Question

You work for the city water department and need to pump 3400 liters/minute of water from a tank at ground level into a vented (i.e., open-top) water tower, 75 meters above ground. The pump can be considered adiabatic, and the combined efficiency of pump and its drive motor is 90 percent. Determine: the electric power required (kW) to drive the pump if friction losses are neglected and the average velocity (m/s) of the water if a 20-cm diameter (ID) pipe is used.

Answer 1

Answer:

$P_E=46.2778\ kW$

$v=1.804\ m.s^{-1}$

Explanation:

Given:

• flow rate of water, $\dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}$

∵Density of water is 1 kg per liter

∴mass flow rate of water, $\dot{m}=3400\ kg.min^{-1}$

• height of pumping, $h=75\ m$

• efficiency of motor drive, $\eta=0.9$

• diameter of pipe, $D =0.2\ m$

Now the power required for pumping the water at given conditions:

$P=\dot{m}.g.h$

$P=\frac{3400}{60} \times 9.8\times 75$

$P=41650\ W$

Hence the electric power required:

$P_E \times \eta=P$

$P_E \times 0.9=41650$

$P_E=46.2778\ kW$

Flow velocity is given as:

$v=\dot{V}\div a$

where: a = cross sectional area of flow through the pipe

$v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )$

$v=1.804\ m.s^{-1}$