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
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.
Related Documents

Air  SCFM versus ACFM and ICFM
Actual air compressor capacity (ACFM) vs. standard air capacity (SCFM) and inlet air capacity (ICFM). 
Efficiency in Pumps or Fans
The overall pump and fan efficiency is the ratio power gained by the fluid to the shaft power supplied. 
Isentropic Flow
Fluid flow with constant entropy is also called isentropic flow. 
Pump Power Calculator
Calculate pumps hydraulic and shaft power. 
Pumping Water  Required Horsepower
Horsepower required to pump water. 
Pumps  Specific Speed
Characterizing of impeller types in pumps in a unique and coherent manner. 
Pumps  Suction Specific Speed
Suction Specific Speed can be used to determine stable and reliable operations for pumps with max efficiency without cavitation. 
Pumps vs. Compressors, Blowers and Fans
The difference between pumps, compressors, blowers and fans. 
Types of Air Compressors
Reciprocating, rotary screw and rotary centrifugal air compressors. 
Work done by Force
Work done by a force acting on an object.