Quiz2 - flow rate in lb m/s Pressure Ratio 2.5 =(5.67472 x 10-5 2 m& –(2.66996 x 10-2 m& 1.6147 Pressure Ratio 3.0 =(3.5414 x 10-5

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Fall 2007 1 ME 416 Computer Assisted Design of Thermal Systems Quiz #2 Closed Book, Closed Notes An actual axial compressor has inlet air at 270 K and 80 kPa applies a pressure ratio of 3.64. For a mass flow rate of 23 kg/s, determine the required actual power for the compressor and the actual exit temperature. The following properties of air may prove useful: density ( ρ ): 1.21 kg/m 3 specific heat (c P ): 1.004 kJ/(kg · K) specific heat ratio (k): 1.4
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ME 416 CAD of Thermal Systems Fall 2007 2 Equations for Quiz #2 The required power for any compressor is given by ( 29 T - T c m W in out P = in kW` For an ideal or isentropic fan the exit temperature is given by P P T T 1)/k - (k in out in ideal out, = Using the isentropic efficiency we can write W W ideal actual η = The following operating equations apply for an actual axial compressor for a mass
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Unformatted text preview: flow rate in lb m /s. Pressure Ratio: 2.5 η = (5.67472 x 10-5 ) 2 m & – (2.66996 x 10-2 ) m & + 1.6147 Pressure Ratio: 3.0 η = (3.5414 x 10-5 ) m & 3 – (4.9514 x 10-3 ) 2 m & + (2.0582 x 10-1 ) m &- 1.7747 Pressure Ratio: 3.5 η = (1.0412 x 10-4 ) m & 3 – (1.5478 x 10-2 ) 2 m & + (7.3349 x 10-1 ) m &- 10.324 Pressure Ratio: 4.0 η = (1.2500 x 10-4 ) m & 3 – (1.9468 x 10-2 ) 2 m & + (9.7623 x 10-1 ) m &- 14.9977 Pressure Ratio: 4.5 η = (8.6313 x 10-6 ) m & 4- (1.579 x 10-3 ) m & 3 + (1.0377 x 10-1 ) 2 m &- (2.8805) m & + 28.894 Unit conversion factor: 1 kg = 2.204 lb m Linear interpolation formula ) T-)/(T T-(T )-( + = L H L L H L ⋅ Ψ Ψ Ψ Ψ...
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This note was uploaded on 07/25/2008 for the course ME 416 taught by Professor Somerton during the Fall '07 term at Michigan State University.

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Quiz2 - flow rate in lb m/s Pressure Ratio 2.5 =(5.67472 x 10-5 2 m& –(2.66996 x 10-2 m& 1.6147 Pressure Ratio 3.0 =(3.5414 x 10-5

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