--- title: "Water hammer pressure and stress, 2300 m pipeline" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:water-hammer-2300m-pressure-stress-1725m" source: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-2300m-pressure-stress-1725m site: BCT Engineering --- # Water hammer pressure and stress, 2300 m pipeline Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2080 Chaitra, 2+3+3 marks - Video: [Water Hammer - The Joukowsky Equation (3/8)](https://www.youtube.com/watch?v=GfCCsdHKrF8) — EngineeringTrainer - 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-pressure-stress-1725m - 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 thickness of 2.8 mm. When a discharge of 2200 liters per minute of water was flowing, the valve was suddenly closed completely. What water hammer pressure and stress would develop at this condition? Also sketch the variation of the water hammer pressure with time at a distance of 1725 m downstream from reservoir end. Take $K = 2\times10^9$ Pa for water and $E = 2.08\times10^{11}$ Pa for steel. ## Short answer Water hammer pressure rise Δp = 2.38 MPa, hoop stress = 63.9 MPa. At 1725 m downstream from reservoir (575 m from valve): pressure steps: 0 to 0.50 s static, +2.38 MPa from 0.50 to 3.50 s, 0 from 3.50 to 4.50 s, -2.38 MPa from 4.50 to 7.51 s, 0 thereafter; cycle 8.01 s. ## Answer **Data.** $L = 2300$ m, $D = 0.15$ m, $t = 2.8$ mm, $Q = 2200$ l/min $= 0.03667$ m³/s, $K = 2\times10^9$ Pa, $E = 2.08\times10^{11}$ Pa. Instantaneous closure. #### Velocity and wave speed $$ V_0 = \frac{Q}{\pi D^2/4} = \frac{0.03667}{0.017671} = 2.075\ \text{m/s} $$ $$ c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}} = \frac{\sqrt{2\times10^{6}}}{\sqrt{1 + \dfrac{2\times10^9(0.15)}{2.08\times10^{11}(0.0028)}}} = \frac{1414.2}{\sqrt{1 + 0.5151}} = 1148.9\ \text{m/s} $$ Check closure type: $2L/c = 4.00$ s, and the valve closes instantaneously, so the closure is rapid (full Joukowsky rise). #### Water hammer pressure $$ \Delta p = \rho cV_0 = 1000(1148.9)(2.075) = 2.384\times10^6\ \text{Pa} = 2.38\ \text{MPa}\quad(\Delta H = 243.0\ \text{m of water}) $$ #### Stress developed Hoop (circumferential) stress due to the pressure rise, thin-walled pipe: $$ \sigma = \frac{\Delta p\,D}{2t} = \frac{(2.384\times10^6)(0.15)}{2(0.0028)} = 63.9\ \text{MPa} $$ (The longitudinal stress is about half of this, $\Delta p D/4t$.) #### Pressure-time variation at 1725 m from the reservoir The point is $d = 2300 - 1725 = 575$ m from the valve ($x = 1725$ m from the reservoir). Wave arrival times ($c = 1148.9$ m/s): - $t_1 = d/c = 0.500$ s: wave from the valve arrives; pressure rises by $+2.38$ MPa - $t_2 = (2L - d)/c = 3.503$ s: wave reflected from the reservoir arrives (negative); pressure falls back to the static value - $t_3 = (2L + d)/c = 4.504$ s: negative wave reflected from the valve arrives; pressure falls to $-2.38$ MPa - $t_4 = (4L - d)/c = 7.507$ s: positive wave reflected from the reservoir arrives; pressure returns to the static value - The cycle repeats with period $4L/c = 8.007$ s. | Time interval (s) | Pressure change at the point | |---|---| | 0 to 0.50 | 0 | | 0.50 to 3.50 | $+2.38$ MPa | | 3.50 to 4.50 | 0 | | 4.50 to 7.51 | $-2.38$ MPa | | 7.51 to 8.01 | 0 (then the cycle repeats) | ```text ****************** ***------------------******------------------**** ****************** (top = 1.0, bottom = -1.0; '-' is zero line) ``` (Pressure change relative to static pressure: +1 = $+\Delta p$, $-1$ = $-\Delta p$; time from 0 to $4L/c = 8.0$ s.) **Answer:** $\Delta p = 2.38$ MPa (243 m of water); hoop stress $\sigma = 63.9$ MPa; the pressure at 1725 m from the reservoir is a stepped wave with the changes at $t$ = 0.50, 3.50, 4.50 and 7.51 s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the full Joukowsky pressure rise applied here? A. Because closure is instantaneous, so it is faster than 2L/c = 4.0 s B. Because the pipe is steel C. Because the discharge is large D. Because friction is high Answer: A. A closure time not exceeding 2L/c is rapid, so the full delta p = rho c V0 develops. (Key point: Closure type) 2. What is the water hammer pressure rise (c = 1148.9 m/s, V0 = 2.075 m/s)? A. 2.38 MPa B. 0.24 MPa C. 63.9 MPa D. 11.5 MPa Answer: A. delta p = rho c V0 = 1000 x 1148.9 x 2.075 = 2.38 MPa (243 m of water). (Key point: Water hammer pressure) 3. How is the stress produced in the thin pipe wall found? A. sigma = delta p D / (2t), giving about 63.9 MPa B. sigma = delta p t / D, giving about 0.1 MPa C. sigma = delta p D / (4t) only, giving 32 MPa as the main stress D. sigma = rho c V0, giving 2.38 MPa Answer: A. Hoop stress is the largest: 2.384e6 x 0.15 / (2 x 0.0028) = 63.9 MPa; the longitudinal stress is half. (Key point: Hoop stress) 4. At 575 m from the valve, what happens to the pressure between t = 0.50 s and 3.50 s? A. It is at static value B. It is raised by 2.38 MPa C. It is lowered by 2.38 MPa D. It oscillates every 0.1 s Answer: B. The wave from the valve arrives at 0.50 s and the reservoir's negative reflection arrives at 3.50 s, so the point is +2.38 MPa between them. 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