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# 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)

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

w = g h                                        (6)

where

g= acceleration of gravity (m/s2)

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.

## Related Documents

### Air - SCFM versus ACFM and ICFM

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### 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.

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