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Air Motors

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The application of the gas laws to Air Motors

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Air Motors
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Air Motors

13108/img_therm24.jpg

Work done by Air

$= \frac{n}{n\;-\;1}\left(P_1V_1\;-\;P_2V_2 \right)$

$=\frac{n}{n-1}wR\left(T_1-T_2 \right)$

$=\;\frac{n}{n-1}\;wRT_1\;\left[1-\left(\frac{P_1}{P_2} \right)^{\frac{n-1}{n}} \right]$

Isothermal Operation:

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

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.

13108/img_therm25.jpg

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

Example 1:

A reciprocating Air Motor of bore 13.54 in. and Stroke 18 in. is supplied with Air at $180^0F$ 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.

13108/img_therm26.jpg

Bore = 13.54 in. and Stroke = 18 in.

$V_2=\frac{\pi (13.54)^2}{4\times 144}\times 1.5=1.5\;ft^3$

$P_1V_1^n\;=\;P_2V_2^n$

$V_1=V_2\left(\frac{P_2}{P_1} \right)^{\frac{1}{1.25}}\;=\;1.5\div 6^{\frac{1}{1.25}}\;=\;0.357\;ft^3$

$Work\;out put\;per\;cycle\;=\;\frac{n}{n\;-\;1}\left(P_1V_1\;-\;P_2V_2 \right)$

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

The Stroke is now reduced to 12 ins.

$V_2'\;=\;\frac{\pi (13.54)^2\times 1.0}{4\times 144}\;=\;1\.ft^3\;=\;V_3$

$V_1\;=\;0.357\;ft^3$

$P_1V_1^{1.25}\;=\;P_2V_2^{1.25}$

$\therefore\;\;\;\;\;P_2\;=\;P_1\left(\frac{V_1}{V_2} \right)^{1.25}$

$\therefore\;\;\;\;\;P_2\;=\;90\left(\frac{0.357}{1} \right)^{1.25}\;=\;25\;psi.$

Work out put per cycle is:

$P_1V_1+\left(\frac{P_1V_1-P_2V_3}{n-1 }\right)-P_3V_3$

$=\;(90\times 144\times 0.35)\;+\left(\frac{90\times 144\times 0.357\;-\;25\times 144\times 1.0}{0.25} \right)\;-\;(15\times 144\times 1.0)$

$=\;6560\;ft\;lbs.$

Therefore the loss of Power is:

$\frac{6910\;-\;6560}{6910}\times 100\;\approx 5\%$

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Last Modified: 2009-02-12 00:23:58 Page Rendered: 2010-09-24 11:42:58