--- title: "Flow rate 2 s after pump failure" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:pump-failure-flow-rate-after-2s" source: https://www.bctengineer.com/q/semester_4/hydraulics/pump-failure-flow-rate-after-2s site: BCT Engineering --- # Flow rate 2 s after pump failure Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2076 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/pump-failure-flow-rate-after-2s - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question A pump draws water from a reservoir and delivers it at a steady rate of 115 L/s to a tank in which the free surface level is 12 m higher than that in the reservoir. The pipe system consists of 30 m of 225 mm diameter pipe (f = 0.028) and 100 m of 150 mm diameter pipe (f = 0.032) arranged in series. Determine the flow rate 2 s after a failure of the power supply to the pump, assuming that the pump stops instantaneously. Neglect minor losses in the pipes and in the pump, and assume an incompressible fluid in rigid pipes with f independent of Reynolds number. ## Short answer After instantaneous pump stop, the water column decelerates under static head and friction. Using rigid-column theory, the flow rate at t=2 s is 29.3 L/s. ## Answer **Data.** Static lift $H_s = 12$ m. Pipe 1: $L_1 = 30$ m, $D_1 = 0.225$ m, $f = 0.028$. Pipe 2: $L_2 = 100$ m, $D_2 = 0.15$ m, $f = 0.032$. Steady flow $Q_0 = 115$ l/s. After the pump stops instantaneously, the water column keeps moving for a short time, decelerating under the static head and friction. Treat the water as incompressible and the pipes as rigid (rigid-column theory). #### Constants - Areas: $A_1 = 0.03976$ m², $A_2 = 0.01767$ m². - Inertia term: $I = \sum\dfrac{L}{A} = \dfrac{30}{0.03976} + \dfrac{100}{0.01767} = 6413.4\ \text{m}^{-1}$ - Friction coefficients, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$: $r_1 = 120.4$, $r_2 = 3481.9$, so $r = r_1 + r_2 = 3602.3$ s²/m⁵ (check: the steady friction head at 115 l/s is $r Q_0^2 = 47.64$ m). #### Equation of motion After the pump fails, two effects decelerate the whole water column: the static head $H_s$ (pushing the water back) and friction (opposing the forward motion). For the pipes in series: $$ \frac{\sum (L/A)}{g}\frac{dQ}{dt} = -\left(H_s + rQ^2\right) \;\Rightarrow\; \frac{dQ}{dt} = -\frac{g}{I}\left(H_s + rQ^2\right) $$ Separate the variables and integrate from $Q_0$: $$ \int_{Q_0}^{Q}\frac{dQ}{H_s + rQ^2} = -\frac{g}{I}t \;\Rightarrow\; \frac{1}{\sqrt{H_sr}}\left[\tan^{-1}\left(Q\sqrt{\frac{r}{H_s}}\right) - \tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)\right] = -\frac{g}{I}t $$ $$ Q = \sqrt{\frac{H_s}{r}}\tan\left[\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right) - \frac{g\sqrt{H_sr}}{I}\,t\right] $$ #### Numbers $$ \sqrt{\frac{H_s}{r}} = \sqrt{\frac{12}{3602.3}} = 0.05772,\qquad Q_0\sqrt{\frac{r}{H_s}} = 0.115\sqrt{\frac{3602.3}{12}} = 1.9925,\qquad \tan^{-1}(1.9925) = 1.1056\ \text{rad} $$ $$ \frac{g\sqrt{H_sr}}{I} = \frac{9.81\sqrt{12(3602.3)}}{6413.4} = 0.31803\ \text{s}^{-1} $$ At $t = 2$ s: $$ Q = 0.05772\tan\left(1.1056 - 0.31803(2)\right) = 0.05772\tan(0.4696) = 29.29\times10^{-3}\ \text{m}^3/\text{s} $$ (The flow stops completely at $t = 1.1056/0.31803 = 3.48$ s.) **Answer:** The flow rate 2 s after the power failure is about $29.3$ l/s (still forward), down from 115 l/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After the pump stops, what decelerates the forward-moving water column? A. Only friction B. The static head H_s plus friction, both opposing the flow C. Only the static head D. The pump's reverse thrust Answer: B. The 12 m static lift pushes the water back and friction opposes forward motion, so dQ/dt = -(g/I)(H_s + rQ^2). (Key point: Forces after pump failure) 2. How is the inertia constant I of the two pipes in series found? A. I = L1 + L2 B. I = sum of L/A = 30/A1 + 100/A2 C. I = average of the areas D. I = sum of D Answer: B. Rigid-column inertia of each pipe is L/A, which add for pipes in series (6413.4 per m). (Key point: Inertia term for series pipes) 3. Which form of solution results from integrating dQ/(H_s + rQ^2)? A. A tan^-1 function, so Q = sqrt(H_s/r) tan[tan^-1(Q0 sqrt(r/H_s)) - (g sqrt(H_s r)/I) t] B. A tanh law C. A linear decrease D. A pure exponential decay Answer: A. The integral of 1/(a + bQ^2) is an inverse tangent, giving a tangent expression for Q(t). (Key point: Solution form) 4. What is the flow 2 s after the failure, and when does it stop? A. About 29.3 l/s; stops at about 3.5 s B. About 115 l/s; never stops C. About 58 l/s; stops at 10 s D. About 5 l/s; stops at 2.1 s Answer: A. Q = 0.05772 tan(0.4696) = 29.3 l/s and flow ceases at 1.1056/0.31803 = 3.48 s. (Key point: Flow at 2 s) ## Related in Hydraulics - [Duration of flow after pump stops](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-stop-duration-of-forward-flow) — Hydraulics, chapter 4 - [Pump power and pipe length with Colebrook](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-power-pipe-length-colebrook) — Hydraulics, chapter 2 - [Three reservoirs with pump in pipe 2](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apump-three-reservoir-flow-and-head-j) — Hydraulics, chapter 3 - [Flow rate from drop in surface over hump](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acontraction-3m-to-2m-floor-raised-drop-0-2) — Hydraulics, chapter 7 - [Branching pipe system and three-reservoir cases](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abranching-pipe-system-three-cases) — Hydraulics, chapter 3 ## Related in other subjects - [Flow between reservoirs (20 cm and 30 cm pipe)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Areservoirs-pipeline-20-30cm) — Hydro Power, chapter 1 - [Reservoir level change, water balance, 1600 ha](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-level-change-1600ha) — Engineering Hydrology, chapter 1, asked 2× - [Reservoir level change, water balance, 15.5 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Areservoir-level-change-15-5km2) — Engineering Hydrology, chapter 1, asked 2× - [Pipe outlet minimum water level, 50 lps](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Apipe-outlet-min-water-level-50lps) — Irrigation and Drainage Engineering, chapter 7