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

`W_s =` 283.181 hp

Step-by-step 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)+ (1)/(2)(v^2_2-v_1^2) = (Q_(cv))/(m)- (W_s)/(m)`

where;

`g(z_2-z_1) =0` and
`(1)/(2)(v^2_2-v_1^2) = 0`

Then; we have :

`h_2 - h_1 = (Q_(cv))/(m)- (W_s)/(m)`

`h_2 - h_1 = (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 = \frac { 1216235.664 N/m^2 * 0.2000856 m^3 /sec } {287 J/kg K * 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