Answer to Question #230026 in Molecular Physics | Thermodynamics for Anuj

Question

 In an air compressor air flows steadily at the rate of 0.5 kg/s through an

air compressor. It enters the compressor at 6 m/s with a pressure of 1 bar and a specific volume

of 0.85 m3/kg and leaves at 5 m/s with a pressure of 7 bar and a specific volume of 0.16 m3/kg. The

internal energy of the air leaving is 90 kJ/kg greater than that of the air entering. Cooling water

in a jacket surrounding the cylinder absorbs heat from the air at the rate of 60 kJ/s. Calculate :

(i) The power required to drive the compressor ;

(ii) The inlet and output pipe cross-sectional areas

Answer 1

Mass flow rate of air m = 0.5 kg/s

Velocity at inlet $V_1 = 6 \; m/s$

Pressure $P_1 = 1\; bar = 10^5 \;Pa$

Specific volume $v_1 = 0.85 \;m^3/kg$

Outlet velocity $V_2 = 5 \;m/s$

Pressure $P_2 = 7 \;bar = 7 \times 10^5 \;Pa$

Specific volume $v_2 = 0.16 \;m^3/kg$

Internal energy U = 90 kJ/kg

Heat absorbed Q = -60 kJ/s

Power required W =?

$∆Pv = P_2v_2 – P_1v_1 \\ = 7 \times 10^5 \times 0.16 - 10^5 \times 0.85 \\ = 27000 \; J/kg \\ = 27 \; kJ/kg \\ ∆h = ∆(U + Pv) \\ = 90 + 27 \\ = 117 \; kJ/kg$

Change in kinetic energy

$∆KE = 0.5(V_2^2 - V_1^2) \\ = 0.5(5^2 - 6^2) \\ = - 5.5 \; J/kg$

Energy balance equation

$Q - W = m(∆h + ∆KE) \\ -60 - W = 0.5(117 - \frac{5.5}{1000}) \\ W = - 118.5 \; kW$

Negative sign shows that work is done on compressor.

Inlet and outlet pipe cross-sectional areas, $A_{1}$ and $A_{2}$ :

Using the relation,

$m = \frac{VA}{v} \\ A_1 = \frac{mv_1}{V_1} = \frac{0.5 \times 0.85 }{6}=0.0708 \;m^2 \\ A_2 = \frac{mv_2}{V_2}= \frac{0.5 \times 0.16}{5}= 0.016 \;m^2$