HW9_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 #9 - Solutions 28 total points, each section of each problem is worth 2 points HW#9 topics covered: entropy change and application of the variation of entropy 1. 5 kg of R-134a is initially at 550 kPa with a quality of 15%. The temperature of the refrigerant is increased by 35°C by an isobaric process. Use the Kleinstreuer tables for your analysis. a) What is the total entropy change of the process? b) Can the process occur adiabatically? Solutions a) State 1 from Table B.10: s 1 = s f + xs fg = 0.29461 kJ/kgK + 0.15*0.62821 kJ/kgK = 0.3888 kJ/kgK From the problem statement, we know P 2 = P 1 = 550 kPa. Since it’s saturated R-134a, we know the temperature for state 1 is T 1 = T sat @ 550 kPa = 18.73 °C T 2 = T 1 + 35 °C = 53.73 °C Comparing P2 and T2 with the tables, we see state 2 is superheated vapor. Since 550 kPa (0.55 MPa), you can either assume P = 0.5 MPa OR P = 0.6 MPa and interpolate with T 2 to find s 2 P = 0.5 MPa: s 2 = 1.0309 + (1.0599 - 1.0309) 53 . 73 50 60 50 = 1.0417 kJ/kgK s 2 - s 1 = 1.0417 - 0.3888 = 0.6529 kJ/kgK ΔS = m(s 2 - s 1 ) ΔS = m( s 2 - s 1 ) = 5 kg(0.6529 kJ/kgK) = 3.26 kJ/K OR P = 0.6 MPa: s 2 = 1.0121 + (1.0417 - 1.0121) 53 . 73 50 60 50 = 1.0231 kJ/kgK s 2 - s 1 = 1.0231 - 0.3888 = 0.6343 kJ/kgK ΔS = m(s 2 - s 1 ) ΔS = m(s 2 - s 1 ) = 5 kg(0.6343 kJ/kgK) = 3.17 kJ/K b) For a closed system, VOS is
ΔS = Q/Tb+Sgen For an adiabatic process, Q = 0. and Sgen = ΔS = 3.3 or 3.2 kJ/K from part a). Since Sgen>0, yes the process can occur adiabatically.
2. You are a hungry student trying to improve your cooking skills. You use a pressure cooker to make a meal while you do your thermodynamics homework. The cooker is insulated, with a volume of 2 m 3 , 80°C, 500 kPa. Your food is finished, so you open the valve to release the pressure. The internal pressure of the cooker drastically decreases to 50 kPa and the valve is closed. Assume the working fluid is air and that the process is reversible and adiabatic. Using the ideal gas law and associated state relations, determine the mass released during the process. You can assume constant specific heats. Solutions C.V. the volume inside cooker Since the working fluid is air, we can use ideal gas relations: T 2 = T 1 (P 2 /P 1 ) k-1/k = (273.15+80 K)(50/500 kPa) 1.4-1/1.4 = 183 K m 1 = P 1 V/RT 1 = (500 kPa)(2 m 3 ) / (0.287 kJ/kgK)(273.15+80 K) = 9.87 kg m 2 = P 2 V/RT 2 = (50 kPa)(2 m 3 ) / (0.287 kJ/kgK)(183 K) = 1.90 kg m RELEASED = 9.87 - 1.90 = 7.97 kg 3. A piston/cylinder contains air at 1500 kPa and 750 K, which expands to a pressure of 200 kPa. Assuming the process is a reversible, isothermal process: a) Find the final temperature b) The specific work Solutions a) C.V. air, mass unknown We know it’s an isothermal process, so T 1 = T 2 = 750 K b) Additionally, u 2 = u 1 , s 𐩑𐩑 T2 = s 𐩑𐩑 T1 energy: 1 q 2 = 1 w 2 Energy equation: u 2 - u 1 = 1 q 2 - 1 w 2 Entropy equation: s 2 - s 1 = 𝑑𝑑𝑑𝑑 𝑇𝑇 + 1 S 2, gen Process: reversible 1 S 2, gen = 0, T = constant 𝑑𝑑𝑑𝑑 𝑇𝑇 = 1 q 2 / T 1 w 2 = T(s 2 - s 1 ) = T(s 𐩑𐩑 T2 - s 𐩑𐩑 T1 - R ln 𝑃𝑃 2 𝑃𝑃 1 ) = -RTln 𝑃𝑃 2 𝑃𝑃 1 = ( -0.287 kJ/kgK)(750 K)(ln 200 𝑘𝑘𝑘𝑘𝑘𝑘 1500 𝑘𝑘𝑘𝑘𝑘𝑘 ) 1 w 2 = 433.7 kJ/kg 4. 0.5 kg of steam is heated in a piston/cylinder from a saturated liquid at 150 °C to a saturated vapor in a reversible constant pressure process.
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