Question 2.69: Water is pumped from a lake to a storage tank 20 m above at ......
Water is pumped from a lake to a storage tank 20 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.
Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pumpmotor unit and the pressure difference between the inlet and the exit of the pump are to be determined. √
Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible.
Properties We take the density of water to be \rho=1000 \mathrm{~kg} / \mathrm{m}^3
Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level \left(z_1=0\right), and thus the potential energy at points 1 and 2 are \mathrm{pe}_1=0 and \mathrm{pe}_2=g z_2. The flow energy at both points is zero since both 1 and 2 are open to the atmosphere \left(P_1=P_2=P_{\text {atm }}\right). Further, the kinetic energy at both points is zero \left(\mathrm{ke}_1=\mathrm{ke}_2=0\right) since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are
\displaystyle \dot{m}=\rho \dot{\boldsymbol{V}}=\left(1000 \mathrm{~kg} / \mathrm{m}^3\right)\left(0.070 \mathrm{~m}^3 / \mathrm{s}\right)=70 \mathrm{~kg} / \mathrm{s}\displaystyle p e_2=g z_2=\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)(20 \mathrm{~m})\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=0.196 \mathrm{~kJ} / \mathrm{kg}
Then the rate of increase of the mechanical energy of water becomes
The overall efficiency of the combined pump-motor unit is determined from its definition,
(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 \mathrm{~kW} :
Solving for \Delta P and substituting,
Therefore, the pump must boost the pressure of water by 196 \mathrm{~kPa} in order to raise its elevation by 20 \mathrm{~m}.
Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.
