## Pump power calculations - problem statement

Calculate the pump power and motor power requirement to pump 200,000 kg/hr of water available at 25^{0}C and atmospheric pressure from a storage tank. The rated differential head requirement is 30 m.

Assume the mechanical efficiency of the pump to be 70%.

Assume the motor efficiency to be 90%.

## Solution

First we will calculate the theoretical power requirement using the pump power equation. This is the * power required by the pump and provided by the motor*. Next, we will divide this power requirement by the efficiency of the motor to

*.*

**calculate power required by the motor**Let's go step by step through these pump power calculations.

### Step1

The first step is to determine the important physical properties of water at given conditions. The only important physical property for solving this problem is the mass density of water.

Using EnggCyclopedia’s Liquid Density Calculator, water density at 25^{0}C =994.72 kg/m^{3}

Using water density, the mass flow rate is converted to volumetric flow rate.

Volumetric flow = 200,000 / 994.72 = 201.06 m^{3}/hr

Also the differential pressure is determined using differential head as,

ΔP = ρgΔh = 994.72 × 9.81 × 30/10^{5} = 2.93 bar

### Step2

The next step is to calculate the theoretical pump power requirement. As per the pump power equation, power requirement is the product of volumetric flow (Q) and differential pressure (ΔP).

Power requirement = Q × ΔP = 201.06/3600 m^{3}/s × 2.93 × 10^{5} N/m^{2}

Theoretical power requirement = 16350 Watt = 16.35 kW

### Step3

Pump shaft power requirement = Theoretical power requirement / pump efficiency.

For a pump that has been already purchased or has been ordered for manufacturing, the efficiency can be determined using the pump performance curves provided by pump manufacturer. Here the problem statement has specified pump efficiency to be 70%.

Hence, pump shaft power requirement = 16.35 kW / 0.7 = 23.36 kW

Similarly, motor power requirement = Pump shaft power requirement / motor efficiency

Similar to pump efficiency, electric motor efficiency for motors already purchased or ordered, can be provided by the manufacturer of motor. However for purpose of this sample problem the efficiency is to be taken as 90% as per problem statement.

Motor power requirement = 23.36 / 0.9 = 25.95 kW = 25.95 × 1.3596 HP = 35.28 HP

Electric motors are available for following standard **Horsepower** ratings.

1 | 1.5 | 2 | 3 | 5 | 7.5 | 10 | 15 | 20 | 25 | 30 | 40 | 50 |

60 | 75 | 100 | 125 | 150 | 200 | 250 | 300 | 350 | 400 | 450 | 500 | 600 |

700 | 800 | 900 | 1000 | 1250 | 1500 | 1750 | 2000 | 2250 | 2500 | 3000 | 3500 | 4000 |

Hence in order to have satisfy the minimum power requirement, the motor to be purchased has to have power rating of 40 HP or higher.