--- title: "Pressure variation after sudden valve closure" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:water-hammer-pressure-time-valve-and-midpoint" source: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-pressure-time-valve-and-midpoint site: BCT Engineering --- # Pressure variation after sudden valve closure Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2082 Kartik, 8 marks - Video: [Sudden closure of valve, when pipe is rigid](https://www.youtube.com/watch?v=aU-7L2Vnz2k) — Pavika fluid says Tawi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-pressure-time-valve-and-midpoint - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question Describe with sketches the variation of pressure with time (at valve and at midpoint of pipe) in a long pipeline from the reservoir with valve at the downstream, when the valve is instantaneously closed. ## Short answer At valve, pressure jumps by ΔH = cV0/g at closure, stays at H0+ΔH for 2L/c, drops to H0-ΔH for the next 2L/c, repeating every 4L/c. At midpoint, pressure steps: H0, H0+ΔH for L/c, H0 for L/c, H0-ΔH for L/c, H0, same period. ## Answer **Case.** A long pipeline of length $L$ from a reservoir (upstream) to a valve (downstream) carries water at velocity $V_0$. The valve is closed instantaneously at $t = 0$. Neglect friction. Let $\Delta H = \dfrac{cV_0}{g}$ (pressure rise $\Delta p = \rho cV_0$) and let the static head be $H_0$ (shown as the zero line below). #### At the valve end The pressure wave starts at the valve at $t = 0$, so the pressure there rises at once by $\Delta H$ and stays at $H_0 + \Delta H$ until the negative wave reflected from the reservoir returns at $t = 2L/c$. The pressure then drops by $2\Delta H$ to $H_0 - \Delta H$ (the closed valve prevents backward flow) and stays there until $t = 4L/c$, when the positive wave from the reservoir restores it to $H_0 + \Delta H$. The variation is a square wave of period $4L/c$. Pressure at the valve over two cycles ($+\Delta H$ above and $-\Delta H$ below the zero line): ```text ************ ************ * ------------------------------------------------- ************ ************ (top = 1.0, bottom = -1.0; '-' is zero line) ``` #### At the mid-point of the pipe ($L/2$ from the valve) - $0 < t < L/2c$: no change (the wave has not arrived yet); pressure $H_0$. - $L/2c < t < 3L/2c$: the wave reaches the mid-point; pressure $H_0 + \Delta H$ (duration $L/c$). - $3L/2c < t < 5L/2c$: the negative wave reflected from the reservoir cancels it; pressure $H_0$. - $5L/2c < t < 7L/2c$: the negative wave reflected from the valve arrives; pressure $H_0 - \Delta H$. - $7L/2c < t < 9L/2c$: the positive wave from the reservoir restores the pressure to $H_0$; the cycle then repeats from $t = 9L/2c$. ```text ****** ****** ***------******------******------******------**** ****** ****** (top = 1.0, bottom = -1.0; '-' is zero line) ``` The mid-point pressure has the same amplitude ($\pm\Delta H$) but is out of phase with the valve. It stays for $L/c$ at each extreme level, with $H_0$ in between, giving a "stepped" (rectangular pulse) shape. **Summary.** At the valve the pressure alternates between $H_0 + \Delta H$ and $H_0 - \Delta H$ every $2L/c$; at the mid-point it follows the sequence $H_0 \to H_0 + \Delta H \to H_0 \to H_0 - \Delta H \to H_0$, each stage lasting $L/c$ (except the first, which lasts $L/2c$). Both have a period of $4L/c$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the pressure at the valve vary after instantaneous closure (friction neglected)? A. A square wave between +delta H and -delta H changing every 2L/c B. A gradual exponential rise to +delta H C. A short pulse lasting L/c each cycle D. Constant at +delta H forever Answer: A. The pressure rises at once, holds until the reflected wave returns at 2L/c, then drops to -delta H until 4L/c. (Key point: Valve-end pressure trace) 2. At t = 2L/c the valve-end pressure drops by 2 delta H. Why? A. The reservoir sends a negative wave that arrives and the closed valve prevents backward flow B. The valve opens at this instant C. Friction removes the pressure D. The pipe wall collapses Answer: A. The reflected -delta H wave arrives at the closed valve, taking pressure from +delta H to -delta H. (Key point: Why the valve pressure drops at 2L/c) 3. At the mid-point, when does the pressure first rise to H0 + delta H? A. At t = 0 B. At t = L/2c, when the wave from the valve reaches it C. At t = L/c D. At t = 2L/c Answer: B. The mid-point is L/2 from the valve, so the wave takes L/2c to arrive. (Key point: Mid-point first change) 4. Which sequence describes the mid-point pressure over one period? A. H0, H0+dH, H0, H0-dH, H0 B. H0, H0+dH, H0-dH, H0+dH, H0 C. H0+dH, H0-dH alternately every L/2c D. H0, H0+dH, H0+dH, H0, H0 Answer: A. Stepped pulses: +dH lasts L/c, then reservoir wave cancels it, then -dH for L/c, then back to H0; period 4L/c. 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