Question

Maximum pressure rise due to water hammer in a pipeline (a = area of the pipe; V₀ = velocity, g = acceleration due to gravity; t = time period; L = length of the pipeline) is
1. \(\frac{{a{V_0}}}{{2g}}\)
2. \(\frac{{aV_0^2}}{{2g}}\)
3. \(\frac{{L{V_0}}}{{gT}}\)
4. Independent of the dimensions of the pipe

Answer 1

Correct Answer - Option 3 : \(\frac{{L{V_0}}}{{gT}}\)

For instantaneous closure of valve, the pressure head at valve end is

\({\rm{\Delta }}h = \frac{{C{V_0}}}{g}\)

Time period for travel of pressure wave from one end to the other is

\(t = \frac{L}{C}\)

Therefore velocity of pressure wave

\(C = \frac{L}{t}\)

and

\({\rm{\Delta }}h = \frac{{L{V_0}}}{{gt}}\)