Air enters a compressor operating at steady state with pressure of 1 atm, at a t
ID: 2994919 • Letter: A
Question
Air enters a compressor operating at steady state with pressure of 1 atm, at a temperature of 80 oC, and a volumetric flow rate of 0.6 m3/s. The air exits the compressor at a pressure of 6 atm. Heat transfer from the compressor to its surrounding occurs at a rate of 20 kJ/kg of air flowing. The compressor power input is 65 kW. Neglecting kinetic and potential energy effects and assuming air to be an ideal gas, please determine the exit temperature. Air enters a compressor operating at steady state with pressure of 1 atm, at a temperature of 80 oC, and a volumetric flow rate of 0.6 m3/s. The air exits the compressor at a pressure of 6 atm. Heat transfer from the compressor to its surrounding occurs at a rate of 20 kJ/kg of air flowing. The compressor power input is 65 kW. Neglecting kinetic and potential energy effects and assuming air to be an ideal gas, please determine the exit temperature. Air enters a compressor operating at steady state with pressure of 1 atm, at a temperature of 80 oC, and a volumetric flow rate of 0.6 m3/s. The air exits the compressor at a pressure of 6 atm. Heat transfer from the compressor to its surrounding occurs at a rate of 20 kJ/kg of air flowing. The compressor power input is 65 kW. Neglecting kinetic and potential energy effects and assuming air to be an ideal gas, please determine the exit temperature. Air enters a compressor operating at steady state with pressure of 1 atm, at a temperature of 80 oC, and a volumetric flow rate of 0.6 m3/s. The air exits the compressor at a pressure of 6 atm. Heat transfer from the compressor to its surrounding occurs at a rate of 20 kJ/kg of air flowing. The compressor power input is 65 kW. Neglecting kinetic and potential energy effects and assuming air to be an ideal gas, please determine the exit temperature.Explanation / Answer
T1 = 80 deg C = 80+273 K = 353 K
P1 = 1 atm = 101.325 kPa
P2 = 6 atm = 607.95 kPa
For isentropic compression
T2_is / T1 = (P2 / P1)^((n-1)/n)
T2_is / 353 = (6/1)^((1.4 - 1)/1.4)
T2_is = 589 K
Inlet density = P1 / (R*T1)
= 101.325*10^3 / (287 * 353)
= 1 kg/m^3
Inlet mass flowrate = volume flowrate *density
= 0.6*1
= 0.6 kg/s
Ideal work input = m*(h2_is - h1) = m*Cp*(T2_is - T1)
= 0.6*1.005*(589 - 353)
= 237.18 kW
Actual work input = 237.18 + 20 = 257.18 kW
m*Cp*(T2 - T1) = 257.18
1*1.005*(T2 - 353) = 257.18
T2 = 609 K = 336 deg C
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