--- title: "Pump power and pipe length with Colebrook" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:pump-power-pipe-length-colebrook" source: https://www.bctengineer.com/q/semester_4/hydraulics/pump-power-pipe-length-colebrook site: BCT Engineering --- # Pump power and pipe length with Colebrook Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2078 Chaitra, 3+3 marks - Video: [Energy Equation with a Pump – Example Problem](https://www.youtube.com/watch?v=Ima9w2eK8Dg) — Brian Bernard - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/pump-power-pipe-length-colebrook - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question If a flow of 0.25 m³/s of water is to be maintained in the system shown below, the power required by the pump to maintain the water levels is 54642 watts. The pipe is made of iron (e = 0.26 mm) and is 250 mm in diameter. Take $\mu$ = 0.0021 N·s/m². Take all losses and use Colebrook-White equation, if needed. (i) Find the length of the pipe. (ii) For the same power if the loss of head is doubled, find the discharge. Write your comment on the result. [Figure: two reservoirs with water surface elevations 12 m (left) and 19 m (right) connected by a pipe of length L with a pump in the line] ## Short answer Pipe length is 124 m using f=0.0203 from Colebrook. If head loss doubles, discharge reduces to 204 L/s, about an 18% drop because friction loss falls with the square of flow. ## Answer **Data.** $Q = 0.25$ m³/s, pump power $P = 54\,642$ W, $D = 0.25$ m, $e = 0.26$ mm, $\mu = 0.0021$ N s/m², water $\rho = 1000$ kg/m³. Reservoir levels 12 m and 19 m, so the static lift is $19 - 12 = 7$ m. All losses are considered: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$ and friction. #### (i) Length of the pipe Pump head: $P = \gamma Q H_p$ $$ H_p = \frac{54\,642}{9810(0.25)} = 22.280\ \text{m} $$ Energy equation between the reservoir surfaces: $H_p = 7 + h_L$, so the head loss is $h_L = 15.280$ m. Velocity and Reynolds number: $$ V = \frac{0.25}{\pi(0.25)^2/4} = 5.093\ \text{m/s},\quad \frac{V^2}{2g} = 1.3220\ \text{m},\quad Re = \frac{\rho VD}{\mu} = \frac{1000(5.093)(0.25)}{0.0021} = 606305 $$ Relative roughness $e/D = 0.00104$. Colebrook-White (solved by iteration): $$ \frac{1}{\sqrt f} = -2\log_{10}\left(\frac{e/D}{3.7} + \frac{2.51}{Re\sqrt f}\right) \;\Rightarrow\; f = 0.02031 $$ Head loss: $h_L = \left(0.5 + f\dfrac{L}{D} + 1.0\right)\dfrac{V^2}{2g}$ $$ 15.280 = \left(1.5 + f\frac{L}{D}\right)(1.3220) \;\Rightarrow\; f\frac{L}{D} = 10.058 $$ $$ L = \frac{10.058\times 0.25}{0.02031} = 123.8\ \text{m} $$ #### (ii) Same power, loss of head doubled "Loss of head doubled" means that, at the same flow, the system now loses twice as much head (for example a longer or rougher pipe). The head loss varies as $Q^2$, so for the original pipe $$ h_L = KQ^2,\qquad K = \frac{15.280}{0.25^2} = 244.5\ \text{s}^2/\text{m}^5 $$ and for the new system $h_L' = 2KQ^2 = 489\,Q^2$ (taking $f$ as constant, since it changes very little). The pump head is then $$ H_p' = 7 + 489\,Q^2 $$ For the same power $P = \gamma Q H_p'$: $$ 9810\,Q\,(7 + 489\,Q^2) = 54\,642 \;\Rightarrow\; Q\,(7 + 489\,Q^2) = 5.570 $$ Solving by trial: $Q = 0.200$ gives 5.31, $Q = 0.205$ gives 5.65, and $Q = 0.204$ gives 5.58. So $$ Q' \approx 0.204\ \text{m}^3/\text{s} = 204\ \text{l/s} $$ At this flow the head loss is $489(0.204)^2 = 20.3$ m and the pump head is $27.3$ m. (Allowing $f$ to vary with $Re$ in the Colebrook-White equation gives 203.6 l/s, practically the same.) **Comment.** Doubling the loss cuts the discharge from 250 l/s to about 204 l/s, a fall of about 18%, for the same power. The 7 m static lift is a fixed requirement, but the pump must also supply the friction head. When friction rises, more of the same power is spent overcoming it and less water is delivered. The discharge falls by much less than the loss rises, because the head loss falls with the square of the flow. It would be wrong to keep the doubled loss of 30.56 m at the old flow and divide the power by $\gamma(7 + 30.56)$. That gives 148 l/s, which is too low, because the friction loss itself drops as the flow drops. **Answer:** (i) $L \approx 124$ m ($f = 0.0203$); (ii) $Q' \approx 204$ l/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the pump head obtained from the given power? A. H_p = P/(gamma Q) = 22.28 m B. H_p = P/Q = 218 m C. H_p = P x gamma x Q D. H_p = P/gamma = 5.57 m Answer: A. Power P = gamma Q H_p, so H_p = P/(gamma Q). (Key point: Pump head from power) 2. The pump head is 22.28 m and the static lift is 7 m. What is the head loss in the system? A. 22.28 + 7 = 29.28 m B. 22.28 m C. 22.28 - 7 = 15.28 m D. 7 m Answer: C. H_p = static lift + h_L, so h_L = 22.28 - 7 = 15.28 m. (Key point: Head loss from energy equation) 3. How is the friction factor found and which losses are included for length L? A. From the Moody chart using e/D only; friction only B. Colebrook-White by iteration using e/D and Re; loss = (0.5 + fL/D + 1.0) V^2/2g including entrance and exit C. f = 64/Re since the flow is laminar D. f = 0.02 assumed; exit loss only Answer: B. Re is about 6x10^5 (turbulent), so f comes from Colebrook-White, and all losses (entrance 0.5, exit 1.0, friction) are considered. (Key point: Friction factor and loss terms) 4. For the same power and doubled head loss, why does the discharge fall to 148 l/s? A. Pump head becomes 7 + 2(15.28) = 37.56 m and Q = P/(gamma H_p) B. Q is halved because loss is doubled C. Pump head stays 22.28 m so Q is unchanged D. Static lift also doubles to 14 m Answer: A. A larger pump head needed for the same power means a smaller Q since Q = P/(gamma H_p). 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