Ridge is n efficiency of 73 is installed at 25 pts The actua
Solution
The ideal hydraulic power required to drive a pump is given by -
Ph = Flow discharge x fluid density x acceleration due to gravity x differential head
Assuming flow discharge as 1 m3/ sec,
Ph(kW) = 1 x 1000 kg/m3 x 9.81 m/s2 x 20.5 = 201105 W = 201.105kW
Now, the shaft power i.e. the power required transferred from the motor to the shaft of the pump depends upon the efficiency of the pump and is equal to Ph/ efficiency of the pump
here, efficiency of the pump = 73%
therefore, the shaft power= 201.105/ 0.73 = 275.486 kW
Now, the efficiency of the motor for powering the pump is 85%,
The power requirement of the motor = 275.486/0.85 = 324.101 kW
Thus, the power requirement of the system is 324.101 kW
Number of 60W lightbulbs demanding an equivalent amount of power = 324.101 x 1000 / 60 = 5401.68 or 5402 lightbulbs.
