HW9_ME235

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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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a) Based on the expression for expansion and compression work, what is the work transfer [kJ] done by the process? b) Based on the conservation of energy, what is the heat transfer of the process [kJ]? c) Based on the variation of entropy (2nd law of thermodynamics), what is the heat transfer of the process [kJ]? d) Are the heat transfer results from c) and d) consistent? Solutions using McGraw Hill Table B.4 Energy equation: m(u 2 - u 1 ) = 1 Q 2 - 1 W 2 Entropy equation: m(s 2 - s 1 ) = ∫ 𝑑𝑑𝑑𝑑 𝑇𝑇 + 0 = 1 𝑄𝑄 2 𝑇𝑇 Process: P = C ⇒ 1 W 2 = mP(v 2 - v 1 ) = mPv fg 1 W 2 = (0.5 kg)(476.16 kPa)(0.39248 - 0.001091) m 3 /kg = 93.2 kJ Energy equation: 1 Q 2 = m(u 2 - u 1 ) + 1 W 2 = mu fg + 1 W 2 1 Q 2 = (0.5 kg)(1927.4 kJ/kg) + 93.182 kJ = 1057 kJ Entropy equation: 1 Q 2 = mT(s 2 - s 1 ) = mTs fg = (0.5 kg)[(150 + 273.15) K](4.9953 kJ/kgK) = 1057 kJ Yes, the terms are reasonably the same, differences due to rounding. 5. You are a hungry student on a different night trying to improve your cooking skills. You have a recipe containing 100 g beef, 20 g broccoli, and 50 g potatoes. You put them all in a pot on an electrical stove and see that the temperature (in Kelvin) doubles. Assuming all ingredients were thawed (above freezing) before cooking, what is the change in entropy of the food due to heating? Assume no heat loss from the ingredients. Solutions using McGraw Hill Appendices C.V. cooking pot and stove VOS: S 2 - S 1 = ∑m i (s 2 - s 1 ) i = ∫ 𝑑𝑑𝑑𝑑 𝑇𝑇 + 1 S 2, gen = 1 S 2, gen Specific heats found from Appendix B.3 ∑m i (s 2 - s 1 ) i = ln ( 𝑇𝑇 2 𝑇𝑇 1 ) ∑m i C i ∑m i C i = (0.1 kg)(3.08 kJ/kgK) + (0.02 kg)(3.86 kJ/kgK) + (0.05 kg)(3.45 kJ/kgK) = 0.558kJ/K S 2 - S 1 = ln ( 𝑇𝑇 2 𝑇𝑇 1 ) ∑m i C i = (0.558 kJ/K) ln(2) = 0.387 kJ/K

Problem 6. Silane (SiH 4 , molecular weight = 32.12) gas is used to manufacture semiconductor materials in the microelectronics industry. It is a highly explosive gas that will spontaneously ignite if exposed to air at a temperature above a critical value of T ignition = 325 o C. However, the silane needs to be pressurized in order to deliver the gas to the reaction chamber. In this problem, you will examine how to safely pressurize SiH 4 . ***********Throughout your analysis treat silane as an ideal gas!!*********** T sur = 298 K a) The silane is initially at a pressure of P 1 =1.5 MPa and occupies a volume of V 1 = 0.5 m 3 . If the silane is compressed isothermally at T = 52 o C (to keep the temperature well below T ignition ) to a volume of V 2 = 0.2 m 3 , what is the pressure in the chamber at the end of the process? You can neglect kinetic and potential energy effects throughout your analysis of parts a)-d). Solution: PV=mRT = const. So P 2 =P 1 ( 1 / 2 ) = (1.5 MPa)( 1 = 0.5 m 3 / 2 = 0.2 m 3 ) = 3.75 MPa b) For the process described in part a), what is the heat transfer required [kJ] and is it into or out of the system? c) What is the specific entropy generation [kJ/(kgK)] associated with the process?

d) Sketch the process on P-V and T-S diagrams. Label the initial and final states. Include the vapor dome on your diagrams. On the T-S diagram, draw two lines of constant pressure that pass through the dome and label the isobaric lines P high and P low . On the P-V diagram, draw two lines of constant temperature that pass through the dome and label the isothermal lines T high and T low .

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