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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 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

    Design of pumping systems and pipelines. With centrifugal pumps, displacement pumps, cavitation, fluid viscosity, head and pressure, power consumption and more.

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