# Power Gained by Fluid from Pump or Fan

## The power gained by fluid from an operating pump or fan

Sponsored Links

### 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*. Density of water is

^{5}N/m^{2})*1000 kg/m*. The volume flow is measured to

^{3}*1 10*.

^{-3}m^{3}/sThe 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

## Related Topics

Sponsored Links

## Related Documents

Sponsored Links