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Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from control volume inlet to exit. The liquid specific enthalpy at the inlet is 40.09 Btu/lb and at the exit is 40.94 Btu/lb. The pump requires 3 Btu/s of power to operate. If kinetic energy effects are negligible and gravitational acceleration is 32.174 ft/s2, the heat transfer rate associated with this steady state process is most closely:

A) 2.02 Btu/s from the liquid to the surroundings.

B) 3.98 Btu/s from the surroundings to the liquid.

C) 4.96 Btu/s from the surroundings to the liquid.

D) 1.04 Btu/s from the liquid to the surroundings.

Respuesta :

raphealnwobi

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet ([tex]h_{1}=40.09[/tex] Btu/lb)

liquid specific enthalpy at the exit ([tex]h_{2}=40.94[/tex] Btu/lb)

initial elevation ([tex]z_1=0ft[/tex])

final elevation ([tex]z_2=100ft[/tex])

acceleration due to gravity (g) = 32.174 ft/s²

[tex]W_{cv}[/tex] = 3 Btu/s

The energy balance equation is given as:

[tex]Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0[/tex]

Since  kinetic energy effects are negligible, the equation becomes:

[tex]Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0[/tex]

Substituting values:

[tex]Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\[/tex]

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

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