Turbo Machines - Specific Work done by Pumps, Compressors or Fans
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 = m2/s2
Specific Work of a Pump or Fan
Specific work of a pump or fan working with an incompressible fluid can be expressed as:
w = (p2 - p1) / ρ (1)
where
w= specific work (Nm/kg, J/kg, m2/s2)
p= pressure (N/m2)
ρ= density (kg/m3)
Specific Work of a Turbine
Specific work of a turbine with an incompressible fluid can be expressed as:
w = (p1 - p2) / 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
p1 v1κ = p2 v2κ (3)
where
v= volume (m3)
κ=cp / cv- ratio of specific heats (J/kg K)
Specific work:
w=κ / (κ -1) R T1 [( p2 / p1)((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 T1 [1 - ( p2 / p1)((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/s2)
Transformed to express head:
h = w / g (7)
Example - Specific Work of a Water Pump
A water pump works between 1 bar (105 N/m2) and 10 bar (10 105 N/m2). The specific work can be calculated with (1):
w = (p2 - p1) / ρ
=((10 105 N/m2) - (105 N/m2)) / (1000 kg/m3)
= 900 Nm/kg
Dividing by acceleration of gravity the head can be calculated using (7):
hwater= (900 Nm/kg) / (9.81 kg/s2)
= 91.74 (m) water column
Example - Specific Work of an Air Compressor
An air compressor works with air at 20 oC compressing the air from 1 bar absolute (105 N/m2) to 10 bar (10 105 N/m2). The specific work can be expressed with (4):
w=κ / (κ -1) R T1 [( p2 / p1)((κ-1)/κ) - 1]
= ((1.4 J/kg K) / ((1.4 J/kg K) - 1 )) (286.9 J/kg K) ((273 K) + (20 K)) [((10 105 N/m2) / (105 N/m2))(((1.4 J/kg K) - 1)/(1.4 J/kg.K)) - 1 ]
= 273826 Nm/kg
where
κair= 1.4 (J/kg K) - ratio of specific heat air
Rair= 286.9 (J/kg K) - individual gas constant air
Dividing by acceleration of gravity the head can be calculated using (7):
hair= (274200 N m/kg) / (9.81 kg/s2)
= 27951 (m) air column
Related Topics
• Fluid Mechanics
The study of fluids - liquids and gases. Involving velocity, pressure, density and temperature as functions of space and time.
• Pumps
Piping systems and pumps - centrifugal pumps, displacement pumps - cavitation, viscosity, head and pressure, power consumption and more.
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