Question 11.9: A pump requires 50 kW to supply water at a rate of 0.2 m³/s ...
A pump requires 50 kW to supply water at a rate of 0.2 m^{3}/s to an overhead tank. The pipe connecting the delivery end of the pump to the overhead tank is 120 m long and 300 mm in diameter and has a friction factor f = 0.02. A valve is inserted in the delivery pipe to control the flow rate. The loss coefficient of the valve under wide open condition is 5.0. Water is supplied from a reservoir 2 m below the horizontal level of the pump though a suction pipe 6 m long and 400 mm in diameter having f = 0.03. Determine the maximum height from the plane of the pump at which the overhead tank can be placed under this situation. (Take the efficiency of the pump η = 80%).
Let H be the height of the overhead tank from the pump
p_{d} be the pressure at the delivery side of the pump
p_{s} be the pressure at the suction side of the pump.
The average velocity of flow in the delivery pipe
V_{d}=\frac{4 \times 0.2}{\pi \times(0.3)^{2}}=2.83 m/s
The average velocity of flow in the suction pipe
V_{s}=\frac{4 \times 0.2}{\pi \times(04)^{2}}=1.59 m/s
Applying Bernoulli’s equation between a point at the inlet to the delivery pipe and a point at the water surface in the overhead tank where the pressure is atmospheric, we have,
\frac{p_{d}}{p g}+\frac{(2.83)^{2}}{2 g}=\frac{p_{ atm }}{\rho g}+H+\left(0.02 \times \frac{120}{0.3}+5.0+1.0\right) \times \frac{(2.83)^{2}}{2 g}
or \frac{p_{d}-p_{ atm }}{\rho g}=H+5.31 (11.49)
Applying Bernoulli’s equation between a point on the water surface in the supply reservoir and a point at the end of the suction pipe connecting the pump, we can write,
\frac{p_{ atm }}{\rho g}=\frac{p_{s}}{\rho g}+2+\left(1+0.5+0.03 \times \frac{6.0}{0.4}\right) \times \frac{(1.59)^{2}}{2 g}
or \frac{p_{ atm }-p_{s}}{\rho g}=2.25 (11.50)
From Eqs (11.49) and (11.50), we get
\frac{p_{d}-p_{s}}{\rho g}=H+7.56
Power delivered by the pump to water = 50 × 0.8 = 40 kW
Therefore, we can write,
0.2 \times\left(p_{d}-p_{c}\right)=40 \times 10^{3}
or 0.2 \times 10^{3} \times 9.81(H+7.56)=40 \times 10^{3}
which gives, H = 12.83 m