Please provide explanation and steps where possible Air flow

**Please provide explanation and steps where possible

Air flows at a rate of 2.5 kg/s and is compressed from 100 kPa and 25°C. The compressor delivers 440 kW of power and operates isentropically (adiabatically and reversibly). The air is then cooled to 50°C at constant pressure using cooling water at 20°C and 2.0 bar. The cooling water leaves the cooler at a temperature of 50°C. The cooled air is then compressed to a final pressure of 1200 kPa and the temperature rises to 150°C. The second compressor loses heat to the surroundings at a rate of 5 kW. Assume that the system is operating under steady flow conditions. For air, use cp = 1.01 kJ kg-1 K -1 , k = 1.4, and R = 0.2871 kJ kg-1 K -1 .

a) What is the temperature (T2) and pressure (P2) of the air leaving the first compressor?

b) What is the required mass flow rate of cooling water (mwater in kg/s)? (ANS: 3.005 kg/s)

c) What is the required power input to the second compressor (W2 in kW)?

mwater = ? woter P,voter,in = 2.0 bar Twater,in-20-C woter,in-2 Q:5kW woter,in- mair 2.5 kg/s P1 100 kPa T1 = 25°C T2-? 50°C Compressor Ta = 150°C Pa = 1200 kPa Compressor Pwoter.out 2.0 bar T 50 woter,out 440 kw water,out 2

Solution

a)

For first compressor, by applying 1st law:

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

0 - (-440) = 2.5*1.01*(T2 - 25)

T2 = 199.26 deg C = 472.4 K

For isentropic compression in first compressor we have,

T2 / T1 = (P2 / P1)^(k-1)/k

472.4 / (273+25) = (P2 / 100)^(1.4-1)/1.4

P2 = 500.7 kPa

b)

Heat gained by water = Heat lost by air

m_water*Cp_water*(T_water,out - T_water,in) = m_air*Cp_air*(T₂ - T₃)

m_water*4.186*(50 - 20) = 2.5*1.01*(199.26 - 50)

m_water = 3.001 kg/s

c)

Applying 1st law of thermodynamics to 2nd compressor:

Q₂ - W₂ = m_air*Cp*(T₄ - T₃)

-5 - W = 2.5*1.01*(150 - 50)

W₂ = -257.5 kW

Hence, required input power 257.5 kW