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Pressure variation after sudden valve closure

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.

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.

Case. A long pipeline of length LL from a reservoir (upstream) to a valve (downstream) carries water at velocity V0V_0. The valve is closed instantaneously at t=0t = 0. Neglect friction. Let ΔH=cV0g\Delta H = \dfrac{cV_0}{g} (pressure rise Δp=ρcV0\Delta p = \rho cV_0) and let the static head be H0H_0 (shown as the zero line below).

At the valve end

The pressure wave starts at the valve at t=0t = 0, so the pressure there rises at once by ΔH\Delta H and stays at H0+ΔHH_0 + \Delta H until the negative wave reflected from the reservoir returns at t=2L/ct = 2L/c. The pressure then drops by 2ΔH2\Delta H to H0−ΔHH_0 - \Delta H (the closed valve prevents backward flow) and stays there until t=4L/ct = 4L/c, when the positive wave from the reservoir restores it to H0+ΔHH_0 + \Delta H. The variation is a square wave of period 4L/c4L/c. Pressure at the valve over two cycles (+ΔH+\Delta H above and −ΔH-\Delta H below the zero line):

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(top = 1.0, bottom = -1.0; '-' is zero line)

At the mid-point of the pipe (L/2L/2 from the valve)

  • 0<t<L/2c0 < t < L/2c: no change (the wave has not arrived yet); pressure H0H_0.
  • L/2c<t<3L/2cL/2c < t < 3L/2c: the wave reaches the mid-point; pressure H0+ΔHH_0 + \Delta H (duration L/cL/c).
  • 3L/2c<t<5L/2c3L/2c < t < 5L/2c: the negative wave reflected from the reservoir cancels it; pressure H0H_0.
  • 5L/2c<t<7L/2c5L/2c < t < 7L/2c: the negative wave reflected from the valve arrives; pressure H0−ΔHH_0 - \Delta H.
  • 7L/2c<t<9L/2c7L/2c < t < 9L/2c: the positive wave from the reservoir restores the pressure to H0H_0; the cycle then repeats from t=9L/2ct = 9L/2c.
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(top = 1.0, bottom = -1.0; '-' is zero line)

The mid-point pressure has the same amplitude (±ΔH\pm\Delta H) but is out of phase with the valve. It stays for L/cL/c at each extreme level, with H0H_0 in between, giving a "stepped" (rectangular pulse) shape.

Summary. At the valve the pressure alternates between H0+ΔHH_0 + \Delta H and H0−ΔHH_0 - \Delta H every 2L/c2L/c; at the mid-point it follows the sequence H0→H0+ΔH→H0→H0−ΔH→H0H_0 \to H_0 + \Delta H \to H_0 \to H_0 - \Delta H \to H_0, each stage lasting L/cL/c (except the first, which lasts L/2cL/2c). Both have a period of 4L/c4L/c.

✓ Answer checked by Claude Opus 5.5 (high effort)

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