M.tech Thermal Advanced RAC Chapter 1 VCC Cycles

07-08-2012

1

Basic Gas Cycles and Vapour Compression Refrigeration

M. Tech. Thermal Engg.

Basic Gas Cycles &

Vapour Compression

Refrigeration

Advanced Refrigeration & Air-Conditioning

M. Tech. Thermal Engineering

ME 0611 SEM - I

Dept. of Mechanical Engineering

ME 0611 SEM-I Advanced Refrigeration and Air-Conditioning



Outline

• Applications of Refrigeration.

• Bell – Coleman Cycle.

• COP and Power Calculations

• Vapour – Compression Refrigeration System.

• Presentation on T-S and P-h diagram.

• Cascade Refrigeration System.

Advanced Refrigeration and Air-Conditioning ME 0611 SEM-I

07-08-2012

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Refrigeration

REFRIGERATION – Science of producing and maintaining temperature below that of

surrounding / atmosphere.

REFRIGERATION – Cooling of or removal of heat from a system.

Refrigerating System – Equipment employed to maintain the system at a low temperature.

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.


Basic Vapour Compression and Vapour Absorption Refrigeration


Refrigeration Circuit

Refrigeration Circuit

Evaporator Compressor

Condenser Expansion

Valve


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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 - Applications

1. Ice making.

2. Transportation of food items above and below freezing.

2. Industrial Air – Conditioning.

4. Comfort Air – Conditioning.

5. Chemical and related industries.

6. Medical and Surgical instruments.

7. Processing food products and beverages.

8. Oil Refining.

9. Synthetic Rubber Manufacturing.

10. Manufacture and treatment of metals.

11. Freezing food products.

12. Manufacturing Solid Carbon Dioxide.

13. Production of extremely low temperatures (Cryogenics)

14. Plumbing.

15. Building Construction.

Applications :


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M. Tech. Thermal Engg. Applied Thermodynamics & Heat Engines

Refrigeration Systems

1. Ice Refrigeration System.

2. Air Refrigeration System.

2. 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 :

ME 0611 SEM-I



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

RCOP n

COPlTheoretica

COPActualCOPlative 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.

ME 0611 SEM-I

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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.




Air Refrigeration System

One of the earliest method.

Obsolete due to low COP and high operating cost.

Preferred in Aircraft Refrigeration due to its low weight.

Characteristic :

- Throughout the cycle, Refrigerant remains in gaseous state.

Air Refrigeration

Closed System Open System

• Air refrigerant contained within

piping or components of system.

• Pressures above atm. Pr.

• Refrigerator space is actual room to be cooled.

• Air expansion to atm. Pr. And then

compressed to cooler pressure.

• Pressures limited to near atm. Pr. levels..


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Air Refrigeration System

1. Suction to compressor in Closed System may be at high pressures. Hence,

the size of Expander and Compressor can be kept small.

Closed System Vs. Open System :

2. In Open Systems, air picks up the moisture from refrigeration chamber. This

moisture freezes and chokes the valves.

3. Expansion in Open System is limited to atm. Pr. Level only. No such restriction

to Closed System.




Reverse Carnot Cycle

3

2

1 4

Isotherms

Adiabatic

T2

Expansion

Compression

T1

Pre

ssu

re

Volume

P –V Diagram

3 2

1 4 T1

Expansion Compression

T2

Tem

per

atu

re

Entropy

1’ 4’

T –s Diagram


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3 2

1 4 T1


T2

Tem

per

atu

re

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

1 4 T1


T2

Tem

per

atu

re

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

ExtractedHeatCOP



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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

absorbedHeatCOPrefrig ….ANS

KTKT

K

TT

TCOPrefrig 6.313

235

23599.2 1

212

1

….ANS

Heat Delivered as Heat Pump ;

sec/189.53.13600

/000,143.11 kJ

hrkJkWtonne

doneWorkabsorbedHeat

….ANS

99.33.1

sec/189.5

kW

kJ

doneWork

deliveredHeatCOPHP ….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.

18.5258308

258

12

1

KK

K

TT

TCOPrefrig

….ANS

hrkJInputWork

InputWork

hrkJX

InputWork

tonne

InputWork

EffectfrigCOPrefrig

/32558

/000,14121216.5

.Re

….ANS kWhrkJhrInputWork

Power 04.93600

/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


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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.

46.9265293

265

12

1

KK

K

TT

TCOPrefrig

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.

kgmkJ

kgkJXkgm

kJdoneWork

kJEffectfrig

W

RCOP

iceice

nrefrig

35.813600

)/(6.418)(46.9

)(

)(.Re

….ANS




Bell – Coleman / Reverse Bryaton Cycle

Elements of this system :

1. Compressor.

2. Heat Exchanger.

3. Expander.

4. Refrigerator.

Work gained from Expander is used

to drive Compressor.

Hence, less external work is required.

Heat Exchanger Cooling

Water

Refrigerator

Compressor Expander

Cold Air

Very Cold Air Warm Air

Hot Air


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3 2

1 4

Isobars

Adiabatic

Expansion

Compression

Pre

ssu

re

Volume

P –V Diagram

3

2

1

4

Expansion

Compression

Tem

per

atu

re

Entropy

Isobars

Adiabatic

T –s Diagram





3

2

1

4

Expansion

Compression

Tem

per

atu

re

Entropy

Isobars

Adiabatic

Heat Absorbed in Refrigerator :

)( 41 TTCmQ Padded

Heat Rejected in Heat Exchanger :

)( 32 TTCmQ Prejected

If process changes from Adiabatic to Polytropic;

11221

VPVPn

nQcomp

4433exp1

VPVPn

nQ n

We know,

1PCR


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Net Work Done :

1234

4312

44331122

exp

1

1

1

1

TTTTCmn

n

TTTTRmn

n

VPVPVPVPn

n

WWW

P

ncomp

For Isentropic Process :

1234

exp

TTTTCm

WWW

P

ncomp





COP :

1234

41

1

1

)(

TTTTCmn

n

TTCm

W

Q

QQ

AddedWorkCOP

P

P

net

added

addedrejected

1234

41

1

1

)(

TTTTn

n

TTCOP


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Air Refrigeration Cycle - Merits / Demerits

Merits :

1. No risk of fire (as in case of NH3); as air is non – flammable.

2. Cheaper (than other systems); as air is easily available.

3. Weight per tonne of refrigeration is quite low (compared to other systems).

Demerits :

1. Low COP (compared with other systems).

2. Weight of air (as Refrigerant) is more (compared to other systems).

ME 0611 SEM-I



Example 4

A Bell – Coleman refrigerator operates between pressure limits of 1 bar and 8 bar. Air is

drawn from the cold chamber at 9 ºC, compressed and then cooled to 29 ºC before

entering the expansion cylinder. Expansion and compression follow the law PV1.35 = Const.

Calculate the theoretical COP.

For air, take γ = 1.4 and Cp = 1.003 kJ/kg.

Polytropic Compression 1-2 :

Kbar

barK

P

PTT

n

n

2.4821

8)282(

35.1

135.11

1

212

Polytropic Expansion 3-4 :

KT

bar

barTK

P

PTT

n

n

6.176

1

8)302(

4

35.1

135.1

4

1

4

343

3 2

1 4

PV1.35=C

Pre

ssu

re

Volume

P2

= 8 bar

P1

= 1 bar

282 K

302 K

ME 0611 SEM-I

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Heat Extracted from Cold Chamber :

kgkJKKXkgkJTTCP /7.105)6.176282()/(003.1)( 41

Example 4….cntd

Heat Rejected to Heat Exchanger :

kgkJKKXkgkJTTCP /7.180)3022.482()/(003.1)( 32

Net Work Done :

kgkJW

KKKKkgkJW

TTTTCmn

nW

net

net

Pnet

/8.82

2822.4823026.176)/003.1(4.1

14.1

135.1

35.1

1

11234

27.1/8.82

/7.105

kgkJ

kgkJ

doneWork

absorbedHeatCOPrefrig ….ANS

ME 0611 SEM-I



Example 5

An air refrigeration open system operating between 1 MPa and 100 kPa is required to

produce a cooling effect of 2000 kJ/min. temperature of the air leaving the cold chamber is

-5 ºC, and at leaving the cooler is 30 ºC. Neglect losses and clearance in the compressor

and expander. Determine :

i) Mass of air circulated per min. ii) Compressor Work, Expander Work, Cycle Work.

ii) COP and Power in kW required.

3 2

1 4

PVγ=C

Pre

ssu

re

Volume

P2

= 1 MPa

P1

= 100 kPa 268 K

303 K

Polytropic Expansion 3-4 :

KT

MPa

MPaTK

P

PTT

9.156

1.0

1)302(

4

4.1

14.1

4

1

4

343

Refrig. Effect per kg :

kgkJ

KKXkgkJ

TTCP

/66.111

)9.156268()/(003.1

)( 41

ME 0611 SEM-I

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Example 5….cntd

min/91.17/66.111

min/2000

.Re

.Rekg

kgkJ

kJ

kgperEffectfrig

Effectfrig

Mass of air circulated per min :

….ANS

Polytropic Compression 1-2 : KkPa

kPaK

P

PTT 4.517

100

1000)268(

4.1

14.11

1

212

….ANS

Compressor Work :

min/85.4486

2684.517)/287.0(min)/91.17(14.1

4.1

112

kJW

KKkgkJkgW

TTRmW

comp

comp

comp

….ANS




Expander Work :

min/42.2628

9.156303)/287.0(min)/91.17(14.1

4.1

1

exp

exp

43exp

kJW

KKkgkJkgW

TTRmW

….ANS

Example 5….cntd

Cycle Work = Wcycle = Wcomp – Wexp

= 4486.85 kJ/min – 2628.42 kJ/min = 1858.43 kJ/min…ANS

076.1min/43.1858

min/2000.Re

kJ

kJ

requiredWork

EffectfrigCOPrefrig

….ANS

Power required :

kWkJ

time

WP

cycle97.30

minsec/60

min/43.1858 ….ANS


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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

2 3

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.


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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

Tem

per

atu

re,

T Condensation

Compression

Evaporation

Expansion

T2

T1

3

4

2

1

Compressor Work,

(W)

Net Refrig. Effect,

(Rn)

Sat. Vapour Line

Sat. Liq. Line

f g

12

41

14321

141

hh

hh

Area

fgArea

DoneWork

AbsorbedHeatCOP





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

AbsorbedHeatCOP

Entropy, s

Tem

per

atu

re,

T Condensation

Compression

Evaporation

Expansion

T2

T1

3

4

2

1

Compressor Work,

(W)

Net Refrig. Effect,

(Rn)

Sat. Vapour Line

Sat. Liq. Line

f g

2’

NOTE : h2 = h2’ + Cp (Tsup – Tsat)


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Case C. Wet Vapour after Compression :


= W = Area 1-2-3-4-1

Heat Absorbed

= W = Area 1-4-g-f-1

Entropy, s

Tem

per

atu

re,

T Condensation

Compression

Evaporation

Expansion

T2

T1

3

4

2

1

Compressor Work,

(W)

Net Refrig. Effect,

(Rn)

Sat. Vapour Line

Sat. Liq. Line

f g

12

41

14321

141

hh

hh

Area

fgArea

DoneWork

AbsorbedHeatCOP

NOTE : h2 = (hf + x.hfg)2




Vapour Compression System : P-h Diagram

Enthalpy, h

Pre

ssu

re,

Pr

Iso

therm

al,

T =

Co

nst

Isen

thalp

ic,

h =

Co

nst.

Isobaric,

P = Const

Sub-cooled

Liq. region 2 – phase

region

Superheated

region


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Vapour Compression System : P-h Diagram

Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

4

Evaporation

Condensation E

xp

an

sion

12

41

hhW

hhRn

12

41

hh

hh

W

RCOP n




Factors Affecting Vapour Compression System

A. Effect of Suction Pressure :

Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

4

1’ 4’

2’

P1

P2

12

41

hh

hh

W

RCOP n

COP of Original Cycle :

COP when Suction Pr. decreased :

2'2'1112

'1141

'1'2

'4'1

hhhhhh

hhhh

hh

hh

W

RCOP n

Thus,

Refrig. Effect ↓

Work Input ↑

COP ↓


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B. Effect of Delivery Pressure :

12

41

hh

hh

W

RCOP n


COP when Delivery Pr. increased :

2'212

4'441

1'2

'41

hhhh

hhhh

hh

hh

W

RCOP n

Thus,

Refrig. Effect ↓

Work Input ↑

COP ↓

Enthalpy, h

Pre

ssu

re,

Pr

1

2

3

4

3’

4’

2’

P1

P2





C. Effect of Superheating :

12

41

hh

hh

W

RCOP n


Thus,

Refrig. Effect ↑

Work Input ↑ or ↓

COP ↓ or ↑

Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

4

1’

2’

P1

P2

1'12'212

1'141

'1'2

4'1

hhhhhh

hhhh

hh

hh

W

RCOP n



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D. Effect of Sub-cooling :

12

41

hh

hh

W

RCOP n


Thus,

Refrig. Effect ↑

Work Input : SAME

COP ↑

12

'4441

12

'41

hh

hhhh

hh

hh

W

RCOP n


Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

4

1’

3’

P1

P2

4’





E. Effect of Suction & Condenser Temperatures :

Now, Condenser Temp. ↓

Evaporator Temp. ↑

12

41

'1'4'3'2'1

1'44'11

hh

hh

Area

fgArea

DoneWork

AbsorbedHeatCOP

COP of Modified Cycle :


12

41

14321

141

hh

hh

Area

fgArea

DoneWork

AbsorbedHeatCOP

COP ↑

Entropy, s

Tem

per

atu

re,

T Condensation

Compression

Evaporation

Expansion

T2

T1

3

4

2

1’

f g

1

4’

2’ 3’


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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

tonnekghh

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

141

tonnemvhh

DisplPistonTh g




Vapour Compression System – Mathematical Analysis

)(

)/(

12

12

kWhhmP

kgkJhhW

theor

comp

a) Isentropic Compression :

D. Theoretical Power Required :

)(1

)/(1

1122

1122

kWVPVPn

nmP

kgkJVPVPn

nW

theor

comp

a) Polytropic Compression :

E. Heat removed through Condenser :

)/(32 kgkJhhmQcond


07-08-2012

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Example 7 Example 6 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 6….cntd

Entropy, s

Tem

per

atu

re,

T

298 K

268 K

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

kgkJhxhh fgf /8.1230.154)6.0(4.31111

kgkJhxhh fgf /2.1330.138)5325.0(7.5922 22

5325.0

268

0.1546.01251.0

298

0.1382232.0

2

2

1

1

11

2

2

22

111222

12

x

x

T

hxs

T

hxs

sxssxs

ss

fg

f

fg

f

fgffgf

Isentropic Compression : 1-2

kgkJhh f /7.5934


82.6/8.1232.133

/7.598.123

12

41

kgkJ

kgkJ

hh

hh

W

RCOP n


07-08-2012

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Example 6….cntd

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

Tem

per

atu

re,

T

298 K

268 K

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

Now;

….ANS

hrsintonneXX

kg

kgkJ

kgkJXkgm

hhm

Xm

W

RCOP

ice

iceactualn

actual

24661.01000

2460459.0

min/459.0

41.3/74.418

)/(8.1232.133)(6

74.41841.3

12

)(




Example 7

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 25 ºC to -15oC. The vapour 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 :


07-08-2012

24



91.823.119622.1319

04.10023.1196

12

41

hh

hhCOP ltheoreticaCOP of the Cycle :

Entropy, s

Tem

per

atu

re,

T

298 K

258 K

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

Example 7….cntd

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

2

1

1112

12

x

x

sxss

ss

fgfg


kgkJhh g /22.131922




Example 7….cntd

Actual COP = ηrel X COPtheor

= 0.62 X 8.91

= 5.52

Entropy, s

Tem

per

atu

re,

T

298 K

258 K

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

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 : kgkgkJ

kJ1599.0

)/(9.678

sec)/(56.108

Total Work done by Compressor :

)(sec/67.19

/)23.119622.1319()(1599.012

kWkJ

kgkJXkghhXmrefrig

….ANS


07-08-2012

25



Example 8

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 8….cntd


94.2/1.1830.220

/53.741.183

12

41

kgkJ

kgkJ

hh

hh

W

RCOP 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

Tem

per

atu

re,

T

40 ºC

-10 ºC

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

2’


07-08-2012

26



Example 9

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 9….cntd

Refrigerating Capacity :

sec/192.0

/53.7461.1782041

kgm

kgkJXmkWhhm

KT

T

T

TCss

ss

P

2.324

313ln747.0682.07082.0

ln

2

2

'2

2'21

21


kgkJ

KKkgkJkgkJ

TTChh P

/4.211

0.3132.324./747.0)/(05.203

'22'22

Entropy, s

Tem

per

atu

re,

T

313 K

253 K

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

2’


07-08-2012

27



Example 9

….ANS

Power Required :

kW

kgkJXkghhm

29.6

/61.1784.211sec)/(192.012

Vol. Efficiency :

%6.87

509.1

607.903.003.01

1

13.1/1

/1

bar

bar

P

Pkk

n

S

dvol

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

Tem

per

atu

re,

T

313 K

253 K

3

4

2

1 Sat. Vapour Line

Sat. Liq. Line

f g

2’




Example 10

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 :


07-08-2012

28



Example 10….cntd

kgkJ

kgkJkgkJ

TTChhh subcoolsatliqPf

/29.343

)/(30330856.405)/(07.366

34'3

KT

T

T

TCssss P

8.371

308ln903.22086.5755.5

ln

2

2

'2

2'2121


kgkJ

KKkgkJkgkJ

TTChh P

/8.1673

0.3088.371./903.2)/(57.1488

'22'22

Entropy, s

Tem

per

atu

re,

T

308 K

263 K

3

4

2

1 Sat. Vapour Line

Sat. Liq. Line

f g

2’

3’ 303 K




Example 10….cntd

sec/04517.0

/29.34322.1450

)(50

/

)(50

41

kg

kgkJ

kW

kgkJhh

kWm

Mass of Refrigerant :

Compressor Power :

kW

kgkJXkg

hhm

1.10

/22.14508.1673)(04517.0

12

….ANS

Cylinder Dimensions :

mmL

mD

kgm

rpmDD

v

NLD

kgmg

228.0)19.0(2.1

19.0

/417477.0

60

)(1000)2.1(

4604sec)/(04517.0

3

22

….ANS

….ANS

Entropy, s

Tem

per

atu

re,

T

308 K

263 K

3

4

2

1 Sat. Vapour Line

Sat. Liq. Line

f g

2’

3’ 303 K


07-08-2012

29



Comparison of Carnot Vs. VCC

Assumption : STATE 1 and STATE 3 are

common for both cycles.

A1 A1 : Additional Work required due to

superheated section

A2 A2 : Additional Work required, as no work

is recovered during expansion. A3

A3 : Loss in cooling effect due to throttling

as compared to isentropic expansion.

Advanced Refrigeration and Air-Conditioning

Entropy, s

Tem

per

atu

re,

T

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

2’

4’ 4”

ME 0611 SEM-I



Comparison of Carnot Vs. VCC

on J/kg basis :

Entropy, s

Tem

per

atu

re,

T

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

2’

4’ 4”

'213'221 ssThhA

A1

'432 hhA

A2

'443 hhA

A3

Areas A2 and A3 are EQUAL.

Throttling causes identical dual loss..!!!.

2121 AAWW comp

314 AQW comp


C

C

rev

R

W

AA

QA

COP

COP

21

3

1

/1

ME 0611 SEM-I

07-08-2012

30


M. Tech. Thermal Engg. Advanced Refrigeration and Air-Conditioning

Example 11

A refrigerant 12 theoretical single stage cycle operates between a condensing temperature

of 32.2 ºC and an evaporating temperature of -17.78 ºC. Assume a Carnot cycle operating

between the same temperatures. Determine :

1. Carnot cycle work of compression.

2. Carnot cycle refrigerating effect.

3. Excess work of compression due to superheat section.

4. Excess work of compression due to throttling.

5. Loss in refrigeration effect due to throttling.

6. Refrigeration efficiency.

Entropy, s

Tem

per

atu

re,

T

3

4

2

1

Sat. Vapour Line

Sat. Liq. Line

f g

2’ A1

A2

A3 4’

4”

ME 0611 SEM-I



Example 11….cntd

1.

kgkJ

ssTTWC

/986.22

033.2477656.70678.1722.32

3112

…ANS.

2.

kgkJ

ssTQC

/332.117

033.2477656.70622.255

311

…ANS.

kgkJ

ssThhA

/1483.0

667.6847656.70622.30543.20033.207

'213'221

3.

…ANS.

ME 0611 SEM-I

07-08-2012

31



Example 11….cntd

2653.0889.807656.706

889.80033.247

"41

"43'4

"41'4"4'43

ss

ssx

ssxsss

4. To evaluate A2, the value of h4’ is to be calculated.

…ANS.

kgkJ

hhxhh

/242.62

817.1973.1792653.0817.19

"41'4"4'4

kgkJ

hhA

/438.4

242.6278.66

'432

…ANS.

5. Since, A3 = A2, we have A3 = 4.438 kJ/kg.

ME 0611 SEM-I



6. Refrigerating efficiency can be calculated in 2 different ways :

Example 11….cntd

8.0

22.25561.27

5095.112

112

1341

Thh

TThhR

…ANS.

8.0

986.22

438.41483.01

332.117/438.41

1

/1

21

3

C

CR

W

AA

QA

…ANS.

OR by Equation :

ME 0611 SEM-I

07-08-2012

32



Multistage Vapour Compression Cycle

To overcome the limitations of the single – stage cycle.

At low Evaporator Temperatures, staging of Compressor is necessary

due to Vol. Efficiency in single – stage cycles.

Compressor staging and vapour intercooling

to excessive discharge temperatures.

Also helps for COP.

ME 0611 SEM-I




Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

9

Evaporation

Condensation

Exp

an

sion

4

5 6

7 8

Evaporator

LP

Compressor

1

2

HP

Compressor

4

5

Condenser 6

7

8

9

NOTE : Typical for AMMONIA and R12

__________

Flash

Intercooler

3 Water

Intercooler

ME 0611 SEM-I

07-08-2012

33



Example 12

Calculate the power needed to compress 10 kg/min of Ammonia from saturated vapour at

1.4 bar to a condensing pressure of 10 bar :

(a) by single – stage compression

(b) by two – stage compression with intercooling by liquid refrigerant at 4 bar.

Assume saturated liquid to leave condenser and dry saturated vapour to leave evaporator.


4 7

8

__________

2

Enthalpy, h

Pre

ssu

re,

Pr

1

2

3

8

Exp

an

sion

4

5 6

7

-27 ºC

-2 ºC

25 ºC

1.4 bar

10 bar

ME 0611 SEM-I



Example 12….cntd

From P – h Chart for Ammonia, we get :

SINGLE – STAGE :

h6= h7 = -645.0 kJ/kg h1 = 464.5 kJ/kg h3 = 753.5 kJ/kg

h5 = 623.5 kJ/kg.K h2 = 602.5 kJ/kg h4 = 498.0 kJ/kg

Higher stage compressor must compress 10 kg/min plus

the quantity of liquid which evaporates to de-superheat

the gas at 2.

The rate of Ammonia compressed can be computed by

making a heat and mass balance about intercooler.

4

(h4 = 498 kJ/kg)

(0.9142 kg/min) 7

(h6 = h7 = -645 kJ/kg)

8

__________

2

(h2=602.5 kJ/kg)

(10 kg/min)


07-08-2012

34



1. Heat Balance : 4985.602min/10645 46 mkgm

2. Mass Balance : 46 10 mm

49804985.6026025645 66 mm

Example 12….cntd

Property W/o Intercooling With Intercooling

1 – 2, 2 – 3 1 – 2, 2 – 4, 4 – 5

h2 – h1, kJ/kg 602.5 – 464.5 = 138 602.5 – 464.5 = 138

h3 – h2, kJ/kg 753.5 – 602.5 = 151 ---

h5 – h4, kJ/kg --- 623.5 – 498.0 = 125.5

Kg/min from 1 – 2 10 10

Kg/min from 2 – 3 10 ---

Kg/min from 4 – 5 --- 10.91426

Power (kJ/min), 1 – 2 10 X 138 = 1380 10 X 138 = 1380

Power (kJ/min), 2 – 3 10 X 151 = 1510 ---

Power (kJ/min), 4 – 5 --- 125.5 X 10.91426 = 1369.7

Total Power, (kJ/min) 2890 2747.7




Example 12….cntd

Thus,

Power requirement without liquid refrigerant intercooling = 2890 kJ/min

Power requirement with liquid refrigerant intercooling = 2747.7 kJ/min


07-08-2012

35




Evaporator

LP

Compressor

Water

Intercooler

__________

HP

Compressor

Condenser

1

2

3

4

5 6

7

8

9

Flash

Intercooler

Flash Intercooler :

Tank with fixed liquid level.

Liquid level maintained by a Float Valve

= Expansion Valve also.

Sat. Liquid at intermediate pressure is

expanded to Evaporator pressure.

Flash vapour @ Throttling is given as

suction to HP Compressor.

Vapour @ 3 is bubbled through orifices in

the Flash Intercooler and de-superheated

by evaporation of liquid.

Vapour @ suction to HP Compressor :

1. Flash Vapour.

2. Refrigerant in evaporator

3. Vapour due to evaporation of

liquid in intercooler.

ME 0611 SEM-I




Evaporator

LP

Compressor

Water

Intercooler

__________

HP

Compressor

Condenser

1

2

3

4

5 6

7

8

9

Flash

Intercooler

Disadvantages of Flash Intercooler :

Liquid refrigerant in tank is at

intermediate pressure and saturated.

1. Evaporator is above Intercooler.

2. Heat is absorbed in liquid line.

1. Some liquid evaporates ahead

of Expansion Valve.

2. Operation of Expansion Valve

is sluggish due to low pressure

differential.

ME 0611 SEM-I

07-08-2012

36




NOTE : Typical for AMMONIA


Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

9

Evaporation

Condensation

Exp

an

sion

4

5 6

7

8

Evaporator

LP

Compressor

Water

Intercooler

HP

Compressor

Condenser

1

2

3

4

5 6

7

8

9

_______

Shell-&-Coil

Intercooler

ME 0611 SEM-I




Evaporator

LP

Compressor

Water

Intercooler

HP

Compressor

Condenser

1

2

3

4

5 6

7

8

9

_______

Shell-&-Coil

Intercooler

Advantages of Shell-&-Coil Intercooler :

Subcools the liquid refrigerant.

Eliminates possibility of flash

liquid ahead the Expansion Valve.

Large pressure differential, as the

liquid is at Condenser pressure.

Limitation of Shell-&-Coil Intercooler :

Low COP, as not possible to

intercool the liquid as much as in

Flash Intercooler.

ME 0611 SEM-I

07-08-2012

37




NOTE : Typical for R12 and R22


Evaporator

LP

Compressor

HP

Compressor

Condenser

1

2

4

5 6

7

8

9

_______

Shell-&-Coil

Liquid

Intercooler

3

Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

9

Evaporation

Condensation

Exp

an

sion

4

5 6

7

8

ME 0611 SEM-I





Evaporator

LP

Compressor

HP

Compressor

Condenser

1

2

4

5 6

7

8

9

_______

Shell-&-Coil

Liquid

Intercooler

3

Vapour from LP Compressor is not

intercooled.

Vapour from LP Compressor is

mixed with refrigerant from

Intercooler.

In stead of Float Valve, a Thermostatic

Expansion Valve is used.

ME 0611 SEM-I

07-08-2012

38





Enthalpy, h

Pre

ssu

re,

Pr

1

2 3

9

Evaporation

Condensation

Exp

an

sion

4,10,11

5 6

7,12

8

Low Temp

Evaporator

LP

Compressor

HP

Compressor

Condenser

1

2

4

5 6

7

8

9

_______

Shell-&-Coil

Liquid

Intercooler

3 Water

Intercooler

High Temp

Evaporator 10

11

12

ME 0611 SEM-I





No. of stages of Compression :

Economic as well as Practical Considerations.

R12, R22 and Ammonia : Single – Stage

Evaporator Temp above – 30 ºC.

R12, R22 : Single – Stage more successful at lower temp than that for Ammonia.

Two – Stage : Evaporator Temp between – 60 ºC to – 30 ºC.

Three – Stage : Evaporator Temp below – 65 ºC.

ME 0611 SEM-I

07-08-2012

39



Cascade Refrigeration System

Entropy, s

Tem

per

atu

re,

T

3a

4a

2a

1a

2’a

3b

4b

2b

1b

2’b

Evaporator

LP

Compressor

1a

2a 3a

4a

HP

Compressor

1b

2b

Condenser 3b

4b

Cascade

Condenser

ME 0611 SEM-I




Evaporator

LP

Compressor

1a

2a 3a

4a

HP

Compressor

1b

2b

Condenser 3b

4b

Temperatures below -65 °C

Two Independent Systems.

Cascade Condenser : Heat Exchanger

Advantage :

Possibility of Multistaging each stage…!!!

Limitation :

Efficiency loss due to Temperature

Overlap in Cascade Condenser.

Applications :

1. Liquefaction of Petroleum Vapour.

2. Liquefaction of Air / Atm. Gases.

3. Manufacture of Dry Ice.

ME 0611 SEM-I

07-08-2012

40


M. Tech. Thermal Engg. ME 0611 SEM-I Advanced Refrigeration and Air-Conditioning

It is one of the oldest Refrigeration System.

Production of Low ( Cryogenic) Temperature Applications.

Two or more Refrigeration Cycles operate in series.



M. Tech. Thermal Engg. ME 0611 SEM-I

Thermodynamic Cycle


07-08-2012

41



Experimental Setup for Cascade System




Experimental Setup for Cascade System


07-08-2012

42



Performance of Cascade System


Sr. No. Parameter Value Parameter Value

1. Te,CO2 -50 0C Te,NH3 -20.96 0C

2. Tc,CO2 -17.48 0C Tc,NH3 29.72 0C

3. COP,CO2 2.40 COP,NH3 2.14



Comparison – Multistage VCC

07-08-2012

43



Precooled Linde – Hampson Cycle

(1)

Makeup Gas

Heat Exchanger Heat Exchanger

J – T Valve

J – T Valve

Main

Compressor

Refrigerant

Compressor

Cooling

Water

Liquid

(2) (3)

(4)

(5)

(f)

(g)

(6) (1)

(a)

(a) (b) (c)

(d)

3

4

Tem

p, T

Entropy, s

1 2

5

6

f g

Refrigerant

Boiling

Point



Cascade System for Air Liquefaction

Makeup Gas

Liquid N2

Cooling Water

Ammonia, NH3

Ethylene, C2H4

Methane, CH4

07-08-2012

44



Cascade System for LNG Liquefaction




Advantages / Limitations

Temperature Overlap in Cascade Condenser .

Limitation :

Advantages :

It permits the use of Two / more different Refrigerants, depending on

their Boiling Points.

Inclusion of Intercooling between every two stages is possible, so it

improve performance of system.

Each cycle is operating separately ( i.e. Pr, Freq, compressor Input

Power, etc).

07-08-2012

45



Thank You !




References

Randall Barron, ‘Cryogenic Systems’, Oxford University Press, 2nd ed.,

1981, pp. 69 – 85.

Roy J. Dossat, ‘Principles of Refrigeration’, Thomson Press, 2nd SI ed., pp.

542 – 548.

Aurora S. C. and Domkundwar S. , ‘Refrigeration and Air Conditioning’,

Dhanpat Rai & Sons, Delhi, 8th ed. pp. 1 – 4.

Cengel Y. & Boles C. , ‘Thermodynamics’, Tata McGraw – Hill Publication,

Sixth ed., pp. 636-638.

ME 0611 SEM-I

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