50 kmol per hour of air is compressed from P1= 1.2 bar toP2= 6.0 bar in a steady

Question

50 kmol per hour of air is compressed from P1= 1.2 bar toP2= 6.0 bar in a steady flow compressor. delivered mechanical poweris 98.8 kW. temperatures and velocities are: T1= 300K         T2= 520K U1= 10m/s         U2= 3.5m/s Estimate the rate of heat transfer from the compressor. assumefor air that Cp= 7/2R and that enthalpy is dependent frompressure 50 kmol per hour of air is compressed from P1= 1.2 bar toP2= 6.0 bar in a steady flow compressor. delivered mechanical poweris 98.8 kW. temperatures and velocities are: T1= 300K         T2= 520K U1= 10m/s         U2= 3.5m/s Estimate the rate of heat transfer from the compressor. assumefor air that Cp= 7/2R and that enthalpy is dependent frompressure

Explanation / Answer

Steady state equation: H1 + KE1 + Q = H2 + KE2 + W Q - W = (H2 - H1) + (KE2 - KE1) where KE is the kinetic energy: KE1 = 0.5*m*u1^2 m = n*MW MW = mol weight of air = 29 kg/kmol m = 50*29 = 1450 kg/hr KE1 = 0.5*1450*10^2/3600/1000 = 0.020 kW KE2 = 0.5*1450*3.5^2/3600/1000 = 0.002 kW H is the enthalpy H1 = m*Cp*T1 R = specific gas constant of air = 0.287 kJ/kg-K = 8.3145/29 (univ. gas constant/mol weight) Cp = (7/2)*0.287 = 1.0045 kJ/kg-K H1 = 1450*1.0045*300/3600 = 121.377 kW H2 = m*Cp*T2 = 1450*1.0045*520/3600 = 210.387 kW Q - 98.8 = (210.387 - 121.377) + (0.002 - 0.020) Q = 187.8 kW

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