Question

Water is pumped at the rate of 0.075 m3/s from a reservoir 20 m above a pump to a free discharge 35 m above the pump. The pressure on the intake side of the pump is 150 kPa and the pressure on the discharge side is 450 kPa. All pipes are commercial steel of 15 cm diameter. Determine (a) the head supplied by the pump and (b) the total head loss between the pump and point of free discharge.

Answer 1

Answer:

(a) 408.16 m

(b) 10.87 m

Solution:

As per the question:

Height of the reservoir, h = 20 m

Pressure on the intake, $P_{i} = 150\ kPa$

Pressure on the discharge, $P_{d} = 450\ kPa$

Diameter of the pipes, d = 15 cm = 0.15 m

Now,

(a) The head supplied by the pump can be calculated as:

Applying Bernoulli's theorem:

$\frac{P_{i}}{\rho g} + \frac{V_{i}^{2}}{2g} + z_{i} + H_{pump} = \frac{P_{d}}{\rho g} + \frac{V_{d}^{2}}{2g} + z_{d}$

where

$V_{i} = V_{d}$

$z_{i} = z_{d}$

Now, the above eqn becomes:

$\frac{P_{i}}{\rho g} + H_{pump} = \frac{P_{d}}{\rho g}$

$H_{pump} = \frac{P_{d} - {P_{i}}{\rho g}$

$H_{pump} = \frac{450 - 150}{0.075\time 9.8} = 408.16\ m$

(b) To calculate the total Head loss between the pump and free discharge point:

$\frac{P_{d}}{\rho g} + \frac{V_{d}^{2}}{2g} + z_{d} + H_{pump} = \frac{P'}{\rho g} + \frac{V'^{2}}{2g} + z' + H_{Loss}$

Since,

$V_{d} = V'$

Thus

$H_{Loss} = \frac{(450 - 0)\times 10^{3}}{9.8\times 1000} - 35 = 10.87 m$