# Specific Work done by Turbo Machines - Pumps, Compressors and Fans

## Specific work from pumps, fans, compressors and turbines

Specific work is work per unit weight. Specific work in turbo machines as fans, pumps, compressors or turbines has the SI-units

*Nm/kg = J/kg = m*^{2}/s^{2}

### Specific Work of a Pump or Fan

Specific work of a pump or fan working with an incompressible fluid can be expressed as:

w = (p_{2}- p_{1}) / ρ(1)

where

w= specific work (Nm/kg, J/kg, m^{2}/s^{2})

p= pressure (N/m^{2})

ρ= density (kg/m^{3})

### Specific Work of a Turbine

Specific work of a turbine with an incompressible fluid can be expressed as:

w = (p_{1}- p_{2}) / ρ(2)

### Specific Work of a Compressor

A compressor works with compressible fluids and the specific work for an isentropic compressor process can be expressed with the help of

p_{1}v_{1}^{κ}= p_{2}v_{2}^{κ}(3)

where

v= volume (m^{3})

κ=c_{p}/ c_{v}- ratio of specific heats (J/kg K)

Specific work:

w=κ / (κ -1) R T_{1}[( p_{2}/ p_{1})^{((κ-1)/κ)}- 1](4)

where

R= individual gas constant (J/kg K)

T= absolute temperature (K)

### Specific Work of a Gas Turbine

A gas turbine expands a compressible fluid and the specific work can be expressed as

w=κ / (κ -1) R T_{1}[1 - ( p_{2}/ p_{1})^{((κ-1)/κ)}](5)

### Head in Turbomachines

The specific work can on basis of the energy equation be expressed with the head as:

w = g h(6)

where

h= head (m)

g= acceleration of gravity (m/s^{2})

Transformed to express head:

h = w / g(7)

### Example - Specific Work of a Water Pump

A water pump works between *1 bar (10 ^{5} N/m^{2})* and

*10 bar (10 10*. The specific work can be calculated with

^{5}N/m^{2})*(1)*:

w = (p_{2}- p_{1}) / ρ

=((10 10^{5}N/m^{2}) - (10^{5}N/m^{2})) / (1000 kg/m^{3})

= 900 Nm/kg

Dividing by acceleration of gravity the head can be calculated using *(7)*:

h_{water}= (900 Nm/kg) / (9,81 kg/s^{2})

= 91,74 (m) water column

### Example - Specific Work of an Air Compressor

An air compressor works with air at *20 ^{o}C* compressing the air from

*1 bar*absolute

*(10*to

^{5}N/m^{2})*10 bar (10 10*. The specific work can be expressed with

^{5}N/m^{2})*(4)*:

w=κ / (κ -1) R T_{1}[( p_{2}/ p_{1})^{((κ-1)/κ)}- 1]

= ((1.4 J/kg K) / ((1.4 J/kg K) - 1 )) (286.9 J/kg K) ((273 K) + (20 K)) [((10 10^{5}N/m^{2}) / (10^{5}N/m^{2}))^{(((1.4 J/kg K) - 1)/(1.4 J/kg.K))}- 1 ]

= 273826 Nm/kg

where

κ_{air}= 1.4 (J/kg K) - ratio of specific heat air

R_{air}= 286.9 (J/kg K) - individual gas constant air

Dividing by acceleration of gravity the head can be calculated using (7):

h_{air}= (274200 N m/kg) / (9.81 kg/s^{2})

= 27951 (m) air column

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