Chapter 5 Fundamentals of Refrigeration

AE 2031 APPLIED THERMODYNAMICS S. Y. B. Tech.
FUNDAMENTALS OF
REFRIGERATIONS
Prof. Aniket Suryawanshi
Asst. Prof. Automobile Engg. Dept.
R. I. T. Rajaramnagar

Outline
• Applications of Refrigeration.
• Reversed Carnot Cycle.
• COP and Power Calculations
• Vapour – Compression Refrigeration System.
• Presentation on T-S and P-h diagram.
• Vapour – Absorption System.

Refrigeration
REFRIGERATION – Cooling of or removal of heat from a system.
Refrigerated System – System which is kept at lower temperature.
Refrigeration – 1) By melting of a solid,
2) By sublimation of a solid,
3) By evaporation of a liquid.
Most of the commercial refrigeration production : Evaporation of liquid.
This liquid is known as Refrigerant.

Refrigeration Circuit
Refrigeration Circuit
Evaporator Compressor
CondenserExpansion
Valve

Refrigeration - Elements
Compressor
Condenser
Evaporator
Expansion
Valve
Wnet, in
Surrounding Air
Refrigerated Space
QH
QL
High Temp
Source
Low Temp
Sink
QH
QL
Wnet, in

Refrigeration Systems
1. Ice Refrigeration System.
2. Air Refrigeration System.
3. Vapour Compression Refrigeration System.
4. Vapour Absorption Refrigeration System.
5. Adsorption Refrigeration System.
6. Cascade Refrigeration System.
7. Mixed Refrigeration System.
8. Thermoelectric Refrigeration System.
9. Steam Jet Refrigeration System.
10. Vortex Tube Refrigeration System.
Refrigeration Systems :

Refrigeration - Applications
1. Ice making.
2. Transportation of food items above and below freezing.
3. Chemical and related industries.
4. Medical and Surgical instruments.
5. Processing food products and beverages.
6. Oil Refining.
7. Synthetic Rubber Manufacturing.
8. Freezing food products.
9. Manufacturing Solid Carbon Dioxide.
Applications :

Performance - COP
COP – Ratio of Heat absorbed by the Refrigerant while passing through the Evaporator
to the Work Input required to compress the Refrigerant in the Compressor.
Performance of Refrigeration System :
- Measured in terms of COP (Coefficient of Performance).
If; Rn = Net Refrigerating Effect. W = Work required by the machine.
Then;
W
R
COP n

COPlTheoretica
COPActual
COPlative Re
Actual COP = Ratio of Rn and W actually measured.
Theoretical COP = Ratio of Theoretical values of Rn and W obtained by applying
Laws of Thermodynamics to the Refrigerating Cycle.

Performance - Rating
Rating of Refrigeration System :
- Refrigeration Effect / Amount of Heat extracted from a body in a given time.
Unit :
- Standard commercial Tonne of Refrigeration / TR Capacity
Definition :
- Refrigeration Effect produced by melting 1 tonne of ice from and at 0 ºC in 24 hours.
Latent Heat of ice = 336 kJ/kg.

Reverse Carnot Cycle
3
2
1
4
Isotherms
Adiabatic
T2
Expansion
Compression
T1
Pressure
Volume
P –V Diagram
3 2
14T1
Expansion Compression
T2
Temperature Entropy
1’4’
T –s Diagram

3 2
14T1
T2
Temperature
Entropy
1’4’
Operation :
3 – 4 : Adiabatic Expansion.
Temp. falls from T2 to T1.
Cylinder in contact with Cold Body at T1.
4 – 1 : Isothermal Expansion.
Heat Extraction from Cold Body.
1 – 2 : Adiabatic Compression.
Requires external power.
Temp. rises from T1 to T2.
Cylinder in contact with Hot Body at T2
2 – 3 : Isothermal Compression.
Heat Rejection to Hot Body.

3 2
14T1
T2
Temperature
Entropy
1’4’
Heat extracted from cold Body : Area 1-1’-4’-4
= T1 X 1-4
Work done per cycle : Area 1-2-3-4
= (T2 – T1) X 1-4
12
1
12
1
)41()(
)41(
4321
4'4'11
TT
T
XTT
XT
Area
Area
DoneWork
ExtractedHeat
COP










Example 1
A Carnot Refrigerator requires 1.3 kW per tonne of refrigeration to maintain a region at
low temperature of -38 ºC. Determine:
i) COP of Carnot Refrigerator.
ii) Higher temperature of the cycle.
iii) Heat delivered and COP, if the same device is used Heat Pump.
99.2
)sec/3600()3.1(
/000,14
3.1
1

hrkW
hrkJ
kW
tonne
doneWork
absorbedHeat
COPrefrig ….ANS
KT
KT
K
TT
T
COPrefrig 6.313
235
235
99.2 1
212
1




 ….ANS
Heat Delivered as Heat Pump ;
sec/189.53.1
3600
/000,14
3.11 kJ
hrkJ
kWtonne
doneWorkabsorbedHeat


….ANS
99.3
3.1
sec/189.5

kW
kJ
doneWork
deliveredHeat
COPHP ….ANS

Example 2
A refrigerating system works on reverse Carnot cycle. The higher temperature in the
system is 35 ºC and the lower temperature is -15 ºC. The capacity is to be 12 tonnes.
Determine :
i) COP of Carnot Refrigerator.
ii) Heat rejected from the system per hour.
iii) Power required.
18.5
258308
258
12
1





KK
K
TT
T
COPrefrig ….ANS
hrkJInputWork
InputWork
hrkJX
InputWork
tonne
InputWork
Effectfrig
COPrefrig
/32558
/000,141212
16.5
.Re


….ANSkW
hrkJhrInputWork
Power 04.9
3600
/32558
3600
/

Heat Rejected / hr = Refrig. Effect / hr + Work Input / hr
= 12 x 14,000 (kJ/hr) + 32,558 (kJ/hr) = 2,00,558 kJ/hr. ….ANS

Example 3
Ice is formed at 0 ºC from water at 20 ºC. The temperature of the brine is -8 ºC. Find out
the kg of ice per kWh. Assume that the system operates on reversed Carnot cycle. Take
latent heat of ice as 335 kJ/kg.
46.9
265293
265
12
1





KK
K
TT
T
COPrefrig
Heat to be extracted per kg of water ( to from ice at 0 ºC)
Rn = 1 (kg) x Cpw (kJ/kg.K) x (293– 273) (K) + Latent Heat (kJ/kg) of ice
= 1 (kg) x 4.18 (kJ/kg.K) x 20 (K) + 335 (kJ/kg)
= 418.6 kJ/kg.
Also, 1 kWh = 1 (kJ) x 3600 (sec/hr) = 3600 kJ.
kgm
kJ
kgkJXkgm
kJdoneWork
kJEffectfrig
W
R
COP
ice
ice
n
refrig
35.81
3600
)/(6.418)(
46.9
)(
)(.Re


….ANS

Vapour Compression System
Elements of this system :
1. Compressor.
2. Condenser.
3. Expansion Valve.
4. Evaporator.
Vapour @ ↓ Pr. and ↓ Temp. (State 1)
Isentropic Compression :
↑ Pr. and ↑ Temp. (State 2)
Condenser : ↑ Pr. Liquid (State 3)
Throttling : ↓ Pr. ↓ Temp. (State 4)
Evaporator : Heat Extraction from surrounding;
↓ Pr. vapour (State 1).
1
2
3
4
1
23
4

Vapour Compression System
Merits :
1. High COP; as very close to Reverse Carnot Cycle.
2. Running Cost is 1/5th of that of Air Refrigeration Cycle.
3. Size of Evaporator is small; for same Refrigeration Effect.
Demerits :
1. Initial cost is high.
2. Inflammability.
4. Evaporator temperature adjustment is simple; by adjusting Throttle Valve.
3. Leakage.
4. Toxicity.

Vapour Compression System : T-s Diagram
Case A. Dry and Saturated Vapour after Compression :
Work done by Compressor
= W = Area 1-2-3-4-1
Heat Absorbed
= W = Area 1-4-g-f-1
Entropy, s
Temperature,T
Condensation
Compression
Evaporation
Expansion
T2
T1
3
4
2
1
Compressor Work,
(W)
Net Refrig. Effect,
(Rn)
Sat. Vapour Line
Sat. Liq. Line
fg
12
41
14321
141
hh
hh
Area
fgArea
DoneWork
AbsorbedHeat
COP








Case B. Superheated Vapour after Compression :
= W = Area 1-2-2’-3-4-1
Heat Absorbed
= W = Area 1-4-g-f-1
12
41
143'221
141
hh
hh
Area
fgArea
DoneWork
AbsorbedHeat
COP







Entropy, s
Temperature,T
Condensation
Compression
Evaporation
Expansion
T2
T1
3
4
2
1
Compressor Work,
(W)
Net Refrig. Effect,
(Rn)
Sat. Vapour Line
Sat. Liq. Line
fg
2’
NOTE : h2 = h2’ + Cp (Tsup – Tsat)

Case C. Wet Vapour after Compression :
= W = Area 1-2-3-4-1
Heat Absorbed
= W = Area 1-4-g-f-1
Entropy, s
Temperature,T
Condensation
Compression
Evaporation
Expansion
T2
T1
3
4
2
1
Compressor Work,
(W)
Net Refrig. Effect,
(Rn)
Sat. Vapour Line
Sat. Liq. Line
fg
12
41
14321
141
hh
hh
Area
fgArea
DoneWork
AbsorbedHeat
COP







NOTE : h2 = (hf + x.hfg)2

Vapour Compression System : P-h Diagram
Enthalpy, h
Pressure,Pr
Isothermal,
T=Const
Isenthalpic,
h=Const.
Isobaric,
P = Const
Sub-cooled
Liq. region
2 – phase
region
Superheated
region

Vapour Compression System : P-h Diagram
Enthalpy, h
Pressure,Pr
1
23
4
Evaporation
Condensation
Compression
12
41
hhW
hhRn


12
41
hh
hh
W
R
COP n





Vapour Compression System – Mathematical Analysis
A. Refrigerating Effect :
  )/(41 kgkJHeatdSuperheateHeatLatenthhQevap 
= Amount of Heat absorbed in Evaporator.
B. Mass of Refrigerant :
= Amount of Heat absorbed / Refrigerating Effect.
 
)sec/(
3600
000,14
41
tonnekg
hh
m 


1 TR = 14,000 kJ/hr
C. Theoretical Piston Displacement :
= Mass of Refrigerant X Sp. Vol. of Refrigerant Gas (vg)1.
 
  )sec/(
3600
000,14
.. 3
1
41
tonnemv
hh
DisplPistonTh g 



Vapour Compression System – Mathematical Analysis
 
  )(
)/(
12
12
kWhhmP
kgkJhhW
theor
comp


a) Isentropic Compression :
D. Theoretical Power Required :
 
  )(
1
)/(
1
1122
1122
kWVPVP
n
n
mP
kgkJVPVP
n
n
W
theor
comp






a) Polytropic Compression :
E. Heat removed through Condenser :
  )/(32 kgkJhhmQcond 

Example 4
A refrigeration machine is required to produce ice at 0º C from water at 20 ºC. The
machine has a condenser temperature of 298 K while the evaporator temperature is 268
K. The relative efficiency of the machine is 50 % and 6 kg of Freon-12 refrigerant is
circulated through the system per minute. The refrigerant enters the compressor with a
dryness fraction of 0.6. Specific heat of water is 4.187 kJ/kg.K and the latent heat of ice is
335 kJ/kg. Calculate the amount of ice produced on 24 hours. The table of properties if
Freon-12 is given below:
Temperature
(K)
Liquid Heat
(kJ/kg)
Latent Heat
(kJ/kg)
Entropy of Liquid
(kJ/kg)
298 59.7 138.0 0.2232
268 31.4 154.0 0.1251
hf1 = 31.4 kJ/kg
hfg1 = 154.0 kJ/kg
hf2 = 59.7 kJ/kg
hfg2 = 138.0 kJ/kg
hf3 = h4 = 59.7 kJ/kg
m = 6 kg/min
ηrel = 50 %
x2 = 0.6
Cpw = 4.187 kJ/kg.K
Latent Heat of ice = 335.7 kJ/kg
Given :

Example 4….contd
Entropy, s
Temperature,T
298 K
268 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
kgkJhxhh fgf /8.1230.154)6.0(4.31111 
kgkJhxhh fgf /2.1330.138)5325.0(7.5922 22 
 
5325.0
268
0.154
6.01251.0
298
0.138
2232.0
2
2
1
1
11
2
2
22
111222
12





























x
x
T
h
xs
T
h
xs
sxssxs
ss
fg
f
fg
f
fgffgf
Isentropic Compression : 1-2
kgkJhh f /7.5934 
COP of Original Cycle :
 
 
82.6
/8.1232.133
/7.598.123
12
41







kgkJ
kgkJ
hh
hh
W
R
COP n

Example 4….contd
Actual COP = ηrel X COPtheor = 0.5 X 6.82 = 3.41
Heat extracted from 1 kg of water at 20 ºC to form 1 kg of ice at 0 ºC :
kgkJ
kgkJ
CXKkgkJXkg
/74.418
)/(335
)()020()./(187.4)(1



Entropy, s
Temperature,T
298 K
268 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
Now;
….ANS
 
 
hrsintonne
XX
kg
kgkJ
kgkJXkg
m
hhm
Xm
W
R
COP
ice
iceactualn
actual
24661.0
1000
2460459.0
min/459.0
41.3
/74.418
)/(8.1232.133)(6
74.418
41.3
12
)(








Example 5
28 tonnes of ice from and at 0 ºC is produced per day in an ammonia refrigerator. The
temperature range in the compressor is from 25ºC to -15ºC. The vapor is dry and
saturated at the end of compression and an expansion valve is used. Assuming a
co-efficient of performance of 62% of the theoretical, calculate the power required to
drive the compressor. Take latent heat of ice = 335 kJ/kg.
Temp
(ºC)
Enthalpy (kJ/kg) Entropy of
Liquid
(kJ/kg.K)
Entropy of Vapour
(kJ/kg.K)Liquid Vapour
25 100.04 1319.22 0.3473 4.4852
-15 -54.56 1304.99 -2.1338 5.0585
hf1 = -54.56 kJ/kg
hg1 = 1304.99kJ/kg
hf2 = 100.04 kJ/kg
hg2 = 1319.22 kJ/kg
hf3 = h4 = 100.04 kJ/kg
Tcond = 25 ºC
Tevap = -15 ºC
x2 = 1….dry saturated vapour
COPactual = 0.62 (COPtheor)
Latent Heat of ice = 335.7 kJ/kg
Given :

 
 
91.8
23.119622.1319
04.10023.1196
12
41







hh
hh
COP ltheoreticaCOP of the Cycle :
Entropy, s
Temperature,T
298 K
258 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
Example 5….contd
processcIsenthalpikgkJhh ...../04.10043 
 
kgkJ
hxhh fgf
/23.1196
)56.54(99.1304)92.0()56.54(
)( 1111



    
92.0
1338.20585.5)1338.2(4852.4
1
1
1112
12




x
x
sxss
ss
fgfg
kgkJhh g /22.131922 

Example 5….contd
Actual COP = ηrel X COPtheor
= 0.62 X 8.91
= 5.52
Entropy, s
Temperature,T
298 K
258 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
Actual Rn = COPactual X Work done
= 5.52 X (h2 – h1)
= 5.52 X (1319.22 – 1196.23)
= 678.9 kJ/kg
Heat extracted from 28 tonnes of water at 0 ºC to form ice at 0 ºC :
)(sec/56.108
)(sec/3600)(24
)/(335)/(1000)(28
kWkJ
hrXhr
kgkJXtonnekgXkg


Mass of refrigerant : kg
kgkJ
kJ
1599.0
)/(9.678
sec)/(56.108

Total Work done by Compressor :
 
)(sec/67.19
/)23.119622.1319()(1599.012
kWkJ
kgkJXkghhXmrefrig


….ANS

Example 6
In a standard vapour compression refrigeration cycle, operating between an evaporator
temperature of -10 ºC and a condenser temperature of 40 ºC, the enthalpy of the
refrigerant, Freon-12, at the end of compression is 220 kJ/kg. Show the cycle diagram on
T-s plane. Calculate:
1. The C.O.P. of the cycle.
2. The refrigerating capacity and the compressor power assuming a refrigerant flow
rate of 1 kg/min.
You may use the extract of Freon-12 property table given below:
Temp (ºC) Pr (MPa) hf (kJ/kg) hg (kJ/kg)
-10 0.2191 26.85 183.1
40 0.9607 74.53 203.1
hf1 = 26.85 kJ/kg
hg1 = h1 = 183.1 kJ/kg
hf2 = 74.53 kJ/kg
hg2 = 203.1 kJ/kg
hf3 = h4 = 74.53 kJ/kg
Tcond = 40 ºC
Tevap = -10 ºC
h2 = 220 kJ/kg
Given :

Example 6….contd
COP of Original Cycle :
 
 
94.2
/1.1830.220
/53.741.183
12
41







kgkJ
kgkJ
hh
hh
W
R
COP n
….ANS
Refrigerating Capacity :
   
min/57.108
/53.741.183)(141
kJ
kgkJXkghhm


….ANS
Compressor Power :
   
kW
kJ
kgkJXkghhm
615.0
min/9.36
/1.1830.220)(112



….ANS
Entropy, s
Temperature,T
40 ºC
-10 ºC
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
2’

Example 7
A Freon-12 refrigerator producing a cooling effect of 20 kJ/sec operates on a simple
cycle with pressure limits of 1.509 bar and 9.607 bar. The vapour leaves the evaporator
dry saturated and there is no undercooling. Determine the power required by the
machine. If the compressor operates at 300 rpm and has a clearance volume of 3% of
stroke volume, determine the piston displacement of the compressor. For compressor
assume that the expansion following the law PV1.3 = Constant.
Temp
(oC)
Ps
(bar)
vg
(m3/kg)
Enthalpy
hf
(kJ/kg)
Enthalpy
hg
(kJ/kg)
Entropy
sf
(kJ/kg)
Entropy
sg
(kJ/kg)
Specific
heat
(kJ/kg.K)
-20 1.509 0.1088 17.8 178.61 0.073 0.7082 ---
40 9.607 --- 74.53 203.05 0.2716 0.682 0.747
hf1 = 17.8 kJ/kg
hg1 = h1 = 178.61 kJ/kg
hf2 = 74.53 kJ/kg
hg2’ = 203.05 kJ/kg
hf3 = h4 = 74.53 kJ/kg
Tcond = 40 ºC
Tevap = -20 ºC
h2 = 220 kJ/kg
Given :

Example 7….contd
Refrigerating Capacity :    
sec/192.0
/53.7461.1782041
kgm
kgkJXmkWhhm




 
KT
T
T
T
Css
ss
P
2.324
313
ln747.0682.07082.0
ln
2
2
'2
2
'21
21
















 
  
kgkJ
KKkgkJkgkJ
TTChh P
/4.211
0.3132.324./747.0)/(05.203
'22'22



Entropy, s
Temperature,T
313 K
253 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
2’

Example 7
….ANS
Power Required :    
kW
kgkJXkghhm
29.6
/61.1784.211sec)/(192.012



Vol. Efficiency :
%6.87
509.1
607.9
03.003.01
1
13.1/1
/1















bar
bar
P
P
kk
n
S
d
vol
Vol of Refrigerant
at Intake :
sec/02089.0
)/(1088.0sec)/(192.0
3
3
m
kgmXkg
vm g




Piston Displ. Vol. :
  3
3
00477.0
)(300876.0
min)(sec/60sec)/(02089.0
)(
.
m
rpm
m
rpm
VolActual
vol






 ….ANS
Entropy, s
Temperature,T
313 K
253 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
2’

Example 8
A food storage locker requires a refrigeration capacity of 50 kW. It works between a
condenser temperature of 35 ºC and an evaporator temperature of -10 ºC. The refrigerator
is ammonia. It is sub-cooled by 5 ºC before entering the expansion valve. By the dry
saturated vapour leaving the evaporator. Assuming a single-cylinder, single-acting
compressor operating at 1000 rpm with stroke equal to 1.2 times the bore, determine :
1. The power required.
2. The cylinder dimensions.
Properties of ammonia are :
Sat.
Temp.
(oC)
Pr.
(bar)
Enthalpy
(kJ/kg)
Entropy
(kJ/kg)
Sp. Vol.
(m3/kg)
Sp. Heat
(kJ/kg.K)
Liquid Vapour Liquid Vapour Liquid Vapour Liquid Vapour
-10 2.9157 154.056 1450.22 0.82965 5.7550 --- 0.417477 --- 2.492
35 13.522 366.072 1488.57 1.56605 5.2086 1.7023 0.095629 4.556 2.903
h1 = 1450.22 kJ/kg
h2’ = 1488.57 kJ/kg
hf3 = 366.072 kJ/kg
Tcond = 35 ºC
Tevap = -10 ºC
State 3 = Sub-cooled by 5 ºC
Given :

Example 8….contd
 
 
kgkJ
kgkJkgkJ
TTChhh subcoolsatliqPf
/29.343
)/(30330856.405)/(07.366
34'3



 
KT
T
T
T
Cssss P
8.371
308
ln903.22086.5755.5
ln
2
2
'2
2
'2121















 
  
kgkJ
KKkgkJkgkJ
TTChh P
/8.1673
0.3088.371./903.2)/(57.1488
'22'22



Entropy, s
Temperature,T
308 K
263 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
2’
3’
303 K

Example 8….contd
   
sec/04517.0
/29.34322.1450
)(50
/
)(50
41
kg
kgkJ
kW
kgkJhh
kW
m






Mass of Refrigerant :
Compressor Power :
 
 
kW
kgkJXkg
hhm
1.10
/22.14508.1673)(04517.0
12




….ANS
Cylinder Dimensions :
 
mmL
mD
kgm
rpm
DD
v
N
LD
kgm
g
228.0)19.0(2.1
19.0
/417477.0
60
)(1000
)2.1(
4604
sec)/(04517.0 3
22


















….ANS
….ANS
Entropy, s
Temperature,T
308 K
263 K
3
4
2
1
Sat. Vapour Line
Sat. Liq. Line
fg
2’
3’
303 K

Simple Theoretical Vapour Absorption System
Evaporator Absorber
GeneratorCondenser
Weak
Solution
Refrig.
Vapour Strong
Solution
PUMP
QE
QA
WP
Expn.
Valve
QC
QG
(2)
(1)
(3)
(4)
(5)
(6)
(7)
(8)
Assumption :
Absorbent does not vaporise in the
Generator.
State (1) : Vapour leaving
the Evaporator
+ Weak Liq. Solution in
the Absorber
Removal of Heat :
Refrig. Vapour be absorbed
by Weak Liq. Solution.
Strong Liq. Solution.
Pressure of Liq. Solution ↑ by pump to
Generator Pressure.

Evaporator Absorber
GeneratorCondenser
Weak
Solution
Refrig.
Vapour Strong
Solution
PUMP
QE
QA
WP
Expn.
Valve
QC
QG
(2)
(1)
(3)
(4)
(5)
(6)
(7)
(8)
Heat Addition in Generator :
Refrig. Vapour be driven out
of the Liq. Solution.
Heat Exchanger in Liq. Solution
Circuit, between Generator and
Absorber.
Generator + Condenser : HP Side.
Evaporator + Absorber : LP Side.

Absorption
Fluid
System
TE
TG
Pump
TO
WP
QO
QG
QE
Generator
Refrig. Space Environment
Energy Transfer to & from
the Fluid of Absorption System :
Generator : Heating Medium
Heat Addition, QG
Pump :
Work Addition, WP
Substance to be cooled :
Heat Addition, QE
Absorber :
Heat Rejection,
QO = QA (Absorber) + QC (Condensor)
to environment through cooling water.

Absorption
Fluid
System
TE
TG
Pump
TO
WP
QO
QG
QE
Generator
Refrig. Space Environment
CAO QQQ 
First Law of Thermodynamics :
PEGO WQQQ 
Assumptions :
Generator Heating Med. Temp, TG
Refrig. Substance Temp, TE
Environmental Temp, TO
Const.

Vapour Absorption System
1
2
4
3
Solubility of NH3 in water @ ↓ Temp and Pr. is MORE than that @ ↑ Temp. and Pr.
NH3 vapour from Evaporator (State 1) is readily absorbed in Absorber. Heat Rejection
This solution is then pumped to ↑ Temp. and Pr. @ Generator.
Reduction in stability of solution Vapour removed from Solution.
Vapour passes to Condenser.
Weak Solution returns to Absorber.

Vapour Absorption System
Merits :
2. Work done on Compression is LESS.
1. Pumping work is much less than work for Compressing vapour.
Demerits :
2. Low COP.
1. Heat input to the Generator is required.
COP :
PumpLiquidondoneWorkGeneratorinpliedHeat
EvaporatorinextractedHeat
COP


sup

Vapour Compression Vs. Vapour Absorption
Sr.
No.
Particulars Vapour Compression Systems Vapour Absorption Systems
1. Type of Energy Supplied Mechanical – High Grade Heat – Low Grade
2. Energy Supply Rate Low High
3. Wear & Tear More Less
4. Performance of Part Load Poor Not affected at Part Load
5. Suitability Used where High Grade Mechanical
Energy is available
Can be used at Remote Places, as
can be used with simple Kerosene
lamp
6. Charging of Refrigerant Simple Difficult
7. Leakage More chances No chances, as no Compressor or
Reciprocating Part
8. Damage Liquid traces in Suction Line may
damage Compressor
No danger

Thermodynamic Properties
A. Critical Temp. & Pressure
1. Highest Temp at which liquid can be condensed.
2. Condenser Pressure : well below Critical Pressure.
3. Otherwise ↑ Power Consumption…!!!

1. High Vaporisation Enthalpy.
Enthalpy, hPressure,Pr
Vaporization
Enthalpy
B. Enthalpy of Vaporization
2. More Refrigerating Effect
per unit mass of refrigerant.
3. ↓ Size of Equipment.

1. As HIGH as possible.
Enthalpy, h
Pressure,Pr
1
23
4
Evaporation
Condensation
Expansion
C. Specific Heat
Vapour : ↑ value is desired.
Liquid : ↓ value is desired.
D. Thermal Conductivity
2. ↑ Heat Transfer Rates.
3. Use of Low Conductivity Materials.

• Both should be + ve.
E. Evaporator & Condenser Pressures
• + ve Pressure :
Refrigerant flows out , if leak…!!
• – ve Pressure :
Air and Moisture enters into system…!!! Enthalpy, h
Pressure,Pr
1
23
4
Evaporation
Condensation
Expansion

1. Chemical Stability – Important as Working Substance.
A. Stability & Inertness
2. Should Not decompose in operating range of temperatures.
3. Inertness – Should Not react with equipment material.
e.g. Ammonia (NH3) attacks on Copper and Cuprous alloys.
Chemical Properties

1. As low as possible
B. Flammability
C. Water Solubility
1. Less solubility for water.
inflammable
2. Straight Hydrocarbons : ↑ flammable
2. Presence of moisture can be detected…!!!
Chemical Properties

1. Ability to mix with oil.
D. Miscibility
E. Toxicity
2. Need Oil Separators.
2. Toxic Refrigerants :
1. As low as possible Non - Toxic
DANGER for Domestic and Comfort Air Conditioning
Chemical Properties

1. Non – Corrosive.
F. Corrosive Properties
G. Viscosity
1. Low Viscosity
2. Better Refrigerant Flow
↑ Heat Transfer
H. Dielectric Strength
1. High Dielectric Strength for Hermetically Sealed Compressors.
Chemical Properties

• Boiling Temperature of refrigerant should be quite low at atmospheric conditions for effective
refrigeration. For refrigerants having higher boiling temperatures at atmospheric
conditions the compressor is run at higher vacuum.
• For an ideal refrigerant the freezing temperature of refrigerant should be quite low so as to
prevent its freezing at evaporative temperature. Freezing point temperature should be less
than evaporative temperature. For example, R-22 has freezing point of -160ºC and normally
most of refrigerants have freezing point below -30ºC.
• Critical temperature of the ideal refrigerant should be higher than the condenser temperature
for the ease of condensation.
• Refrigerant should have large latent heat at evaporative temperature as this shall increase the
refrigerant capacity per kg of refrigerant.
• Refrigerant should have small specific volume at inlet to compressor as this reduces
compressor size for the same refrigeration capacity.
Desired Properties of Refrigerants

• Specific heat of refrigerant in liquid form should be small and it should large for refrigerant in
vapor form, since these increase the refrigerating capacity per kg of refrigerant.
• Thermal conductivity of refrigerant should be high.
• Viscosity of refrigerant should be small for the ease of better heat transfer and small
pumping work requirement.
• Refrigerant should be chemically inert and non-toxic.
• Refrigerant should be non-flammable, non- explosive and do not have any harmful effect upon
coming in contact with material stored in refrigeration space.
• Refrigerant may have pleasant distinct odour so as to know about its leakage.
• Refrigerant should be readily available at lesser price.
Desired Properties of Refrigerants

Classification
Refrigerants
Primary Secondary
1. Halocarbon Refrigerants
2. Inorganic Refrigerants
3. Azeotrope Refrigerants
4. Hydrocarbon Refrigerants
 Usually liquids.
 Transfer heat from the substance
being cooled to a Heat Exchanger
where the heat is absorbed by a
Primary Refrigerant.
 e.g. In an air conditioning system,
air acts as a secondary refrigerant.

Halocarbon Refrigerants
42 identified halocarbons by ASHRAE
• Designated by ‘R’ followed by a Numerical.
• First digit from right ≡ No. of C atoms -1
• Second digit from right ≡ No. of H atoms +1
• Third digit from right ≡ No. of Fluorine (F) atoms
Nomenclature :
e.g. : R 114 – dichloro-tetrafluoro-ethane

R – 11 ( C-Cl3-F )
• Operating Pressures – 0.3 bar (Evaporator)
– 1.25 bar (Condenser)
• Suitable with Centrifugal Compressor.
• Also used as Secondary Refrigerant.
• Safe and Non – Toxic
• One of the Highest Ozone Destruction Potential .
C
Cl
Cl
Cl
F

R – 12 ( C-Cl2-F2 )
• Most widely used.
• Non – Toxic, Non – Flammable.
• Miscible with oil.
• Comparably Lower Latent Heat.
• High Vapour Density.
• High Ozone Depletion Potential .
C
F
Cl
Cl
F

R – 12 ( C-Cl-F3 )
• Used in Ultra Low Temperature applications.
• Boiling Temp. = (-98) ˚C
• Suitable for all types of compressors.
• High Ozone Depletion Potential.
C
F
F
Cl
F

R – 22 ( C-H-Cl-F2 )
• High Discharge Temp. Thus suction superheat should be minimum.
• Smaller Compressor Displacement is required than R 12.
• Low Ozone Depletion Potential.
• Similar Properties as of R 12.
C
F
Cl
H
F

R – 134a
• Leading candidate to Replace R 12.
• Zero Ozone Depletion Potential.
• Non – Flammable, Non – Toxic.
• High Heat Transfer Coefficient.
• Miscibility Problems.
C
F
F
F C
F
H
H

Inorganic Refrigerants
Before Halocarbons, these were used extensively…!!
• Designated by ‘R’ followed by a Numerical.
• Number ≡ 700 + Mole. Weight of the compound.
Nomenclature :
e.g. : Ammonia ≡ 700 + 17 ≡ R 717.

Ammonia ( NH3 – R 717 )
• Extensively used other than Halocarbons.
• Excellent Thermal Properties.
• Highest Refrigerating Effect per unit mass.
• Operating Pressures – 2.37 bar (Evaporator)
– 11.67 bar (Condenser)
• Not Miscible with oil.
• It attacks on Copper & Cuprous alloys.

Carbon Dioxide ( CO2 – R 744 )
• Excellent for Low Temperature Refrigeration.
• Boiling Point = (-73.6) ˚C.
• Operating Pressures – 20.7 bar (Evaporator, -15 ºC)
– 11.67 bar (Condenser, 70 ºC)
• Low Efficiency.
• Immiscible with oil, Non – Flammable.
• High Global Warming Potential.

Azeotropic Refrigerants
• Azeotrope :
• Designated by 500 Series.
Mixture of two / more components in exact proportion.
To form Low / High Pressure Boiling Mixture .

R 500
• Blend of R 12 + R 152a
: R 12 (73.8 %) + R 152a (26.2 %) by weight.
• Good substitute for R 12.
• Compressor Capacity ↑ by 18 %.

R 502
• Blend of R 22 + R 115
: R 22 (48.8 %) + R 115 (51.2 %) by mass.
• Low Compressor Discharge Temp. than R 22.
• Non – Flammable, Non – Toxic.
• Low Oil Miscibility.
• Replacement of R 22.

Straight Hydrocarbons
• Highly Flammable.
• Miscible with oil.
• Ethane ( C2H6 ) , Methane ( CH4 ) & Ethylene ( C2H5 )
: Low Temperature Refrigeration

ASHRAE Designation for Refrigerants
No. Chemical Name Formulae Mole. Weight
Halocarbons :
11 Trichloro-monofloro methane C-Cl3-F 137.4
12 Dichloro-difloro methane C-Cl2-F2 120.9
13 Monochloro-trifloro methane C-Cl-F3 104.5
14 Carbon tetrachloride C-F4 88.0
21 Dichloro-monofloro methane C-H-Cl2-F 102.9
22 Monochloro-difloro methane C-H-Cl-F2 86.5
113 Trichloro-trifloro ethane C-Cl2-F-C-Cl-F2 187.4
114 Dichloro-tetrafloro ethane C-Cl-F2-C-Cl-F2 170.9
Hydrocarbons :
50 Methane Ch4 16.0
170 Ethane C-H3-C-H3 30.0
290 Propane C-H3-C-H2-C-H3 44.0
Inorganic Compounds :
717 Ammonia NH3 17.0
718 Water H2O 18.0
729 Air --- 29.0
744 Carbon dioxide CO2 44.0
764 Sulphur dioxide SO2 64.0

Comparative Refrigerant Data
Refrigerant Boiling Temp.@
1 atm ( ºC)
Freezing
Temp. (ºC)
Critical
Temp. (ºC)
Critical Pr.
(bar)
11 23.83 -111.11 198.00 44.09
12 -29.78 -157.78 112.00 41.15
13 -81.40 -181.10 28.83 38.68
14 -128.00 -183.89 -45.50 37.37
21 8.89 -135.00 177.94 51.71
22 -40.78 -160.00 96.00 49.77
113 47.55 -35.00 214.11 34.13
114 3.56 -94.27 145.72 32.68
717 -33.33 -77.72 133.00 114.25
718 100.00 0.00 374.50 222.42
729 -194.33 --- -140.55 37.71
744 -78.50 -56.61 31.00 73.85
764 -10.00 -75.50 157.11 78.70

Thank You !

Chapter 5 Fundamentals of Refrigeration

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Chapter 5 Fundamentals of Refrigeration