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7-198 7-148 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of air and the rate of entropy generation are to be determined for the cases of an insulated and uninsulated evaporator. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 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 A-1). The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The properties of R-134a at the inlet and the exit states are (Tables A-11 through A-13) P1 = 120 kPa h1 = h f + x1 h fg = 22.49 + 0.3 214.48 = 86.83 kJ/kg x1 = 0.3 s1 = s f + x1 s fg = 0.09275 + 0.3(0.85503) = 0.3493 kJ/kg K T2 = 120 kPa h2 = hg @ 120 sat. vapor s2 = hg @ 120 P V& 3 kPa kPa = 236.97 kJ/kg = 0.9478 kJ/kg K R-134a 6 m3/min AIR Analysis (a) The mass flow rate of air is & mair = 3 3 3 RT3 = (100 kPa )(6 m /min ) (0.287 kPa m3/kg K )(300 K ) = 6.97 kg/min 1 2 kg/min 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): & & & min - mout = msystem 0 (steady) 2 4 sat. vapor & & & & & & & & = 0 min = mout m1 = m2 = mair and m3 = m4 = mR Energy balance (for the entire heat exchanger): Rate of net energy transfer by heat, work, and mass & & Ein - Eout 1 24 4 3 = Rate of change in internal, kinetic, potential, etc. energies & Esystem 0 (steady) 1442443 =0 & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = ke pe 0) Combining the two, Solving for T4, Substituting, & & & mR (h2 - h1 ) = mair (h3 - h4 ) = mair c p (T3 - T4 ) T4 = T3 - & m R (h2 - h1 ) & m air c p (2 kg/min)(236.97 - 86.83) kJ/kg = -15.9 C = 257.1 K (6.97 kg/min)(1.005 kJ/kg K) T4 = 27 C - Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as Rate of net entropy transfer by heat and mass & & Sin - Sout 1 24 4 3 + Rate of entropy generation & Sgen { & = Ssystem 0 (steady) 144 44 2 3 Rate of change of entropy & & & & & m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 (since Q = 0) & & & & & mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 or, 7-149 & & & Sgen = mR (s2 - s1 ) + mair (s4 - s3 ) where s 4 - s 3 = c p ln T4 P T 257.1 K - R ln 4 = c p ln 4 = (1.005 kJ/kg K) ln = -0.1551 kJ/kg K T3 P3 T3 300 K 0 Substituting, & Sgen = (2 kg/min )(0.9478 - 0.3493 kJ/kg K ) + (6.97 kg/min)(-0.1551 kJ/kg K) = 0.116 kJ/min K = 0.00193 kW/K (b) When there is a heat gain from the surroundings at a rate of 30 kJ/min, the steady-flow energy equation reduces to & & & Qin = mR (h2 - h1 ) + mair c p (T4 - T3 ) & & Qin - mR (h2 - h1 ) & air c p m Solving for T4, Substituting, T4 = T3 + T4 = 27 C + (30 kJ/min) - (2 kg/min)(236.97 - 86.83) kJ/kg = -11.6 C = 261.4 K (6.97 kg/min)(1.005 kJ/kg K) The entropy generation in this case is determined by applying the entropy balance on an extended system that includes the evaporator and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surrounding air at all times. The entropy balance for the extended system can be expressed as Rate of net entropy transfer by heat and mass & & Sin - Sout 1 24 4 3 + Rate of entropy generation & Sgen { & = Ssystem 0 (steady) 144 44 2 3 Rate of change of entropy Qin & & & & & + m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 Tb,out Qin & & & & & + mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 Tsurr & Q & & & Sgen = mR (s2 - s1 ) + mair (s4 - s3 ) - in T0 s 4 - s 3 = c p ln T4 P 261.4 K - R ln 4 = (1.005 kJ/kg K) ln = -0.1384 kJ/kg K T3 P3 300 K 0 or where Substituting, 30 kJ/min & Sgen = (2 kg/min )(0.9478 - 0.3493) kJ/kg K + (6.97 kg/min )(- 0.1384 kJ/kg K ) - 305 K = 0.1340 kJ/min K = 0.00223 kW/K Discussion Note that the rate of entropy generation in the second case is greater because of the irreversibility associated with heat transfer between the evaporator and the surrounding air. 7-150 7-199 A room is to be heated by hot water contained in a tank placed in the room. The minimum initial temperature of the water needed to meet the heating requirements of this room for a 24-h period and the entropy generated are 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 container itself is negligible relative to the energy stored in water. 4 The room is maintained at 20 C at all times. 5 The hot water is to meet the heating requirements of this room for a 24-h period. Properties The specific heat of water at room temperature is c = 4.18 kJ/kg C (Table A-3). Analysis Heat loss from the room during a 24-h 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 Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 = Change in internal, kinetic, potential, etc. energies Esystem 123 4 4 - Qout = U = (U )water + (U )air 0 10,000 kJ/h or -Qout = [mc(T2 - T1)]water Substituting, -240,000 kJ = (1500 kg)(4.18 kJ/kg C)(20 - T1) It gives T1 = 58.3 C where T1 is the temperature of the water when it is first brought into the room. (b) We take the house as the system, which is a closed system. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the house and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for the extended system can be expressed as Net entropy transfer by heat and mass 20 C water Sin - Sout 1 24 4 3 - + Sgen = Ssystem { 123 4 4 Entropy generation Change in entropy 0 Qout + Sgen = S water + Sair Tb,out = S water since the state of air in the house (and thus its entropy) remains unchanged. Then the entropy generated during the 24 h period becomes Sgen = S water + Qout T Q = mc ln 2 + out Tb,out T1 water Tsurr 293 K 240,000 kJ = (1500 kg )(4.18 kJ/kg K ) ln + 331.3 K 278 K = -770.3 + 863.3 = 93.0 kJ/K 7-151 7-200 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 and the entropy generated are 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 well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K, cv = 0.743 kJ/kg C and cp =1.039 kJ/kg C for N2, and R = 2.0769 kPa.m3/kg.K, cv = 3.1156 kJ/kg C, and cp = 5.1926 kJ/kg C for He (Tables A-1 and A-2) Analysis The mass of each gas in the cylinder is PV (500 kPa ) 1 m3 mN 2 = 1 1 = = 4.77 kg RT 0.2968 kPa m3/kg K (353 K ) 1 N2 PV mHe = 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 80 C He 1 m3 500 kPa 25 C Taking the entire contents of the cylinder as our system, the 1st law relation can be written as Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 Esystem 123 4 4 0 = U = (U )N 2 + (U )He 0 = [mcv (T2 - T1 )]N 2 + [mcv (T2 - T1 )]He Substituting, (4.77 kg ) 0.743 kJ/kgo C T f - 80 C + (0.808 kg )(3.1156 kJ/kg C) T f - 25 C = 0 ( )( ) ( ) It gives Tf = 57.2 C 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. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Net entropy transfer by heat and mass Sin - Sout 1 24 4 3 + Sgen = Ssystem { 123 4 4 Entropy generation Change in entropy 0 + Sgen = S N 2 + S He But first we determine the final pressure in the cylinder: 4.77 kg 0.808 kg m m N total = N N 2 + N He = + = + = 0.372 kmol M N 2 M He 28 kg/kmol 4 kg/kmol P2 = N total RuT V total = (0.372 kmol)(8.314 kPa m3/kmol K )(330.2 K ) = 510.6 kPa 2 m3 Then, T P S N 2 = m c p ln 2 - R ln 2 T1 P N 1 2 330.2 K 510.6 kPa = (4.77 kg )(1.039 kJ/kg K )ln - (0.2968 kJ/kg K )ln 353 K 500 kPa = -0.361 kJ/K 7-152 P T S He = m c p ln 2 - R ln 2 P He T1 1 330.2 K 510.6 kPa = (0.808 kg )(5.1926 kJ/kg K )ln - (2.0769 kJ/kg K )ln 298 K 500 kPa = 0.395 kJ/K Sgen = S N 2 + S He = -0.361 + 0.395 = 0.034 kJ/K If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas would remain constant in this case: T V S N 2 = m cv ln 2 - R ln 2 T1 V1 V T S He = m cv ln 2 - R ln 2 V1 T1 0 = (4.77 kg )(0.743 kJ/kg K )ln 330.2 K = -0.237 kJ/K 353 K N 2 = (0.808 kg )(3.1156 kJ/kg K )ln 330.2 K = 0.258 kJ/K 298 K He 0 Sgen = S N 2 + S He = -0.237 + 0.258 = 0.021 kJ/K 7-153 7-201 EES Problem 7-200 is reconsidered. The results for constant specific heats to those obtained using variable specific heats are to be compared using built-in EES or other functions. Analysis The problem is solved using EES, and the results are given below. "Knowns:" R_u=8.314 [kJ/kmol-K] V_N2[1]=1 [m^3] Cv_N2=0.743 [kJ/kg-K] "From Table A-2(a) at 27C" R_N2=0.2968 [kJ/kg-K] "From Table A-2(a)" T_N2[1]=80 [C] P_N2[1]=500 [kPa] Cp_N2=R_N2+Cv_N2 V_He[1]=1 [m^3] Cv_He=3.1156 [kJ/kg-K] "From Table A-2(a) at 27C" T_He[1]=25 [C] P_He[1]=500 [kPa] R_He=2.0769 [kJ/kg-K] "From Table A-2(a)" Cp_He=R_He+Cv_He "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, allowing the piston to move." "Conservation of Energy for the closed system:" "E_in - E_out = DELTAE, we neglect DELTA KE and DELTA PE for the cylinder." E_in - E_out = DELTAE E_in =0 [kJ] E_out = 0 [kJ] "At the final equilibrium state, N2 and He will have a common temperature." DELTAE= m_N2*Cv_N2*(T_2-T_N2[1])+m_He*Cv_He*(T_2-T_He[1]) "Total volume of gases:" V_total=V_N2[1]+V_He[1] MM_He = 4 [kg/kmol] MM_N2 = 28 [kg/kmol] N_total = m_He/MM_He+m_N2/MM_N2 "Final pressure at equilibrium:" "Allowing the piston to move, the pressure on both sides is the same, P_2 is:" P_2*V_total=N_total*R_u*(T_2+273) S_gen_PistonMoving = DELTAS_He_PM+DELTAS_N2_PM DELTAS_He_PM=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1])) DELTAS_N2_PM=m_N2*(Cp_N2*ln((T_2+273)/(T_N2[1]+273))-R_N2*ln(P_2/P_N2[1])) "The final temperature of the system when the piston does not move will be the same as when it does move. The volume of the gases remain constant and the entropy changes are given by:" S_gen_PistNotMoving = DELTAS_He_PNM+DELTAS_N2_PNM DELTAS_He_PNM=m_He*(Cv_He*ln((T_2+273)/(T_He[1]+273))) DELTAS_N2_PNM=m_N2*(Cv_N2*ln((T_2+273)/(T_N2[1]+273))) 7-154 "The following uses the EES functions for the nitrogen. Since helium is monatomic, we use the constant specific heat approach to find its property changes." E_in - E_out = DELTAE_VP DELTAE_VP= m_N2*(INTENERGY(N2,T=T_2_VP)INTENERGY(N2,T=T_N2[1]))+m_He*Cv_He*(T_2_VP-T_He[1]) "Final Pressure for moving piston:" P_2_VP*V_total=N_total*R_u*(T_2_VP+273) S_gen_PistMoving_VP = DELTAS_He_PM_VP+DELTAS_N2_PM_VP DELTAS_N2_PM_VP=m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_VP)ENTROPY(N2,T=T_N2[1],P=P_N2[1])) DELTAS_He_PM_VP=m_He*(Cp_He*ln((T_2+273)/(T_He[1]+273))-R_He*ln(P_2/P_He[1])) "Fianl N2 Pressure for piston not moving." P_2_N2_VP*V_N2[1]=m_N2*R_N2*(T_2_VP+273) S_gen_PistNotMoving_VP = DELTAS_He_PNM_VP+DELTAS_N2_PNM_VP DELTAS_N2_PNM_VP = m_N2*(ENTROPY(N2,T=T_2_VP,P=P_2_N2_VP)ENTROPY(N2,T=T_N2[1],P=P_N2[1])) DELTAS_He_PNM_VP=m_He*(Cv_He*ln((T_2_VP+273)/(T_He[1]+273))) SOLUTION Cp_He=5.193 [kJ/kg-K] Cp_N2=1.04 [kJ/kg-K] Cv_He=3.116 [kJ/kg-K] Cv_N2=0.743 [kJ/kg-K] DELTAE=0 [kJ] DELTAE_VP=0 [kJ] DELTAS_He_PM=0.3931 [kJ/K] DELTAS_He_PM_VP=0.3931 [kJ/K] DELTAS_He_PNM=0.258 [kJ/K] DELTAS_He_PNM_VP=0.2583 [kJ/K] DELTAS_N2_PM=-0.363 [kJ/K] DELTAS_N2_PM_VP=-0.3631 [kJ/K] DELTAS_N2_PNM=-0.2371 [kJ/K] DELTAS_N2_PNM_VP=-0.2372 [kJ/K] E_in=0 [kJ] E_out=0 [kJ] MM_He=4 [kg/kmol] MM_N2=28 [kg/kmol] m_He=0.8079 [kg] m_N2=4.772 [kg] N_total=0.3724 [kmol] P_2=511.1 [kPa] P_2_N2_VP=467.7 P_2_VP=511.2 P_He[1]=500 [kPa] P_N2[1]=500 [kPa] R_He=2.077 [kJ/kg-K] R_N2=0.2968 [kJ/kg-K] R_u=8.314 [kJ/kmol-K] S_gen_PistMoving_VP=0.02993 [kJ/K] S_gen_PistNotMoving=0.02089 [kJ/K] S_gen_PistNotMoving_VP=0.02106 [kJ/K] S_gen_PistonMoving=0.03004 [kJ/K] T_2=57.17 [C] T_2_VP=57.2 [C] T_He[1]=25 [C] T_N2[1]=80 [C] V_He[1]=1 [m^3] V_N2[1]=1 [m^3] V_total=2 [m^3] 7-155 7-202 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 and the entropy generated are 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 well-insulated 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, cv = 0.743 kJ/kg C and cp =1.039 kJ/kg C for N2, and R = 2.0769 kPa.m3/kg.K, cv = 3.1156 kJ/kg C, and cp = 5.1926 kJ/kg C for He (Tables A-1 and A-2). The specific heat of the copper at room temperature is c = 0.386 kJ/kg C (Table A-3). Analysis The mass of each gas in the cylinder is PV (500 kPa ) 1 m3 mN 2 = 1 1 = = 4.77 kg RT 0.2968 kPa m3/kg K (353 K ) 1 N2 PV mHe = 1 1 RT 1 He 3 3 ( ) ( ) (500 kPa )(1m ) = (2.0769 kPa m /kg K )(298 K ) = 0.808 kg = Change in internal, kinetic, potential, etc. energies N2 1 m3 500 kPa 80 C He 1 m3 500 kPa 25 C Copper Taking the entire contents of the cylinder as our system, the 1st law relation can be written as Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 Esystem 123 4 4 0 = U = (U )N 2 + (U )He + (U )Cu 0 = [mcv (T2 - T1 )]N 2 + [mcv (T2 - T1 )]He + [mc(T2 - T1 )]Cu where T1, Cu = (80 + 25) / 2 = 52.5 C Substituting, (4.77 kg )(0.743 kJ/kg C)(T f - 80 C + (0.808 kg )(3.1156 kJ/kg C ) T f - 25 C ) ) + (5.0 kg )(0.386 kJ/kg C )(T f - 52.5) C = 0 ( It gives Tf = 56.0 C 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. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Net entropy transfer by heat and mass Sin - Sout 1 24 4 3 + Sgen = Ssystem { 123 4 4 Entropy generation Change in entropy 0 + Sgen = S N 2 + S He + S piston But first we determine the final pressure in the cylinder: 4.77 kg 0.808 kg m m N total = N N 2 + N He = + = + = 0.372 kmol M N 2 M He 28 kg/kmol 4 kg/kmol P2 = N total RuT V total = (0.372 kmol)(8.314 kPa m3/kmol K )(329 K ) = 508.8 kPa 2 m3 7-156 Then, T P S N 2 = m c p ln 2 - R ln 2 T1 P N 1 2 329 K 508.8 kPa = (4.77 kg )(1.039 kJ/kg K )ln - (0.2968 kJ/kg K )ln 353 K 500 kPa = -0.374 kJ/K T P = m c p ln 2 - R ln 2 T1 P He 1 329 K 508.8 kPa = (0.808 kg )(5.1926 kJ/kg K )ln - (2.0769 kJ/kg K )ln 298 K 500 kPa = 0.386 kJ/K 329 K T = mc ln 2 = (5 kg )(0.386 kJ/kg K )ln = 0.021 kJ/K 325.5 K T1 piston S He S piston Sgen = S N 2 + S He + S piston = -0.374 + 0.386 + 0.021 = 0.033 kJ/K If the piston were not free to move, we would still have T2 = 329 K but the volume of each gas would remain constant in this case: V T S N 2 = m cv ln 2 - R ln 2 V1 T1 V T S He = m cv ln 2 - R ln 2 V1 T1 0 = (4.77 kg )(0.743 kJ/kg K ) ln 329 K = -0.250 kJ/K 353 K N2 0 = (0.808 kg )(3.1156 kJ/kg K ) ln 329 K = 0.249 kJ/K 298 K He Sgen = S N 2 + S He + S piston = -0.250 + 0.249 + 0.021 = 0.020 kJ/K 7-157 7-203 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 a specified value. The amount of electrical work done during this process and the total entropy change 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 uniform-flow 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 is R = 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17) Te = 330 K he = 330.34 kJ/kg T1 = 330 K u1 = 235.61 kJ/kg T2 = 330 K u 2 = 235.61 kJ/kg 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 uniform-flow system can be expressed as Mass balance: min - mout = msystem me = m1 - m2 Energy balance: Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 = Change in internal, kinetic, potential, etc. energies Esystem 123 4 4 We,in - me he = m2u2 - m1u1 (since Q ke pe 0) The initial and the final masses of air in the tank are (500 kPa ) 5 m3 PV = 26.40 kg m1 = 1 = RT1 0.287 kPa m3/kg K (330 K ) PV m2 = 2 RT2 ) ( ) (200 kPa )(5 m ) = (0.287 kPa m /kg K )(330 K ) = 10.56 kg 3 3 ( = (15.84 kg )(330.34 kJ/kg ) + (10.56 kg )(235.61 kJ/kg ) - (26.40 kg )(235.61 kJ/kg ) = 1501 kJ (b) The total entropy change, or the total entropy generation within the tank boundaries is determined from an entropy balance on the tank expressed as Sin - Sout + Sgen = Ssystem 1 24 4 3 { 123 4 4 Net entropy transfer by heat and mass Entropy generation Change in entropy Then from the mass and energy balances, me = m1 - m2 = 26.40 - 10.56 = 15.84 kg We,in = me he + m2u2 - m1u1 AIR 5 m3 500 kPa 57 C We - me se + Sgen = S tank = (m1 - m 2 )s e + (m 2 s 2 - m1 s1 ) = m 2 (s 2 - s e ) - m1 (s1 - s e ) Assuming a constant average pressure of (500 + 200)/2 = 350 kPa for the exit stream, the entropy changes are determined to be T 0 P P 200 kPa s 2 - s e = c p ln 2 - R ln 2 = - R ln 2 = -(0.287 kJ/kg K )ln = 0.1606 kJ/kg K Te Pe Pe 350 kPa T1 0 P P 500 kPa - R ln 2 = - R ln 1 = -(0.287 kJ/kg K )ln = -0.1024 kJ/kg K Te Pe Pe 350 kPa Therefore, the total entropy generated within the tank during this process is Sgen = (10.56 kg )(0.1606 kJ/kg K ) - (26.40 kg )(- 0.1024 kJ/kg K ) = 4.40 kJ/K s1 - s e = c p ln or, S gen = m e s e + S tank = m e s e + (m 2 s 2 - m1 s1 ) 7-158 7-204 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 and the entropy generation are 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/kg C, and the specific heat of ice at about 0 C is c = 2.11 kJ/kg C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0 C and 333.7 kJ/kg.. Analysis (a) We take the ice and the water as the system, and disregard any heat transfer between the system and the surroundings. Then the energy balance for this process can be written as Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 = Change in internal, kinetic, potential, etc. energies Esystem 123 4 4 ice -5 C 80 kg 0 = U 0 = U ice + U water [mc(0 C - T1 ) solid + mhif + mc(T2 - 0 C) liquid ] ice + [mc(T2 - T1 )] water = 0 WATER 1 ton Substituting, (80 kg){(2.11 kJ/kg C)[0 - (-5)] C + 333.7 kJ/kg + (4.18 kJ/kg C)(T2 - 0) C} + (1000 kg )(4.18 kJ/kg C)(T2 - 20) C = 0 It gives T2 = 12.42 C which is the final equilibrium temperature in the tank. (b) We take the ice and the water as our system, which is a closed system. Considering that the tank is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Net entropy transfer by heat and mass Sin - Sout 1 24 4 3 + Sgen = Ssystem { 123 4 4 Entropy generation Change in entropy 0 + Sgen = Sice + S water where 285.42 K T = (1000 kg )(4.18 kJ/kg K ) ln = -109.6 kJ/K S water = mc ln 2 293 K T1 water Sice = Ssolid + S melting + Sliquid ice ( ) Tmelting mhif T + + mc ln 2 = mc ln T1 solid Tmelting T1 liquid ice 273 K 333.7 kJ/kg 285.42 K = (80 kg ) (2.11 kJ/kg K )ln + + (4.18 kJ/kg K ) ln 268 K 273 K 273 K = 115.8 kJ/K Then, Sgen = S water + Sice = -109.6 + 115.8 = 6.2 kJ/K 7-159 7-205 An insulated cylinder initially contains a saturated liquid-vapor 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 added and the entropy generation are to be determined. Assumptions 1 Thermal properties of the ice are constant. 2 The cylinder is well-insulated 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 0 C is c = 2.11 kJ/kg C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0 C 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 Net energy transfer by heat, work, and mass Ein - Eout 1 24 4 3 = Change in internal, kinetic, potential, etc. energies Esystem 123 4 4 Wb,in = U H = 0 H ice + H water = 0 or [mc(0 C - T1 ) solid + mhif + mc(T2 - 0 C) liquid ] ice + [m(h2 - h1 )] water = 0 The properties of water at 100 C are (Table A-4) v f = 0.001043, v g = 1.6720 m 3 /kg h f = 419.17, h fg = 2256.4 kJ.kg s f = 1.3072 s fg = 6.0490 kJ/kg.K ice -18 C v1 = v f + x1v fg = 0.001043 + (0.1)(1.6720 - 0.001043) = 0.16814 m3/kg h1 = h f + x1h fg = 419.17 + (0.1)(2256.4) = 644.81 kJ/kg h2 = h f @100o C = 419.17 kJ/kg s2 = s f @100o C = 1.3072 kJ/kg K msteam = s1 = s f + x1s fg = 1.3072 + (0.1)(6.0470 ) = 1.9119 kJ/kg K WATER 0.02 m3 100 C V1 0.02 m3 = = 0.119 kg v1 0.16814 m3/kg Noting that T1, ice = -18 C and T2 = 100 C and substituting gives m{(2.11 kJ/kg.K)[0-(-18)] + 333.7 kJ/kg + (4.18 kJ/kg C)(100-0) C} +(0.119 kg)(419.17 644.81) kJ/kg = 0 m = 0.034 kg = 34.0 g ice (b) We take the ice and the steam as our system, which is a closed system. Considering that the tank is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Sin - Sout + Sgen = Ssystem 1 24 4 3 { 123 4 4 Net entropy transfer by heat and mass Entropy generation Change in entropy 0 + Sgen = Sice + Ssteam Ssteam = m(s2 - s1 ) = (0.119 kg )(1.3072 - 1.9119 )kJ/kg K = -0.0719 kJ/K Sice = Ssolid + S melting + Sliquid ( )ice = mc ln Tmelting T 1 solid + mhif Tmelting T + mc ln 2 T1 liquid ice 273.15 K 333.7 kJ/kg 373.15 K = 0.0907 kJ/K = (0.034 kg ) (2.11 kJ/kg K )ln + + (4.18 kJ/kg K )ln 255.15 K 273.15 K 273.15 K Then, Sgen = Ssteam + Sice = -0.0719 + 0.0907 = 0.0188 kJ/K 7-160 7-206 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 and the amount of entropy generated 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 uniform-flow 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). Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). 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 uniform-flow system can be expressed as Mass balance: min - mout = msystem mi = m2 Energy balance: Net energy transfer by heat, work, and mass (since mout = minitial = 0) Ein - Eout 1 24 4 3 = Change in internal, kinetic, potential, etc. energies Esystem 123 4 4 Qin + mi hi = m2u2 (since W Eout = Einitial = ke pe 0) Combining the two balances: Qin = m2 (u2 - hi ) where 10 kPa 17 C m2 = P2V (100 kPa ) 0.005 m 3 = = 0.0060 kg RT2 0.287 kPa m 3 /kg K (290 K ) ( ( ) ) 17 Ti = T2 = 290 K Table A - hi = 290.16 kJ/kg u 2 = 206.91 kJ/kg Qout = 0.5 kJ 5L Evacuated Substituting, Qin = (0.0060 kg)(206.91 - 290.16) kJ/kg = - 0.5 kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the bottle and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be expressed as Net entropy transfer by heat and mass Sin - Sout 1 24 4 3 + Sgen = Ssystem { 1 23 4 4 Entropy generation Change in entropy 0 mi si - Qout + Sgen = S tank = m2 s2 - m1s1 Tb,in Qout = m2 (s2 - si ) Tb,out = m2 s2 Therefore, the total entropy generated during this process is Sgen = -mi si + m2 s2 + 0 + 0.5 kJ Qout Q = out = = 0.0017 kJ/K Tb, out Tsurr 290 K 7-161 7-207 Water is heated from 16 C to 43 C 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 rate of entropy generation are to be determined. The reduction in power input and entropy generation as a result of installing a 50% efficient regenerator are also to be determined. Assumptions 1 This is a steady-flow 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 of water is given to be = 1 kg/L. The specific heat of water at room temperature is c = 4.18 kJ/kg C (Table A-3). Analysis (a) 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 steady-flow system can be expressed in the rate form as & & & & & Ein - Eout = Esystem 0 (steady) = 0 Ein = Eout 1 24 4 3 1442443 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & & We,in + mh1 = mh2 (since ke pe 0) & & & We,in = m(h2 - h1 ) = mc(T2 - T1 ) where Substituting, WATER 16 C 43 C & m = V& = (1 kg/L)(10 L/min) = 10 kg/min & W = (10/60 kg/s )(4.18 kJ/kg C )(43 - 16 ) C = 18.8 kW e,in The rate of entropy generation in the heating section during this process is determined by applying the entropy balance on the heating section. Noting that this is a steady-flow process and heat transfer from the heating section is negligible, & & & & Sin - Sout + Sgen = Ssystem 0 = 0 1 24 4 3 { 1 24 4 3 Rate of net entropy transfer by heat and mass Rate of entropy generation Rate of change of entropy & & & & ms1 - ms2 Sgen + = 0 Sgen = m( s2 - s1 ) & Noting that water is an incompressible substance and substituting, 316 K T & & = 0.0622 kJ/K Sgen = mc ln 2 = (10/60 kg/s )(4.18 kJ/kg K )ln 289 K T1 (b) The energy recovered by the heat exchanger is & & & Qsaved = Qmax = mC (Tmax - Tmin ) = 0.5(10/60 kg/s )(4.18 kJ/kg C )(39 - 16 ) C = 8.0 kJ/s = 8.0 kW 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 Taking the cold water stream in the heat exchanger as our control volume (a steady-flow system), the temperature at which the cold water leaves the heat exchanger and enters the electric resistance heating section is determined from & & Q = mc(Tc, out - Tc,in ) Substituting, It yields 8 kJ/s = (10/60 kg/s )( 4.18 kJ/kg o C)(Tc, out - 16o C) Tc, out = 27.5o C = 300.5 K The rate of entropy generation in the heating section in this case is determined similarly to be T 316 K & & = 0.0350 kJ/K Sgen = mcln 2 = (10/60 kg/s )(4.18 kJ/kg K ) ln T1 300.5 K Thus the reduction in the rate of entropy generation within the heating section is & S reduction = 0.0622 - 0.0350 = 0.0272 kW/K 7-162 7-208 EES Using EES (or other) software, the work input to a multistage compressor is to be determined for a given set of inlet and exit pressures for any number of stages. The pressure ratio across each stage is assumed to be identical and the compression process to be polytropic. The compressor work is to be tabulated and plotted against the number of stages for P1 = 100 kPa, T1 = 17 C, P2 = 800 kPa, and n = 1.35 for air. Analysis The problem is solved using EES, and the results are tabulated and plotted below. GAS$ = 'Air' Nstage = 2 "number of stages of compression with intercooling, each having same pressure ratio." n=1.35 MM=MOLARMASS(GAS$) R_u = 8.314 [kJ/kmol-K] R=R_u/MM k=1.4 P1=100 [kPa] T1=17 [C] P2=800 [kPa] R_p = (P2/P1)^(1/Nstage) W_dot_comp= Nstage*n*R*(T1+273)/(n-1)*((R_p)^((n-1)/n) - 1) Nstage 1 2 3 4 5 6 7 8 9 10 230 220 210 Wcomp [kJ/kg] 229.4 198.7 189.6 185.3 182.8 181.1 179.9 179 178.4 177.8 W comp [kJ/kg] 200 190 180 170 1 2 3 4 5 6 7 8 9 10 Nstage 7-163 7-209 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed of three processes. The work and heat transfer for each process are to be determined. Assumptions 1 All processes are reversible. 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 = 0.287 kPa.m3/kg.K (Table A-1). Analysis Using variable specific heats, the properties can be determined using the air table as follows u1 = u2 = 214.07 kJ/kg 0 0 s1 = s2 = 1.70203 kJ/kg.K T1 = T2 = 300 K Pr1 = Pr 2 = 1.3860 Pr 3 = u = 283.71 kJ/kg P3 400 kPa (1.3860) = 3.696 Pr 2 = 3 T3 = 396.6 K P2 150 kPa P = const. 1 T = const. 3 s = const. 2 The mass of the air and the volumes at the various states are m= P1V1 (400 kPa)(0.3 m 3 ) = = 1.394 kg RT1 (0.287 kPa m 3 /kg K)(300 K) mRT2 (1.394 kg)(0.287 kPa m 3 /kg K)(300 K) = = 0.8 m 3 P2 150 kPa mRT3 (1.394 kg)(0.287 kPa m 3 /kg K)(396.6 K) = = 0.3967 m 3 P3 400 kPa P2 150 kPa = (1.394 kg)(0.287 kJ/kg.K)ln = 0.3924 kJ/kg.K 400 kPa P1 V2 = V3 = Process 1-2: Isothermal expansion (T2 = T1) S1- 2 = -mR ln Qin,1- 2 = T1 S1- 2 = (300 K)(0.3924 kJ/K) = 117.7 kJ Wout,1- 2 = Qin,1- 2 = 117.7 kJ Process 2-3: Isentropic (reversible-adiabatic) compression (s2 = s1) Win,2 -3 = m(u3 - u2 ) = (1.394 kg)(283.71 - 214.07) kJ/kg = 97.1 kJ Q2-3 = 0 kJ Process 3-1: Constant pressure compression process (P1 = P3) Win,3-1 = P3 (V3 - V1 ) = (400 kg)(0.3924 - 0.3) kJ/kg = 37.0 kJ Qout,3-1 = Win,3-1 - m(u1 - u3 ) = 37.0 kJ - (1.394 kg)(214.07 - 283.71) kJ/kg = 135.8 kJ 7-164 7-210 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust gases and a compressor driven by the turbine is considered. The air temperature at the compressor exit and the isentropic efficiency of the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Exhaust gases have air properties and air is an ideal gas with constant specific heats. Properties The specific heat of exhaust gases at the average temperature of 425 C is cp = 1.075 kJ/kg.K and properties of air at an anticipated average temperature of 100 C are cp = 1.011 kJ/kg.K and k =1.397 (Table A-2). Analysis (a) The turbine power output is determined from & & W = m c (T - T ) T exh p 400 C Turbine Air, 70 C 95 kPa 0.018 kg/s Compressor 1 2 Exh. gas 450 C 0.02 kg/s 135 kPa = (0.02 kg/s)(1.075 kJ/kg. C)(450 - 400) C = 1.075 kW For a mechanical efficiency of 95% between the turbine and the compressor, & & WC = mWT = (0.95)(1.075 kW) = 1.021 kW Then, the air temperature at the compressor exit becomes & & W = m c (T - T ) C air p 2 1 1.021 kW = (0.018 kg/s)(1.011 kJ/kg. C)(T2 - 70) C T2 = 126.1 C (b) The air temperature at the compressor exit for the case of isentropic process is T2 s P = T1 2 P 1 ( k -1) / k 135 kPa = (70 + 273 K) 95 kPa (1.397 -1)/1.397 = 379 K = 106 C The isentropic efficiency of the compressor is determined to be C = T2 s - T1 106 - 70 = 0.642 = T2 - T1 126.1 - 70 7-165 7-211 Air is compressed in a compressor that is intentionally cooled. The work input, the isothermal efficiency, and the entropy generation are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K and the specific heat of air at an average temperature of (20+300)/2 = 160 C = 433 K is cp = 1.018 kJ/kg.K (Table A-2). Analysis (a) The power input is determined from an energy balance on the control volume & & & W = mc (T - T ) + Q C p 2 1 out Q 300 C 1.2 MPa Compressor Air 20 C, 100 kPa = (0.4 kg/s)(1.018 kJ/kg. C)(300 - 20) C + 15 kW = 129.0 kW (b) The power input for a reversible-isothermal process is given by P 1200 kPa & & WT =const. = mRT1 ln 2 = (0.4 kg/s)(0.287 kJ/kg.K)(20 + 273 K)ln = 83.6 kW P1 100 kPa Then, the isothermal efficiency of the compressor becomes & W 83.6 kW T = T =const. = = 0.648 & 129.0 kW WC (c) The rate of entropy generation associated with this process may be obtained by adding the rate of entropy change of air as it flows in the compressor and the rate of entropy change of the surroundings & T P Q & & & S gen = S air + S surr = c p ln 2 - R ln 2 + out T1 P1 Tsurr = (1.018 kJ/kg.K)ln = 0.0390 kW/K 300 + 273 K 1200 kPa 15 kW - (0.287 kJ/kg.K)ln + 20 + 273 K 100 kPa (20 + 273) K 7-166 7-212 Air is allowed to enter an insulated piston-cylinder device until the volume of the air increases by 50%. The final temperature in the cylinder, the amount of mass that has entered, the work done, and the entropy generation are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K and the specific heats of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K (Table A-2). Analysis The initial pressure in the cylinder is P1 = m2 = m1 RT1 (0.7 kg)(0.287 kPa m 3 /kg K)(20 + 273 K) 0.25 m 3 Air 0.25 m3 0.7 kg 20 C Air 500 kPa 70 C V1 = = 235.5 kPa P2V 2 (235.5 kPa)(1.5 0.25 m 3 ) 307.71 = = RT2 T2 (0.287 kPa m 3 /kg K)T2 307.71 - 0.7 T2 A mass balance on the system gives the expression for the mass entering the cylinder mi = m 2 - m1 = (c) Noting that the pressure remains constant, the boundary work is determined to be W b,out = P1 (V 2 -V1 ) = (235.5 kPa)(1.5 0.25 - 0.5)m 3 = 29.43 kJ (a) An energy balance on the system may be used to determine the final temperature mi hi - Wb, out = m2u2 - m1u1 mi c pTi - Wb, out = m2cv T2 - m1cv T1 307.71 307.71 - 0.7 (1.005)(70 + 273) - 29.43 = T (0.718)T2 - (0.7)(0.718)(20 + 273) T 2 2 There is only one unknown, which is the final temperature. By a trial-error approach or using EES, we find T2 = 308.0 K (b) The final mass and the amount of mass that has entered are m2 = 307.71 = 0.999 kg 308.0 mi = m 2 - m1 = 0.999 - 0.7 = 0.299 kg (d) The rate of entropy generation is determined from S gen = m 2 s 2 - m1 s1 - mi s i = m 2 s 2 - m1 s1 - (m 2 - m1 ) s i = m 2 ( s 2 - s i ) - m1 ( s1 - s i ) T P T P = m 2 c p ln 2 - R ln 2 - m1 c p ln 1 - R ln 1 Ti Pi Ti Pi 308 K 235.5 kPa = (0.999 kg) (1.005 kJ/kg.K)ln - (0.287 kJ/kg.K)ln 343 K 500 kPa 293 K 235.5 kPa - (0.7 kg) (1.005 kJ/kg.K)ln - (0.287 kJ/kg.K)ln 343 K 500 kPa = 0.0673 kJ/K 7-167 7-213 A cryogenic turbine in a natural gas liquefaction plant produces 350 kW of power. The efficiency of the turbine is to be determined. Assumptions 1 The turbine operates steadily. 2 The properties of methane is used for natural gas. Properties The density of natural gas is given to be 423.8 kg/m3. Analysis The maximum possible power that can be obtained from this turbine for the given inlet and exit pressures can be determined from & m & Wmax = ( Pin - Pout ) = (55 kg/s) 423.8 kg/m3 3 bar Cryogenic turbine LNG, 40 bar -160 C, 55 kg/s (4000 - 300)kPa = 480.2 kW Given the actual power, the efficiency of this cryogenic turbine becomes & 350 kW W = & = = 0.729 = 72.9% 480.2 kW W max This efficiency is also known as hydraulic efficiency since the cryogenic turbine handles natural gas in liquid state as the hydraulic turbine handles liquid water. 7-168 Fundamentals of Engineering (FE) Exam Problems 7-214 Steam is condensed at a constant temperature of 30 C as it flows through the condenser of a power plant by rejecting heat at a rate of 55 MW. The rate of entropy change of steam as it flows through the condenser is (a) 1.83 MW/K (b) 0.18 MW/K (c) 0 MW/K (d) 0.56 MW/K (e) 1.22 MW/K Answer (b) 0.18 MW/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 "C" Q_out=55 "MW" S_change=-Q_out/(T1+273) "MW/K" "Some Wrong Solutions with Common Mistakes:" W1_S_change=0 "Assuming no change" W2_S_change=Q_out/T1 "Using temperature in C" W3_S_change=Q_out/(T1+273) "Wrong sign" W4_S_change=-s_fg "Taking entropy of vaporization" s_fg=(ENTROPY(Steam_IAPWS,T=T1,x=1)-ENTROPY(Steam_IAPWS,T=T1,x=0)) 7-215 Steam is compressed from 6 MPa and 300 C to 10 MPa isentropically. The final temperature of the steam is (a) 290 C (b) 300 C (c) 311 C (d) 371 C (e) 422 C Answer (d) 371 C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=6000 "kPa" T1=300 "C" P2=10000 "kPa" s2=s1 s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) T2=TEMPERATURE(Steam_IAPWS,s=s2,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant" W2_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P2" W3_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P1" 7-169 7-216 An apple with an average mass of 0.15 kg and average specific heat of 3.65 kJ/kg. C is cooled from 20 C to 5 C. The entropy change of the apple is (a) 0.0288 kJ/K (b) 0.192 kJ/K (c) -0.526 kJ/K (d) 0 kJ/K (e) 0.657 kJ/K Answer (a) 0.0288 kJ/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). C=3.65 "kJ/kg.K" m=0.15 "kg" T1=20 "C" T2=5 "C" S_change=m*C*ln((T2+273)/(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_S_change=C*ln((T2+273)/(T1+273)) "Not using mass" W2_S_change=m*C*ln(T2/T1) "Using C" W3_S_change=m*C*(T2-T1) "Using Wrong relation" 7-217 A piston-cylinder device contains 5 kg of saturated water vapor at 3 MPa. Now heat is rejected from the cylinder at constant pressure until the water vapor completely condenses so that the cylinder contains saturated liquid at 3 MPa at the end of the process. The entropy change of the system during this process is (a) 0 kJ/K (b) -3.5 kJ/K (c) -12.5 kJ/K (d) -17.7 kJ/K (e) -19.5 kJ/K Answer (d) -17.7 kJ/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=3000 "kPa" m=5 "kg" s_fg=(ENTROPY(Steam_IAPWS,P=P1,x=1)-ENTROPY(Steam_IAPWS,P=P1,x=0)) S_change=-m*s_fg "kJ/K" 7-218 Helium gas is compressed from 1 atm and 25 C to a pressure of 10 atm adiabatically. The lowest temperature of helium after compression is (a) 25 C (b) 63 C (c) 250 C (d) 384 C (e) 476 C Answer (e) 476 C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 P1=101.325 "kPa" T1=25 "C" P2=10*101.325 "kPa" 7-170 "s2=s1" "The exit temperature will be lowest for isentropic compression," T2=(T1+273)*(P2/P1)^((k-1)/k) "K" T2_C= T2-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant" W2_T2=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W3_T2=(T1+273)*(P2/P1)-273 "Assuming T is proportional to P" W4_T2=T1*(P2/P1) "Assuming T is proportional to P, using C" 7-219 Steam expands in an adiabatic turbine from 8 MPa and 500 C to 0.1 MPa at a rate of 3 kg/s. If steam leaves the turbine as saturated vapor, the power output of the turbine is (a) 2174 kW (b) 698 kW (c) 2881 kW (d) 1674 kW (e) 3240 kW Answer (a) 2174 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=8000 "kPa" T1=500 "C" P2=100 "kPa" x2=1 m=3 "kg/s" h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,x=x2,P=P2) W_out=m*(h1-h2) "Some Wrong Solutions with Common Mistakes:" s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) h2s=ENTHALPY(Steam_IAPWS, s=s1,P=P2) W1_Wout=m*(h1-h2s) "Assuming isentropic expansion" 7-220 Argon gas expands in an adiabatic turbine from 3 MPa and 750 C to 0.2 MPa at a rate of 5 kg/s. The maximum power output of the turbine is (a) 1.06 MW (b) 1.29 MW (c) 1.43 MW (d) 1.76 MW (e) 2.08 MW Answer (d) 1.76 MW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=0.5203 k=1.667 P1=3000 "kPa" T1=750 "C" m=5 "kg/s" P2=200 "kPa" 7-171 "s2=s1" T2=(T1+750)*(P2/P1)^((k-1)/k) W_max=m*Cp*(T1-T2) "Some Wrong Solutions with Common Mistakes:" Cv=0.2081"kJ/kg.K" W1_Wmax=m*Cv*(T1-T2) "Using Cv" T22=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W2_Wmax=m*Cp*(T1-T22) W3_Wmax=Cp*(T1-T2) "Not using mass flow rate" T24=T1*(P2/P1) "Assuming T is proportional to P, using C" W4_Wmax=m*Cp*(T1-T24) 7-221 A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining heat from the surroundings at temperature T in the amount of q. If the entropy of the substance is s1 at state 1, and s2 at state 2, the entropy change of the substance s during this process is (b) s > s2 s1 (c) s = s2 s1 (d) s = s2 s1 + q/T (a) s < s2 s1 (e) s > s2 s1 + q/T Answer (c) s = s2 s1 7-222 A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas constant of the gas is R, the entropy change of the gas s during this process is (b) s = R ln(P2/P1)- q/T (c) s =R ln(P1/P2) (d) s =R ln(P1/P2)-q/T (a) s =R ln(P2/P1) (e) s= 0 Answer (c) s =R ln(P1/P2) 7-223 Air is compressed from room conditions to a specified pressure in a reversible manner by two compressors: one isothermal and the other adiabatic. If the entropy change of air is sisot during the reversible isothermal compression, and sadia during the reversible adiabatic compression, the correct statement regarding entropy change of air per unit mass is (b) sisot= sadia>0 (c) sadia> 0 (d) sisot < 0 (e) sisot= 0 (a) sisot= sadia=0 Answer (d) sisot < 0 7-224 Helium gas is compressed from 15 C and 5.4 m3/kg to 0.775 m3/kg in a reversible adiabatic manner. The temperature of helium after compression is (a) 105 C (b) 55 C (c) 1734 C (d) 1051 C (e) 778 C Answer (e) 778 C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). 7-172 k=1.667 v1=5.4 "m^3/kg" T1=15 "C" v2=0.775 "m^3/kg" "s2=s1" "The exit temperature is determined from isentropic compression relation," T2=(T1+273)*(v1/v2)^(k-1) "K" T2_C= T2-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant" W2_T2=T1*(v1/v2)^(k-1) "Using C instead of K" W3_T2=(T1+273)*(v1/v2)-273 "Assuming T is proportional to v" W4_T2=T1*(v1/v2) "Assuming T is proportional to v, using C" 7-225 Heat is lost through a plane wall steadily at a rate of 600 W. If the inner and outer surface temperatures of the wall are 20 C and 5 C, respectively, the rate of entropy generation within the wall is (a) 0.11 W/K (b) 4.21 W/K (c) 2.10 W/K (d) 42.1 W/K (e) 90.0 W/K Answer (a) 0.11 W/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q=600 "W" T1=20+273 "K" T2=5+273 "K" "Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives" Q/T1-Q/T2+S_gen=0 "W/K" "Some Wrong Solutions with Common Mistakes:" Q/(T1+273)-Q/(T2+273)+W1_Sgen=0 "Using C instead of K" W2_Sgen=Q/((T1+T2)/2) "Using avegage temperature in K" W3_Sgen=Q/((T1+T2)/2-273) "Using avegage temperature in C" W4_Sgen=Q/(T1-T2+273) "Using temperature difference in K" 7-226 Air is compressed steadily and adiabatically from 17 C and 90 kPa to 200 C and 400 kPa. Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is (a) 0.76 (b) 0.94 (c) 0.86 (d) 0.84 (e) 1.00 Answer (d) 0.84 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=1.005 "kJ/kg.K" k=1.4 P1=90 "kPa" T1=17 "C" 7-173 P2=400 "kPa" T2=200 "C" T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 Eta_comp=(Cp*(T2s-T1))/(Cp*(T2-T1)) "Some Wrong Solutions with Common Mistakes:" T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K in finding T2s" W1_Eta_comp=(Cp*(T2sW1-T1))/(Cp*(T2-T1)) W2_Eta_comp=T2s/T2 "Using wrong definition for isentropic efficiency, and using C" W3_Eta_comp=(T2s+273)/(T2+273) "Using wrong definition for isentropic efficiency, with K" 7-227 Argon gas expands in an adiabatic turbine steadily from 500 C and 800 kPa to 80 kPa at a rate of 2.5 kg/s. For an isentropic efficiency of 80%, the power produced by the turbine is (a) 194 kW (b) 291 kW (c) 484 kW (d) 363 kW (e) 605 kW Answer (c) 484 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=0.5203 "kJ/kg-K" k=1.667 m=2.5 "kg/s" T1=500 "C" P1=800 "kPa" P2=80 "kPa" T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 Eta_turb=0.8 Eta_turb=(Cp*(T2-T1))/(Cp*(T2s-T1)) W_out=m*Cp*(T1-T2) "Some Wrong Solutions with Common Mistakes:" T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K to find T2s" Eta_turb=(Cp*(T2W1-T1))/(Cp*(T2sW1-T1)) W1_Wout=m*Cp*(T1-T2W1) Eta_turb=(Cp*(T2s-T1))/(Cp*(T2W2-T1)) "Using wrong definition for isentropic efficiency, and using C" W2_Wout=m*Cp*(T1-T2W2) W3_Wout=Cp*(T1-T2) "Not using mass flow rate" Cv=0.3122 "kJ/kg.K" W4_Wout=m*Cv*(T1-T2) "Using Cv instead of Cp" 7-228 Water enters a pump steadily at 100 kPa at a rate of 35 L/s and leaves at 800 kPa. The flow velocities at the inlet and the exit are the same, but the pump exit where the discharge pressure is measured is 6.1 m above the inlet section. The minimum power input to the pump is (a) 34 kW (b) 22 kW (c) 27 kW (d) 52 kW (e) 44 kW Answer (c) 27 kW 7-174 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=0.035 "m^3/s" g=9.81 "m/s^2" h=6.1 "m" P1=100 "kPa" T1=20 "C" P2=800 "kPa" "Pump power input is minimum when compression is reversible and thus w=v(P2-P1)+Dpe" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) m=V/v1 W_min=m*v1*(P2-P1)+m*g*h/1000 "kPa.m^3/s=kW" "(The effect of 6.1 m elevation difference turns out to be small)" "Some Wrong Solutions with Common Mistakes:" W1_Win=m*v1*(P2-P1) "Disregarding potential energy" W2_Win=m*v1*(P2-P1)-m*g*h/1000 "Subtracting potential energy instead of adding" W3_Win=m*v1*(P2-P1)+m*g*h "Not using the conversion factor 1000 in PE term" W4_Win=m*v1*(P2+P1)+m*g*h/1000 "Adding pressures instead of subtracting" 7-229 Air at 15 C is compressed steadily and isothermally from 100 kPa to 700 kPa at a rate of 0.12 kg/s. The minimum power input to the compressor is (a) 1.0 kW (b) 11.2 kW (c) 25.8 kW (d) 19.3 kW (e) 161 kW Answer (d) 19.3 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=1.005 "kJ/kg.K" R=0.287 "kJ/kg.K" Cv=0.718 "kJ/kg.K" k=1.4 P1=100 "kPa" T=15 "C" m=0.12 "kg/s" P2=700 "kPa" Win=m*R*(T+273)*ln(P2/P1) "Some Wrong Solutions with Common Mistakes:" W1_Win=m*R*T*ln(P2/P1) "Using C instead of K" W2_Win=m*T*(P2-P1) "Using wrong relation" W3_Win=R*(T+273)*ln(P2/P1) "Not using mass flow rate" 7-230 Air is to be compressed steadily and isentropically from 1 atm to 25 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be (a) 3 atm (b) 5 atm (c) 8 atm (d) 10 atm (e) 13 atm Answer (b) 5 atm 7-175 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1 "atm" P2=25 "atm" P_mid=SQRT(P1*P2) "Some Wrong Solutions with Common Mistakes:" W1_P=(P1+P2)/2 "Using average pressure" W2_P=P1*P2/2 "Half of product" 7-231 Helium gas enters an adiabatic nozzle steadily at 500 C and 600 kPa with a low velocity, and exits at a pressure of 90 kPa. The highest possible velocity of helium gas at the nozzle exit is (a) 1475 m/s (b) 1662 m/s (c) 1839 m/s (d) 2066 m/s (e) 3040 m/s Answer (d) 2066 m/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=5.1926 "kJ/kg.K" Cv=3.1156 "kJ/kg.K" T1=500 "C" P1=600 "kPa" Vel1=0 P2=90 "kPa" "s2=s1 for maximum exit velocity" "The exit velocity will be highest for isentropic expansion," T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "C" "Energy balance for this case is h+ke=constant for the fluid stream (Q=W=pe=0)" (0.5*Vel1^2)/1000+Cp*T1=(0.5*Vel2^2)/1000+Cp*T2 "Some Wrong Solutions with Common Mistakes:" T2a=T1*(P2/P1)^((k-1)/k) "Using C for temperature" (0.5*Vel1^2)/1000+Cp*T1=(0.5*W1_Vel2^2)/1000+Cp*T2a T2b=T1*(P2/P1)^((k-1)/k) "Using Cv" (0.5*Vel1^2)/1000+Cv*T1=(0.5*W2_Vel2^2)/1000+Cv*T2b T2c=T1*(P2/P1)^k "Using wrong relation" (0.5*Vel1^2)/1000+Cp*T1=(0.5*W3_Vel2^2)/1000+Cp*T2c 7-232 Combustion gases with a specific heat ratio of 1.3 enter an adiabatic nozzle steadily at 800 C and 800 kPa with a low velocity, and exit at a pressure of 85 kPa. The lowest possible temperature of combustion gases at the nozzle exit is (a) 43 C (b) 237 C (c) 367 C (d) 477 C (e) 640 C Answer (c) 367 C 7-176 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 T1=800 "C" P1=800 "kPa" P2=85 "kPa" "Nozzle exit temperature will be lowest for isentropic operation" T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T2=T1*(P2/P1)^((k-1)/k) "Using C for temperature" W2_T2=(T1+273)*(P2/P1)^((k-1)/k) "Not converting the answer to C" W3_T2=T1*(P2/P1)^k "Using wrong relation" 7-233 Steam enters an adiabatic turbine steadily at 400 C and 3 MPa, and leaves at 50 kPa. The highest possible percentage of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is (a) 5% (b) 10% (c) 15% (d) 20% (e) 0% Answer (b) 10% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=3000 "kPa" T1=400 "C" P2=50 "kPa" s2=s1 s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) x2=QUALITY(Steam_IAPWS,s=s2,P=P2) misture=1-x2 "Checking x2 using data from table" x2_table=(6.9212-1.091)/6.5029 7-234 Liquid water enters an adiabatic piping system at 15 C at a rate of 8 kg/s. If the water temperature rises by 0.2 C during flow due to friction, the rate of entropy generation in the pipe is (a) 23 W/K (b) 55 W/K (c) 68 W/K (d) 220 W/K (e) 443 W/K Answer (a) 23 W/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=4180 "J/kg.K" m=8 "kg/s" T1=15 "C" T2=15.2 "C" S_gen=m*Cp*ln((T2+273)/(T1+273)) "W/K" 7-177 "Some Wrong Solutions with Common Mistakes:" W1_Sgen=m*Cp*ln(T2/T1) "Using deg. C" W2_Sgen=Cp*ln(T2/T1) "Not using mass flow rate with deg. C" W3_Sgen=Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with deg. C" 7-235 Liquid water is to be compressed by a pump whose isentropic efficiency is 75 percent from 0.2 MPa to 5 MPa at a rate of 0.15 m3/min. The required power input to this pump is (a) 4.8 kW (b) 6.4 kW (c) 9.0 kW (d) 16.0 kW (e) 12.0 kW Answer (d) 16.0 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=0.15/60 "m^3/s" rho=1000 "kg/m^3" v1=1/rho m=rho*V "kg/s" P1=200 "kPa" Eta_pump=0.75 P2=5000 "kPa" "Reversible pump power input is w =mv(P2-P1) = V(P2-P1)" W_rev=m*v1*(P2-P1) "kPa.m^3/s=kW" W_pump=W_rev/Eta_pump "Some Wrong Solutions with Common Mistakes:" W1_Wpump=W_rev*Eta_pump "Multiplying by efficiency" W2_Wpump=W_rev "Disregarding efficiency" W3_Wpump=m*v1*(P2+P1)/Eta_pump "Adding pressures instead of subtracting" 7-236 Steam enters an adiabatic turbine at 8 MPa and 500 C at a rate of 18 kg/s, and exits at 0.2 MPa and 300 C. The rate of entropy generation in the turbine is (a) 0 kW/K (b) 7.2 kW/K (c) 21 kW/K (d) 15 kW/K (e) 17 kW/K Answer (c) 21 kW/K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=8000 "kPa" T1=500 "C" m=18 "kg/s" P2=200 "kPa" T2=300 "C" s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) s2=ENTROPY(Steam_IAPWS,T=T2,P=P2) S_gen=m*(s2-s1) "kW/K" 7-178 "Some Wrong Solutions with Common Mistakes:" W1_Sgen=0 "Assuming isentropic expansion" 7-237 Helium gas is compressed steadily from 90 kPa and 25 C to 600 kPa at a rate of 2 kg/min by an adiabatic compressor. If the compressor consumes 70 kW of power while operating, the isentropic efficiency of this compressor is (a) 56.7% (b) 83.7% (c) 75.4% (d) 92.1% (e) 100.0% Answer (b) 83.7% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=5.1926 "kJ/kg-K" Cv=3.1156 "kJ/kg.K" k=1.667 m=2/60 "kg/s" T1=25 "C" P1=90 "kPa" P2=600 "kPa" W_comp=70 "kW" T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 W_s=m*Cp*(T2s-T1) Eta_comp=W_s/W_comp "Some Wrong Solutions with Common Mistakes:" T2sA=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W1_Eta_comp=m*Cp*(T2sA-T1)/W_comp W2_Eta_comp=m*Cv*(T2s-T1)/W_comp "Using Cv instead of Cp" 7-238 ... 7-241 Design and Essay Problems

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