Air enters a compressor operating at steady state at 50 degr

Air enters a compressor operating at steady state at 50 degree C, 0.9 bar, 70% relative humidity with a volumetric flow rate of 0.8 m^3/s. The moist air exits the compressor at 195 degree C, 1.5 bar. Assuming the compressor is well insulated, determine the relative humidity at the exit the power input, in kW the rate of entropy production, in kW/K

Solution

a)

At 50 C, we get from water properties, saturation pressure Psat1 = 12.4 kPa

Rel Humidity = Pv / Psat

0.7 = Pv / 12.4

Pv = 8.68 kPa = 0.0868 bar

Pair = P - Pv

= 0.9 - 0.0868

Pair = 0.8132 bar

Humidity ratio w1 = 0.622*Pv / Pair

= 0.622*0.0868 / 0.8132

= 0.0664

During compression, w2 = w1

0.0664 = 0.622*Pv2 / (1.5 - Pv2)

Pv2 = 0.1447 bar

From water properties at 195 C, we get Psat2 = 14 bar

RH2 = Pv2 / Psat2

= 0.1447 / 14

= 0.01 or 1%

b)

Q - W = h2 - h1

0 - W = m*Cp*(T2 - T1)

-W = rho*V*Cp*(T2 - T1)

W = 1.2*0.8*1.005*(195 - 50)

W = 174.87 kW

c)

At 195 C, 1.5 bar we get s2 = 7.62 kJ/kg-K

At 50 C, 0.9 bar, we get s1 = 0.704 kJ/kg-K

Entropy production = m*(s2 - s1)

= 1.2*0.8*(7.62 - 0.704)

= 6.64 kW/K