Question 5 of 13: Temperature Rise in a Closed Pump Circuit

Question 5 — Temperature Rise in a Closed Pump Circuit (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The rise in water temperature after 1 hour of operation.

Approach. In a closed circuit with no elevation change and no heat escaping the system, every watt of shaft power the pump draws must end up as heat in the water — the useful hydraulic power overcomes friction head (which dissipates as heat as the fluid recirculates), and the inefficient 10% appears as heat too, since bearing loss is stated negligible so the loss is internal/hydraulic. Convert the total energy input over 1 h into a temperature rise via Qheat=mcpΔT.

1. Shaft power input to the pump. $$P_{{hyd}} = \rho g Q H = 1000\times9.81\times0.125\times100 = 122,625\text{{ W}}$$ $$P_{{shaft}} = \frac{{P_{{hyd}}}}{{\eta}} = \frac{{122,625}}{{0.90}} = 136,250\text{{ W}}$$
2. Total energy delivered to the water in 1 h. $$E = P_{{shaft}}\times3600\text{{ s}} = 136,250\times3600 = 490.5\times10^6\text{{ J}}$$
3. Temperature rise. Mass of water in the circuit: m = ρV = 1000×40 = 40,000 kg. $$\Delta T = \frac{{E}}{{mc_p}} = \frac{{490.5\times10^6}}{{40,000\times4190}} = \boxed{{2.93^\circ\text{{C}}}}$$