Question 7 of 13: Efficiency of a Submersible Sump Pump

Question 7: Efficiency of a Submersible Sump Pump (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 overall (wire-to-water) efficiency of the pump, and a comment on the result.

Approach. The useful hydraulic power delivered to the water equals ρgQH, where the total head H the pump must supply is the elevation lift plus the friction head loss; efficiency compares this to the stated power input.

1. Total head added by the pump. $$ H = \Delta z + h_L = 2.10 + 1.03 = 3.13\ \text{m} $$
2. Hydraulic power delivered to the water. With $Q = 96/60/1000 = 1.6\times10^{-3}$ m³/s: $$ P_{water} = \rho g Q H = (1000)(9.81)(1.6\times10^{-3})(3.13) $$ $$ P_{water} \approx 49.1\ \text{W} $$
3. Overall efficiency. $$ \eta = \frac{P_{water}}{P_{in}} = \frac{49.1}{186} $$ $$ \boxed{\eta \approx 26.4\%} $$

A 26% overall efficiency is low but not unreasonable for a small consumer-grade submersible pump at a modest flow: the stated 186 W is the electrical (or nameplate) input to the whole unit, so this figure already bundles motor electrical losses, bearing/seal friction, and impeller hydraulic losses into one number, and small pumps rarely exceed 30–40% overall for this reason. Several factors could further affect the calculated efficiency: the nameplate ¼-HP rating may not equal the pump's actual electrical draw at this operating point; the measured head loss (1.03 m) itself depends on an assumed pipe roughness and flow regime; a portion of the discharge kinetic energy at the pipe exit was neglected in H; and impeller wear, entrained air, or a partially clogged sump strainer would all reduce measured output without appearing anywhere in the input-power figure.