HW10_ME235

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University of Michigan, Dearborn *

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ME 230

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Mechanical Engineering

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Apr 3, 2024

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THE UNIVERSITY OF MICHIGAN Department of Mechanical Engineering ME 235: Thermodynamics I Section 002 Winter 2022 © Wooldridge 2022 Homework Set #10 - Solutions 18 total points, each section of each problem is worth 2 points HW#10 topics covered: second law of thermodynamics and Rankine problems 1. In a piston/cylinder, 0.1 kg of water is at 400 °C, 100 kPa. In an isobaric process, the water is cooled to 100 °C. The heat transfer out of the water is used as a heat source into a heat engine. The heat engine rejects heat to the atmosphere at 25 °C. The entire system (piston/cylinder and heat engine) can be considered reversible. a) What is the total heat transfer out of the water [kJ]? b) What is the work out of the heat engine [kJ]? Solutions using CATT3 a) Energy equation: m(u 2 - u 1 ) = 1 Q 2 - 1 W 2 Entropy equation: m(s 2 - s 1 ) = + 0 𝑑𝑞 𝑇 Process: P = Constant 1 W 2 = P(V 2 - V 1 ) = mP(v 2 - v 1 ) ∫ ?𝑑𝑉 = C.V. of piston/cylinder State 1 of water in cylinder using T1 and P1: s 1 = 9.608 kJ/kgK, h 1 = 3280 kJ/kg State 2 of water using T2 and P1=P2: s 2 = 8.448 kJ/kgK, h 2 = 2687 kJ/kg 1 Q 2 = m(u 2 - u 1 ) + 1 W 2 = m(h 2 - h 1 ) = (0.1 kg)(2687 kJ/kg - 3280 kJ/kg) = -59.3 kJ b) C.V. of system (piston/cylinder and heat engine) Energy equation: m(u 2 - u 1 ) = Q L - 1 W 2 - W HE Entropy equation: m(s 2 - s 1 ) = + 0 −? 𝐿 𝑇 ?𝑚? Q L = mT amb (s 2 - s 1 ) = (0.1 kg)(298.15 K)(8.448 kJ/kgK - 9.608 kJ/kgK) = -34.585 kJ W HE = (u 2 - u 1 ) = - Q L - 1 Q 2 = 34.585 + 59.3 = 24.7 kJ
2. Two tanks containing air are connected with a valve. Tank A has 1 kg at 250 kPa, 500 K. Tank B has 2.5 kg at 150 kPa, 350 K. Assume the tanks are insulated and rigid. The valve opens and the air equilibrates to a final uniform state without external heat or work transfer. a) What is the final T [K]? b) What is the final P [kPa]?0. c) What is the entropy generated [kJ/K]? Solutions C.V. Tank A + Tank B COM: m tot = m 2 = m A + m B V tot = V 2 = V A + V B Energy eq: U 2 - U 1 = m 2 u 2 - m A u A1 - m B u B1 = 1 Q 2 - 1 W 2 = 0 Entropy eq: S 2 - S 1 = m 2 s 2 - m A s A1 - m B s B1 = 1 Q 2 /T + 1 S 2gen Process eq: V = constant 1 W 2 = 0, adiabatic 1 Q 2 = 0 State A 1 : V A = m A RT A1 /P A1 = (1 kg)(0.287 kJ/kgK)(500 K)/(250 kPa) = 0.574 m 3 State B 1 : V B = m B RT B1 /P B1 = (2.5 kg)(0.287 kJ/kgK)(350 K)/(150 kPa) = 1.674 m 3 State 2: m 2 = m A + m B = 1 + 2.5 = 3.5 kg V 2 = V = V A + V B = 0.574 + 1.674 = 2.248 m 3 Using energy eq and constant specific heats to find T 2 : u 2 = (m A u A1 + m B u B1 )/m 2 T 2 = (m A T A1 + m B T B1 )/m 2 = ((1 kg)(500 K) + (2.5 kg)(350 K))/3.5 kg = 393 K = 120 °C Ideal gas to find P 2 given T 2 : P 2 = m 2 RT 2 /V = (3.5 kg)(0.287 kJ/kgK)(393 K)/(2.248 m 3 ) = 175.6 kPa Entropy eq to find total entropy generated: S 2 - S 1 = m A [C p ln(T 2 /T A1 ) - Rln(P 2 /P A1 )] + m B [C p ln(T 2 /T B1 - Rln(P 2/ P B1 )] = (1)[(1.004)ln(393/500) - (0.287)ln(175.6/250)] + (2.5)[(1.004)ln(393/350) - (0.287)ln(175.6/150)] = 0.037 kJ/K
3. A bronze sphere weighs 25 kg, 5 kg of which is water. The water and sphere are initially equilibrated and the water inside the sphere is at 3 MPa, 300°C. The entire system cools to an ambient temperature of 25 °C. What is the entropy generated by cooling the bronze and the water [kJ/K] ? c bronze = 0.4 kJ/(kgK) Solutions using CATT3 C.V. bronze sphere and water (controlled mass) Energy eq: U 2 - U 1 = 1 Q 2 - 1 W 2 , If we assume the sphere doesn't change shape/volume, so there is no expansion or compression work and V = constant 1 W 2 , = 0 U 2 - U 1 = 1 Q 2 Entropy eq: S 2 - S 1 = dQ /T + 1 S 2gen = 1 Q 2 /T 0 + 1 S 2gen For water, From CATT3 output for T1 = 300 oC and P1 = 3 MPa: v 1 = 0.08114 m3/kg, u 1 = 2750 kJ/kg, s 1 = 6.539 kJ/kgK From CATT3 output for T2 = 25 oC and v 2 = v 1 = 0.08114 x 2 = 0.001848, u 2 = 109.1 kJ/kg, s 2 = 0.3825 kJ/kgK 1 Q 2 = m bronze (u 2 - u 1 ) bronze + m H20 (u 2 - u 1 ) H2O 1 Q 2 = m bronze c bronze *(T 2 - T 1 ) + m H20 (u 2 - u 1 ) = (20 kg)(0.4 kJ/kgK)(25 - 300 °C) + (5 kg)(109.1 - 2750 kJ/kg) = -15,405 kJ S 2 - S 1 = m bronze (s 2 - s 1 ) + m H20 (s 2 - s 1 ) = (20 kg)(0.4 kJ/kgK)ln(298.15/573.15 K) + (5 kg)(0.3825 - 6.539) = -36.0 kJ/K 1 S 2gen = S 2 - S 1 - 1 Q 2 /T 0 = -36.0 +15,405/298K = +15.7 kJ/K
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4. An ideal Rankine cycle uses water as the working fluid. Saturated water vapor leaves the boiler at 100 °C and saturated liquid water leaves the condenser at 50 °C. What is the thermal efficiency of the cycle? Solutions using CATT3 C.V. turbine State 3: x = 1, T = 100°C h = 2676 kJ/kg, s = 7.355 kJ/kgK, P = 101.3 kPa State 4: T = 50°C, s = s 3 = 7.355 kJ/kgK h = 2359 kJ/kg w T,s = h 3 - h 4 = 2676 - 2359 = 317 kJ/kg State 1: x = 0, T = 50°C P 1 = 12.35 kPa, v = 0.001012 m 3 /kg, s = 0.7037 kJ/kgK, h = 209.3 kg/kg State 2: P 3 = P 2 w P, s = v 1 (P 2 - P 1 ) = -(0.001012)(101.3 - 12.35) = -0.09 kJ/kg h 2 = h 1 + w P, s = 209.3 + 0.09 = 209.39 kJ/kg C.V. boiler q H = h 3 - h 2 = 2676 - 209.39 = 2467 kJ/kg 𝝶 CYLCE = (w TURBINE - w PUMP )/q H = (317 - 0.09)/2467 = 0.128 5. Steam exits the boiler of a power plant at 2.5 MPa, 500 °C. Coolant fluid cools the condenser via heat exchange. If the steam flows through the plant at 35 kg/s, what is the net power output? Assume no pressure losses through the condenser and boiler and treat the turbine and pump as reversible and adiabatic. Solutions using CATT3 Known from problem statement: State 1: , T = 75 °C State 3: P 3 = P 2 = 2.5 MPa, 500 °C C.V. pump State 1: x = 0, T 1, = 75 °C P = 38.58 kPa, h 1 = 313.9 kJ/kg, v = 0.001026 m 3 /kg w P, s = v 1 (P 2 - P 1 ) = (0.001026)(2500 - 38.58) = 2.525 kJ/kg C.V. turbine State 3: P = 2500 kPa, T = 500 °C h = 3462 kJ/kg, s = 7.323 kJ/kgK State 4: T 1 = T 4 = 75 °C h = 2510 kJ/kg w T,s = h 3 - h 4 = 3462 - 2510 = 952 kJ/kg w NET = w TURBINE - w PUMP = 949.475 kJ/kg W NET = ṁw NET = (35 kg/s)(949.475 kJ/kg) = 33,231 kJ/kg = 33.2 MJ/kg
6. A steam power plant has a high pressure of 2 MPa. The water exits the boiler at 750 °C and exits the condenser at 75 °C. The turbine has an efficiency of 90% and the pump has an efficiency of 75%. What is the thermal efficiency of the cycle? Solutions using CATT3 C.V. pump State 1: into the pump = saturated liquid (x = 0), T 1, P 1 = 38.58 kPa, h 1 = 313.9 kJ/kg, v = 0.001026 m 3 /kg w P, s = v 1 (P 2 - P 1 ) = (0.001026)(2000 - 38.58) = 2.0124 kJ/kg h 2 = h 1 + w P, s = 313.9 + 2.012 = 315.91 kJ/kg 𝝶 PUMP = w s / w a w a = w s /0.75 = 2.0124/0.75 = 2.683 kJ/kg C.V. turbine State 3: P = 2000 kPa, T = 750 °C h = 4033 kJ/kg, s = 8.065 kJ/kgK Find State 4 using s 3 = s 4 and P 4 = P 1 and CATT3: h 4 = 2783 kJ/kg w T,s = h 3 - h 4 = 4033 - 2783 = 1250 kJ/kg 𝝶 TURBINE = w a / w s w a = (w s )(0.90) = (1394)(0.9) = 1125 kJ/kg C.V. boiler q H = h 3 - h 2 = 4033 - 316 = 3717 kJ/kg 𝝶 CYLC = (w TURBINE - w PUMP )/q H = (1125 - 2.68)/3717 = 30.2%