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Unformatted text preview: 1 A heatpowered reversible portable air compressor consists of three components: (a) an adiabatic compressor; (b) a constant pressure heater (heat supplied from an outside source); and (c) an adiabatic turbine. Ambient air enters the compressor at 100 kPa, 300 K, and is compressed to 600 kPa. All of the power from the turbine goes into the compressor, and the turbine exhaust is the supply of compressed air. If this pressure is required to be 200 kPa, what must the temperature be at the exit of the heater? Heater C T q H 1 2 3 4 P 2 = 600 kPa, P 4 = 200 kPa Adiabatic and reversible compressor: q = 0 and s gen = 0 Energy Eq.(1 st Law): h 1 w c = h 2 Entropy Eq.(2 nd Law): s 2 = s 1 Assuming constant specific heat the isentropic relation become C =1 004; k=1 4 Assuming constant specific heat the isentropic relation becomes (C P0 =1.004; k=1.4): T 2 = T 1 P 2 P 1 k1 k = 300(6) 0.2857 = 500.8 K w c = C P (T 2 T 1 ) = 1.004(500.8 300) = 201.5 kJ/kg Adiabatic and reversible turbine: q = 0 and s gen = 0 Energy Eq.(1 st Law): h 3 = w T + h 4 ; Entropy Eq.(2 nd Law): s 4 = s 3 For constant specific heat the isentropic relation gives: T 4 = T 3 (P 4 /P 3 ) k1 k = T 3 (200/600) 0.2857 = 0.7304 T 3 4 3 4 3 3 3 Energy Eq. for shaft: w c = w T = C P (T 3 T 4 ) 201.5 = 1.004 T 3 (1 0.7304) => T 3 = 744.4 K T 3 P 2 3 200 kP 600 kPa 2 1 v s 1 300 100 kPa 4 200 kPa 4 2 . . : . . = + = gen surr m c net and heat both exchanges V C S dS dS dS that seen have we mass control a For Principle of Increase of Entropy: Control volume dS cv dt Q c.v....
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 Fall '07
 Borgnakke

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