--- title: "One cycle of water hammer pressure wave" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:water-hammer-one-cycle-wave-sketch" source: https://www.bctengineer.com/q/semester_4/hydraulics/water-hammer-one-cycle-wave-sketch site: BCT Engineering --- # One cycle of water hammer pressure wave Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2073 Magh, 8 marks; 2069 Poush, 8 marks (asked 2 times) - Video: [Water Hammer - What is Water Hammer? (1/8)](https://www.youtube.com/watch?v=1E26PgIq9JM) — 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-one-cycle-wave-sketch - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question Discuss water hammer phenomenon. Describe with neat sketches the one cycle of pressure wave propagation in a pipe connected to a reservoir, when the valve at the end of the pipe is closed suddenly (showing flow velocity direction and wave celerity at the specified times). One cycle represents t = 0 to t = 4L/C. ## Short answer Water hammer is a pressure surge Δp=ρcV0 from sudden velocity change. After sudden valve closure, one cycle: 0–L/c compression wave stops flow; L/c–2L/c wave reflects, restoring pressure with reverse flow −V0; 2L/c–3L/c negative wave stops reverse flow and drops pressure; 3L/c–4L/c wave restores forward flow +V0. Period T=4L/c. ## Answer #### Water hammer When the velocity of flowing water in a closed pipe is changed suddenly (valve closed quickly, pump stopped), the kinetic energy of the moving column is converted into pressure energy, producing a pressure wave that travels along the pipe at a speed $c$ (celerity) and is reflected at the ends. The pressure change is $\Delta p = \rho cV_0$. This repeated pressure rise and fall is called **water hammer** (it produces noise and vibration like hammering). #### One cycle of the wave after sudden valve closure Consider a pipe of length $L$ with a reservoir at the upstream end and a valve at the downstream end, initial velocity $V_0$ towards the valve. The valve is closed suddenly at $t = 0$. (Friction neglected; the wave speed $c$ is constant.) **Stage 1: $0 < t < L/c$** (wave travels from the valve to the reservoir) ```text reservoir |<-------------- L -------------->| valve |=====[ V=V0 ]=====|====[ V=0 ]===X closed <---- c <----- pressure wave +dp ``` The layer at the valve is stopped at once and compressed; pressure rises by $\Delta p$ and the pipe expands. The wave front moves upstream at $c$; the water behind it ($V = 0$) is at high pressure; ahead of it, the water still moves at $V_0$. At $t = L/c$ the whole pipe is at $p_0 + \Delta p$ with $V = 0$. **Stage 2: $L/c < t < 2L/c$** (wave travels from the reservoir to the valve) ```text reservoir |=====[ V=-V0 ]=====|====[ V=0, +dp ]===X ----> c -----> (reflected wave, pressure back to p0) ``` At the reservoir the pressure cannot stay above the reservoir pressure, so the water flows back into the reservoir with $V = -V_0$ and a wave $-\Delta p$ travels back towards the valve, restoring the pressure to normal. At $t = 2L/c$ the whole pipe is at normal pressure with $V = -V_0$ (flow towards the reservoir). **Stage 3: $2L/c < t < 3L/c$** (wave travels from the valve to the reservoir) ```text reservoir |=====[ V=-V0, p0 ]=====|====[ V=0, p0-dp ]===X <---- c <----- negative wave ``` The valve prevents any further backward flow, so $V$ drops to 0 at the valve and the pressure falls by $\Delta p$ below normal (possible column separation if the pressure falls below vapour pressure). The negative wave travels towards the reservoir. At $t = 3L/c$ the pipe is at $p_0 - \Delta p$, $V = 0$. **Stage 4: $3L/c < t < 4L/c$** (wave travels from the reservoir to the valve) ```text reservoir |=====[ V=+V0, p0 ]=====|====[ V=0, p0-dp ]===X ----> c -----> ``` The low pressure in the pipe draws water from the reservoir at $V = +V_0$; the wave restores normal pressure as it moves to the valve. At $t = 4L/c$ conditions are the same as at $t = 0$ just before closure (flow $V_0$ towards the valve with pressure $p_0$), and the cycle repeats. In reality friction damps the oscillations. | Time | Wave direction | Velocity | Pressure in pipe | |---|---|---|---| | 0 to $L/c$ | valve to reservoir | $V_0 \to 0$ behind the front | $p_0 + \Delta p$ behind the front | | $L/c$ to $2L/c$ | reservoir to valve | $-V_0$ behind the front | back to $p_0$ behind the front | | $2L/c$ to $3L/c$ | valve to reservoir | $0$ behind the front | $p_0 - \Delta p$ behind the front | | $3L/c$ to $4L/c$ | reservoir to valve | $+V_0$ behind the front | back to $p_0$ behind the front | The period of one cycle is $T = 4L/c$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What converts into pressure energy to produce water hammer when a valve is closed suddenly? A. Potential energy of the reservoir B. Kinetic energy of the moving water column C. Heat energy of the pipe wall D. Elastic energy of the valve spindle Answer: B. The moving column is stopped, and its kinetic energy appears as a pressure wave (delta p = rho c V0). (Key point: Cause of water hammer) 2. At t = L/c after instantaneous valve closure (pipe connected to a reservoir), what is the condition in the whole pipe? A. V = V0 and pressure p0 B. V = 0 and pressure p0 + delta p C. V = -V0 and pressure p0 D. V = 0 and pressure p0 - delta p Answer: B. The positive wave has travelled to the reservoir, leaving all water at rest and compressed. (Key point: Stage 1 state) 3. Why does water flow back into the reservoir between L/c and 2L/c? A. The valve reopens automatically B. The high pressure cannot be maintained at the reservoir, so a negative wave returns and water flows out with V = -V0 C. Friction pushes the water back D. The pipe contracts and expels water Answer: B. The reservoir holds pressure at p0, so a -delta p wave reflects and restores normal pressure while the flow reverses. (Key point: Reflection at the reservoir) 4. What are the pressure at 3L/c and the period of one cycle? A. p0 + delta p, period 2L/c B. p0 - delta p, period 4L/c C. p0, period L/c D. p0 - delta p, period 2L/c Answer: B. The valve stops the reverse flow, giving a negative wave so the pipe is at p0 - delta p at 3L/c; the cycle repeats after 4L/c. 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