Skip to main content

Peak pressure for 4.5 s closure

Short answer

Closure time 4.5 s is less than 2L/c = 6.4 s, so rapid closure applies. Peak pressure rise Δp = ρcV = 2.70 MPa (275.2 m head). The peak pressure acts over the first 950 m from the valve.

A valve is closed in 4.5 s at the down stream end of a 3200 m pipeline carrying water at 2.7 m/s. What is the peak pressure developed by the closure, if the wave travels with velocity of 1000 m/s? Determine the length of pipe subjected to the peak discharge.

The wave from the valve travels to the reservoir and back in time 2L/c2L/c. If the closure finishes before the reflected wave returns to the valve, the full Joukowsky pressure develops at the valve.

Given data

L=3200L=3200 m, V=2.7V=2.7 m/s, c=1000c=1000 m/s, tc=4.5t_c=4.5 s.

Peak pressure

2Lc=2×32001000=6.4 s>tc=4.5 s\frac{2L}{c}=\frac{2\times3200}{1000}=6.4\ \text{s}>t_c=4.5\ \text{s}

The closure is rapid (the valve is completely shut before the negative wave returns), so

Δp=ρcV=1000×1000×2.7=2.70 MPa(ΔH=cVg=275.2 m)\Delta p=\rho cV=1000\times1000\times2.7=2.70\ \text{MPa}\quad\left(\Delta H=\frac{cV}{g}=275.2\ \text{m}\right)

Length of pipe subjected to the peak pressure

The last pressure increment leaves the valve at t=tct=t_c and reaches a point xx from the valve at tc+x/ct_c+x/c. The negative wave reflected from the reservoir reaches the same point at (2L−x)/c(2L-x)/c. The full peak is felt only where the last increment arrives first:

tc+xc≤2L−xc  ⇒  x≤L−ctc2=3200−1000×4.52=950 mt_c+\frac{x}{c}\le\frac{2L-x}{c}\;\Rightarrow\;x\le L-\frac{ct_c}{2}=3200-\frac{1000\times4.5}{2}=950\ \text{m} 
 Reservoir                              Valve
 |<------ 2250 m ------>|<---- 950 m --->|
   reduced pressure        peak pressure

Answer: peak pressure rise = 2.70 MPa (275.2 m of water); length of pipe subjected to the peak pressure = 950 m from the valve.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Hydraulics chapter 4 questions →