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Air Compressors

Examines the application of the gas laws to Air Compressors and Motors.
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Introduction

Single Stage Reciprocating without Clearance:
23287/Air-Compressors-0020.png
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Stage 1 - a
  • Suction Stroke: The inlet valve opens and the cylinder fills with air at Ambient Pressure.

Stage 1 - b
  • Compression Stroke: Both Valves are shut. The Pressure is raised from P_1 to P_2.

Stage 2 - b
  • Delivery Stage: The Exhaust Valve opens. Air at P_2 is delivered to the receiver at Constant Pressure.

The Work Done during compression = \displaystyle\frac{P_2V_2-P_1V_1}{n-1}

The Work Done in the Compressor = \displaystyle \frac{n}{n-1}\;(P_2V_2-P_1V_1)

Work Done In An Isothermal Compression

23287/Air-Compressors-0021.png
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Work Done on Air But,

or,

Work Done In An Adiabatic Compression

Alternative Forms Of The Work Done Expression

The Work Done is given by the following Expressions: where N is the number of Cycles per Min.

where W is the weight Handled per Min.

or,

A Comparison Of The Work Done With Different Indices Of Compression

23287/Air-Compressors-0022.png
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where R =Pressure Ratio.

Points on the Graph:-
  • 2 Isothermal. n = 1
  • 2' Compression when \gamma \geq n\geq 1
  • 2'' n = \gamma - Adiabatic Reversible.
  • 2''' n\geq \gamma

For Reciprocating Compressors:

The efficiency referred to is the Isothermal case since fairly successful cooling can be achieved.

For Rotary Compressors:

The Cooling is very difficult and Indices of less than \gamma are never achieved. It is therefore normal to compare the Performance with the Adiabatic reversible case.

The Overall Isothermal Efficiency Of The Plant:

The Cooling Of Compressors.

It is usually considered that the heat is given up during the Compression:

For a Polytropic Compression.

For an Isothermal Compression.

Example:

[imperial]
Example - Application of the gas laws to Air Compressors and Motors
Problem
An Air Compressor takes in Air at 14 psi and at 20 degrees C. It is compressed in accord to the law P\;V^{1.2}=Constant and delivers it to receiver at 140psi.

Find the Temperature at the end of the Compression and Calculate per pound of Air, the Compressor Work input and the heat rejected during Compression.
Workings
23287/Air-Compressors-0023.png
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=429^{\circ}K

Work input to Compressor per lb.
=96\times \frac{1.2}{0.2}\left(429-293 \right)=78,300\,ft.lb.

Heat Rejected during Compression.
= 23.4\;CHU
Solution
Temperature at the end of the Compression 429^{\circ}K

Compressor Work input 78,300\,ft.lb.

Heat Rejected during Compression 23.4\;CHU