A penstock is 3000 m long. Pressure wave travels in it with a velocity 1500 m/s. If the turbine gates are closed uniformly and completely in 4.5 s, then it is called a

Question

Question

This question was previously asked in

BPSC AE Paper 6 (Civil) 11 Nov 2022 Official Paper

Answer 1

Concept:

Critical Time of closure is given by:

\({t_c} = \frac{{2L}}{C}\)

Where 2L/C = Time taken by pressure wave to travel from the valve to the tank and from tank to valve.

L = Length of pipe, C = Velocity of pressure wave

(i) The valve closure is gradual if, \({t_c} > \frac{{2L}}{C}\)

(ii) The valve closure is sudden if, \({t_c} < \frac{{2L}}{C}\)

Calculation:

Given,

L = 3000 m, C = 1500 m/s, tc = 4.5 second

Critical time of closure, \({t_c} = \frac{{2L}}{C}\)

⇒ \({t_c} = \frac{{2 \ \times\ 3000}}{{1500}} = 4\sec \)

∵ tc (4.5 sec) > 4 sec, So it is the condition of gradual closure (slow closure).