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# Air Motors

The application of the gas laws to Air Motors
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Contents

## Air Motors

##### MISSING IMAGE!

Work done by Air
$=&space;\frac{n}{n&space;-&space;1}\left(P_1V_1&space;-&space;P_2V_2&space;\right)$
$=\frac{n}{n-1}wR\left(T_1-T_2&space;\right)$
$=&space;\frac{n}{n-1}\;wRT_1\;\left[1-\left(\frac{P_1}{P_2}&space;\right)^{\frac{n-1}{n}}&space;\right]$

### Isothermal Operation

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

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.
##### MISSING IMAGE!

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

Example:
[imperial]
##### Example - The application of the gas laws to Air Motors
Problem
A reciprocating Air Motor of bore 13.54 in. and Stroke 18 in. is supplied with Air at $\inline&space;180^0F$ 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.
Workings

Bore = 13.54 in. and Stroke = 18 in.

$V_2=\frac{\pi&space;(13.54)^2}{4\times&space;144}\times&space;1.5=1.5\;ft^3$
$P_1V_1^n&space;=&space;P_2V_2^n$
$V_1=V_2\left(\frac{P_2}{P_1}&space;\right)^{\frac{1}{1.25}}&space;=&space;1.5\div&space;6^{\frac{1}{1.25}}&space;=&space;0.357\;ft^3$
$Work\;out&space;put\;per\;cycle&space;=&space;\frac{n}{n&space;-&space;1}\left(P_1V_1&space;-&space;P_2V_2&space;\right)$
$=&space;\frac{1.25}{0.25}(90\times&space;144\times&space;0.357&space;-&space;15\times&space;144\times&space;1.5)=&space;6910\;ft.lbs./cycle$

The Stroke is now reduced to 12 ins.
$V_2'&space;=&space;\frac{\pi&space;(13.54)^2\times&space;1.0}{4\times&space;144}&space;=&space;1\.ft^3&space;=&space;V_3$
$V_1&space;=&space;0.357\;ft^3$
$P_1V_1^{1.25}&space;=&space;P_2V_2^{1.25}$
$\therefore\;\;\;\;\;P_2&space;=&space;P_1\left(\frac{V_1}{V_2}&space;\right)^{1.25}$
$\therefore\;\;\;\;\;P_2&space;=&space;90\left(\frac{0.357}{1}&space;\right)^{1.25}&space;=&space;25\;psi.$

Work out put per cycle is:
$P_1V_1+\left(\frac{P_1V_1-P_2V_3}{n-1&space;}\right)-P_3V_3$
$=(90\times&space;144\times&space;0.35)\;+\left(\frac{90\times&space;144\times&space;0.357&space;-&space;25\times&space;144\times&space;1.0}{0.25}&space;\right)&space;-&space;(15\times&space;144\times&space;1.0)$
$=&space;6560\;ft\;lbs.$
Therefore the loss of Power is:
$\frac{6910&space;-&space;6560}{6910}\times&space;100\;\approx&space;5\%$
Solution
Work out put per cycle is $\inline&space;6910\;ft.lbs./cycle$

The loss of Power is $\inline&space;\approx&space;5\%$