--- title: "Pump horsepower, TEL and HGL" subject: "Fluid Mechanics" chapter: 5 key: "semester_3/fluid_mechanics:pump-horsepower-tel-hgl" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pump-horsepower-tel-hgl site: BCT Engineering --- # Pump horsepower, TEL and HGL Fluid Mechanics · Chapter 5: Hydrodynamics - Asked in: 2074 Asoj, 5+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pump-horsepower-tel-hgl - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-5 ## Question Water is pumped at 0.12 m³/s from the lower to the upper reservoir as shown in figure below. Pipe friction losses $h_f = 27v^2/2g$, where V is the average velocity in the pipe (diameter = 15 cm). If pump is 75% efficient, what horse power is needed to drive it? Draw TEL and HGL. [Figure: lower reservoir with water surface at 120 m and upper reservoir at 150 m, pump P in the inclined pipe between them] ## Short answer Pipe area 0.01767 m², velocity 6.79 m/s, friction loss 63.46 m, required pump head 93.46 m. Water power 110 kW; input power 146.7 kW (197 hp) at 75% efficiency. TEL and HGL rise abruptly at the pump by 93.46 m then fall uniformly. ## Answer **Given:** $Q = 0.12$ m³/s, $d = 0.15$ m, $h_f = 27V^2/2g$, efficiency 75 %, lower reservoir surface at 120 m, upper reservoir surface at 150 m. The pump delivers to the upper reservoir, and both surfaces are open to the atmosphere with negligible velocity. #### Velocity and losses $$ \begin{aligned} A &= \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2\\ V &= \frac{Q}{A} = \frac{0.12}{0.017671} = 6.791\ \text{m/s}\\ \frac{V^2}{2g} &= \frac{6.791^2}{19.62} = 2.350\ \text{m}\\ h_f &= 27\times 2.350 = 63.46\ \text{m} \end{aligned} $$ #### Pump head (energy equation between the two surfaces) $$ \begin{aligned} z_1 + H_p &= z_2 + h_f\\ H_p &= (150 - 120) + 63.46 = 93.46\ \text{m} \end{aligned} $$ #### Power $$ \begin{aligned} P_{water} &= \gamma Q H_p = 9.81\times 0.12\times 93.46 = 110.02\ \text{kW}\\ P_{input} &= \frac{110.02}{0.75} = 146.69\ \text{kW}\\ &= \frac{146.69\times 1000}{746} = 196.6\ \text{hp} \end{aligned} $$ **Answer: pump head 93.46 m; power needed to drive the pump $\approx$ 146.7 kW, i.e. about 197 hp (water power 110 kW, or 147.5 hp).** #### TEL and HGL The pump's position along the pipe is not given, so take the suction pipe as short, with all of the loss $h_f = 63.46$ m (pipe friction plus the exit loss) in the delivery pipe. ```text El. (m) 213.46 + T T = TEL 211.11 + H . H = HGL (2.35 m below TEL) | | ' . | | ' . both fall uniformly by friction 152.35 + | ' . T TEL at the pipe outlet 150.00 + | H========= upper reservoir | | 120.00 +=T=P pump P: TEL and HGL rise by Hp = 93.46 m 117.65 + H +----------------------------------> distance along pipe ``` - The TEL starts at 120 m at the lower reservoir surface (no velocity). Where the water enters the pipe, the HGL drops below the TEL by $V^2/2g = 2.35$ m (to 117.65 m). - At the pump, both the TEL and the HGL **rise suddenly** by $H_p = 93.46$ m: the TEL to 213.46 m and the HGL to 211.11 m. - In the delivery pipe, the TEL slopes down uniformly with friction, and the HGL runs parallel to it, 2.35 m below (uniform diameter). - At the outlet the HGL equals the upper reservoir level, 150 m. The TEL is 152.35 m there and drops by the exit loss $V^2/2g = 2.35$ m to 150 m. The total fall of the TEL is 63.46 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the pipe velocity? A. 3.4 m/s B. 13.6 m/s C. 6.79 m/s D. 0.12 m/s Answer: C. V = Q/A = 0.12/0.017671 = 6.79 m/s. (Key point: Velocity) 2. What is the pump head? A. 30 m B. 63.46 m C. 213.46 m D. 93.46 m Answer: D. Hp = elevation difference 30 + hf 63.46. (Key point: Pump head) 3. How does efficiency enter? A. Input power = gamma Q Hp / 0.75 B. Input power = 0.75 gamma Q Hp C. Hp divided by 0.75 only D. Not used Answer: A. 110 kW water power divided by 0.75 = 146.7 kW. (Key point: Efficiency) 4. How does TEL behave at the pump? A. Falls by Hp B. Rises suddenly by Hp C. Stays constant D. Equal to HGL Answer: B. Pump adds energy. 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