Reconsidering Case 14 from the Morning of 2011

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14. The boiler feed water at 30°C (with a density of 1000 kg/m3) is pumped from 0.1 MPa to 1.5 MPa using a pump and sent into the boiler; the actual efficiency of the pump is 60% of the efficiency of an adiabatic reversible process. Given that the fundamental thermodynamic relation is dH = TdS + VdP, and assuming that the kinetic and potential energies in this steady-flow adiabatic reversible process can be ignored, what is the shaft power of the pump? (A) 1400 (B) 233 (C) 1485 (D) 2333 Regarding this question, I reviewed multiple posts on forums where people discussed it. It was found that the first step of calculating the adiabatic reversible work was straightforward. The debate is whether to multiply or divide by efficiency. Actually, if you look closely at the description in the question, that’s exactly where the problem lies. You haven’t even provided the efficiency for the adiabatic reversible process – how am I supposed to calculate it? It’s not as simple as just multiplying or dividing by 60%; one also needs to take into account the efficiency mentioned earlier. Just change it to the pump’s efficiency being 60% of the power consumption in an adiabatic reversible process. Firstly, 1400 represents the minimum amount of work required, which is the ideal work, or what we are familiar with as useful work. Dividing wid by ws gives an efficiency of 60%, so the shaft work ws = 1400/0.6; this makes it understandable. Of course, there is a very simple way to solve this problem. ws = (H2 - H1) ÷ 0.6 or (P2 - P1) × V. Of course, the unit for this question is also problematic.