Exam4s - Spring 2006 ME 201 Thermodynamics Exam#4 Solution...

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Spring 2006 1 ME 201 Thermodynamics Exam #4 Solution Problem 1 Often in an ideal jet propulsion cycle a second burner is used after the turbine, as shown in the figure. Consider the following operating conditions: Inlet conditions for the turbine: 900 kPa and 1800 K Required turbine work output: 270.13 kJ/kg Exit temperature of the second burner: 2000 K Nozzle exit pressure: 59.31 kPa Determine a.) Pressure of the second burner b.) Exit velocity of the nozzle Solution: Since the block diagram has been sketched and nodded for us, we begin by setting up our table Node T(K) P(kPa) h(kJ/kg) φ (kJ/kg K) 1 1800 900 2003.3 3.6684 2 1580 515.18 1733.17 3.50829 3 2000 515.18 2252.1 3.7994 4 1200 59.31 1277.79 3.17898 Italicized values from air tables. Bold values calculated State 1 is fixed, so we may go to the air tables and find h 1 = 2003.3 kJ/kg and φ 1 = 3.6684 kJ/(kg K) Now traversing the cycle 1-2 Turbine (Isentropic) The work of the turbine must provide the work of the compressor, so that h 2 -h 1 = -w c Solving for h 2 h 2 = h 1 - w c = 2003.3 - 270.13 = 1733.17 kJ/kg
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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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Exam4s - Spring 2006 ME 201 Thermodynamics Exam#4 Solution...

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