Exam2s - Spring 2006 Thermodynamics Exam #2 Solution...

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Spring 2006 1 ME 201 Thermodynamics Exam #2 Solution Problem 1 Steam at 300 kPa with quality 0.96316 passes through a valve to convert it to saturated vapor. Determine the exit pressure required. Solution: We begin with our template. Substance Type: Compressible System Type: Control Volume Device: Valve Process: Isenthalpic Inlet State: Fixed Outlet State: UNKNOWN q = 0 w sh = 0 1st Law: h 2 - h 1 = 0 State 1 State 2 T 1 = 133.52ºC T 2 = 81.31ºC P 1 = 300 kPa P 2 = 50 kPa h 1 = 2645.2 kJ/kg h 2 = 2645.2 kJ/kg Phase: 2 phase with x 1 = 0.96316 Phase: sat.vap Italicized values are from steam tables. Bold values are calculated. Approach: Go to the steam tables and get h f and h g at 300 kPa. Use the quality to calculate h 1 . Set h 2 = h 1 . Go to the saturation pressure table and find the pressure that gives an h g = h 2 . At 300 kPa, we have h f = 561.43 kJ/kg and h g = 2724.9 kJ/kg Then we can calculate h 1 = (1-x 1 )h f +x 1 h g = (1-0.96316)(561.43) + (0.96316)(2724.9) = 2645.2 kJ/kg Since our valve is isenthalpic h 2 = h 1 = 2645.2 kJ/kg Now going to the saturation pressure table we find P 2 = 50 kPa
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ME 201 Thermodynamics Spring 2006 2 Problem 2 A piston-cylinder device contains 0.001 kg of CO 2 at 750 kPa and 1340 K. The device undergoes a polytropic process with polytropic exponent -0.3 or PV -0.3 = constant which decreases the pressure to 550.2 kPa. Determine the boundary work and heat
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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Exam2s - Spring 2006 Thermodynamics Exam #2 Solution...

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