Refrigeration Cycleszahurul.buet.ac.bd/ME6101/ME6101_RefrigerationCycle.pdf · 2019-07-04 · Refrigeration Capacity/Performance • 1 ton refrigeration: heat absorbed by 1 ton (2000

Refrigeration Cycles

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

http://zahurul.buet.ac.bd/

ME 6101: Classical Thermodynamics

http://zahurul.buet.ac.bd/ME6101/

c

Dr. Md. Zahurul Haq (BUET) Refrigeration Cycles ME6101 (2019) 1 / 24

T266 T273

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Refrigerators, Air-conditioners & Heat Pumps

T270

• COPR =

_QL

_WR: COPAC =

_QL

_WAC: COPHP =

_QH

_WHP

• COPHP = COPR + 1

c


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

c


Basic Refrigeration Cycle Reversed Carnot Cycle

Reversed Carnot Cycle

T639

COP =

_Qin

_Wc − _Wt

=TC∆s

(TH − TC )∆s=

TC

TH − TC

c



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 ◭

c



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.

c



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.

c


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

c



Ideal Vapour Compression Refrigeration Cycle

T641 T640

c



T255

T256

c



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

c



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 ◭

c



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

c



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

0.44R

efri

gera

nt fl

ow r

ate

(kg/

s)

T261

c



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

c


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

c



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.

c



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% ◭

c



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 ◭

c


Miscellaneous Refrigeration Systems

Vapour Absorption Refrigeration System

T272

c



T267

Ammonia absorption refrigeration system

c



Vapour-Compression Heat Pump (HP) System

T268

• _Qout = _Wc + _Qin

• COP |HP =

_Qout

_Wc= 1 + COP |Ref

c



Gas Refrigeration System: Brayton Cycle

T269

COP =

_Qin

_Wc − _Wt

=(h1 − h4)

(h2 − h1) − (h3 − h4)

c


Documents

Refrigeration Cycleszahurul.buet.ac.bd/ME6101/ME6101_RefrigerationCycle.pdf · 2019-07-04 · Refrigeration Capacity/Performance • 1 ton refrigeration: heat absorbed by 1 ton (2000