FinalExamS - Spring 2006 ME 201 Thermodynamics Final Exam...

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Spring 2006 1 ME 201 Thermodynamics Final Exam Solutions Directions: Work all three problems. The exam is open notes and open text book. All problems have equal weight. Note that you may round where appropriate to avoid interpolation. Problem 1 A more realistic model for the compression process in an internal combustion uses a polytropic process followed by an isentropic process instead of a single isentropic process. The engine operates with the following conditions: Six cylinder, four stroke engine with displacement of 2.8 liters Compression ratio: 9.065 Compression ratio for polytropic process: 3 Polytropic exponent for first compression process: 1.203 Initial air conditions: 280 K and 120 kPa Engine speed of 1300 rpm Determine a.) the air temperature at the end of both compression processes b. the heat transfer for the polytropic process b.) the engine power in kW for the two compression processes Solution: We begin with a block diagram of the system nodded out for the two processes of interest. Next we need to obtain our volumes. We are given V disp = 2.8 x 10 -3 m 3 and r = 9.065 then TDC BDC disp V V V - = TDC BDC rV V = 3 V / V 2 BDC = Polytropic Compression Isentropic Compression 1 2 3
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ME 201 Thermodynamics Spring 2006 2 Substituting gives 3 4 - 3 disp TDC m 10 x 3.4718 1 065 . 9 10 x 8 . 2 1 r V V = - = - = - and 3 3 4 BDC m 10 x 1472 . 3 ) 10 x 4718 . 3 )( 065 . 9 ( V - - = = Following the polytropic process we have 3 -3 3 BDC 2 m 10 x 1.0491 3 / ) 10 x 14724 . 3 ( 3 / V V = = = - Now setting up our table and entering our operating information Node T(K) P(kPa) V(m 3 ) v(m 3 /kg) u(kJ/kg) v r 1 280 120 3.1472 x 10 -3 0.6697 199.75 2 350 449.94 1.0491 x 10 -3 0.2232 250.02 422.2 3 540 2098 3.4718 x 10 -4 0.0739 389.34 139.72 Italicized values from ideal gas relations. Bold values calculated. We begin by calculating our specific volume at 1
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FinalExamS - Spring 2006 ME 201 Thermodynamics Final Exam...

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