Power Gained by Fluid from Pump or Fan
Calculate the power gained by fluid from an operating pump or fan.
Power Gained by Fluid
The power gained by the fluid from a pump or fan can be expressed as:
P = m w (1)
where
P = power (W)
m = mass flow rate (kg/s)
w = specific work (Nm/kg, J/kg)
Specific Work
Specific work  w  can be expressed:
w = g h (2)
where
h = head (m)
g = acceleration of gravity (9.81 m/s^{2})
Mass Flow Rate
Mass flow rate  m  can be expressed:
m =ρ Q (3)
where
ρ = density (kg/m^{3})
Q = volume flow rate (m^{3}/s)
Combining (1), (2) and (3) the power gained by the fluid from a pump or fan can be expressed as:
P = ρ Q g h (4)
With specific weight expressed as:
γ = ρ g (5)
where
γ= specific weight (N/m^{3})
equation (4) can be modified so the power gained by the fluid from a pump or fan can be expressed as:
P = γ Q h (6)
Since head can be expressed as
h = (p_{2}  p_{1}) /γ (7)
equation (4) can be modified so the power gained by the fluid from a pump or fan can be expressed as:
P = Q (p_{2}  p_{1}) (8)
Example  Head Rise of a Inline Pump
An inline water pump works between measured pressure 1 bar (1 10^{5} N/m^{2}) and 10 bar (10 10^{5} N/m^{2}). Density of water is 1000 kg/m^{3}. The volume flow is measured to 1 10^{3} m^{3}/s.
The actual water head (water column) can be calculated using (7):
h = (p_{2}  p_{1}) /γ
= (p_{2}  p_{1}) /ρ g
= ((10 10^{5} N/m^{2})  (1 10^{5} N/m^{2})) / (1,000 kg/m^{3}) (9.81 m/s^{2})
= 91.7 m  water column
The power gained by the fluid can be calculated using equation (4):
P = ρ Q g h
= (1,000 kg/m^{3}) (1 10^{3} m^{3}/s) (9.81 m/s^{2}) (91.7 m)
= 899.6 kg.m^{2}/s^{3} (W)
= 0.9 kW
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