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RECIPROCATING COMPRESSORSThere are various compressor designs:Rotary vane; Centrifugal & Axial flow (typically used on gasturbines); Lobe (Roots blowers), andReciprocating.The main advantages of the reciprocating compressor are thatit can achieve high pressure ratios (but at comparatively lowmass flow rates) and is relatively cheap.

It is a piston and cylinder device with (automatic) springcontrolled inlet and exhaust valves. Delivery is usually to areceiver. The receiver is effectively a store of energy used todrive (eg) compressed air tools.

Receiver

Delivery

TDC BDCInlet

Clearance vol.

Swept vol.

2

Reciprocating compressors usually compress air but are alsoused in refrigeration where they compress a superheatedvapour (to which the gas laws strictly do not apply).

In order to be practical there is a clearance between the pistoncrown and the top of the cylinder. Air 'trapped' in this clearancevolume is never delivered, it expands as the piston movesback and limits the volume of fresh air which can be inducedto a value less than the swept volume.

The induced volume flow is an important purchasingparameter. It is called the "Free Air Delivery" (FAD), and itmeasures the capacity of a compressor in terms of the air flowit can handle. It is normally measured at standard sea level(SSL) atmospheric conditions and allows the capacities (size)of compressors to be compared.

N.B. The inducedmass per cycle must equal the deliveredmass per cycle (continuity!), although the induced anddelivered volumes will be different.

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0100200300400500600700800900

1000

0 0.2 0.4 0.6 0.8 1 1.2

41 Induction

12 Compression

23 Delivery

34 Expansion

Cycle Analysis

The cycle may be analysed as two non-flow (compression andexpansion) processes and two flow processes (delivery andinduction)

PROCESS GROSS WORK

p2V2 - p1V1n-1

p2(V2-V3)

p4V4 - p3V3n-1

p1(V4-V1)

Note that we assumepolytropic compressionandexpansion.This is because some degree of cooling is usually attemptedfor reasons we shall see later.If no cooling were attemptedn becomesg.

On p-V co-ordinates:

1

23

4

pres

sure

(kP

a)

Volume (litres)

4

The work per cycle is given by:gross work

p2V2 - p1V1n-1 p2(V2-V3)

p4V4 - p3V3n-1 p1(V4-V1)work per cycle = + + +

p4V4 - p1V1n-1

p2V2 - p3V3n-1p1(V4-V1) p2(V2-V3)

p1(V4-V1)n-1 p1(V4-V1)

p2(V2-V3)n-1 p2(V2-V3)

p1(V4-V1) {1+ } + p2(V2-V3){1+ } 1n-1 1n-1

but mass delivered = mass induced

p1(V1-V4) RT1

p2(V2-V3) RT2

=

p2(V2-V3) p1(V1-V4)T2T1

p1(V1-V4) { } [ -1] nn-1T2T1

for a polytropic process : T2T1p2p1

n-1 n

= ( ) = rpn-1 n

=

Noting that (V1-V4) is the induced volume (Vind), and p1 is theinlet pressure (pin) we may re-arrange and write:

work per cycle = pin Vind { rp -1} nn-1n-1 n

work per cycle =

+ + +=

but p1=p4 & p2=p3

work per cycle = + + +

=

\

NB Power required = work per cyclexcycles per sec

5

00.10.20.30.40.50.60.70.80.9

1

0 10 20 30 40 50

Volumetric Efficiency

The reference conditions (p & T) at which the volumetricefficiency is measured should always be quoted (it wouldnormally be SSL conditions).

[The concept ofhvol applies also to reciprocating engines.]

We have already noted that the induced volume is less thanthe swept volume. To enable this effect to be evaluated wedefine volumetric efficiency (hvol) as:

hvol = Induced volumeSwept volume

V1-V4 Vs

=

but p3V3 = p4V4nn

\ V4 = V3 rp1n

V3 is the clearance volume (Vc), and V1 = Vc + Vs

hvol = Vc + Vs - Vc rp Vs

1n

\

hvol = 1 - ( rp - 1)VcVs1n

hvol

rp

= 0.05

n = 1.27

VcVs

6

Volumetric Efficiency referred to SSL conditions.

In testing a compressor, the measured induced volume flowwill be that of the actual test inlet conditions.

It is unlikely that these inlet conditions will be SSL.

We therefore need to refer our results to SSL conditions.

SSL Inlet >>TsPs

TiPi

The mass flow of gas must be the same both at SSL (s)conditions and at Inlet (i) conditions.

ms = mi

psVsRTs

piViRTi

=

Vspi Tsps Ti

= Vi

dividing both sides by Vswept

hvol(SSL)= hvol(inlet)pi Tsps Ti

. .

. .

. .

.

(Measured).Vi

.

7

0

50

100

150

200

250

1 1.1 1.2 1.3 1.4

Compressor Efficiency

If we plot the specific work (kJ/kg delivered) against thepolytropic index n we obtain:

w = RTin { rp -1} nn-1n-1 n

wkJ/kg

n

rp=4

rp=8

Isot

herm

al

com

pres

sion

Adi

abat

ic

com

pres

sion

It is clear that the closer the compression is to isothermal theless work is required.The savings are greater at higher pressure ratios(eg above 18% @ rp=4 and 27% @ rp=8)

It is also apparent that isothermal compressionrepresentsthe ideal minimum work input for a compression process(where cooling is feasible).

Polytropic

8

We can therefore define compressor efficiency as:

Isothermal work per cycleActual work per cyclehiso =

If we recalculate the work input assuming isothermalcompression [ W12 = p1V1 ln(p1/p2) etc] it is found that:

hiso =

Note that this efficiency is known as theisothermal efficiency.

The degree of cooling possible during a single stagecompression process tends to be limited. It improves at lowspeeds but this limits compressor capacity.

One way of improving efficiency, especially at highercompression ratios and speeds, is to go tomultistagecompression with cooling of the gas between each stage.

{ rp -1} nn-1n-1 n

ln rp

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Multistage compression

To avoid unacceptable reductions in compressor capacity(RPM and volumetric efficiency) and to minimise power inputwith high compression ratios, multistaging with inter-cooling isused.The number of stages will normally be between two and four.

Stage 1 Stage 2 Stage 3

in> >out

intercoolers

Each stage may be treated as a separate compressor,however, with multistaging, all will normally rotate at the samespeed.

The volumetric efficiency of the compressor as a whole isdetermined by thefirststage.

If the inter-cooling is such that the inlet temperature to thefollowing stage(s) is the same as the inlet temperature to thefirst stage we have ideal (or perfect) inter-cooling.

Since the work input to any stage is dependent on thepressure ratio across it, it should be possible to minimise thetotal work input by the correct choice of compression ratioacross each stage.

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Optimum stage pressure ratio

Assume we havetwo stages of compression with idealintercooling and the same index of compression (andexpansion) 'n' in each stage.

Total Work per cycle

= p1in V1ind { rp1 -1} nn-1n-1 n + p2in V2ind { rp2 -1}

n-1 n

Since themass induced by the first stage must be equal tothemass induced by the second stage:

p1in V1ind R Tin

p2in V2ind R Tin=

= p1in V1ind { rp1 -1} nn-1n-1 n + { rp2 -1}

n-1 n

Total Work per cycle

If p1 is the inlet pressure and p2 the final delivery pressure,let pi= the inter-stage pressure:

rp1 =pip1 rp2 =p2pi&

If we substitute the above in the expression for Total Work anddifferentiate wrt pi, we can find pifor minimum Total Work.

pi = [p1 p2]

or rp1 = rp2 =p2p1( )

We could extend the same method to N stages with the resultthat, for minimum work input, the pressure ratio across eachstage must be the same and equal to the Nth root of theoverall pressure ratio.

rp(opt)= rp(overall)1N

\

then

whence