Question

1. Air flow steadily at the rate of 0.95 kg/s through air compressor, entering at 7.5 m/s speed, 95 Kpaa

pressure and 0.95 m 3 /kg specific volume and leaving 20 m/s, 725 kpaa and 0.19 m 3 /kg. the internal

energy of the air leaving is 100 KJ/kg greater than that of the air entering. Cooling water in the

compressor jacket absorbs heat from the air at the rate of 60 kJ/s. compute the steady flow work in kW.

Answer 1

$m= 0.95 \;kg/sec \\ v_1 = 0.95 \;m^3/kg \\ v_2= 0.19 \;m^3/kg \\ V_1 = 7.5 \;m/s \\ V_2 = 20 \;m/s \\ p_1 = 95 \;kPa = 95 \times 10^3 \;Pa \\ p_2 = 725 \;kPa = 725 \times 10^3 \;Pa \\ Q= -60 \;kJ/s \\ Δu=100 \;kJ/kg \\ wf_1 = p_1v_1 \\ = 95 \;kPa \times \frac{0.95 \; m^3}{kg} = 90.25 \;kJ/kg \\ wf_2 = p_2v_2 \\ = 725 \;kPa \times \frac{0.19 \; m^3}{kg} = 137.75 \;kJ/kg \\ KE_1 = \frac{V_1^2}{2k} \\ = \frac{(7.5 \; m/s)^2}{2 \times \frac{1 kg \;m \cdot \;m}{N \cdot s^2}} \\ = 0.028125 \;kJ/kg \\ KE_2 = \frac{V^2_2}{2k} \\ = \frac{(20 \;m/s)^2}{2 \times \frac{1 kg \;m \cdot \;m}{N \cdot s^2}} \\ = 0.2 \; kJ/kg \\ E_{in} = E_{out} \\ u_1 + wf_1 +KE_1+Q = w +u_2 +wf_2 +KE_2 \\ w = Δu +wf_2 +KE_2 - wf_1 -KE_1 -Q \\ = 100 \;kJ/kg + 137.75 \;kJ/kg+ 0.2 \; kJ/kg -90.25 \;kJ/kg -0.028 \;kJ/kg -(-60 \;kJ/s) \\ = 147.672 \;kJ/kg \times 0.95 \;kg/s + 60 \;kJ/s \; \\ = 140.288 \;kJ/s + 60 \;kJ/s \\ = 200.28 \;kJ/s \\ = 200.28 \;kW$