Turbo Machines  Specific Work done by Pumps, Compressors or Fans
Calculate specific work done by pumps, fans, compressors or turbines.
Specific work is work per unit weight. Specific work in turbo machines as fans, pumps, compressors or turbines has the SIunits
 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})^{((κ1)/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})^{((κ1)/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^{5} N/m^{2}). The specific work can be calculated with (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^{5} N/m^{2}) to 10 bar (10 10^{5} N/m^{2}). The specific work can be expressed with (4):
w=κ / (κ 1) R T_{1} [( p_{2} / p_{1})^{((κ1)/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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