--- title: "Pump head and power between two reservoirs" subject: "Fluid Mechanics" chapter: 8 key: "semester_4/fluid_mechanics_mechanical:pumped-system-power-between-reservoirs" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/pumped-system-power-between-reservoirs site: BCT Engineering --- # Pump head and power between two reservoirs Fluid Mechanics · Chapter 8: Flow Measurement - Asked in: year not given, 8 marks - Video: [Fluid Mechanics Final Exam Question: Energy Equation Analysis of Pumped Storage](https://www.youtube.com/watch?v=20xOMDL6pyc) — Fluid Matters - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/pumped-system-power-between-reservoirs - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-8 ## Question A pump delivers water at 0.06 m$^3$/s from a lower reservoir to an upper reservoir whose water surface is 30 m higher. The suction pipe is 20 m long and the delivery pipe is 600 m long, both 200 mm in diameter with Darcy friction factor $f = 0.02$. The total minor losses (entry, bends, valves) are equal to 3 velocity heads, and the exit loss into the upper reservoir is one velocity head. Find the head the pump must develop, the hydraulic power, and the shaft power if the pump efficiency is 70%. ## Short answer Velocity V=1.91 m/s, friction loss 11.53 m, minor+exit loss 0.74 m. Pump head Hp = 30+11.53+0.74 = 42.3 m. Hydraulic power = ρgQHp = 24.9 kW, shaft power = 24.9/0.70 = 35.5 kW. ## Answer #### Given data $Q = 0.06\ \text{m}^3/\text{s}$, $D = 0.2\ \text{m}$, $f = 0.02$, $L_s = 20\ \text{m}$, $L_d = 600\ \text{m}$, static lift $z = 30\ \text{m}$, $\sum K = 3$ (minor), exit $K = 1$, $\eta = 0.70$. #### Velocity and velocity head $$ \begin{aligned} A &= \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2 \\ V &= \frac{Q}{A} = \frac{0.06}{0.03142} = 1.910\ \text{m/s} \\ \frac{V^2}{2g} &= \frac{1.910^2}{19.62} = 0.1859\ \text{m} \end{aligned} $$ #### Energy equation between the two reservoir surfaces Both surfaces are at atmospheric pressure and velocity is nearly zero: $$ 0 + 0 + H_p = z + h_f + h_{minor} + h_{exit} $$ **Friction loss** (total length 620 m): $$ h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02 \times \frac{620}{0.2} \times 0.1859 = 11.53\ \text{m} $$ **Minor losses:** $3 \times 0.1859 = 0.558\ \text{m}$ **Exit loss:** $1 \times 0.1859 = 0.186\ \text{m}$ #### Head developed by the pump $$ H_p = 30 + 11.53 + 0.558 + 0.186 = 42.27\ \text{m} $$ #### Power Hydraulic (water) power: $$ P_w = \rho g Q H_p = 1000 \times 9.81 \times 0.06 \times 42.27 = 24\,880\ \text{W} = 24.9\ \text{kW} $$ Shaft power: $$ P_s = \frac{P_w}{\eta} = \frac{24.88}{0.70} = 35.5\ \text{kW} $$ **Answer: $H_p = 42.3\ \text{m}$; hydraulic power $= 24.9\ \text{kW}$; shaft power $= 35.5\ \text{kW}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Between the two reservoir surfaces, what is H_p? A. Static lift only = 30 m B. Static lift minus all losses C. Static lift plus friction, minor and exit losses D. Friction loss plus velocity head only Answer: C. Both surfaces are at atmospheric pressure with negligible velocity, so H_p = z + h_f + h_minor + h_exit. (Key point: Energy equation) 2. Which length is used in the friction loss h_f = f (L/D)(V^2/2g)? A. Delivery pipe only, 600 m B. Suction pipe only, 20 m C. Total 620 m of suction plus delivery pipe D. Static lift 30 m Answer: C. Friction occurs along the whole pipe, so L = 20 + 600 = 620 m, giving h_f = 11.53 m. (Key point: Friction loss length) 3. After finding hydraulic power 24.9 kW, how is shaft power obtained for 70% efficiency? A. 24.9 x 0.70 = 17.4 kW B. 24.9 / 0.70 = 35.5 kW C. 24.9 + 0.70 kW D. 24.9 / 0.30 = 83 kW Answer: B. Shaft power is hydraulic power divided by pump efficiency: 24.88/0.70 = 35.5 kW. 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