Unformatted text preview: Chapter 7 Exergy: A Measure of Work Potential 715 Windmills are to be installed at a location with steady winds to generate power. The minimum number of windmills that need to be installed is to be determined. Assumptions Air is at standard conditions of 1 atm and 25°C Properties The gas constant of air is 0.287 kPa.3/kg.K. Analysis The exergy or work potential of the blowing air is the kinetic energy it possesses, Exergy = kei = V12 (12 m/s) 2 § 1 kJ/kg · = ¨ ¸ = 0.072 kJ/kg 2 2 2 2 © 1000 m / s ¹ At standard atmospheric conditions (25°C, 101 kPa), the density and the mass flow rate of air are ρ=
and P 101 kPa . = = 118 m 3 / kg RT (0.287 kPa ⋅ m 3 / kg ⋅ K)(298 K) m = ρAV1 = ρ
Thus, π D2 . V1 = (118 kg / m 3 )(π / 4 )(10 m) 2 (12 m / s) = 1,112 kg / s 4 Available Power = m ke1 = (1112 kg / s)(0.072 kJ / kg) = 80.1 kW , The minimum number of windmills that needs to be installed is N= Wtotal 600 kW = = 7.49 ≅ 8 windmills 80.1 kW W 716 Water is to be pumped to a high elevation lake at times of low electric demand for use in a hydroelectric turbine at times of high demand. For a specified energy storage capacity, the minimum amount of water that needs to be stored in the lake is to be determined.
Assumptions The evaporation of water from the lake is negligible. Analysis The exergy or work potential of the water is the potential energy it possesses, Exergy = PE = mgh
Thus, 75 m m= 5 × 10 6 kWh § 3600 s ·§ 1000 m 2 / s 2 PE = ¨ ¸¨ gh (9.8 m/s 2 )(75 m) © 1 h ¹¨ 1 kW ⋅ s/kg © · ¸ = 2.45 × 10 10 kg ¸ ¹ 72 Chapter 7 Exergy: A Measure of Work Potential 729 A cylinder is initially filled with R134a at a specified state. The refrigerant is cooled and condensed at constant pressure. The exergy of the refrigerant at the initial and final states, and the exergy destroyed during this process are to be determined. √
Assumptions The kinetic and potential energies are negligible. Properties From the refrigerant tables (Tables A11 through A13), v1 = 0.02846 m 3 / kg P1 = 0.8 MPa ½ ¾ u1 = 261.62 kJ/kg T1 = 50°C ¿ s = 0.9711 kJ/kg ⋅ K 1
v ≅ v f @ 30°C = 0.0008417 m 3 / kg P2 = 0.8 MPa ½ 2 ¾ u 2 ≅ u f @ 30°C = 90.84 kJ/kg T2 = 30°C ¿s ≅s 2 f @ 30°C = 0.3396 kJ/kg ⋅ K v = 0.24216 m 3 / kg P0 = 0.1 MPa ½ 0 ¾ u 0 = 254.54 kJ/kg T0 = 30°C ¿ s = 1.1122 kJ/kg ⋅ K 0
Analysis (a) From the closed system exergy relation,
R134a 0.8 MPa P = const. Q X 1 = Φ 1 = m{(u1 − u 0 ) − T0 ( s1 − s 0 ) + P0 ( v1 − v 0 )} = (5 kg){(261.62 − 254.54) kJ/kg − (303 K)(0.9711 − 1.1122) kJ/kg ⋅ K + (100 kPa)(0.02846 − 0.24216)m 3 /kg[kJ/m 3 ⋅ kPa]} = 142.3 kJ
and, X 2 = Φ 2 = m{(u 2 − u 0 ) − T0 ( s 2 − s 0 ) + P0 ( v 2 − v 0 )} = (5 kg){(90.84 − 254.54) kJ/kg  (303 K)(0.3396 − 1.1122) kJ/kg ⋅ K + (100 kPa)(0.0008417 − 0.24216)m 3 /kg[kJ/m 3 ⋅ kPa]} = 231.1 kJ (b) The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out
Net exergy tra nsfer by heat, work,and mass − X destroyed ©0 (reversibl e) = ∆X system
Change in exergy Exergy destructio n Wrev,in = X 2 − X 1 = 231.1 −142.3 = 88.8 kJ
Noting that the process involves only boundary work, the useful work input during this process is simply the boundary work in excess of the work done by the surrounding air, Wu ,in = Win − Wsurr ,in = Win − P0 (V1 − V2 ) = P(V1 − V2 ) − P0 m( v1 − v 2 ) = m( P − P0 )( v 1 − v 2 ) § 1 kJ · = (5 kg)(800  100 kPa)(0.02846 − 0.0008417 m 3 / kg)¨ ¸ = 96.7 kJ 3 © 1 kPa ⋅ m ¹
Knowing both the actual useful and reversible work inputs, the exergy destruction or irreversibility that is the difference between the two is determined from its definition to be X destroyed = I = Wu ,in − Wrev,in = 96.7 − 88.8 = 7.9 kJ 77 Chapter 7 Exergy: A Measure of Work Potential 737 An insulated cylinder is initially filled with saturated R134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The change in the exergy of the refrigerant during this process and the reversible work are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is wellinsulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The process is stated to be reversible. Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables, v1 = v g @ 0.8 MPa = 0.0255 m 3 / kg P1 = 0.8 MPa ½ ¾ u1 = u g @ 0.8 MPa = 243.78 kJ/kg sat.vapor ¿s =s 1 g @ 0.8 MPa = 0.9068 kJ/kg ⋅ K
The mass of the refrigerant is m= V 0.08 m 3 = = 3.137 kg v1 0.0255 m 3 / kg
x2 = s2 − s f s fg = 0.9066 − 0.2399 = 0.988 0.9145 − 0.2399 R134a 0.8 MPa Reversible P2 = 0.4 MPa ½ 3 ¾ v 2 = v f + x 2 v fg = 0.0007904 + 0.988 × (0.0509 − 0.0007904) = 0.05030 m /kg s 2 = s1 ¿ u = u + x u = 61.69 + 0.988 × (231.97 − 61.69) = 229.93 kJ/kg 2 f 2 fg The reversible work output, which represents the maximum work output Wrev,out can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out
Net exergy tra nsfer by heat, work,and mass − X destroyed©0 (reversible) = ∆X system
Exergy destructio n Change in exergy  Wrev,out = X 2 − X1 → Wrev,out = X1 − X 2 = Φ1 − Φ2 Therefore, the change in exergy and the reversible work are identical in this case. Using the definition of the closed system exergy and substituting, the reversible work is determined to be Wrev ,out = Φ 1 − Φ 2 = m (u1 − u 2 ) − T0 ( s1 − s 2 ) © + P0 ( v1 − v 2 ) = m[(u1 − u 2 ) + P0 ( v1 − v 2 )]
0 [ = (3.137 kg)[(243.78 − 229.93) kJ/kg + (100 kPa)(0.0255 − 0.05030)m 3 / kg[kJ/kPa ⋅ m 3 ] = 35.7 kJ 713 Chapter 7 Exergy: A Measure of Work Potential 759E Air is compressed steadily by a compressor from a specified state to another specified state. The minimum power input required for the compressor is to be determined.
Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 100 psia 480°F Properties The gas constant of air is R = 0.06855 Btu/lbm.R (Table A1E). From the air table (Table A17E) T1 = 520 R → h1 = 124.16 Btu/lbm s1o = 0.59173 Btu/lbm ⋅ R T2 = 940 R → h2 = 226.11 Btu/lbm
o s 2 = 0.73509 Btu/lbm ⋅ R AIR 15 lbm/min 14.7 psia 60°F Analysis The reversible (or minimum) power input is determined from the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, Xin − Xout
Rate of net exergy transfer by heat, work, and mass − Xdestroyed Ê0 (reversible) = ∆Xsystem Ê0 (steady) =0 Rate of exergy destruction Rate of change of exergy Xin = Xout mψ 1 + Wrev,in = mψ 2 Wrev,in = m(ψ 2 − ψ 1 ) = m[(h2 − h1 ) − T0 (s2 − s1 ) + ∆keÊ0 + ∆peÊ0 ] where
o o ∆s air = s2 − s1 − R ln P2 P 1 = (0.73509 − 0.59173) Btu / lbm ⋅ R − (0.06855 Btu / lbm ⋅ R) ln = 0.01193 Btu / lbm ⋅ R Substituting, 100 psia 14.7 psia Wrev,in = (22/60 lbm/s)[(226.11 − 124.27)Btu/lbm − (520 R)(0.01193 Btu/lbm ⋅ R)] = 35.1 Btu/s = 49.6 hp Discussion Note that this is the minimum power input needed for this compressor. 733 Chapter 7 Exergy: A Measure of Work Potential 771 Steam is accelerated in an adiabatic nozzle. The exit velocity of the steam, the isentropic efficiency, and the exergy destroyed within the nozzle are to be determined.
Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A4 through A6) P1 = 7 MPa ½ h1 = 3410.3 kJ/kg ¾ T1 = 500°C ¿ s1 = 6.7975 kJ/kg ⋅ K P2 = 5 MPa ½ h2 = 3316.2 kJ/kg ¾ T2 = 450°C ¿ s 2 = 6.8186 kJ/kg ⋅ K P2 s = 5 MPa ½ ¾ h2 s = 3301.5 kJ/kg s 2 s = s1 ¿ 7 MPa 500°C 70 m/s STEAM 5 MPa 450°C Analysis We take the nozzle 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
Rate of net energy transfer by heat, work, and mass = ∆Esystem Ê0 (steady) =0 Rate of change in internal, kinetic, potential, etc. energies Ein = Eout m(h1 + V12
2 / 2) = m(h2 + V2 / 2) (since W = Q ≅ ∆pe ≅ 0) 2 V2 − V12 2 0 = h2 − h1 + Then the exit velocity becomes § 1000 m 2 /s 2 V2 = 2(h1 − h2 ) + V12 = 2(3410.3 − 3316.2) kJ/kg¨ ¨ 1 kJ/kg ©
(b) The exit velocity for the isentropic case is determined from · ¸ + (70 m/s) 2 = 439.4 m/s ¸ ¹ § 1000 m 2 /s 2 V2 s = 2(h1 − h2 s ) + V12 = 2(3410.3 − 3301.5) kJ/kg¨ ¨ 1 kJ/kg ©
Thus, · ¸ + (70 m/s) 2 = 471.7 m/s ¸ ¹ ηN = 2 V2 / 2 2 V2 s / 2 = (439.4 m / s) 2 / 2 (471.7 m / s) 2 / 2 = 86.8% (c) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on the actual nozzle. It gives S in − S out
Rate of net entropy transfer by heat and mass + Rate of entropy generation S gen N = ∆S system ©0 =0 Rate of change of entropy ms1 − ms 2 + S gen = 0 → S gen = m(s 2 − s1 ) or s gen = s 2 − s1
Substituting, the exergy destruction in the nozzle on a unit mass basis is determined to be x destroyed = T0 s gen = T0 ( s 2 − s1 ) = (298 K)(6.8186 − 6.7975)kJ/kg ⋅ K = 6.26 kJ/kg 745 Chapter 8 Gas Power Cycles 820 The three processes of an airstandard cycle are described. The cycle is to be shown on Pv and Ts diagrams, and the net work per cycle and the thermal efficiency are to be determined. √
Assumptions 1 The airstandard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A17. Analysis (b) The properties of air at various states are → T1 = 290 K u1 = 206.91kJ/kg h1 = 290.16kJ/kg P
2 P2 v 2 P1v1 P 380kPa (290K ) = 1160K = → T2 = 2 T1 = T2 T1 P1 95kPa → u 2 = 897.91kJ/kg, Pr2 = 207.2 Pr3 = P3 95kPa Pr2 = → (207.2) = 51.8 h3 = 840.38kJ/kg P2 380kPa
T qin
1 3 qout v Qin = m(u 2 − u1 ) = (0.003kg )(897.91 − 206.91)kJ/kg = 2.073kJ Qout = m(h3 − h1 ) = (0.003kg )(840.38 − 290.16 )kJ/kg = 1.651kJ Wnet ,out = Qin − Qout = 2.073 − 1.651 = 0.422kJ
1 2 qin
3 qout s
(c) η th = Wnet ,out Qin = 0.422 kJ = 20.4% 2.073 kJ 87 Chapter 8 Gas Power Cycles 833 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle are to be determined.
Assumptions 1 The airstandard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A17. Analysis (a) Process 12: isentropic compression. T1 = 300 K → u1 = 214.07 kJ / kg vr1 = 621.2 P
3 v r2 v 1 1 = 2 v r1 = v r1 = (621.2 ) = 77.65 8 v1 r → T2 = 673.1K u 2 = 491.2kJ/kg 750 kJ/kg
2 4 1 v § 673.1K · P2 v 2 P1v1 vT = → P2 = 1 2 P1 = (8)¨ ¸ ¨ 300K ¸(95kPa ) = 1705kPa T2 T1 v 2 T1 ¹ © Process 23: v = constant heat addition. q23,in = u3 − u2 → u3 = u2 + q23,in = 491.2 + 750 = 1241.2 kJ / kg →
§ 1539K · P3 v3 P2 v 2 T = → P3 = 3 P2 = ¨ ¨ 673.1K ¸(1705kPa ) = 3898kPa ¸ T3 T2 T2 © ¹ T3 = 1539 K vr 3 = 6.588 (b) Process 34: isentropic expansion. v r4 = T = 774.5K v1 v r3 = rv r3 = (8)(6.588) = 52.70 → 4 u 4 = 571.69kJ/kg v2 Process 41: v = constant heat rejection. qout = u4 − u1 = 571.69 − 214.07 = 357.62 kJ / kg
wnet ,out = qin − qout = 750 − 357.62 = 392.38 kJ / kg (c) η th =
v1 = w net,out q in = 392.38kJ/kg = 52.3% 750kJ/kg (d) RT1 0.287 kPa ⋅ m 3 /kg ⋅ K 300K ) = = 0.906m 3 /kg = v max P1 95kPa v max r = wnet,out v1 (1 − 1 / r ) = ( )( v min = v 2 = MEP = wnet,out v1 − v 2 ( § kPa ⋅ m 3 ¨ kJ 0.906m 3 /kg 1 − 1/8) ¨ © 392.38kJ/kg )( · ¸ = 495.0kPa ¸ ¹ 834 EES solution of this (and other comprehensive problems designated with the computer icon) is available to instructors at the Instructor Manual section of the Online Learning Center (OLC) at www.mhhe.com/cengelboles. See the Preface for access information. 813 Chapter 8 Gas Power Cycles 849 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined.
Assumptions 1 The airstandard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are Cp = 1.005 kJ/kg·K, Cv = 0.718 kJ/kg·K, and k = 1.4 (Table A2). Analysis (a) Process 12: isentropic compression.
P
2
k −1 qin §V T2 = T1 ¨ 1 ¨V ©2 · ¸ ¸ ¹ 3 = (293K )(20 ) 0.4 = 971.1K
4 Process 23: P = constant heat addition. P3 V3 P2 V2 = T3 T2 → V3 T3 2200 K = = = 2.265 V2 T2 971.1 K qout
1 v Process 34: isentropic expansion. q in = h3 − h2 = C p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2200 − 971.1)K = 1235 kJ/kg q out = u 4 − u1 = C v (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(920.6 − 293)K = 450.6 kJ/kg wnet ,out = q in − q out = 1235 − 450.6 = 784.4 kJ/kg §V T4 = T3 ¨ 3 ¨V ©4 · ¸ ¸ ¹ k −1 § 2.265V2 = T3 ¨ ¨V 4 © · ¸ ¸ ¹ k −1 § 2.265 · = T3 ¨ ¸ ©r¹ k −1 § 2.265 · = (2200 K )¨ ¸ © 20 ¹ 0.4 = 920.6 K η th =
(b) v1 = wnet ,out q in = 784.4 kJ/kg = 63.5% 1235 kJ/kg RT1 0.287kPa ⋅ m 3 /kg ⋅ K 293K ) = = 0.885m 3 /kg = v max 95kPa P1 v max r = v1 (1 − 1 / r ) wnet ,out = ( )( v min = v 2 = MEP = wnet ,out v1 − v 2 ( § kPa ⋅ m 3 ¨ kJ 0.885m 3 /kg 1 − 1/20 ) ¨ © 784.4kJ/kg )( · ¸ = 933kPa ¸ ¹ 825 Chapter 8 Gas Power Cycles 873 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The airstandard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are Cp = 1.005 kJ/kg·K and k = 1.4 (Table A2). Analysis (a) Using the isentropic relations,
(k −1) / k T2 s §P = T1 ¨ 2 ¨P ©1 §P = T3 ¨ 4 ¨P ©3 · ¸ ¸ ¹ = (300 K )(12) 0.4/1.4 = 610.2 K = 491.7 K T4 s ws ,C ,in = h2 s − h1 = C p (T2 s − T1 ) = (1.005 kJ/kg ⋅ K )(610.2 − 300 )K = 311.75 kJ/kg ws ,T ,out = h3 − h4 s = C p (T3 − T4 s ) = (1.005 kJ/kg ⋅ K )(1000 − 491.7 )K = 510.84 kJ/kg ws ,net ,out = ws ,T ,out − ws ,C ,in = 510.84 − 311.75 = 199.09 kJ/kg ms = Wnet ,out ws ,net ,out = 90,000 kJ/s = 452.1 kg/s 199.09 kJ/kg · ¸ ¸ ¹ (k −1) / k §1· = (1000 K )¨ ¸ © 12 ¹ 0.4/1.4 T 1000 K
2s 2 4s 4 3 (b) The net work output is determined to be wa ,net ,out = wa ,T ,out − wa ,C ,in = ηT ws ,T ,out − ws ,C ,in / η C Wnet ,out wa ,net ,out = (0.80)(510.84) − 311.75 0.80 = 18.98kJ/kg 90,000 kJ/s = = 4742 kg/s 18.98 kJ/kg
300 K 1 ma = s 839 ...
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 Fall '08
 HAMMOND
 Thermodynamics, Energy, Entropy, kJ/kg, Exergy, potential energy changes

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