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Air Motors

The application of the gas laws to Air Motors

Air Motors


Work done by Air

Isothermal Operation

Work Done = \inline \displaystyle P_1V_1\;Ln\,\frac{V_2}{V_1}

It is not possible to expand the Air down to Atmospheric Pressure. By opening the valve early the small loss in work done is accompanied by a large decrease in the Swept Volume.

Work output = Area of Diagram = \inline \displaystyle P_1V_1+\left(\frac{P_1V_1-P_2V_2}{n\;_\;1} \right) -P_3V_3

Example - The application of the gas laws to Air Motors
A reciprocating Air Motor of bore 13.54 in. and Stroke 18 in. is supplied with Air at \inline 180^0F and 90 psi. Find the Work output per cycle with complete expansion to 15 psi. Find also the percentage loss of output resulting from a decrease of stroke to 12 in. with incomplete expansion. Neglect Clearance and take the Index of expansion as 1.25.

Bore = 13.54 in. and Stroke = 18 in.

= \frac{1.25}{0.25}(90\times 144\times 0.357 - 15\times 144\times 1.5)= 6910\;ft.lbs./cycle

The Stroke is now reduced to 12 ins.
\therefore\;\;\;\;\;P_2 = 90\left(\frac{0.357}{1} \right)^{1.25} = 25\;psi.

Work out put per cycle is:
=(90\times 144\times 0.35)\;+\left(\frac{90\times 144\times 0.357 - 25\times 144\times 1.0}{0.25} \right) - (15\times 144\times 1.0)
= 6560\;ft\;lbs.
Therefore the loss of Power is:
\frac{6910 - 6560}{6910}\times 100\;\approx 5\%
Work out put per cycle is \inline 6910\;ft.lbs./cycle

The loss of Power is \inline \approx 5\%

Last Modified: 26 Sep 11 @ 07:23     Page Rendered: 2022-03-14 15:56:02