REFRIGERATION AND AIR CONDITIONING COURSE LECTURES - Week 2.pdf

REFRIGERATION AND AIR CONDITIONING COURSE LECTURES

1. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
CCSS,NGSS
Refrigeration &
Air-Conditioning
(ME-319)

2. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
2
Ideal and Actual Vapor-Compression
Refrigeration Cycles

3. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Refrigeration and Air Conditioning
3
❖Refrigerationisdefinedastheprocessofextractingheatfroma
lower-temperatureheatsource,substance,orcoolingmediumand
transferringittoahigher-temperatureheatsink.
❖Refrigerationsystemsarealsousedextensivelyforprovidingthermal
comforttohumanbeingsbymeansofairconditioning.
❖AirConditioningreferstothetreatmentofairsoastosimultaneously
controlitstemperature,moisturecontent,cleanliness,odourand
circulation,asrequiredbyoccupants,aprocess,orproductsinthe
space.
❖Arefrigerationsystemisacombinationofcomponentsandequipment
connectedinasequentialordertoproducetherefrigerationeffect.

4. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Refrigeration and Air Conditioning
Refrigerator Air-Conditioner

5. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
5
❖Thecoolingcapacityofarefrigerationsystem—thatis,therateofheatremovalfrom
therefrigeratedspace—isoftenexpressedintermsoftonsofrefrigeration.
❖Thecapacityofarefrigerationsystemthatcanfreeze1ton(2000���)ofliquidwater
at0℃(32℉)intoiceat0℃in24hissaidtobe1ton.
One ton of refrigeration is the amount of heat required to melt 1 ton of ice in a 24-hour
period.
Refrigeration and Air Conditioning
Rating Refrigeration Equipment

6. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
6
Energyrequiredtomelt1��ofice=144���
Energyrequiredtomelt2000��ofice=(2000∗144)���=288000���
1tonofrefrigeration(TR):
Energyrequiredtomelt2000��oficein24hr=288000/24���/ℎ
1TR=12000BTU/hor200BTU/min1tonofrefrigeration(inSI
Units):(1���=1.055��)
�????????????=���????????????/�??????�=�.����????????????
Refrigeration and Air Conditioning
Rating Refrigeration Equipment

7. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
7
Refrigerators and Heat Pumps
Refrigerators
Heat Pumps
���
�=
DesiredOutput
����??????��������
=
����??????��������
���������
=
�
??????
�
���,??????�
���
??????�=
DesiredOutput
����??????��������
=
����??????��������
���������
=
�
??????
�
���,??????�
Coefficient of Performance
���
�+1=
�
??????
�
??????−�
??????
+1
���
�+1=
�
??????+�
??????−�
??????
�
??????−�
??????
���
�+1=
�
??????
�
??????−�
??????
���
�+1=���
??????�
∴�
���,??????�=�
??????−�
??????
Theobjectiveofarefrigeratoristoremoveheat(�
??????)from
thecoldmedium;theobjectiveofaheatpumpistosupply
heat(�
??????)toawarmmedium

8. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
8
Reversed Carnot Cycle
Carnot
Refrigerator
ReversibleIsothermal Heat Addition (process 1-2, T
L=constant)
ReversibleIsentropic Compression(process 2-3, temperature risesfrom T
Lto T
H)
ReversibleIsothermal Heat rejection (process 3-4, T
H=constant)
ReversibleIsentropicExpansion (process 4-1, temperature drops from T
Hto T
L)
���
�??????����,�=
�
??????
�
??????−�
??????
���
�??????����,??????�=
�
??????
�
??????−�
??????

9. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
9
Reversed Carnot Cycle
Carnot
Refrigerator
Process 1-2
andProcess
3-4
Possible to be
achieved
practically
Isothermal Processes heat
transfer processes are not
difficult to achieve since
maintaining a constant
pressure automatically fixes
the temperature.
Process 2-3
Cannot be
approximated
closely in
practice
Requires compression of a
liquid-vapor mixture, which
requires a compressor that
will handle two phases.
Process 4-1
Cannot be
approximated
closely in
practice
Requires expansionof high-
moisture content in turbine.

10. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
10
Ideal and Actual Vapor-Compression
Refrigeration Cycles

11. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
11
Ideal Vapor-Compression Refrigeration Cycle
Ideal Vapor-Compression
Refrigeration Cycle
Removed the impracticalities of reversed Carnot cycle
❖Completevaporizationoftherefrigerantbeforecompression
❖Replacingtheturbinewithathrottlingdevicesuchasan
expansionvalveorcapillarytube
•Isentropic compression in a compressor (process 1-2)
•sat. vapor to superheated vapor
•Constant-pressure heat rejection in a condenser (process 2-3)
•superheated vapor to sat. liquid
•Throttling (irreversible) in an expansion device (process 3-4)
•sat. liquid to sat. mixture
•Constant-pressure heat absorption in an evaporator(process 4-1)
•sat. mixture to sat. vapor

12. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
12
Ideal Vapor-Compression Refrigeration Cycle
Household Refrigerator (example)
(Heat Rejection
to Environment)
(Heat Absorbed From
Refrigerated Space)
OnT-sDiagram:
•Theareaundertheprocesscurve4-1representsheat
absorbedbyrefrigerantintheevaporator.
•Theareaundertheprocesscurve2-3representsheat
rejectedbyrefrigerantinthecondenser.
Replacingtheexpansion
valvebyaturbineisnot
practical,alsotheadded
benefitscannotjustify
theaddedcostand
complexity.

13. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
13
Ideal Vapor-Compression Refrigeration Cycle
Coefficient of Performance
�−ℎdiagramofanidealvapor-
compressionrefrigerationcycle
���
�=
����??????��������
���������
=
�
??????
�
���,??????�
=
ℎ
1−ℎ
4
ℎ
2−ℎ
1
���
??????�=
����??????��������
���������
=
�
??????
�
���,??????�
=
ℎ
2−ℎ
3
ℎ
2−ℎ
1

14. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem #1
Anidealvapor-compressionrefrigerationcycle
thatusesrefrigerant-134aasitsworkingfluid
maintainsacondenserat800kPaandthe
evaporatorat-12°C.Determinethissystem’s
COPandtheamountofpowerrequiredtoservice
a150kWcoolingload.

15. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
�
Problem #1
Solution
Required
(a)COP
(b)Work Inputin
COP
W
Formula( )
21in
Wmhh=− L
in
Q
COP
W
=
Solution
State-11
12TC
Sat. Vapor
=−

 1 @12
1 @12
391.46/
1.7348/
vapor C
vapor C
hh kJkg
ss kJkgK
−
−
= =
= =
State-22
2
21
800
421.87/
PkPa
h kJkg
ss
= 
= 
=

P s h
800 kPa
1.7144415.58
1.7348421.87
1.7437424.61

16. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem #1
Solution
State-33
3
800
243.58/
.
PkPa
h kJkg
SatLiquid
= 
= 

State-443
243.58/hh kJkg== ( )
14L
Qmhh=−
Energy Balance –Evaporator ( )
14
L
Q
m
hh
=
− ( )
150
1.014/
391.46243.58
m kgs= =
−
Part (a)( )
21in
Wmhh=− ( )( )1.014421.87391.46
30.84
in
in
W
W kW
= −
=
Part (b)L
in
Q
COP
W
= 150
4.86
30.84
COP==

17. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
17
Problem #2
AplantusingR134aevaporatesat0℃andcondensesat35℃.
Therefrigerationcapacityoftheplantis352kWanditoperates
ontheidealvaporcompressioncycle.
Determinethefollowing:
a)thedrynessfractionatentrytotheevaporator,
b)therefrigerationeffect(inkJ/kg)
c)themassflowrateofrefrigerant,
d)thevolumetricflowrateatthesuctionstate,
e)thecompressorpower,
f)therateofheatrejectionatthecondenser,and
g)COPoftheplantanditsrefrigeratingefficiency.

18. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
18
Problem #2
Condenser and evaporator operated at constant pressures �
1=�
4;�
2=�
3
Point 1: �
1=0°�; Saturated Vapor=> ℎ
1=398.60��/��
�
1=1.7271��/��
Point 2:�
2=�
1; �
2=35°�; Superheated State
=> �
2=
0.86263+0.91185
2
=0.88724??????��
Solution
P = 1.000 MPa
P = 0.800 MPa
�

19. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
19
Problem #2
Solution
s h
1.7139419.99
1.7271424.19
1.7482430.91
s h
1.7144415.58
1.7271419.49
1.7437424.61
P s h
1.0001.7271424.19
0.887241.7271421.54
0.8001.7271419.49
P=1.000MPa P=0.800MPa
ℎ
2=421.54��/��
�

20. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
20
Problem #2
Point 3: �
3=35°�; Saturated Liquid=> ℎ
3=249.01��/��
Solution
Point 4:ℎ
4=ℎ
3; ℎ
4=249.01��/��
�
Part (a) -the dryness fraction at entry to the evaporator
ℎ=ℎ
�+�ℎ
��
�=0.252
ℎ=ℎ
4,ℎ
�=200��/��;ℎ
��=(398.6−200)��/��
Part (b) the refrigeration effect (in kJ/kg)
���.??????�����=ℎ
1−ℎ
4=398.6−249.01
= 149.59 kJ/kg

21. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem #2
Solution
Part (c) -the mass flow rate of refrigerant
�=�(ℎ
2−ℎ
1)= (2.353)(421.54 –398.6)
�= 54 kW
Part (d) -the volumetric flow rate at the suction state
�
�=��
1= (2.353)(0.06931)
�
�= 0.163.1 m
3
/s
Part (e) -the compressor power
�
??????=�(ℎ
1−ℎ
4)
�= 2.353 kg/s
Part (f) -the rate of heat rejection at the condenser
�
??????=�(ℎ
3−ℎ
2)= (2.353)(421.54 –249.01)
�
??????= 406 kW

22. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem #2
Solution
Part (g) -COP of the plant and its refrigerating efficiency
���=
�
??????
??????
=
352
54
=6.52
���
�??????����=
�??????
�??????−�??????
=
273
308−273
= 7.80
Refrigerating Efficiency = COP / COP
Carnot
Refrigerating Efficiency = 83.6%

23. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
23
Actual Vapor-Compression Refrigeration Cycle
Two Common Sources of Irreversibilitiesin various components of actual cycle
❖FluidFriction(causingpressuredrop)
❖HeatTransfertoandfromtheSurroundings
Pressure drop and heat transfer
in the line connecting the
evaporator and compressor
oPressure drop in the evaporator
oSlight Superheating as saturated vapor condition can’t be
controlled so precisely.
Actual Compression with friction
effects and heat transfer, case
where combined effect increase
the entropy
Actual Compression with friction
effects and heat transfer, case
where combined effect decrease
the entropy
Pressure drop and heat transfer in the line connecting the
compressor and condenser
oPressure drop in the condenser
oSubcooled liquid as saturated liquid
condition can’t be controlled so
precisely.
Pressure drop and heat transfer in the
line connecting the condenser and
throttling valve
Pressure drop and heat transfer in the
line connecting the throttling valve and
evaporator

24. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
24
Actual Vapor-Compression Refrigeration Cycle
Actual Vapor-Compression
Refrigeration Cycle
Two Common Sources of Irreversibilitiesin various components of actual cycle
❖FluidFriction(causingpressuredrop)
❖HeatTransfertoandfromtheSurroundings
ThiswouldresultindecreaseofCOPwhencomparedwithidealcycle
Difference Between Ideal and
Actual Vapor Compression Cycle
oNon-isentropiccompression
oSuperheatedvaporat
evaporatorexit
oSub-cooledliquidat
condenserexit
oPressuredropsincondenser,
evaporatorandconnecting
lines.
oHeatTransferinconnecting
linesandcompressor

25. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
25
Actual Vapor-Compression Refrigeration Cycle
Liquid-to-Suction Heat Exchanger
❖SomeRefrigerationSystemsusealiquid-to-suctionheat
exchanger,whichsubcoolstheliquidfromthecondenser
withsuctionvaporcomingfromtheevaporator.
❖Eventhoughtherefrigeratingeffectisincreased,the
compressionispushedfartheroutintothesuperheat
regionandtheworkofcompressionisalsoincreased
becauseofhigherspecificvolumeatinletofcompressor.
Thepotentialimprovementsinperformanceisthus
counterbalancedandprobablyhasnegligible
thermodynamicadvantage.
❖Heatexchangerhoweverensuretherequireddegreeof
sub-coolingandsuperheating,thereforenoliquidenters
intothecompressoranditalsopreventsanybubblesof
vaportoenterinexpansionvalve.

26. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
26
Actual Vapor-Compression Refrigeration Cycle
Factors Effecting Performance of Vapor Compression Cycle
❖Bypassingtheliquidrefrigerantfromcondenserthroughaheat
exchanger.Thisprocesssubcoolstheliquidbutsuperheatsthe
vapor.Thus,COPisnotimprovedthoughrefrigerationeffectis
increased.
❖Bymakinguseofenoughquantityofcoolingwatersothatthe
liquidrefrigerantisfurthercooledbelowthetemperatureof
saturation.Insomecases,aseparatesub-coolerisalsousedoffor
thispurpose.Inthiscase,COPisimprovedas(ℎ
�−ℎ
�
′)>
(ℎ
�−ℎ
�).
Effect of sub-cooling of refrigerant at condenser exit
Effect of superheating of refrigerant at exit of evaporator
❖Therefrigeratingeffectisincreasedas(ℎ
�
′′−ℎ
�)>(ℎ
�−ℎ
�).
However,COPmayincrease,decreaseorremainunchanged
dependingupontherangeofpressureofthecycle.
Sub-cooling and superheating of refrigerant

27. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
27
Actual Vapor-Compression Refrigeration Cycle
Factors Effecting Performance of Vapor Compression Cycle
❖Letthesuctionpressureortheevaporatingpressureinasimple
refrigerationcyclebereducedfrom�
�to�′
�.
RefrigeratingEffectreducedto:(ℎ
�
′−ℎ
�
′)<(ℎ
&#3627408437;−ℎ
&#3627408436;)
WorkofCompressionincreasedto:ℎ
&#3627408438;
′−ℎ
&#3627408437;
′>(ℎ
&#3627408438;−ℎ
&#3627408437;)
BoththeeffectstendtodecreasetheCOP.
Effect of Change in Suction Pressure (&#3627408451;
&#3627408480;)
Effect of Change in Discharge Pressure (&#3627408451;
&#3627408465;)
❖Letthedischargepressureorthecondensationpressureinasimple
refrigerationcycleisincreasedfrom&#3627408451;
&#3627408465;to&#3627408451;′
&#3627408465;.
RefrigeratingEffectreducedto:(ℎ
&#3627408437;−ℎ
&#3627408436;
′′)<(ℎ
&#3627408437;−ℎ
&#3627408436;)
WorkofCompressionincreasedto:ℎ
&#3627408438;
′′−ℎ
&#3627408437;>(ℎ
&#3627408438;−ℎ
&#3627408437;)
BoththeeffectstendtodecreasetheCOP.
Effect of change in evaporator and
condenser pressure

28. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Fluids Properties for Analysis of Fluids Behavior
28
Actual Vapor-Compression Refrigeration Cycle
Factors Effecting Performance of Vapor Compression Cycle
❖Foragivencondensertemperature&#3627408455;
&#3627408464;theCOPincreasesrapidlywithevaporatortemperature(&#3627408455;
&#3627408466;),particularlyat
lowcondensingtemperatures.
❖Foragivenevaporatortemperature,theCOPdecreasesascondensertemperatureincreases.However,theeffectof
condensertemperaturebecomesmarginalatlowevaporatortemperatures.
Effect of Evaporator (&#3627408455;
&#3627408466;)and Condenser (&#3627408455;
&#3627408464;)Temperature
&#3627408455;
&#3627408464;
&#3627408455;
&#3627408466;

29. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem # 3
Anidealvapor-compressionrefrigeration
cyclethatusesrefrigerant-134aasits
workingfluidmaintainsacondenserat800
kPaandtheevaporatorat-10.07°C.
a)Determinethesystem’sCOPandthe
refrigerationeffect(inkJ/kg).
b)Aliquid-vaporregenerativeheat
exchangerisinstalledinthesystem,with
suctionvaporsuperheatedby20.07°C.
Determinethissystem’sCOPandthe
refrigerationeffect(inkJ/kg).
29

30. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
&#3627408451;
Problem # 3
Solution
Required
(a)COP
(b)Refrigeration Effect
Formula( )
21in
whh=− L
in
q
COP
w
=
Solution
State-11
10.07TC
Sat. Vapor
=− 

 1 @10.07
1 @10.07
392.71/
1.7337/
vapor C
vapor C
hh kJkg
ss kJkgK
−
−
= =
= =
State-22
2
21
800
421.52/
PkPa
h kJkg
ss
= 
= 
=

P s h
800 kPa
1.7144415.58
1.7337421.52
1.7437424.61( )
14L
qhh=−
Part (a)

31. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem # 3
Solution
State-33
3
800
243.58/
.
PkPa
h kJkg
SatLiquid
= 
= 

State-443
243.58/hh kJkg== ( )
14
149.13/
L
qhh kJkg=−=
Refrigeration Effect ( )
21
28.81/whh kJkg=−= 149.13
5.176
28.81
COP==
COP
31

32. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem # 3
Solution
Required
(a)COP
(b)Refrigeration Effect
Formula( )
21in
whh=− L
in
q
COP
w
=
Solution
State-1( )
65L
qhh=−
Part (b) 1 @10.07
11
409.73/&1.7961/
20.07
sat C
PP
h kJkgs kJkgK
SuperheatedbyC
−
= 
==



32

33. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem # 3
Solution
State-22
21
800
1.7961/
PkPa
ss kJkgK
=
==
33
P s h
800 kPa
1.7758434.85
1.7961441.50
1.8067444.98
State-33
3
800
243.58/
.
PkPa
h kJkg
SatLiquid
= 
= 

State-43 4 1 6
hhhh−=−
Energy balance across HX
State-66
10.07TC
Sat. Vapor
=− 

 6 @10.07
392.71/
vapor C
hh kJkg
−
= = 4
226.56/h kJkg=

34. Refrigeration and Air Conditioning (ME-331) Dr. Haider Ali
Problem # 3
Solution( )
65
166.15/
L
qhh kJkg=−=
Refrigeration Effect ( )
21
31.77/whh kJkg=−= 166.15
5.23
31.77
COP==
COP
34