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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:
MISSING IMAGE!

Stage 1 - a
• Suction Stroke: The inlet valve opens and the cylinder fills with air at Ambient Pressure.
$\text{Work&space;done}=P\;dV=P_1V_1$

Stage 1 - b
• Compression Stroke: Both Valves are shut. The Pressure is raised from $\inline&space;P_1$ to $\inline&space;P_2$.

Stage 2 - b
• Delivery Stage: The Exhaust Valve opens. Air at $\inline&space;P_2$ is delivered to the receiver at Constant Pressure.

$\text{Work&space;done}=P_2V_2$

The Work Done during compression = $\inline&space;\displaystyle\frac{P_2V_2-P_1V_1}{n-1}$

The Work Done in the Compressor = $\inline&space;\displaystyle&space;\frac{n}{n-1}\;(P_2V_2-P_1V_1)$
$=\frac{n}{n-1}\left(P_2V_2&space;-&space;P_1V_1&space;\right)$
$\therefore\;\;\;\;\;\;\;\text{Net&space;Work&space;Done}=P_1V_1-P_2V_2+\frac{P_2V_2-P_1V_1}{n-1}$

Work Done In An Isothermal Compression

MISSING IMAGE!

Work Done on Air
$=P_2V_2+P_1V_1\;Ln&space;\frac{V_1}{V_2}-P_1V_1$
But,
$P_1V_1=P_2V_2$

$\therefore\;\;\;\;\;\text{Work&space;Done}=P_1V_1\;Ln\frac{V_1}{V_2}$
or,
$P_1V_1\;Ln\frac{P_2}{P_1}$

Work Done In An Adiabatic Compression

$\text{Work&space;Done}=\frac{\gamma&space;}{\gamma&space;-1}\left(P_2V_2-P_1V_1&space;&space;\right)$
$=\frac{\gamma&space;}{\gamma&space;-1}\;wR\;\left(T_2-T_1&space;&space;\right)$
$=w\;C_P\left(T_2-T_1&space;\right)=\text{The&space;Increase&space;in&space;Enthalpy}$

Alternative Forms Of The Work Done Expression

The Work Done is given by the following Expressions:
$\frac{n}{n-1}\left(P_2V_2-P_1V_1&space;\right)$
$\frac{n}{n-1}\;wR\left(T_2-T_1&space;\right)N$
where N is the number of Cycles per Min.

$\frac{n}{n-1}\;WR\left(T_2-T_1&space;\right)$
where W is the weight Handled per Min.

$\frac{n}{n-1}\;wRT_1\left(\frac{T_2}{T_1}-1\right)$
or,
$\frac{n}{n-1}\;wRT_1\left[\left(\frac{P_2}{P_1}&space;\right)^{\frac{n-1}{n}}-1&space;\right]$

A Comparison Of The Work Done With Different Indices Of Compression

MISSING IMAGE!

$P_1V_1^n=P_2V_2^n$
$V_2=V_1\left(\frac{P_1}{P_2}&space;\right)^{\frac{1}{n}}$
$\therefore\;\;\;\;\;V_2=\frac{V_1}{R^\frac{1}{n}}$
where R =Pressure Ratio.

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

For Reciprocating Compressors:

The efficiency referred to is the Isothermal case since fairly successful cooling can be achieved.
$\text{Isothermal}\;\eta&space;=\frac{\text{Work&space;Input&space;with&space;Isothermal&space;Compression}}{\text{Work&space;Input&space;required&space;with&space;actual&space;index&space;n}}$

For Rotary Compressors:

The Cooling is very difficult and Indices of less than $\inline&space;\gamma$ are never achieved. It is therefore normal to compare the Performance with the Adiabatic reversible case.
$\text{Adiabatic}&space;\eta&space;=\frac{\text{Work&space;Input&space;with&space;Adiabatic&space;Reversible&space;Compression}}{\text{Work&space;Input&space;required&space;with&space;actual&space;index&space;n}}$

The Overall Isothermal Efficiency Of The Plant:

$\text{Transmission&space;Efficiency}&space;\eta_T=\frac{\text{Work&space;In&space;put&space;at&space;Compressor&space;Drive}}{\text{BHP&space;of&space;Motor&space;Supplying&space;Power}}$
$\text{Mechanical}&space;\;\eta_M&space;\;\text{of&space;Compression}&space;=&space;\frac{\text{Work&space;Done&space;on&space;Air&space;in&space;Cylinder}}{\text{Work&space;In&space;put&space;at&space;Compressor&space;Drive}}$
$\text{Overall&space;Mechanical}\;\eta&space;=\eta&space;_T&space;-&space;\eta&space;_m$

$\text{Overall&space;isothermal}\;\eta&space;=\frac{\text{Work&space;Done&space;on&space;Air&space;in&space;Cylinder&space;with&space;isothermal&space;Compression}}{\text{BHP&space;of&space;Motor}}$
$\;=\text{Overall&space;mechanical}\;\eta&space;\times&space;\text{The&space;Isothermal}&space;\;\eta\;\text{of&space;compression}$

The Cooling Of Compressors.

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

For a Polytropic Compression.
$\text{Heat&space;Rejected}&space;=\left(\frac{\gamma&space;-n}{\gamma&space;-1}&space;\right)\times&space;\left(\text{Work&space;done&space;on&space;Air&space;during&space;the&space;compression&space;stroke}\right)$

For an Isothermal Compression.
$\text{Heat&space;Rejected}&space;=&space;\text{Work&space;done&space;on&space;Air}=&space;P_1V_1\;ln\;\frac{P_2}{P_1}$

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 $\inline&space;P\;V^{1.2}=Constant$ 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
$P\;V^{1.2}=\text{Constant}$
$\frac{P_1}{T_1^{\frac{1.2}{0.2}}}=\frac{P_2}{T_2^{\frac{1.2}{0.2}}}$
$\therefore\;\;\;\;\;T_2=T_1\left(\frac{P_2}{P_1}&space;\right)^{\frac{0.2}{1.2}}=293\left(\frac{140}{14}&space;\right)^{\frac{0.2}{1.2}$
$=429^{\circ}K$

Work input to Compressor per lb.
$=w\;R\times&space;\left(\frac{n}{n-1&space;\right)\times&space;\left(T_2-T_1&space;\right)&space;}$
$=96\times&space;\frac{1.2}{0.2}\left(429-293&space;\right)=78,300\,ft.lb.$

Heat Rejected during Compression.
$\text{Heat&space;Rejected}=\frac{\gamma&space;-n}{\gamma&space;-1}\times&space;\text{Work&space;Done&space;in&space;Compression}$
$=\frac{\gamma&space;-n}{\gamma&space;-1}\times&space;\frac{wR(T_2-T_1)}{n-1}$
$=\frac{\gamma&space;-n}{\gamma&space;-1}\times&space;\frac{78,300}{1.2}=\frac{1.4-1.2}{0.4}\times&space;\frac{78,300}{1.2}\times&space;\frac{1}{1400}$
$=&space;23.4\;CHU$
Solution
Temperature at the end of the Compression $\inline&space;429^{\circ}K$

Compressor Work input $\inline&space;78,300\,ft.lb.$

Heat Rejected during Compression $\inline&space;23.4\;CHU$