A Centrifugal pump is used in DWASA which delivers water at a height of 25m. Length of the pipe is 200m and the diameter is 150mm. Flow rate of water through the pipe is 50L/s. Pressure wave velocity is 1200m/s. Darcy weisbach factor is 0.03. Determine-

(a) Total dynamic head 

(b) Total power of the pump if the efficiency is 75%. 

(c) What is the increase in pressure if the valve is closed in 0.02 s?

(d) suggest one mechanical solution to reduce the WASA's pump hammering.

Water velocity in the pipe,

V = $\frac{4Q}{\pi D^2}$

= $\frac{4\times 50}{\pi \times 0.{15}^2\times 1000}$

= 2.83 m/s

(a)

TDH = H_stat + H_f + $\frac{V^2}{2g}$

= H_stat + $\frac{fL{V}^2}{2gd}$ + $\frac{V^2}{2g}$

= 25 + $\frac{0.03 \times 200 \times 2.{83}^2}{2 \times 9.81 \times 0.15}+\frac{2.{83}^2}{2 \times 9.81}$

= 41.74 m

(b)

${\eta }_o=\frac{\gamma Q.\left(TDH\right)}{P}$

⇒ P = $=\frac{\gamma Q.\left(TDH\right)}{{\eta }_o}$

= $\frac{9.81\times 50\times 41.74}{0.75\times 1000}$

= 27.3 kW

(c)

Here, we get,

$t =\frac{2L}{C}$

= $\frac{2\times 200}{1200}$

= 0.33s > 0.02s

So, the closure of the valve is instantaneous.

Hammering pressure,

$p = \rho VC$

$= 1000 \times 2.83 \times 1200$

= 3396000Pa 

= 3.396MPa

(d)

Closing the valve gradually can be a solution of reducing the WASA's pump hammering. Another way can be the installation pressure-relief valves at critical points in the pipe system. These valves are pressure-activated so that water is automatically diverted out of the system when the water-hammer pressure reaches excessive levels.