--- title: "Water hammer sketch, 2300 m pipeline" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:water-hammer-2300m-pipe-sketch-two-points" source: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-2300m-pipe-sketch-two-points site: BCT Engineering --- # Water hammer sketch, 2300 m pipeline Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2081 Chaitra, 8 marks - Video: [Water Hammer in Pipes | Part 1 | Fluid Mechanics](https://www.youtube.com/watch?v=WijY45fmJe0) — The Content Paradox - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-2300m-pipe-sketch-two-points - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question A 2300 m long pipeline leading from a large tank has a diameter of 15 cm and the thickness of 2.8 mm. When a discharge of 2200 l/min of water was flowing the valve was suddenly closed completely. Sketch the variation of the water hammer pressure with time at (i) the valve end (ii) 57.5 m from the upstream tank. Take $K = 2\times10^9$ Pa and $E = 2.08\times10^{11}$ Pa for steel. ## Short answer Wave speed c = 1149 m/s, water hammer pressure rise Δp = 2.38 MPa. At the valve: square wave ±2.38 MPa with half-period 4.00 s (period 8.01 s). At 57.5 m from the upstream tank (2242.5 m from valve): short pulses of ±2.38 MPa lasting 0.10 s each cycle. ## Answer **Data.** $L = 2300$ m, $D = 0.15$ m, wall thickness $t = 2.8$ mm, $Q = 2200$ l/min $= 0.03667$ m³/s, $K = 2\times10^9$ Pa, $E = 2.08\times10^{11}$ Pa. The valve is closed suddenly and completely. #### Wave speed and pressure rise Cross-section area $A = \dfrac{\pi}{4}(0.15)^2 = 0.017671$ m², so $V_0 = \dfrac{Q}{A} = 2.075$ m/s. $$ c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}} = \sqrt{\frac{2\times10^9/1000}{1 + \dfrac{(2\times10^9)(0.15)}{(2.08\times10^{11})(0.0028)}}} = \frac{1414.2}{\sqrt{1 + 0.5151}} = 1148.9\ \text{m/s} $$ $$ \Delta p = \rho cV_0 = 1000(1148.9)(2.075) = 2.384\ \text{MPa},\qquad \Delta H = \frac{\Delta p}{\rho g} = 243.0\ \text{m} $$ Time for the wave to travel the pipe: $L/c = 2.002$ s; $2L/c = 4.004$ s; period $4L/c = 8.007$ s. #### (i) Pressure variation at the valve end $\Delta p$ rises at $t = 0$ and stays at $+2.38$ MPa above the static pressure until $t = 2L/c = 4.00$ s. Then it falls to $-2.38$ MPa until $t = 4L/c = 8.01$ s, then rises again. (Square wave, period 8.01 s.) ```text ************ ************ * ------------------------------------------------- ************ ************ (top = 1.0, bottom = -1.0; '-' is zero line) ``` #### (ii) Pressure variation at 57.5 m from the upstream tank The point is $d = 2300 - 57.5 = 2242.5$ m from the valve. Wave arrival times: - $t_1 = d/c = 1.952$ s: pressure rises by $+2.38$ MPa - $t_2 = (2L - d)/c = 2.052$ s: the wave reflected at the tank (negative) returns, pressure falls back to the static value - $t_3 = (2L + d)/c = 5.956$ s: the negative wave reflected at the valve arrives, pressure falls to $-2.38$ MPa - $t_4 = (4L - d)/c = 6.056$ s: the positive wave reflected at the tank arrives, pressure returns to the static value So at this point the pressure rise is a short pulse: $+2.38$ MPa for only $0.100$ s (from 1.95 to 2.05 s), then the static value (up to 5.96 s), then $-2.38$ MPa for 0.100 s (5.96 to 6.06 s), then static until $t = 4L/c + t_1$, and the cycle repeats. ```text (not to scale in time; pulses are narrow) +2.38 | __ | | | 0 +------+--+----------__----------> t (s) | 1.95 2.05 5.96 6.06 8.0 -2.38 | |__| ``` (The pulses are only about 0.1 s wide because the point is close to the tank.) **Answer:** $c = 1149$ m/s; $\Delta p = 2.38$ MPa. At the valve: $\pm2.38$ MPa square wave of half-period 4.00 s. At 57.5 m from the tank: short $\pm2.38$ MPa pulses of about 0.1 s each cycle of 8.01 s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula is used for the wave speed in an elastic pipe? A. c = sqrt((K/rho) / (1 + KD/(Et))) B. c = sqrt(K rho) C. c = sqrt((K/rho)(1 + KD/(Et))) D. c = K/(rho V) Answer: A. Pipe elasticity reduces the wave speed below the free-water value sqrt(K/rho) by the factor 1/sqrt(1 + KD/Et). (Key point: Wave speed formula) 2. With c = 1149 m/s and V0 = 2.075 m/s, what is the water hammer pressure rise? A. About 0.24 MPa B. About 2.38 MPa C. About 23.8 MPa D. About 1.15 MPa Answer: B. delta p = rho c V0 = 1000 x 1149 x 2.075 = 2.38 MPa. (Key point: Pressure rise) 3. What does the pressure at the valve end look like with time? A. Narrow pulses of 0.1 s B. A +2.38 MPa / -2.38 MPa square wave with half-period 2L/c = 4.0 s C. A steady +2.38 MPa D. A linear rise over 8 s Answer: B. At the valve the pressure stays high for 2L/c, then low for 2L/c, period 8.0 s. (Key point: Valve-end trace) 4. Why is the pressure pulse only about 0.1 s wide at 57.5 m from the tank? A. The wave speed is lower there B. The reflected negative wave from the tank returns very soon after the incident wave passes, since the point is close to the tank C. The valve closes slowly D. The pipe thickness changes there Answer: B. The gap between arrival times d/c and (2L-d)/c is 2 x 57.5/c, about 0.1 s. 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