HW4_ME235

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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 #4 - Solutions 32 total points, each section of each problem is worth 2 points HW#4 topics covered: Thermodynamic properties; P-v-T behavior; states and phases; property data, diagrams, tables; ideal gases, equations of state 1. For water at the following conditions, determine the missing properties from T, P, x and v. Use CATT3 or another digital database for your work. a. T = 150 ℃, P = 450 kPa: b. T = 80 ℃, P = 100 kPa c. P = 300 kPa, v = 0.500 m 3 /kg Solutions using CATT3: a. Temp Pressure Specific Volume Internal Energy Specific Enthalpy Specific Entropy Quality Phase 150°C 0.450 MPa 0.4164 m3/kg 2561 kJ/kg 2749 kJ/kg 6.868 kJ/kg/K Undefin ed Superheated vapor b. Temp Pressure Specific Volume Internal Energy Specific Enthalpy Specific Entropy Quality Phase 80°C 0.100 MPa 0.00102 9 m3/kg 334.8 kJ/kg 334.9 kJ/kg 1.075 kJ/kg/K Undefin ed Compressed liquid

c. Temp Pressure Specific Volume Internal Energy Specific Enthalpy Specific Entropy Quality Phase 133.5°C 0.300 MPa 0.500 m3/kg 2197 kJ/kg 2347 kJ/kg 6.061 kJ/kg/K 0.825 Liquid vapor mixture 2. To design parts of the air-conditioning system for a car we would like to know the following information. a. the pressure when R- 134a is throttled to become a two -phase saturated liquid- vapor mixture at - 30 ℃. b. the temperature of R- 134a if it is condensed at 800 kPa. Use the Kleinstreuer data tables to find the information requested. You can check your answers using CATT3, but use the tables first. Solutions using Tables: a. Using Table B.9, we want to know the saturation pressure for T = - 30 ℃: Psat = 84.43 kPa b. Assuming R134a condenses to a mixed vapor and liquid phase, we want Tsat for P = 800 kPa. From Table B.10, P = 0.8 MPa = 800 kPa. Tsat = 31.31 ℃. Solutions using CATT3: c. Using the “T & X” with X = quality; inputting a temperature of - 30 ℃ and any X value ( 0,1) we can find Psat: Psat = 0.08438 MPa = 84.38 kPa d. Assuming R134a condenses to a mixed vapor and liquid phase, using any value for x and P and input “P & X” into CATT3 where P = 0.8 MPa = 800 kPa, x = 0.0, Tsat 31.33 ℃. 3. A storage facility for methane (main component of natural gas) will store methane as a liquid at 155 K. A vent valve on the tank will open at a pressure of 3,000 kPa in case the cooling system fails and the pressure starts to increase. a. What is the minimum normal storage pressure? b. What is the maximum temperature in the tank while there is some liquid methane inside the tank?

Solutions using CATT3: Using Methane (under the Cryogenics section of CATT3): a. We have a compressed liquid state when the pressure is higher than the saturation pressure. Utilize “T & X” where Temperature = 155 K = - 118.15 ℃ and X is any value (0,1). Psat = 1.293 MPa = 1293 kPa = This is the minimum needed P. Any lower pressure and the methane will vaporize. b. The maximum pressure in the tank while there is some liquid is 3,000 kPa. Using “P & X” to input the maximum pressure = 3,000 kPa = 3 MPa and X is (0,1). The maximum temperature is - 96 ℃ = 177 K. 4. a) What is the internal energy of methane at 155 K and 0.04 m3/kg? Use CATT3 for your work. b) Show the state on a pressure - volume diagram. Label two isotherms Thigh and Tlow on your diagram. Label the three phases on your diagram and the saturation region. c) Show the state on a temperature - volume diagram. Label two isobars Phigh and Plow on your diagram. Label the three phases on your diagram and the saturation region. Solutions using CATT3: a) For cryogenic methane, “T & V” where Temperature = 15 5 K and v = 0.04 m3/kg. u = 237.5 kJ/kg . b) Sketch c) Sketch

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5. Two tanks are connected as shown in the figure, both containing water. Tank A is at 200 kPa, v = 0.5 m3/kg, V = 1 m3 and tank B contains 3.5 kg at 0.5 MPa, 400C. The valve is now opened and the two come to a uniform equilibrated state. Find the final specific volume. Hint: Recall the definition of specific volume is Vtotal/mtotal. We need to find mtotal = mA + mB; and Vtotal = VA+ VB State A1: (P, v) m A = V A /v A = 1/0.5 = 2 kg State B1: (P, T) using CATT3 v B = 0.61732 m 3 /kg V B = m B v B = 3.5 kg * 0.61732 m 3 /kg = 2.161 m 3 Final state: mtot = m A + m B = 3.5 + 2 kg = 5.5 kg vtot = Vtot/mtot = V A + V B = (2.161 + 1) m 3 / 5.5 kg = v 2 0.575 m 3 /kg 6. A closed system is subject to three processes. The working fluid is 2 kg of ammonia. Process 1 to 2 is a constant volume process, with P1 = 10 bar and x1 = 0.6. The constant volume process ends with ammonia as a saturated vapor. Process 2 to 3 is a constant temperature process where heat is added to the ammonia, and the process ends with a pressure, P3, equal to P1. The final process (3 to 1) is a constant pressure process where the final state is the same as the initial state 1. a) Sketch the processes on a P-V diagram. Label the saturation region, liquid and vapor regions, two isotherms (Thigh and Tlow) and all states (1, 2, and 3). b) Sketch the processes on a T-V diagram. Label the saturation region, liquid and vapor regions, two isotherms (Phigh and Plow) and all states (1, 2, and 3). c) What is the pressure at state 2? d) What is the work transfer [kJ] for process 1-2? e) What is the heat transfer [kJ] for process 1-2? Solutions using CATT3: a) Sketch b) Sketch c) To fine P2, we need two independent properties. Based on the processes, v1 = v2, and we are told x2 = 1.0.

Using catt3 P1 = 10 bar = 1 MPa, x = 0.6; v1 = 0.007778 m3/kg and x2 = 1.0, using CATT3 (you have to iterate since there is no v and x option; P2 = 1.66 MPa= 16.5 bar d) W = integral PDV = 0 e) DE = Q-W, assume changes in kinetic and potential energy are negligible. DU = Q-0, so Q = m(u2-u1) Using CATT3, P1 = 10 bar = 1 MPa, x = 0.6; u1 = 919.4 kJ/kg. v1 = v2 (constant volume process) = 0.007778 m3/kg and x2 = 1.0, using CATT3 (you have to iterate since there is no v and x option; P2 = 1.66 MPa= 16.5 bar, u2 = 1341 kJ/kg. Q = 2 kg (1341 - 919.4 kJ/kg) = 843 kJ