Lecture note compressor
description
Transcript of Lecture note compressor
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Reciprocating Air-Compressor
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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
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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
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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)
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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 ( )
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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.
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Receiverpressure
Atmosphericpressure
Single-acting reciprocating air-compressor
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Reciprocating Compressors
• 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.
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Double-acting reciprocating
air-compressor
Animated Double Acting Compressor Cylinder.mp4
Reciprocating Compressors
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
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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
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A pressure-volume (p-V) diagram
for a reciprocating compressor.
Clearance volume is neglected.
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Air-Compressors without Clearance
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.
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A pressure-volume (p-V) diagram for a
reciprocating compressor.
Clearance volume is neglected.
Air-Compressors without Clearance
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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A pressure-volume (p-V) diagram for a
reciprocating compressor.
Clearance volume is neglected.
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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.
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A pressure-volume (p-V) diagram for a
reciprocating compressor.
Clearance volume is neglected.
n = index of polytropic process
( ) ( )1121
12112
/
aVpb
Vpn
n
aVpb
Vpn
aVpb
Vp
adOfareabcOeareaabefarea
dcbaareacycleworkIndicated
−−
=
−+−
−=
−+=
−−−=
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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
Analysis of the Cycle
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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)
Analysis of the Cycle
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,
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•
m
•
m
( ) (5) 1
12 TTRmn
npowerIndicated −
−=
•
Analysis of the Cycle
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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
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(6)
(7)
Analysis of the Cycle
Analysis of the Cycle
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
Analysis of the Cycle
n = index of polytropic process
The Relationship between States
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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%.
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Condition for Minimum Work
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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.
Condition for Minimum Work
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Possible compression process
Condition for Minimum Work
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The indicated work when the air is compressed isothermally,
is given by:
Condition for Minimum Work
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
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Note: When m and Va are the mass and volume of
air induced per unit time, Eqs. (9) and (10) give
the isothermal power.
Condition for Minimum Work
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Isothermal Efficiency:
By definition, based on the indicator diagram,
powerIndicated
powerIsothermalefficiencyIsothermal =
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(11)
Condition for Minimum Work
Calculate the isothermal efficiency of the
compressor given in Example 2.
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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.
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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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Air-Compressors with Clearance
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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
Air-Compressors with Clearance
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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Ideal indicator diagram
Air-Compressors with Clearance
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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.
Ideal indicator diagram
Air-Compressors with Clearance
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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
Air-Compressors with Clearance
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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
Ideal indicator diagram
Air-Compressors with Clearance
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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
Air-Compressors with Clearance
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
Air-Compressors with Clearance
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Analysis of the Cycle
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
Air-Compressors with Clearance
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Comparison between cycles with & without
clearance
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Without clearancevolume
With clearancevolume
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Using the definition given in Eq. (2), the indicated power/cycle,
Air-Compressors with Clearance
( ) ( )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
Air-Compressors with Clearance
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.
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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
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(14)
Air-Compressors with Clearance
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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Air-Compressors with Clearance
Ideal indicator diagram
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From the relationship established before, if p1 = po and
T1 =To, the volume induced becomes,
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( )exbxN
VVVV o
oda
•
==−
Air-Compressors with Clearance
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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Ideal indicator diagram
Air-Compressors with Clearance
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The pressure-volume relationship for the polytropic process,
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=
n
c
d
p
p
V
V1
1
2 i.e.
=
n
cdp
pVV
1
1
2
Ideal indicator diagram
Air-Compressors with Clearance
Therefore,
Volume induced =
=
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n
ccsp
pVVV
1
2
−+
−
− 1
1
2n
csp
pVV
Air-Compressors with Clearance
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
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( )s
dav
V
VV
volumeSwept
inducedVolume −==η
Air-Compressors with Clearance
For this case, if p1 < po and T1 >To, the volume inducedbecomes,
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( )o
ooda
Tp
TVpVV
1
1=−
Air-Compressors with Clearance
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Therefore
(18)
Where
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exbxN
VV o
o
•
=
o
o
s
ov
T
T
p
p
V
V 1
1
⋅⋅=η
Air-Compressors with Clearance
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.
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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.
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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.
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Indicator diagram for
2-stage compression
Multistage 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
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−
−=
−••
11
1
2''
n
n
i
aip
p
pTRm
n
nW
Indicator diagram for 2-stage
compression
Multistage 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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Multistage Compression
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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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Multistage Compression
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Complete Intercooling
The indicated work/cycle saved with
intercooling
The T-s diagram for a complete
intercooling
Multistage Compression
Complete Intercooling
The shaded area representsthe amount of indicated workper cycle that can be saved ifthe intercooling is complete.
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The indicated work/cycle
saved with intercooling
Multistage Compression
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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
−
−
=
=
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Multistage Compression
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 ⋅=
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21 pppi ⋅=
Multistage Compression
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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
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Multistage Compression
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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.
Multistage Compression
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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)
Multistage Compression
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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)
Multistage Compression
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37
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.
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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.
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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.
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11 November 2013 Reciprocating Air-Compressor 76
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.
Multistage Compression
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11 November 2013 Reciprocating Air-Compressor 77
Energy Balance for a 2-stage Compressor With Intercooling
a) For the LP stage, the heat loss rate:
(27)( )1TTcmWQ ipLL −+=•••
Multistage Compression
11 November 2013 Reciprocating Air-Compressor 78
Energy Balance for a 2-stage Compressor With Intercooling
b) For the intercooler, the cooling rate:
(28)
Note: Eq. (28) only valid for complete intercooling
( )1TTcmQ ipi−=
••
Multistage Compression
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40
11 November 2013 Reciprocating Air-Compressor 79
( )12 TTcmWQ pHH−+=
•••
Energy Balance for a 2-stage Compressor With Intercooling
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)
Multistage Compression
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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41
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).
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Example 12 (Final Exam, October 2004)