Lecture note compressor

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Reciprocating Air-Compressor

11 November 2013 Reciprocating Air-Compressor 1

Introduction

• The function of a compressor is to take a definitequantity of fluid and deliver it at a requiredpressure.

• Usually a gas, and most often air.

• To do this, mechanical work must be supplied tothe air compressor, by an electric motor.

• The air compressor does work on the air, calledindicated work.


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Types of Air-Compressor

Can be classified into:

a) Positive displacement

b) Dynamic impulse

Positive displacement air compressors:

a) Reciprocating

b) Rotary or Screw type

Dynamic impulse:

a) Centrifugal

b) Mixed-flow

c) Axial flow11 November 2013 Reciprocating Air-Compressor 3

Types of Air-Compressor


http://www.pneumofore.com/img/technology/operation.swf

Positive displacement type air compressor

Dynamic impulse type air compressor

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Reciprocating

• Low mass flow-rate due to pulsating operation

• High pressure ratio

• High efficiency

• Bigger size and heavy

• Complex mechanical design


Rotary or Screw type

• High mass flow-rate due to

continuous operation

• Low pressure ratios

• Low efficiency

• Smaller in size and light

• Simpler mechanical design

(Out of the scope of SKMM2423)


Screw type

compressor

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The Analysis Objectives

• To calculate air pressure and temperature before

and after compression process

• To calculate indicated work/power

• To obtain compressor efficiency

• To calculate free air delivery rate ( )


smV

skg

m3

or ••

Reciprocating Compressors

Basic Components

A reciprocating air-compressor typically comprises of:

a) A cylinder

b) A piston – moving inside the cylinder

c) A connecting rod arrangement

d) Intake and exhaust valves

Reciprocating air-compressors can be classified into single-acting or

double- acting type.

11 November 2013Reciprocating Air-Compressor

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Receiverpressure

Atmosphericpressure

Single-acting reciprocating air-compressor

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• In a double-acting air-

compressor, the induction

and delivery of

compressed air occurs

during every strokes of the

piston.

• The intake and exhaust

valves are of spring-loaded

type. They operate

automatically.


Double-acting reciprocating

air-compressor

Animated Double Acting Compressor Cylinder.mp4


Assumptions

The working fluid is assumed as a perfect gas and P-v-T can

be calculated by using simple equation of state:

a)

b)

Usually, these assumptions are used to calculate estimate

pressure, P, volume, V, and temperature, T, of the working

fluid.

1KT

PVormRTPV ==

2KPV n =


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Lecture Structure

1. Compressor without clearance volume

2. Compressor with clearance volume

3. Multistage compressors


Air-Compressors without Clearance

The Cycle of Operation

• The cycle of operation of a reciprocating

air-compressor is best shown on a

pressure-volume (p-V) diagram.

• It is known as an indicator diagram for

the compressor.

• The cycle comprises of three processes:

d - a: An induction stroke

a - b: A compression stroke

b - c: A delivery stroke


A pressure-volume (p-V) diagram

for a reciprocating compressor.

Clearance volume is neglected.

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Description of the Processes

d – a: The induction stroke

Intake valve opens, while exhaust valve

closed. Atmospheric air is drawn into the

cylinder at constant pressure p1 and

temperature T1. Ideally, there is no heat loss

to the surrounding from the air.

a – b: The compression stroke

Both intake and exhaust valves closed. The

air is compressed according to a polytropic

law pVn = constant. Its pressure is increased

from p1 to p2. The temperature is also

increased to T2.


A pressure-volume (p-V) diagram for a

reciprocating compressor.



b – c: The delivery stroke

The intake valve closed while

the exhaust valve opens. The

compressed air is pushed out

of the cylinder at constant

pressure p2 and temperature

T2. There is no heat loss from

the air to the surroundings.





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Analysis of the Cycle

The indicated work per cycle:

The area under the p-V diagram

represents the net or indicated work

done on the air per cycle.





n = index of polytropic process

( ) ( )1121

12112

/

aVpb

Vpn

n

aVpb

Vpn

aVpb

Vp

adOfareabcOeareaabefarea

dcbaareacycleworkIndicated

−−

=

−+−

−=

−+=

−−−=


Assuming the air as a perfect gas,

2211 mRTVpandmRTVp ba ==

where m is the mass of air induced and delivered per cycle, R is the

universal gas constant, where R = 0.287 kJ/kgK.

Substituting,

( ) ( )

( )311

/

or

21

/

1

21

12

−

−=

−−

=

T

TmRT

n

ncycleworkIndicated

TTmRn

ncycleworkIndicated


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So, indicated work/cycle can also be expressed as:

−

−=

−

11

/

1

1

21

n

n

p

pTRm

n

ncycleworkIndicated (4)

or

−

−=

−

11

/

1

1

21

n

n

ap

pVp

n

ncycleworkIndicated (5)


Indicated Power:

The indicated power is the work done on the air per unit time.

The mass flow per unit time, is often used to compute the work

done/time or indicated power. Thus, all equations (2) to (4) must

be replaced with to obtain indicated power. As an example,


•

m

•

m

( ) (5) 1

12 TTRmn

npowerIndicated −

−=

•


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Mechanical Efficiency, ηηηηmech :

The mechanical efficiency ηmech is defined as,

where,

Shaft power = Indicated Power + Friction Power

Note: Shaft power is the power supplied by the electric motor to the

compressor. Friction power is the power loss to overcome friction in

moving mechanical parts.

powerShaft

powerIndicatedmech =η


(6)

(7)



Motor Efficiency:

The motor efficiency is defined as,

(8)

where input power is the electric power supplied to

the electric motor.

powerInput

powerShaftmotor =η


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Another perfect gas equation:

2

2

1

1

T

Vp

T

Vp ba =

and For polytropic process:

n

b

n

a VpVp 21 =

Therefore,

( )n

b

a

n

n

n

b

a

V

V

T

T

p

p

T

T

p

p

V

V

−

−

=

=

=

1

1

2

1

1

2

1

2

1

1

2


n = index of polytropic process

The Relationship between States


Free Air BeforeCompression

AfterCompression

o

oo

T

Vp=

1

1

T

Vp a =2

2

T

Vp b

or

o

oo

RT

Vp=

1

1

RT

Vp a =2

2

RT

Vp b

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= =m0 m1 m2

Vo Va Vb

Vo at po To V1 at p1 T1 V2 at p2 T2

n

b

n

a VpVp 21 =

The Relationship between States

Example 1

A single reciprocating compressor takes 1 m3 air per

minute at 1.013 bar and 15o C and delivers it at 7 bar.

Assuming that the law of compression is pV1.35 =

constant, and that clearance is negligible, calculate

the indicated power.


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Example 2

If the compressor of Example 1 is to be driven at 300

rev/min and is a single-acting, single cylinder

machine, calculate the cylinder bore required,

assuming a stroke to bore ratio of 1.5/1. Calculate

the power of the motor required to drive the

compressor if the mechanical efficiency of the

compressor is 85% and that of the motor

transmission is 90%.


Condition for Minimum Work


It is desirable to have work done on the air (indicated power) to be as

minimum as possible.

Recall that the indicated work equals the area under the indicator (p-V)

diagram. The indicated work is minimum if the area under the diagram is

smallest.

The height of the p-V diagram is fixed by

the required pressure ratio. The length

d – a is fixed by the stroke of the piston.

Thus, the area under the diagram will be

smallest only if the compression process

is carried out in reversible isothermal

process according to a law,

pV = constant

where n = 1. Possible compression process

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The area under the diagram will be biggest only if the

compression process is carried out in an isentropic

process where n = γ = 1.4. It also can be called reversible

adiabatic or entropy constant.

The area under the diagram will be moderate when the

compression process is carried out in a polytropic

process. Where n = n.



Possible compression process


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The indicated work when the air is compressed isothermally,

is given by:


Area ab1cda = area ab1efa + area b1c0eb1 – area ad0fa

=

Also, pV = constant or p1Va = p2Vb1

Therefore,

Indicated work per cycle =

or = (9)

or = (10)

Possible compression process

abb VpVpp

pVp 112

1

2

12 ln −+

1

212 ln

p

pVp b

1

21 ln

p

pVp a

1

2lnp

pmRT


Note: When m and Va are the mass and volume of

air induced per unit time, Eqs. (9) and (10) give

the isothermal power.


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Isothermal Efficiency:

By definition, based on the indicator diagram,

powerIndicated

powerIsothermalefficiencyIsothermal =


(11)


Calculate the isothermal efficiency of the

compressor given in Example 2.


Example 3

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Air-Compressors with Clearance

What is clearance volume?

• It is a spacing between the top of

the piston and the valve’s heads

when the piston is at the end of

the delivery stroke.

• Good quality machines has a

clearance volume of about 6%.

But compressors with clearance

of 30 – 35% are also common.


Clearance volume

End of delivery stroke

Why is it required?

Clearance volume is required for the following

reasons:

• Give a mechanical freedom to the moving parts

• Reduce noise and vibration during operation

• Prevent damage to moving components



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Processes

c – d: Expansion process

The piston begins the induction stroke.

The compressed air occupying the

clearance volume Vc expands according

to the polytropic law of pVn = const., until

the pressure and temperature fall to p1

and T1, respectively. Ideally, there is no

heat transfer from the air to the

surroundings.

d – a: Induction process

The inlet valve opens. Fresh atmospheric air is induced into the cylinder

at constant pressure p1 and temperature T1. The volume of air induced is

(Va – Vd). Ideally, there is no heat transfer from the air to the surroundings.

Ideal indicator diagram


a – b: Compression process

Both valves closed. The induced air is

compressed according to the polytropic

law of pVn = const., until the pressure

and temperature increases to p2 and T2,

respectively. Ideally, there is no heat

transfer from the air to the surroundings.

b – c: Delivery process

The exhaust valve opens. The compressed air is delivered out of the

cylinder at constant pressure p2 and temperature T2. Ideally, there is

no heat transfer from the air to the surroundings.




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Note: At the end of the delivery stroke, the clearance

volume Vc is filled with compressed air at pressure p2 and

temperature T2.

The cycle is repeated.




The relationship between states

Before this, the relationship between states for air-compressors

without clearance is given as follows,

Free Air BeforeCompression

AfterCompression

o

oo

T

Vp=

1

1

T

Vp a =2

2

T

Vp b

or

o

oo

RT

Vp=

1

1

RT

Vp a =2

2

RT

Vp b


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However, when clearance volume is considered, it becomes,

( ) ( )2

2

1

1

T

VVp

T

VVp

T

Vp cbda

o

oo −=

−=

or

( ) ( )2

2

1

1

RT

VVp

RT

VVp

RT

Vp cbda

o

oo −=

−=

mo = m1 = m2




Compressor capacity is the rate of volume of air delivered, measured at

ambient conditions and is often expressed as Free Air Delivery (F.A.D),

m3/min.

If the F.A.D, is measured at ambient conditions, i.e., p = po and T = To, with

compressor speed, N (rpm), compressor type, b and number of cylinders,

e, the actual volume delivered per cycle is,

Note:• If single-acting compressor, b = 1• If double- acting compressor, b = 2

•

oV


exbxN

VV o

o

•

=

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Therefore, the relationship between states for air-

compressor with clearance becomes:

( ) ( )o

oo

o

oocbda

T

exbxN

Vp

T

Vp

T

VVp

T

VVp

==−

=−

•

2

2

1

1




Indicated Power:

The indicated work/cycle is given by the area enclosed by

the p-V diagram.

Indicated work = area abcd

= area abef – area cefd


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Comparison between cycles with & without

clearance


Without clearancevolume

With clearancevolume


Using the definition given in Eq. (2), the indicated power/cycle,


( ) ( )121211

TTRmn

nTTRm

n

npowerIndicated da −

−−−

−=

••

( )12

1TTRmm

n

npowerIndicated da −

−

−=

••

( )121

TTRmn

npowerIndicated −

−=

•

(12)

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Where is the mass induced per unit time = •

m

or

−

−=

−•

11

/

1

1

21

n

n

p

pTRm

n

ncyclepowerIndicated (13)

••

− da mm


A single stage, double-acting air

compressor is required to deliver 14 m3

of air per minute measured at 1.013 bar

and 15oC. The delivery pressure is 7 bar

and the speed 300 rev/min. Take the

clearance volume as 5% of the swept

volume with compression and re-

expansion index of n = 1.3. Calculate the

swept volume of the cylinder, the

delivery temperature, and the indicated

power.


Vc = 0.05 Vs

Example 4

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Volumetric Efficiency, ηηηηv:

Volumetric efficiency, ηv is defined as,

ηv = the volume of air delivered, measured

at the free air pressure and temperature,

divided by the swept volume of the

cylinder


(14)


We consider two different cases as follows:

Case 1: When p1 = po, and T1 = To, from Eq. (14),

Where, swept volume = Vs.

( )s

dav

V

VV

volumeSwept

inducedVolume −==η




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From the relationship established before, if p1 = po and

T1 =To, the volume induced becomes,


( )exbxN

VVVV o

oda

•

==−


Therefore, volumetric efficiency, ηv becomes,

s

ov

V

V=η (15)

Or,

ηv = (Va – Vd )/ Vs (16)

Alternatively, we have

Volume induced = Va – Vd = Vs + Vc – Vd




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The pressure-volume relationship for the polytropic process,


=

n

c

d

p

p

V

V1

1

2 i.e.

=

n

cdp

pVV

1

1

2



Therefore,

Volume induced =

=


n

ccsp

pVVV

1

2

−+

−

− 1

1

2n

csp

pVV


Substituting the above expression into Eq. (16), we have:

s

n

cs

s

da

vV

p

pVV

V

VV

−

−

=−

=

1

1

1

2

η

−

−= 11

1

1

2n

s

cv

p

p

V

Vη (17)

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Case 2: When p1 < po, and T1 > To, from Eq. (14),

Where, swept volume = Vs


( )s

dav

V

VV

volumeSwept

inducedVolume −==η


For this case, if p1 < po and T1 >To, the volume inducedbecomes,


( )o

ooda

Tp

TVpVV

1

1=−


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Therefore

(18)

Where


exbxN

VV o

o

•

=

o

o

s

ov

T

T

p

p

V

V 1

1

⋅⋅=η


Example 5 (Final Exam, April 2001)

As a mechanical engineer, you are requested to do analysis ondesigning an air reciprocating compressor. The required parametersare as follows: the pressure and temperature of free air conditions are1.013 bar and 32oC respectively. The pressure ratio is 20:1, thecompression and expansion index, n = 1.32, R = 0.287 kJ/kgK. Byassuming the initial conditions are the same with the free airconditions, for each kg of air calculate,

i. the indicated work if the compression process is done at singlestage, in kJ/kg.

If the indicated power of the compressor is fixed to 10 kW and run at1200 rpm and the clearance volume is 5% of the swept volume,calculate,

ii. the swept volume of the cylinder, in m3,

iii. the volumetric efficiency of the compressor.


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Example 6

A single-stage, double acting air compressor

has a FAD of 14 m3/min measured at 1.013

bar and 15oC. The pressure and

temperature in the cylinder during

induction are 0.95 bar and 32oC. The

delivery pressure is 7 bar and the index of

compression and expansion, n, is equal to

1.3. Calculate the indicated power required

and the volumetric efficiency. The clearance

volume is 5% of the swept volume.


Example 7

A four-cylinder single acting compressor with

cylinder bore 100 mm and stroke 150 mm

runs at 400 rev/min. If the volumetric

efficiency is 70%, find the mass of air

delivered per hour. The ambient pressure and

temperature are 1 bar and 25oC respectively.


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Multistage Compression

What is Multistage Compression?

The air is compressed in more than onecylinder (or stage) to the desired deliverypressure p2, before being delivered for usage.

Need for Multistage Compression

When the delivery pressure is increased to p3,

the volume of the fresh air induced is reduced

from (Va – Vd) to (Va – Vd’), and so on.

Since the volumetric efficiency is given by

ηv = (Va – Vd )/ Vs

the volumetric efficiency decreases with

increasing delivery pressure.


The effect of increasing the delivery

pressure on the volume of fresh air

induced

This situation can be improved by performing

multistage compression process.

Observation

After the first stage compression, the air is passed

into a smaller cylinder, in which it is further

compressed to desired final pressure.

The cycle assumes that the delivery process of the

first stage and the induction process of the second

stage take place at the same pressure pi.

Advantage

• Each cylinder works with lower pressure ratio.

Thus the operational safety of the compressor is

improved.

• The overall volumetric efficiency, ηv increases.


Indicator diagram for

2-stage compression


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Indicated Power:

The indicated power for stage 1

The indicated power for stage 2

−

−=

−••

11

1

1

n

n

iaip

p

pTRm

n

nW


−

−=

−••

11

1

2''

n

n

i

aip

p

pTRm

n

nW

Indicator diagram for 2-stage

compression


Intercooling Process Between Stages

• With multistage compression, the air can be cooled as it isbeing transferred from one cylinder to the next, bypassing it through an intercooler.

• The process of cooling the air is called the intercoolingprocess.

• With intercooling process, a saving in the indicated workcan be achieved. Thus the power supplied to thecompressor can be reduced.



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Intercooling Process Between Stages

Two-stage compression with complete intercooling

Complete Intercooling

When the intercooling process is complete, the inlet air temperature

for first- and second-stage compressors is equivalent i.e.,

Ta’ = Ta

The indicator diagram shows that the two compressor stages share

a common intermediate pressure pi. In real machine, there will be a

small pressure drop between the two stages.



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The indicated work/cycle saved with

intercooling

The T-s diagram for a complete

intercooling



The shaded area representsthe amount of indicated workper cycle that can be saved ifthe intercooling is complete.


The indicated work/cycle

saved with intercooling


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Delivery Temperature for Complete Intercooling

The delivery temperature after each stage are

After 1st stage (19)

After 2nd stage (20)

( )

( )n

n

i

n

n

ii

p

pTT

p

pTT

1

212

1

1

1

−

−

=

=



Ti

Ideal Intermediate Pressure

• The value of intermediate pressure pi affects the amount of indicatedwork/cycle for multistage air compressor.

• The ideal value of intermediate pressure pi is one that gives equalpressure ratio for each stage of the compressor. For two-stagecompression, this means

(21)

or

Hence, (22)

i

i

p

p

p

p 2

1

=

21

2pppi ⋅=


21 pppi ⋅=


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Ideal Intermediate Pressure

In general, for z stages of compressions, with completeintercooling, the pressure ratio for each stage can beexpressed as below,

Where, is the final delivery pressure.

zz

p

p1

1

1

+

1+zp




Conditions for Minimum Work

The indicated work/cycle in a multistage

compression will be minimum when the following

conditions are met:

a) The intercooling is complete, i.e. when Ta’ = Ta

b)The intermediate pressure is the ideal one.


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Total Minimum Indicated Power

In general, if there are z stages of compression, the total minimum

indicated power can be determined using the relation:

−

−=

−

+••

11

11

1

11

n

n

zz

z

ip

p

pTRm

n

nzW (23)

And the pressure ratio of each stage is given by

zz

z

z

p

p

p

p

p

p

p

p1

1

11

2

3

1

2

=

=== ++

⋯ (24)



Total Minimum Indicated Power

T, p and V relationships

−

+

=

n

n

zz

p

p

T

T11

1

1

1

2

+

=

nzz

c

d

p

p

V

V11

1

1

(25)

(26)


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In a single-acting, two-stage reciprocating air

compressor 4.5 of kg/min are compressed

from 1.013 bar and 15oC through a pressure

ratio of 9 to 1. Both stages have the same

pressure ratio, and the law of compression and

expansion in both stages is pV1.3 = constant. If

inter-cooling is complete, calculate the

indicated power and the cylinder swept

volumes required. Assume that the clearance

volumes of both stages are 5% of the swept

volumes and that the compressor runs at 300

rev/min.


Example 8

A three-stage, single-acting air compressor running in an

atmosphere at 1.013 bar and 15oC has a free air delivery of

2.83 m3/min. The suction pressure and temperature are 0.98

bar and 32oC respectively. Calculate the indicated power

required, assuming complete inter-cooling, n = 1.3, and that

the machine is designed for minimum. The delivery pressure

is to be 70 bar.


Example 9

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Example 10 (Final Exam, April, 2001)

As a mechanical engineer, you are requested to do analysis on designing

an air reciprocating compressor. The required parameters are as follows:

the pressure and temperature of free air conditions are 1.013 bar and

32oC, respectively. The pressure ratio is 20:1, the compression and

expansion index, n = 1.32, R = 0.287 kJ/kgK. By assuming the initial

conditions are the same with the free air conditions, for each kg of air

calculate,

i. the indicated work if the compression process is done in two stage

(kJ/kg) with complete inter-cooling and optimum pressure ratio.

If the indicated power of the compressor is fixed to 10 kW and run at 1200

rpm and the clearance volume is 5% of the swept volume, calculate,

ii. the swept volume of the cylinder (m3),

iii. the volumetric efficiency of the compressor.



Energy Balance for a 2-stage Compressor With Intercooling

A steady-flow energy equation (SFEE) can be used to perform an energy

balance analysis on a 2-stage compressor with intercooling.


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a) For the LP stage, the heat loss rate:

(27)( )1TTcmWQ ipLL −+=•••




b) For the intercooler, the cooling rate:

(28)

Note: Eq. (28) only valid for complete intercooling

( )1TTcmQ ipi−=

••


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( )12 TTcmWQ pHH−+=

•••


c) For the HP stage, the heat loss rate:

(29)

With complete intercooling and if the compressor is designed for

minimum work, the indicated power is given by,

( )121

TTRmn

nWW HL −

−==

•••(30)


Example 11

A two stage, single acting compressor handling 12 m3/min ofFAD measured at 1.013 bar and 30oC. The air is compressedfrom the induction conditions of 1 bar and 28oC to thedelivery pressure of 15 bar through a polytropic process ofpV1.32 = constant. The clearance volume is equivalent to 10%of the swept volume in each cylinder, and the compressorspeed is 600 rpm.

(a) If equal pressure ratios in each stage with completeinter-cooling, calculate the total indicated power andthe volumetric efficiency for the first stage,

(b) If the intercooler cools the air to 50oC, calculate theindicated power required, and the heat that has beenrejected.


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A two-stage single acting reciprocating air compressor

with inter-cooling is designed such that the total

compressor work is minimum in which, the inter-

cooling process is complete and the pressure ratio for

both stages is the same. After some time, the intercooler

has performed poorly due to fouling. Consequently, the

inter-cooling process can only accomplish 75% of design

rate of cooling. The compressor is required to deliver air

at rate of 4.6 kg/min from 0.98 bar and 30oC through an

overall pressure ratio of 12. The compression and re-

expansion index 1.25 and the clearance volume is 5% of

the swept volume. For air, R = 0.287 kJ/kgK. If the free

air conditions are 1.013 bar and 25oC and the

compressor runs at 750 rev/min, determine,

i) the indicated power, kW;

ii) the free air delivery (m3/min)

iii) the cylinder swept volumes for each cylinder, (cm3).


Example 12 (Final Exam, October 2004)

Lecture note compressor

Education

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