Refrigeration Cycles
Dr. Md. Zahurul Haq
ProfessorDepartment of Mechanical Engineering
Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh
[email protected]://teacher.buet.ac.bd/zahurul/
ME 203: Engineering Thermodynamics
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 1 / 24
T266 T273
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 2 / 24
Refrigerators, Air-conditioners & Heat Pumps
T270
COPR =
_QL
_WR: COPAC =
_QL
_WAC: COPHP =
_QH
_WHP
COPHP = COPR + 1
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 3 / 24
Refrigeration Capacity/Performance
1 ton refrigeration: heat absorbed by 1 ton (2000 lb) of ice melting
at 0oC in 24 hours.
1 ton refrigeration (TR) = 3.516 kW = 12000 BTU/hr = 200
BTU/min
Coefficient of Performance, COPR = Refrigeration EffectNet Work Required
kW /ton ⇒ power required per ton of refrigeration
kW /ton = 3.516COP
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 4 / 24
Basic Refrigeration Cycle Reversed Carnot Cycle
Reversed Carnot Cycle
T639
COP =
_Qin
_Wc − _Wt
=TC∆s
(TH − TC )∆s=
TC
TH − TC
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 5 / 24
Basic Refrigeration Cycle Reversed Carnot Cycle
⊲ Example:
T271
ηCarnot = 1 − TL
TH= 1 − 293.15
473.15= 0.38 = 38% ◭
COPR/AC = TL
TH−TL= 293.15
473.15−293.15= 1.63 ◭
COPHP = TH
TH−TL= 473.15
473.15−293.15= 2.63 ◭
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 6 / 24
Basic Refrigeration Cycle Reversed Carnot Cycle
Problems: Wet Compression in Carnot Cycle
During compression, droplets in liquid are vaporised by the
internal heat transfer process which requires finite time.
High-speed compressors are susceptible to damage by liquid
because of the short time available.
In wet compression, the droplets of the liquids may wash the
lubricating oil from the walls of the cylinder, accelerating wear.
Dry compression takes place with no droplets and is preferable.
Liquid refrigerants may be trapped in the head of reciprocating
compressor by the rising piston, possibly damaging the valves or
the cylinder head.
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 7 / 24
Basic Refrigeration Cycle Reversed Carnot Cycle
Problems: Expansion Process in Carnot Cycle
Carnot cycle demands that the expansion take place isentropically and
that the resulting work be used to help drive the compressor. Practical
difficulties, however, militate against the expansion engine:
the possible work that can be derived from the engine is small
fraction that must be supplied to the compressor.
practical problems such as lubrication intrude when a fluid of two
phases drives the engine.
the economics of the power recovery have in past not justified the
cost of the expansion engine.⊗
A throttling device, such as a valve or other restriction, is almost
universally used for this purpose.
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 8 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Basic Refrigeration System using 2-Phase Refrigerant
R-134a
-50 -40 -30 -20 -10 0 10 20 30 40 50 0
200
400
600
800
1000
1200
1400
P sa
t (kP
a)
T sat
( o C)
T265 T264
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 9 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Ideal Vapour Compression Refrigeration Cycle
T641 T640
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 10 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
T255
T256
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 11 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Processes of Vapour Compression Refrigeration System
T257
QL = Q41 = _m(h1 − h4)
QH = Q23 = _m(h2 − h3)
Win = W12 = _m(h2 − h1)
COP = QL/Win
1 → 2: Isentropic compression, Pevap → Pcond
2 → 3: Isobaric heat rejection, QH
3 → 4: Isenthalpic expansion, Pcond → Pevap
4 → 1: Isobaric heat extraction, QL
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 12 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
⊲ Example: A theoretical single stage cycle using R134a as refrigerant
operates with a condensing temperature of 30oC and an evaporator
temperature of -20oC. The system produces 50 kW of refrigeration effect.
Estimate:
1 Coefficient of performance, COP
2 Refrigerant mass flow rate, _m
T257
QL = Q41 = _m(h1 − h4) = 50 kW
_m = 0.345 Kg/s ◭
Win = W12 = _m(h2 − h1) = 12.5 kW
COP = QL/Win = 50.0/12.5 = 4.0 ◭
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 13 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Effect of Evaporator Temperature
-50 -45 -40 -35 -30 -25 -20 -15 -10 -5 00
1
2
3
4
5
6
7
8
9
10R134a: RE = 50 kW, T
cond = 30oC
Ref. flow rate
COP
CO
P
Tevap
(oC)
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
Ref
rige
rant
flow
rat
e (k
g/s)
T259
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 14 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Effect of Condenser Temperature
0 5 10 15 20 25 30 35 40 45 502
3
4
5
6
7
8
9
10
11
12R134a: RE = 50 kW, T
evap = -20oC
Ref. flow rate
COP
CO
P
Tcond
(oC)
0.24
0.26
0.28
0.30
0.32
0.34
0.36
0.38
0.40
0.42
0.44
Ref
rige
rant
flow
rat
e (k
g/s)
T261
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 15 / 24
Basic Refrigeration Cycle Ideal Vapour Compression Refrigeration Cycle
Effect of Evaporator & Condenser Temperatures
-50 -45 -40 -35 -30 -25 -20 -15 -10 -5 00
1
2
3
4
5
6
7
8
9
10
11
12
Tcond
= 40oC
Tcond
= 30oC
Tcond
= 20oC
CO
P
Tevap
(oC)T260
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 16 / 24
Deviation from Simple Cycle
Deviations from Ideal Cycle
1 Refrigerant pressure drop in piping, evaporator, condenser,
receiver tank, and through the valves and passages.
2 Sub-cooling of liquid leaving the condenser.
3 Super-heating of vapour leaving the evaporator.
4 Compression process is not isentropic.
T262
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 17 / 24
Deviation from Simple Cycle
Super-heating & Sub-cooling
T263
Sub-cooling of liquid serves a desirable function of ensuring that
100% liquid will enter the expansion device.
Super-heating of vapour ensures no droplets of liquid being carried
over into the compressor.
Even through refrigeration effect is increased, compression work is
greater & probably has negligible thermodynamic advantages.
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 18 / 24
Deviation from Simple Cycle
⊲ Example:
T258
h1 = h(P = 0.14 MPa,T = −10oC)
h2 = h(P = 0.80 MPa,T = 50oC)
h3 ≃ hf ,26oC
h4 = h3
QL = (h1 − h4) = 158.53 kJ/kg
Win = (h2 − h1) = 40.33 kJ/kg
COPR = QL
Win= 3.93 ◭
ηc = h2s−h1
h2−h1
= 93.6% ◭
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 19 / 24
Deviation from Simple Cycle
⊲ Example: A dual-evaporator refrigeration system: TF = -18.0oC, TR =
4.0oC, Tc = 30.0oC & ηc = 80%.
T274
COPR = ? x5 = ?
_m =
_QR+ _QF
h1−h4h= 1.62 × 10−3 kg/s ◭
_Wc =
_m(h2s−h1)
ηc
COPR =
_QR+ _QF
_Wc= 3.37 ◭
_QR = _m(h5 − h4h) → h5
h5 =⇒ x5 = 0.56 ◭
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Miscellaneous Refrigeration Systems
Vapour Absorption Refrigeration System
T272
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Miscellaneous Refrigeration Systems
T267
Ammonia absorption refrigeration system
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Miscellaneous Refrigeration Systems
Vapour-Compression Heat Pump (HP) System
T268
_Qout = _Wc + _Qin
COP |HP =
_Qout
_Wc= 1 + COP |Ref
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Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME 203 (2017) 23 / 24
Miscellaneous Refrigeration Systems
Gas Refrigeration System: Brayton Cycle
T269
COP =
_Qin
_Wc − _Wt
=(h1 − h4)
(h2 − h1) − (h3 − h4)
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