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

283.181 hp

Explanation:

Given that:

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

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure  = 15.4 lbf/in.^2 and Temperature 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:

= -6800 Btu/h  = - 1.9924 kW

Using the steady  state  energy in the process;

where;

 and  

Then; we have :

----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure =  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 = 2.121 kg/sec

The change in enthalpy:

= 213.1605 kW

From (1)

 - 1.9924 kW + 213.1605 kW

211.1681  kW

283.181 hp

The power input is 283.181 hp

answer:

an example of solubility is salt or sugar dissolving in water.

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