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 closely1.04 Btu/s from the liquid to the surroundings. 3.98 Btu/s from the surroundings to the liquid. 4.96 Btu/s from the surroundings to the liquid. 2.02 Btu/s from the liquid to the surroundings.

Question

Kian Maynard

Engineering

20 March, 23:58

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 closely1.04 Btu/s from the liquid to the surroundings. 3.98 Btu/s from the surroundings to the liquid. 4.96 Btu/s from the surroundings to the liquid. 2.02 Btu/s from the liquid to the surroundings.

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Davis Aguirre · Answer 1

1.04Btu/s from the liquid to the surroundings

Explanation:

Given

m = 2lb/s

V1 = 40.09Btu/lb

V2 = 40.94Btu/lb

z1 = 0ft, z2 = 100ft

g = 32.174 ft/s²

Wc = 3Btu/s

The balance on the pump is given as

Qc - Wc + m[ (h1 - h2) + ½ (V1² - V2²) + g (z1 - z2) ] = 0

By neglecting kinetic energy; we substitute the above values:

This is as follows;

Qc - (-3) + 2[ (40.09 - 40.94) + 32.174 (0-100) / (778*32.174) ] = 0

Qc + 3 + 2 (40.09 - 40.94) + 2 * 32.174 (0-100) / (778*32.174) = 0

Qc + 3 - 1.7 - 0.257 = 0

Qc = - 1.04Btu/s