglPist= m_N2*Cv_N2*(T_2_neglPistT_N2[1])+m_He*Cv_He*(T_2_neglPistT_He[1])"[kJ]" "The entire cylinder is considered to be a closed system, including the piston." "Conservation of Energy for the closed system:" "E_in  E_out = DELTAE_withPist, we neglect DELTA KE and DELTA PE for the cylinder." E_in  E_out = DELTAE_withPist "At the final equilibrium state, N2 and He will have a common temperature." DELTAE_withPist= m_N2*Cv_N2*(T_2_withPistT_N2[1])+m_He*Cv_He*(T_2_withPistT_He[1])+m_Pist*Cv_Pist*(T_2_withPistT_Pist[1])"[kJ]" T_Pist[1]=(T_N2[1]+T_He[1])/2"[C]" "Total volume of gases:" V_total=V_N2[1]+V_He[1]"[m^3]" "Final pressure at equilibrium:" "Neglecting effect of piston, P_2 is:" P_2_neglPist*V_total=N_total*R_u*(T_2_neglPist+273)"[kPa]" "Including effect of piston, P_2 is:" N_total=m_N2/molarmass(nitrogen)+m_He/molarmass(Helium)"[kmol]" P_2_withPist*V_total=N_total*R_u*(T_2_withPist+273)"[kPa]" 5217 Chapter 5 The First Law of Thermodynamics mPist [kg] 1 2 3 4 5 6 7 8 9 10 T2,neglPist [C] 57.17 57.17 57.17 57.17 57.17 57.17 57.17 57.17 57.17 57.17 T2,withPist [C] 56.89 56.64 56.42 56.22 56.04 55.88 55.73 55.59 55.47 55.35 60 W ithout Piston
58 T 2 [C] 56 54 W ith Piston 52 50 1 2 3 4 5 6 7 8 9 10 Mass of Piston [kg] 5218 Chapter 5 The First Law of Thermodynamics 5229 A relation for the explosive energy of a fluid is given. A relation is to be obtained for the explosive energy of an ideal gas, and the value for air at a specified state is to be evaluated. Analysis The explosive energy per unit volume is given as u u eexplosion = 1 2 v1 For an ideal gas, u1  u2 = Cv(T1  T2) C p  Cv = R v1 = and thus Cv Cv 1 1 = = = R C p  Cv C p / Cv  1 k  1 Substituting, e explosion = which is the desired result. Using the relation above, the total explosive energy of 20 m of air at 5 MPa and 100C when the surroundings are at 20C is determined to be E explosion = Veexplosion = P1V1 T2 1  k  1 T1 (5000kPa ) 20m 3 = 1.4  1 C v (T1  T2 ) P T = 1 1  2 RT1 / P1 k  1 T1 RT1 P 1 ( ) 1  293K 1kJ 1kPa m 3 373K = 53,619kJ 5219 Chapter 5 The First Law of Thermodynamics 5230 Using the relation for explosive energy given in the previous problem, the explosive energy of steam and its TNT equivalent at a specified state are to be determined. Assumptions Steam condenses and becomes a liquid at room temperature after the explosion. Properties The properties of steam at the initial and the final states are (Table A4 through A6) P1 = 10MPa v1 = 0.03279m 3 /kg T1 = 500 o C u1 = 3045.8kJ/kg T2 = 25 o C u 2 u f @ 25o C = 104.88kJ/kg Comp.liquid Analysis The mass of the steam is m= V 20 m = = 609.9 kg v1 0.03279 m 3 / kg
3 25C STEAM
10 MPa 500C Then the total explosive energy of the steam is determined from E explosive = m(u1  u 2 ) = (609.9kg )(3045.8  104.88)kJ/kg = 1,793,667kJ which is equivalent to 1,793,667 kJ = 552 kg of TNT 3250 kJ / kg of TNT 5220 Chapter 5 The First Law of Thermodynamics 5231 Solar energy is to be stored as sensible heat using phasechange materials, granite rocks, and water. The amount of heat that can be stored in a 5m3 = 5000 L space using these materials as the storage medium is to be determined. Assumptions 1 The materials have constant properties at the specified values. 2 No allowance is made for voids, and thus the values calculated are the upper limits. Analysis The amount of energy stored in a medium is simply equal to the increase in its internal energy, which, for incompressible substances, can be determined from U = mC (T2  T1 ) . (a) The latent heat of glaubers salts is given to be 329 kJ/L. Disregarding the sensible heat storage in this case, the amount of energy stored is becomes Usalt = mhif = (5000 L)(329 kJ/L) = 1,645,000 kJ This value would be even larger if the sensible heat storage due to temperature rise is considered. (b) The density of granite is 2700 kg/m3 (Table A3), and its specific heat is given to be C = 2.32 kJ/kg.C. Then the amount of energy that can be stored in the rocks when the temperature rises by 20C becomes Urock = VCT = (2700 kg/m3 )(5 m3)(2.32 kJ/kg.C)(20C) = 626,400 kJ (c) The density of water is about 1000 kg/m3 (Table A3), and its specific heat is given to be C = 4.0 kJ/kg.C. Then the amount of energy that can be stored in the water when the temperature rises by 20C becomes Urock = VCT = (1000 kg/m3 )(5 m3)(4.0 kJ/kg.C)(20C) = 400,00 kJ Discussion Note that the greatest amount of heat can be stored in phasechange materials essentially at constant temperature. Such materials are not without problems, however, and thus they are not widely used. 5221 Chapter 5 The First Law of Thermodynamics 5232 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratio of the mass flow rates of the extracted seam the feedwater are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The enthalpies of steam and feedwater at are (Tables A4 through A6) P = 1.2MPa 1 h1 = 2827.9kJ/kg o T1 = 250 C P = 1MPa h2 = h f @1MPa = 762.81kJ/kg 1 sat.liquid T2 = 179.91o C and P3 = 2.5MPa h3 h f @50o C = 209.33kJ/kg T3 = 50 o C P4 = 2.5MPa h4 h f @170o C = 719.2kJ/kg o T4 = T2  10 170 C STEAM 1
Feedwater 3 4 Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance (for each fluid stream):
0 & & & min  mout = msystem (steady) 2 & & & & & = 0 min = mout m1 = m2 = ms and & & & m3 = m 4 = m fw Energy balance (for the heat exchanger): & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, & & m s (h 2  h1 ) = m fw (h3  h4 ) & Dividing by m fw and substituting, & ms h  h4 (719.2  209.33)kJ/kg = = 0.247 = 3 & m fw h2  h1 (2827.9  762.81)kJ/kg 5222 Chapter 5 The First Law of Thermodynamics 5233 A building is to be heated by a 30kW electric resistance heater placed in a duct inside. The time it takes to raise the interior temperature from 14C to 24C, and the average mass flow rate of air as it passes through the heater in the duct are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is negligible. 5 No air leaks in and out of the building. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A1). The specific heats of air at room temperature are Cp = 1.005 and Cv = 0.718 kJ/kgK (Table A2). Analysis (a) The total mass of air in the building is m= P1V1 (95 kPa ) 400 m 3 = = 461.3 kg . RT1 0.287 kPa m 3 /kg K (287 K ) ( ( ) ) We first take the entire building as our system, which is a closed system since no mass leaks in or out. The time required to raise the air temperature to 24C is determined by applying the energy balance to this constant volume closed system: E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies & & & t We,in + W fan ,in  Qout = mC v, ave (T2  T1 ) Solving for t gives ( We,in + W fan,in  Qout = U ) (since KE = PE = 0) t = mC v ,ave (T2  T1 ) (461.3 kg )(0.718 kJ/kg o C)(24  14) o C = = 146 s & & (30 kJ/s) + (0.25 kJ/s)  (450/60 kJ/s) We,in + W fan ,in  Qout (b) We now take the heating duct as the system, which is a control volume since mass crosses the boundary. & & & There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this adiabatic steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies 0 (steady) =0
450 kJ/min T2 = T1 + 5C V = 400 m3 P = 95 kPa & & Ein = Eout & & & & & We,in + W fan,in + mh1 = mh2 (since Q = ke pe 0) & & & & We,in + W fan,in = m(h2  h1 ) = mC p (T2  T1 ) Thus, & m= & & We,in + W fan,in C p T = (30 + 0.25) kJ / s (1.005 kJ / kgo C)(5 o C) = 6.02 kg / s 14C 24C T1 We 250 W 5223 Chapter 5 The First Law of Thermodynamics 5234 [Also solved by EES on enclosed CD] An insulated cylinder equipped with an external spring initially contains air. The tank is connected to a supply line, and air is allowed to enter the cylinder until its volume doubles. The mass of the air that entered and the final temperature in the cylinder are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasiequilibrium. 3 Kinetic and potential energies are negligible. 4 The spring is a linear spring. 5 The device is insulated and thus heat transfer is negligible. 6 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kJ/kgK (Table A1). The specific heats of air at room temperature are Cv = 0.718 and Cp = 1.005 kJ/kgK (Table A2a). Also, u = CvT and h = CpT. Analysis We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass mi = m2  m1 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies mi hi = Wb,out + m2 u2  m1u1 (since Q ke pe 0) Combining the two relations, or, Fspring (m 2  m1 )hi = Wb,out + m 2 u 2  m1u1 (m2  m1 )C p Ti = Wb,out + m2 Cv T2  m1Cv T1 P1V1 (200kPa ) 0.2m 3 = = 0.472kg RT1 0.287 kPa m 3 /kg K (295K ) P2V2 RT2
3 3 The initial and the final masses in the tank are m1 = m2 = ) ( ) (600kPa )(0.4m ) = (0.287kPa m /kg K )T ( P = 200 kPa T1 = 22C V1 = 0.2 m 3 Air Pi = 0.8 MPa Ti = 22C = 2 836.2 T2 Then from the mass balance becomes 836.2  0.472 T2 The spring is a linear spring, and thus the boundary work for this process can be determined from mi = m2  m1 = Wb = Area = P1 + P2 (V2  V1 ) = (200 + 600)kPa (0.4  0.2)m 3 = 80kJ 2 2 Substituting into the energy balance, the final temperature of air T2 is determined to be 836.2 836.2  80 =  T  0.472 (1.005)(295) + T (0.718)(T2 )  (0.472 )(0.718)(295 ) 2 2 It yields Thus, and T2 = 344.1 K m2 = 836.2 836.2 = = 2.430 kg T2 344.1 mi = m2  m1 = 2.430  0.472 = 1.958 kg 5224 Chapter 5 The First Law of Thermodynamics 5235 Pressurized air stored in a large cave is to be used to drive a turbine. The amount of work delivered by the turbine for specified turbine exit conditions is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The system is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ/kgK (Table A1). The specific heats of air at room temperature are Cv = 0.718 and Cp = 1.005 kJ/kgK (Table A2a). Also, u = CvT and h = CpT. Analysis We take the cave as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem E in  E out 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = E system 1 24 4 3
Change in internal, kinetic, potential, etc. energies  m e he = m 2 u 2  m1u1 (since Q W ke pe 0) Combining the two: or, (m1  m2 )he + m2u2  m1u1 = 0
CAVE 10,000 m3 500 kPa 400 K AIR P1V P2V T1 + T2 P2V PV C v T2  1 C v T1 = 0 + RT  RT C p 2 RT2 RT1 2 1 P1 P2  T T 2 1 T1 + T2 k + P2  P1 = 0 2 Multiply by R/VCv: Substituting, 500 300 400 + T2 + 300  500 = 0 400  T (1.4 ) 2 2 T22  68.57T2  96,000 = 0 It yields T2 = 346 K The initial and the final masses of air in the cave are determined to be m1 = m2 = P1V (500kPa ) 10 4 m 3 = = 43,554kg RT1 0.287kPa m 3 /kg K (400K ) turbine Air W P2V (300kPa ) 10 4 m 3 = = 30,211kg RT2 0.287kPa m 3 /kg K (346K ) ( ( ( ( ) ) ) ) 100 kPa 300 K Then from the mass balance we get me = m1  m2 = 43,554  30,211 = 13,343 kg The average temperature at the turbine inlet is (400 + 346)/2 = 373 K. Taking the turbine as system and assuming the air properties at the turbine inlet to be constant at the average temperature, the & & turbine work output is determined from the steadyflow energy balance Ein = Eout to be mi hi  me he  Wout = 0 or Wout = m e (hi  he ) turbine = (13,343 kg )(373.7  300.19 )kJ/kg = 981MJ 5225 Chapter 5 The First Law of Thermodynamics 5236E Steam is decelerated in a diffuser from a velocity of 500 ft/s to 100 ft/s. The mass flow rate of steam, the rate of heat transfer, and the inlet area of the diffuser are to be determined. Assumptions 1 This is a steadyflow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. Properties From the steam tables (Tables A4E through A6E) P1 = 14.7 psia v1 = 31.36ft 3 /lbm o T1 = 320 F h1 = 1202.1Btu/lbm and T2 = 240 o F v 2 = 16.327ft 3 /lbm sat.vapor h2 = 1160.7Btu/lbm Q=? P1 =14.7 psia T1 = 320F V1 = 500 ft/s Steam T2 = 240F Sat. vapor V2 = 100 ft/s A2 = 120 in 2 Analysis (a) The mass flow rate of the steam can be determined from its definition to be & m= 1 1 V2 A2 = (100ft/s ) 120/144ft 2 = 5.104lbm/s v2 16.327ft 3 /lbm ( ) (b) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & Ein  Eout 1 24 4 3
Rate of net energy transfer by heat, work, and mass = & Esystem 0 (steady) 1442443 4 4
Rate of change in internal, kinetic, potential, etc. energies =0 & & Ein = Eout & & & & Qin + m(h1 + V12 / 2) = m(h 2 + V12 /2) (since W pe 0) V 2  V12 & & Qin = m h2  h1 + 2 2 Substituting, (100ft/s )2  (500ft/s )2 & Qin = (5.104lbm/s )1160.7  1202.1 + 2 1Btu/lbm 25,037ft 2 /s 2 = 235.8Btu/s & & & (c) There is only one inlet and one exit, and thus m1 = m2 = m . Then the inlet area of the diffuser becomes & m= & mv1 (5.104lbm/s ) 31.36ft 3 /lbm 1 V1 A1 A1 = = = 0.320ft 2 v1 V1 500ft/s ( ) 5226 Chapter 5 The First Law of Thermodynamics 5237 20% of the volume of a pressure cooker is initially filled with liquid water. Heat is transferred to the cooker at a rate of 400 W. The time it will take for the cooker to run out of liquid is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of water are (Tables A4 through A6) P1 = 200 kPa v f = 0.001061 m 3 /kg, v g = 0.8857 m 3 /kg u f = 504.49 kJ/kg, u g = 2529.5 kJ/kg P2 = 200 kPa v 2 = v g @ 200 kPa = 0.8857 m 3 /kg sat.vapor u 2 = u g @ 200 kPa = 2529.5 kJ/kg Pe = 200 kPa he = h g @ 200 kPa = 2706.7 kJ/kg sat.vapor Pressure Cooker 5L 200 kPa & Q in = 400 W Analysis We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Qin  me he = m2 u2  m1u1 (since W ke pe 0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + m g = Vf vf + Vg vg = 0.001 m 3 0.004 m 3 + = 0.9425 + 0.0045 = 0.9470 kg 0.001061 m 3 /kg 0.8857 m 3 /kg U 1 = m1u1 = m f u f + m g u g = (0.9425)(504.49 ) + (0.0045)(2529.5) = 486.86 kJ m2 = 0.005 m 3 V = = 0.0056 kg v 2 0.8857 m 3 /kg
m e = m1  m 2 = 0.9470  0.0056 = 0.9414 kg
and From mass and energy balances, (0.4 kJ/s )t = (0.9414 kg )(2706.7 kJ/kg ) + (0.0056 kg )(2529.5 kJ/kg )  486.86 kJ
It yields t = 5188 s = 1.44 h & Qt = me he + m2 u 2  m1u1 5227 Chapter 5 The First Law of Thermodynamics 5238 A balloon is initially filled with pressurized helium gas. Now a valve is opened, and helium is allowed to escape until the pressure inside drops to atmospheric pressure. The final temperature of helium in the balloon and the mass of helium that has escaped are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniformflow process by assuming the properties of helium that escape to be constant at average conditions. 2 Kinetic and potential energies are negligible. 3 There are no work interactions other than boundary work. 4 Helium is an ideal gas with constant specific heats. 5 Heat transfer is negligible. Properties The gas constant of helium is R =2.0769 kPa.m3/kg.K (Table A1). The specific heats of helium are Cp = 5.1926 and Cv = 3.1156 kJ/kgK (Table A2). Analysis (a) The properties of helium leaving the balloon are changing during this process. But we will treat them as a constant at the average temperature. Thus Te (T1 + T2)/2. Also h = CpT and u = CvT. We take the balloon as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as Mass balance: Energy balance: min  mout = msystem Ein  Eout 1 24 4 3
Net energy transfer by heat, work, and mass me = m1  m2 = Esystem 1 24 4 3
Change in internal, kinetic, potential, etc. energies Wb,in  me he = m2 u2  m1u1 (since Q ke pe 0) or Wb,in = me C p T1 + T2 + m2 Cv T2  m1Cv T1 2 He 20C 150 kPa 25 m3 The final volume of helium is P1 = 100 + bV1 b = (P1 + 100 ) / V1 = (150 + 100 ) / 25 = 10 P2 = 100 + 10V2 V2 = (P2 + 100 ) / 10 = (100 + 100 ) / 10 = 20m 3 The initial and the final masses of helium in the balloon are m1 = m2 = P1V1 (150kPa ) 25m 3 = = 6.162kg RT1 2.0769kPa m 3 /kg K (293K ) P2V2 (100kPa ) 20m 3 962.974 = = 3 RT2 T2 2.0769kPa m /kg K T2 ( ( ( ( ) ) ) ) Then from the mass balance we have mi = m2  m1 = 6.162  962.974 T2 The boundary work done during this process is Wb ,in = P1 + P2 (V1  V2 ) = (150 + 100 )kPa (25  20)m 3 = 625kJ 2 2 Then from the energy balance, T +T Wb,in = me C p 1 2 + m2 Cv T2  m1Cv T1 2 293 + T2 962.974 962.974 (5.1926 ) (3.1156 )(T2 )  (6.162 )(3.1156 )(293) 625 = 6.162  + T 2 T2 2 It yields Solving for T2 yields T22  66.37T2  45,789 = 0 T2 = 249.7 K 5228 Chapter 5 The First Law of Thermodynamics (b) The amount of helium that has escaped is . me = m1  m2 = 6162  962.974 962.974 . = 6162  = 2.306 kg 249.7 T2 5239 Problem 5238 is to be solved using a stepwise approach. Use (a) 5, (b) 20, and (c) 50 increments for pressure between the initial value of 150 kPa and the final value of 100 kPa. Take the starting point of the first step to be the initial state of the helium (150 kPa, 20C, and 25 m^3). The starting point of the second step is the state of the helium at the end of the first step, and so on. Compare your results with those obtained by using the uniformflow approximation (i.e., a onestep solution). Procedure EnergyFunc(P1,T1, V1,P2,T2,a,b:V2,W_b,mout,F,T_out) C_P = 5.1926"[kJ/kgK ]" C_V = 3.1156 "[kJ/kgK ]" R=2.0769 "[kPam^3/kgK]" V2 = (P2a)/b"[m^3]" Tout=(T1+T2)/2 T_out=Tout "Analysis:" "Mass balance:" m1=P1*V1/(R*T1) m2=P2*V2/(R*T2) min = 0"[kg]" mout = min  m2 + m1 "Energy balance:" "Boundary Work: Due to pressure changing linearly with volume. Note the minus sign for work in" W_b = (P1+P2)/2*(V2V1) u1=C_V*T1"[kJ/kg]" u2= C_V*T2"[kJ/kg]" hout = C_P*Tout"[kJ/kg]" Ein= W_b"[kJ]" Eout = mout*hout"[kJ]" DELTAEsys = m2*u2m1*u1 F=Ein  Eout  DELTAEsys End Procedure StepSolution(P_1,T_1, V_1,P_2,DELTAP_step,a,b:V_2_step,T_2_step,W_b_step,m_out_step,T_out_step) R=2.0769 "[kPam^3/kgK]" P_1_step=P_1 T_1_step=T_1 V_1_step=V_1 DELTAT=1 W_b_step =0 m_out_step=0 Repeat P_2_step=P_1_stepDELTAP_step T_2_step=T_1_stepDELTAT T2=T_2_step T_out_step=(T_1_step + T2)/2 V_1_step_a=V_1_step T_1_step_a=T_1_step 5229 Chapter 5 The First Law of Thermodynamics V_2_step = (P_2_stepa)/b"[m^3]" Call EnergyFunc(P_1_step,T_1_step, V_1_step,P_2_step,T2,a,b:V2,W_b,mout,F,Tout) F1=F T22 = 0.99*T2 ICT = 0 Repeat ICT = ICT +1 Call EnergyFunc(P_1_step,T_1_step, V_1_step,P_2_step,T22,a,b:V2,W_b,mout,F,Tout) F2=F T_2_step = T22F2/(F2F1)*(T22T2) T2=T22 F1=F2 T22=T_2_step Until (abs(F2)<1E4) or (ICT>100) T_out_step = Tout V_2_step = V2 W_b_step=W_b_step+ W_b m_out_step=m_out_step+(P_1_step*V_1_step/(R*T_1_step)P_2_step*V_2_step/(R*T_2_step)) V_1_step=V_2_step T_1_step=T_2_step P_1_step=P_2_step Until (P_2_step<=P_2) END "UNIFORM_FLOW SOLUTION:" "Knowns:" C_P = 5.1926"[kJ/kgK ]" C_V = 3.1156 "[kJ/kgK ]" R=2.0769 "[kPam^3/kgK]" P_1= 150"[kPa]" T_1 = 293"[K]" V_1 = 25"[m^3]" "P = a + b*V, where" a = 100 "[kPa]" b=(P_1a)/V_1"[kPa/m^3]" T_out = (T_1 + T_2)/2"[K]" {DELTAP_step=50"[kPa]"} P_2= 100"[kPa]" V_2 = (P_2a)/b"[m^3]" "Analysis: We take the balloon as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, the mass and energy balances for this uniformflow system can be expressed as" "Mass balance:" m_in = 0"[kg]" m_in  m_out = m_2  m_1 5230 Chapter 5 The First Law of Thermodynamics "Energy balance:" E_in  E_out = DELTAE_sys E_in = W_b_in"[kJ]" E_out = m_out*h_out"[kJ]" h_out = C_P*T_out"[kJ/kg]" DELTAE_sys = m_2*u_2m_1*u_1 u_1=C_V*T_1"[kJ/kg]" u_2= C_V*T_2"[kJ/kg]" "The volume flow rates of air are determined to be:" P_1*V_1=m_1*R*T_1 P_2*V_2=m_2*R*T_2 "Boundary Work: Due to pressure changing linearly with volume. Note the minus sign for work in" W_b_in = (P_1+P_2)/2*(V_2V_1) "The minus sign is for work in." Call StepSolution(P_1,T_1, V_1,P_2,DELTAP_step,a,b : V_2_step,T_2_step,W_b_step,m_out_step,T_out_step) SOLUTION Variables in Main a=100 [kPa] b=10 [kPa/m^3] C_P=5.193 [kJ/kgK ] C_V=3.116 [kJ/kgK ] DELTAE_sys=2625 [kJ] DELTAP_step=50 [kPa] E_in=625 [kJ] E_out=3250 [kJ] h_out=1409 [kJ/kg] m_1=6.162 [kg] m_2=3.856 [kg] m_in=0 [kg] m_out=2.307 [kg] m_out_step=2.307 [kg] P_1=150 [kPa] P_2=100 [kPa] R=2.077 [kPam^3/kgK] T_1=293 [K] T_2=249.7 [K] T_2_step=249.7 [K] T_out=271.4 [K] T_out_step=271.4 [K] u_1=912.9 [kJ/kg] u_2=778.1 [kJ/kg] V_1=25 [m^3] V_2=20 [m^3] V_2_step=20 W_b_in=625 [kJ] W_b_step=625 [kJ] 5 240 ... 2246 Design and Essay Problems hg 5231 ...
View
Full
Document
This note was uploaded on 04/02/2008 for the course MEEN 227 taught by Professor Notsure during the Spring '06 term at Texas A&M.
 Spring '06
 Notsure
 Heat Transfer

Click to edit the document details