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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pdf

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5

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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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