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Air Motors

The application of the gas laws to Air Motors

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Air Motors
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Air Motors

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Work done by Air

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

(2)

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

(3)

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

(4)

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.

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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:

[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 $180^0F$ 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.