. A centrifugal pump has the following specifications:
Power = 4 H.P.; Speed = 800 rpm
Head = 8 metres
Flow = 1000 litres/minutes.
If its speed is halved, then the new head will be __________ metres.

ANS:A - 2

To determine the new head of the centrifugal pump when its speed is halved, we can use the affinity laws that relate the pump performance to changes in speed. The affinity laws relevant for this scenario are:

1. Head Affinity Law: The head developed by a centrifugal pump is proportional to the square of the speed (RPM). H1H2=(N1N2)2\frac{H_1}{H_2} = \left( \frac{N_1}{N_2} \right)^2H2​H1​​=(N2​N1​​)2
2. Flow Affinity Law: The flow rate through a centrifugal pump is directly proportional to the speed (RPM). Q1Q2=N1N2\frac{Q_1}{Q_2} = \frac{N_1}{N_2}Q2​Q1​​=N2​N1​​

Given:

• Initial head, H1=8H_1 = 8H1​=8 metres
• Initial speed, N1=800N_1 = 800N1​=800 rpm
• New speed, N2=8002=400N_2 = \frac{800}{2} = 400N2​=2800​=400 rpm (since speed is halved)

To find the new head H2H_2H2​: (H1H2)=(N1N2)2\left( \frac{H_1}{H_2} \right) = \left( \frac{N_1}{N_2} \right)^2(H2​H1​​)=(N2​N1​​)2 H1H2=(800400)2\frac{H_1}{H_2} = \left( \frac{800}{400} \right)^2H2​H1​​=(400800​)2 H1H2=4\frac{H_1}{H_2} = 4H2​H1​​=4 Now, solving for H2H_2H2​: H2=H14H_2 = \frac{H_1}{4}H2​=4H1​​ H2=84H_2 = \frac{8}{4}H2​=48​ H2=2 metresH_2 = 2 \text{ metres}H2​=2 metres

Conclusion:

Therefore, when the speed of the centrifugal pump is halved, the new head will be 2 metres. Hence, the answer is 2 metres.