Question

Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp

Answer 1

Answer:

$W_s =$ 283.181 hp

Explanation:

Given that:

Air enters a compressor operating at steady state at a pressure $P_1$ =  176.4 lbf/in.^2  and Temperature $T_1$ at 260°F

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure $P_2$  = 15.4 lbf/in.^2 and Temperature $T_2$ at 80°F.

Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:

$Q_{cv}$ = -6800 Btu/h  = - 1.9924 kW

Using the steady  state  energy in the process;

$h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}$

where;

$g(z_2-z_1) =0$  and  $\dfrac{1}{2}(v^2_2-v_1^2) = 0$

Then; we have :

$h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}$

$h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}$

${m}(h_2 - h_1) ={Q_{cv} - W_s}$

$W_s ={Q_{cv} + {m}(h_2 - h_1)$ ----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure $P_1$ =  176.4 lbf/in.^2   = ( 176.4 ×  6894.76 ) N/m² = 1216235.664 N/m²

Volumetric flow rate V = 424 ft^3/min = (424 ×  0.0004719) m³  /sec

= 0.2000856 m³  /sec

Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K

Gas constant R=287 J/kg K

Then;

1216235.664 N/m² × 0.2000856 m³  /sec = m × 287 J/kg K × 399.817 K

$m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }$

m = 2.121 kg/sec

The change in enthalpy:

$m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)$

$= 2.121* 1.005* ( 399.817 -299.817)$

= 213.1605 kW

From (1)

$W_s ={Q_{cv} + {m}(h_2 - h_1)$

$W_s =$  - 1.9924 kW + 213.1605 kW

$W_s =$ 211.1681  kW

$W_s =$ 283.181 hp

The power input is $W_s =$ 283.181 hp