Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] 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:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]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}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 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
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
