1

Applied Thermodynamics

2

6. REFRIGERATION Definition Refrigeration is the process of providing and

maintaining temperature of the system below that of the surrounding atmosphere.The refrigeration effect can be accomplished by non – cyclic processes, making use of substances at temperature well below the temperature of the surroundings – e.g., ice, snow, dry ice (solid CO2) etc.

However, of greater importance are cyclic refrigeration systems, wherein the cooling substance (called refrigerant) is not consumed and discarded, but used again and again in a thermodynamic cycle.

3

REFRIGERATION

4

A ton of refrigeration is defined as the quantity of heat required to be removed to produce one ton (1000kg) of ice within 24 hours when the initial condition of water is 00C.

Consider a refrigerator of T tons capacity, Refrigeration capacity = 3.5 kJ/sHeat removed from refrigerator = Refrigeration effect

=R.E. kJ/sPower of the compressor =work/kg of refrigerant x mass flow rate

kJ/s 5.3360024

3351000ionrefrigerat of

x

xTon

5

Reversed Heat Engine Cycle A reversed heat engine is a potential refrigerating machine. It receives heat from a low temperature region at T2, discharge heat to a high temperature region at T1, and requires a net inflow of work. Removal of heat from a low temperature region reduces the temperature of that region below the temperature of the surroundings, thus producing refrigeration.

6

According to First Law Q2 – Q1 = -W i.e., Q1 = Q2 + W

Such a device is called a Refrigerator or Heat Pump, depending on whether the focus is on heat received from the low temperature region Q2 or the heat discharged to the high temperature region Q1. Q2 is known as the refrigeration effect.

The performance of a refrigerator/heat pump is measured by means of its coefficient of performance (COP). COP of a refrigeration/heat pump is defined as

The working fluid in a refrigeration cycle is called a Refrigerant.

7

Important application of Refrigeration1. Ice plants2. Food processing units and transportation, including

dairies3. Industrial air – conditioning4. Comfort air – conditioning5. Chemical and related industries.6. Hospitals.7. Laboratories.8. Domestic applications

Basic processes (operations) in a Refrigeration CycleSince a refrigeration cycle is essentially a reversed heat engine cycle, the working substance (refrigerant) will undergo the following basic operations.

1. Compression - resulting in increase in pressure and temperature.

2. Heat rejection at high temperature.3. Expansion – resulting in reduction in pressure and

temperature and4. Heat addition at low temperature – during which heat

is transferred from the body to be cooled to the refrigerant.

8

Vapour Compression Refrigeration CycleIn this, the refrigerant used is a vapour (e.g., ammonia, Freon-22, Freon-11, Freon -12 etc). The refrigerant undergoes the following operations in a cyclic manner.

1. Compression in a compressor (Usually reciprocating), with work input.

2. Condensation of the vapour into liquid in a condenser, wherein heat is rejected to a cooling medium (air, water) at high pressure and temperature.

3. Expansion of the liquid refrigerant in a suitable device (engine, expansion valve, capillary etc). There may or may not be work output. The liquid may evaporate partially.

4. Evaporation of the mixture of liquid and vapour in an evaporator where heat is added to the refrigerant from the substance to be cooled, producing the necessary refrigeration effect.

9

4-1: Reversible adiabatic (isentropic) compression, with work

input WC.

1-2: Condensation at constant pressure and temperature with heat Q1 rejected to some cooling medium.

2-3: Reversible adiabatic expansion, with work output WE.

3-4: Evaporation at constant pressure and temperature wherein heat Q2 is absorbed from the substance to be cooled.

10

Reversed Carnot Cycle as a Refrigeration Cycle

11

Q1 = area under 2 – 3 = Tmax (s2 – s3)

Q2 = area under 4 – 1 = Tmin (s1 – s4)

= Tmin (s2 – s3), s1 = s2 & s3 = s4

Wnet = WC – WE = Q1 – Q2

= (Tmax-Tmin) (s2-s3)

12

These are the maximum values for any refrigerator or heat pump operating between two fixed temperatures Tmax and Tmin. In other words, no refrigerator/ heat pump has a COP greater than that of a Carnot refrigerator/heat pump, operating between the same maximum and minimum temperatures.

When the refrigerator/heat pump operates on a cycle other than a Carnot cycle, the heat rejection (condensation) and heat addition (evaporation) process may not be isothermal. Then the COPs are given by

13

Where Tcond = average temperature during condensation.

Tevap = average temperature during evaporation.

It can be seen that the closer the temperatures Tcond and Tevap,

the higher the COP.

14

In practice, an expansion engine is not used in a vapour compression refrigeration unit.

This is because; the power output of such an engine is too small to justify its cost. Instead, some kind of expansion device – like a throttling valve or a capillary tube – is used to reduce the pressure and temperature of the refrigerant.

15

• The most convenient property diagram.

16

17

Process 1-2 or 1’–2’: Reversible adiabatic compression. Process 1–2, starting with saturated vapour (state 1) and ending in the superheated region (state 2) is called Dry compression. Process 1’-2’, starting with wet vapour (state 1’) and ending as saturated vapour (state 2’) is called wet compression. Dry compression is always preferred to wet compression. .

18

Process 2-3 (or 2’–3): Reversible constant pressure heat rejection, at the end of which the refrigerant is in saturated liquid state. 2–2’ is desuperheating, and 2’-3 is condensation.

19

Process 3-4: Adiabatic throttling process, for which enthalpy before is equal to enthalpy after throttling. This process is adiabatic but not isentropic. Since it is irreversible, it cannot be shown on a property diagram. States 3 and 4 are equilibrium points and are simply joined by a dotted line following a constant enthalpy line.

20

Process 3-4: Adiabatic throttling process, for which enthalpy before is equal to enthalpy after throttling. This process is adiabatic but not isentropic. Since it is irreversible, it cannot be shown on a property diagram. States 3 and 4 are equilibrium points and are simply joined by a dotted line following a constant enthalpy line.

21

Analysis:The compressor, the condenser and the evaporator can be treated as steady–flow devices, governed by the Steady Flow Energy Equation. Application of S.F.E.E. to these devices results in:

Compressor: Process 1–2 isentropic

Q1-2 = 0 W1-2 = - ∆h

W1-2 = - (h2 - h1)

Compressor work WC = (h2-h1) kJ/kg, on a unit mass basis.

If mr is the mass flow rate of the refrigerant in kg/sec, then the power input in the compressor is given by

Power input = mr (h2 – h1) kW

22

Condenser: Process 2–3: reversible constant pressure process W2-3 = 0

Q2-3 = ∆h = (h3 – h2) kJ/kg

This is negative i.e., heat rejected.Heat rejected per unit mass of the refrigerant isQ1 = (h2 - h3) kJ/kg.

Rate of heat rejection Q1 = mr (h2 - h3) kJ/sec

23

Evaporator: Process 4 – 1: reversible constant pressure process , W4-1 = 0

Q4-1 = ∆h = (h1 - h4) kJ/kg

Heat received by unit mass of the refrigerant= heat received from the substance being cooled= Q2 = (h1 – h4) kJ/kg of refrigerant.

Rate of heat removed = Refrigerating effect = Q2 = m r(h1 - h4) kJ/sec

Refrigerating Effect in terms of refrigeration

24

Expansion: for process 3 – 4, h3 = h4

but, it is not a constant enthalpy process.Note: Values of enthalpy h1, h2, h3 & h4 can be obtained

from property Tables or Property Charts (Diagrams).

25

Actual Vapour Compression Refrigeration Cycle:A constant amount of superheating of the vapour before it enters the compressor is recommended. This is to ensure that no liquid refrigerant droplets enter the compressor. Further, a small degree of sub cooling (under cooling) of the liquid refrigerant at the condenser exit is desirable, in order to reduce the mass of vapour formed during expansion. Excessive formation of vapour bubbles may obstruct the flow of liquid refrigerant through the expansion valve.

26

Both the superheating at the evaporator outlet and the subcooling at the condenser outlet contribute to an increase in the refrigerating effect. However, the load on the condenser also increases. There will be an increase in the compressor discharge temperature. Since the compressor input more or less remains unchanged, the COP of the cycle appears to increase due to this superheating/subcooling.

However, for a fixed temperature of the refrigerated space, the evaporation temperature must be lowered (i.e., Tevap is reduced). Further, for a fixed temperature of the cooling medium, the condensation temperature must be raised (i.e., Tcond will be higher). Hence COP will reduce.

27

Refrigerants and desirable properties:

The most commonly used refrigerants are a group of

halogenated hydrocarbons, marketed under various

proprietary names of freon, genetron, arcton etc.

Among them Freon–22 (Mono-chloro Difluoro Methane),

Freon–11 (Tri-chloro – mono-fluoro methane) & Freon–12

(Dichloro Difluoro methane) are extensively used.

Ammonia is another commonly used refrigerant. Other

refrigerants include CO2, SO2, Methyl chloride, Methylene

chloride, Ethyl chloride etc.

28

Desirable properties of a good refrigerant:Thermodynamic properties1. Low boiling point2. Low freezing point3. Positive gauge pressure in condenser and evaporator,

but not very high4. High latent heat of vaporization Chemical properties5. Non–toxic6. Non–inflammable & non–explosive7. Non–corrosive8. Chemically stable9. No effect on quality of stored products

29

Desirable properties of a good refrigerant: Physical properties.1. Low specific volume of vapour2. Low specific heat3. High thermal conductivity4. Low viscosity

Other properties5. Ease of leakage detection6. Cost7. Ease of handling

30

Ammonia is a good refrigerant with the highest refrigerating effect per unit mass.It is relatively cheap. But it is toxic and corrosive. Leakage can be easily detected because if its pungent odour. Freons are Non–toxic & non–inflammable. Leakage cannot be detected easily as they are odour less and colour less. Some coloured additives are sometimes mixed with Freons to facilitate detection of leakage.

31

Gas (Air) Cycle Refrigeration:Refrigeration can also be accomplished by means of

a gas cycle, the most common being the one using air as a refrigerant. In such a cycle, a throttle valve cannot be used for expansion of the working fluid. During the throttling process, enthalpy at the beginning is equal to enthalpy at the end. For an ideal gas, (all gases including air are assumed to be ideal), enthalpy is a function of temperature only. Hence, during throttling temperature at the beginning will be equal to temperature at the end.

32

Gas (Air) Cycle Refrigeration contin…..:Since there is no cooling of air during expansion,

refrigeration is not possible. In place of a throttle valve, an expander is used. Work output obtained from the expander can be utilized for compression, thus decreasing the net work input. In a gas refrigeration cycle, the refrigerant (gas/air) remains in a gaseous state throughout the cycle. Since there is no phase change, the terms ‘condenser’ and ‘evaporator’ are not appropriate. The device in which heat is rejected at a higher temperature can be called a cooler, while the device in which heat is absorbed at a lower temperature is called the ‘refrigerator’.

33

Reversed Carnot CycleA reversed Carnot cycle

using air as the working substance can be a Refrigeration cycle, through it is not practicable.1 – 2: isentropic compression.2 – 3: heat rejection at constant temperature.3 – 4: Expansion4 – 1: heat addition at constant temperature (refrigeration)

34

35

36

Heat rejected during process 2 – 3= Q1 = Tmax (s2-s3)

= Tmax (s1-s4)

Heat received during process 4 – 1= Q2 = Tmin (s1-s4)

Wnet = WC - WE = Q1 - Q2 (First Law)

= (Tmax - Tmin) (s1-s4)

These COPs are the maximum possible COPs for given maximum and minimum temperatures.

37

Reversed Brayton Cycle.A reversed Brayton cycle with air as the working substance is a more practical refrigeration cycle.1 – 2: isentropic compression2 – 3: constant pressure heat rejection3 – 4: isentropic expansion4 - 1: constant pressure heat addition.

38

39

On a unit mass basis,Compressor work input = WC = h2 - h1 = Cp (T2 - T1)

Expansion work output = WE = h3 - h4 = Cp (T3 - T4)

Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)

Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)

40

On a unit mass basis,Compressor work input = WC = h2 - h1 = Cp (T2 - T1)

Expansion work output = WE = h3 - h4 = Cp (T3 - T4)

Heat rejected at constant pressure = Q1 = h2 - h3 = Cp (T2 - T3)

Heat received at constant pressure = Q2 = h1 - h4 = Cp (T1 - T4)

41

For the isentropic process 1 – 2,

For the isentropic process 3-4,

42

The COP of a gas cycle refrigeration system is low. The power required per unit capacity is high. Its main application is in aircrafts and missiles, where a vapour compression refrigeration system becomes heavy and bulky. Another application of gas cycle refrigeration is in the liquefaction of gases.

Shown below is a schematic flow diagram of an open cycle air refrigeration system.

43

A small amount of compressed air is blend from the main compressor of a turbojet or a supercharged aircraft engine, and is cooled by rejecting heat to large amounts of cooler ambient air.

The cooled compressed air expands in an expander, and as a result cools further.

The cool air enters the cabin. The output of the expander is used to run a blower which sucks the ambient air in.

In addition to cooling, replacement of stale air in the cabin is possible. At high altitudes the pressurization of the cabin is also possible. Because of this consideration air cycle refrigeration is extensively used in aircrafts.

44

Vapour Absorption RefrigerationA vapour absorption refrigeration system uses a refrigerant as well as an absorbent which can be a liquid or solid.

Possibly the best known combination is ammonia as the refrigerant and water as the absorbent.

A vapour absorption refrigeration system does not have a compressor.

The compressor is replaced by a combination of generator, an absorber and a pump.

45

46

Working:The generator acts as a reservoir for the solution of ammonia in water.

Heat from an external source QG is supplied to the solution, leading to evaporation of ammonia and water.

The mixture of ammonia vapour and water vapour rises through the analyzer, where most of the water vapour condenses, gets separated from NH3 and drops back into the generator.

The analyzer is a direct–contact heat exchanger consisting of a series of trays mounted above the generator.

The strong solution of NH3 from the absorber flows down over the trays, comes into contact with and cools the rising vapours. Since the saturation temperature of water is higher than that of NH3 at a given pressure, water vapour will condense first.

47

As the vapour passes upward through the analyzer, it is cooled and enriched by ammonia.

The ammonia vapour leaving the analyzer may still contain traces of water vapour.

If allowed to flow through the condenser and expansion valve, the water vapour will freeze and block the expansion valve.

Traces of water vapour are separated from ammonia vapour in the rectifier.

The rectifier is a water cooled heat exchanger, wherein all of the remaining water vapour and some ammonia vapour condense and return to the generator through the drip line.

48

The net result is that pure ammonia vapour flows into the condenser and condenses to form saturated or slightly under cooled liquid.

The refrigerant then expands through the valve, resulting in a drop in its pressure and temperature.

The cold refrigerant then flows through the evaporator, extracting heat from the substance to be cooled.

Saturated or slightly saturated ammonia vapour from evaporator flows into the absorber.

The weak solution of ammonia (with low concentration of ammonia in water) coming from the generator is sprayed into the absorber.

The ammonia vapour comes into contact with the weak solution, and gets readily absorbed, releasing the latent heat of condensation.

49

This heat QA taken away by cooling water, thereby maintaining the temperature in the absorber constant.

The resulting strong NH3 solution is pumped to the generator, where heat Qa is supplied to it from an external source.

The weak solution leaving the generator and the pressurized strong solution going to the generator flow through a heat exchanger.

In this heat exchanger, the strong solution is preheated while the weak solution is pre-cooled, reducing both Qa, the heat to be supplied in the generator and QA the heat to be removed in the absorber.

50

The combination of the generator and absorber is equivalent to a heat engine, which does the job of the compressor, namely, receiving from the evaporator at low pressure, comparatively low temperature ammonia vapour and delivering high pressure, higher temperature ammonia vapour to the condenser. This is shown in the diagram.

51

Advantages of vapour absorption Refrigeration System:1. No moving parts (in some cases, there is a small pump) Less

wear and tear. Less maintenance cost.2. Low grade fuels can be used. Waste heat can be used.3. System not affected by variation of loads.4. No electricity required.5. No chance of leakage.Disadvantages6. Low COP7. Bulky8. Higher cost initially Vapour absorption refrigeration systems using solar energy as

the heat source to the generator, hold a lot of promise for the future, in the areas of food preservation and comfort cooling.

52

Steam jet refrigeration: It operates on the principle of reducing the boiling point

of water below 100OC by reducing the pressure on the surface of water below the atmospheric pressure. This low pressure or high vacuum is maintained by throttling the steam through jet or nozzles.

53

Working: The system consists of an evaporator, one or more booster ejectors, a surface type of barometric condenser and two stage ejector air pump.

54

Boiler supplies high pressure steam to the nozzle inlet where steam expands.

The warm water returning from the A/C plant is sprayed through the nozzle in the flash chamber to ensure maximum surface area for evaporation.

The water vapor leaving the flash chamber mixes with the high velocity steam from the nozzle and gets further compressed in thermo compressor.

The K.E is converted into pressure energy and the mass discharged into the condenser returns as condensate to the boiler.

About 1% evaporation of water in evaporator is sufficient to reduce the temperature of water to 6OC. the chilled water is circulated to the A/C plant, which returns as warm water into the flash chamber.

55

The water lost due to evaporation in the flash chamber and any loss of cold water is compensated by make-up water.

Air dissolved in feed water is released in the condenser of the system and covers the condensing surfaces along with other gases and increases the condenser pressure.

This air is removed by using small two stage air ejector to maintain high vacuum.

In order to maintain temperature in the evaporator below 0OC it is necessary to add antifreeze or brine.

56

Limitations:1. It requires very high vacuum and cannot be used if

temperature in the evaporator is below 0OC.2. Size of the compressor should be larger to handle

larger volume.3. The compression ratio used in the thermo compressor

is limited to 8 only.4. Heat removed in the condenser is almost double as

compared to vapor compression system.

57

Advantages:1. Due to no moving parts it is noiseless.2. Flexible in operation as cooling capacity can be quickly

changed.3. Weight/Ton of Ref is low and plant life is more.4. Used in cold water processing of rubber mills, chemical

and food processing plants, breweries, refineries etc.5. Safer in operation and absolutely no hazard from

leakage.6. Cheaper in operation and easy to maintain.7. Useful in comfort air conditioning, but not suitable if

water temperature is below 4OC.

58

Prob 1. A cold storage is to be maintained at -5°C (268k) while the surroundings are at 35°C. the heat leakage from the surroundings into the cold storage is estimated to be 29kW. The actual C.O.P of the refrigeration plant is one third of an ideal plant working between the same temperatures. Find the power required to drive the plant. VTU Jan 2007.

7.6682308

268

TT

T ideal C.O.P

cycle.carnot on based C.O.P

but nothing isplant ideal theof C.O.P

2685T 30835T

-:Solution

21

2

1 2

kCkC -

59

233.27.63

1

..3

1 ..

x

POidealCPOCActual

Q2 = The heat removed from low temperature reservoir (cold storage) must be equal to heat leakage from surroundings to the cold storage(which is 29kw)

kW 12.98required 233.2

29

.. W

..

29

2

2

2

PowerPOCActual

QW

QPOCActual

kWQ

60

2. A refrigeration machine of 6 tones capacity working on Bell coleman cycle has an upper limit pressure of 5.2 bar. The pressure and temperature at the start of the compression are 1 bar and 18°C respectively. The cooled compressed air enters the expander at 41°C, assuming both expansion and compression to be adiabatic with an index of 1.4.

Calculate:-Co-efficient of performance.Quantity of air circulated per minute.Piston displacement of compressor and expanderBore of compression and expansion cylinder when the

unit runs at 240 rpm and is double acting with stroke length =200 mm , Power required to drive the unit

61

bar2.5P 41T

1bar P 18T

-:Solution

23

11

C

C

kgkJ

TTTTWork

/57196314291466005.1

Cinput 4312p

21kJ/s6x3.5 tons6capacity Re

67.157

95.42

inputWork

Re..

nfrigeratio

effectngriferatioPOC

62

13.2kg/min0.22x60air/min of Mass

12.54kW0.22 x 57

air/sec of Massair x of /kgrequired

/22.095.42

21

R.E

apacity Reair/sec of

workdonePower

skg

cngriferatioMass

min/42.7Vexpander ofnt displaceme

min/42.7101

196287.02.13

min/11V compressor ofnt displaceme

min/11101

291287.02.13

34

32

4

44

31

32

1

11

mPiston

mx

xx

P

mRTV

mPiston

mx

xx

P

mRTV

63

31.3cm0.313mcylinder expander of

2402.04

242.7

42

38cm0.38mcylinder compressor of

2402.04

211

42

1

21

224

1

21

211

diameterd

xxd

LNdV

diameterd

xxd

LNdVBut

64

Problem3 An air refrigerator system operating on Bell Column cycle, takes in air from cold room at 268 K and compresses it from 1 bar to 5.5 bar the index of compression being 1.25. the compressed air is cooled to 300 K. the ambient temperature is 200C. Air expands in expander where the index of expansion is 1.35.

Calculate:C.O.P of the systemQuantity of air circulated per minute for production of

1500 kg of ice per day at 0°C from water at 20ºC.Capacity of the plant.

65

K8.376

5.5268 25.1

125.11

1

212

P

PTT

Solution

K83.1925.5

1300

35.1

135.11

3

434

P

PTT

kgkJ

TTCn

np

/2.1562688.376005.14.1

14.1

125.1

25.1

1

1W 12C

66

kgkJ

TTCn

np

/69.11883.192300005.14.1

14.1

135.1

35.1

1

1W 43E

25.37

54.75..

/54.75)83.192268(005.1)(.

/5.3769.1182.156

41

work

REPOC

skJTTCER

kgkJWWNetwork

p

EC

skg

heatLatentHeat

/0173.024x3600

1500ecproduced/s ice of Mass

g418.74kJ/k3354.187(20)

_)020(Cice of kgextracted/ pw

67

skg

Mass

tons

heatActual

/096.0

54.75

26.7

efection Refrigerat

Capacityion Refrigeratrate flow

02.25.3

7.267.26kJ/scapacityion Refrigerator

173418.74x0.0secextracted/

68

Problem 4An air refrigeration system is to be designed according to the

following specificationsPressure of air at compressor inlet=101kPaPressure of work at compressor outlet=404kPaPressure loss in the inter cooler=12kPaPressure loss in the cold chamber=3kPaTemperature of air at compressor inlet=7°Temperature of air at turbine inlet=27° Isentropic efficiency of compressor =85%Isentropic efficiency of turbine =85%DetermineC.O.P of cyclePower required to produce 1 ton of refrigerationMass flow rate of air required for 1 ton of refrigeration

69

85.0;85.0 27T

101kPaP 7T

-:Solution

CT3

11

C

C

K4.395101

404266

'Hence ,isentropic is 2-1 Pr

4.1

14.1

1

1

212

P

PTTocess

kPaxPPPPP

kPaxPPPPP

kT

TTTT

TTC

39240497.097.0 03.0

10410103.103.1 03.0

2.418'

88.0

2664.395'or

'

23232

14114

2

1212

12

70

202.3K392

104300

P

PTT ,isentropic is 4-3 Process

1.4

11.4

γ

1γ

3

434

g152.96kJ/k266]-.21.005x[418

'Cair of work/kgompressor

kJ/kg47.05216.53]-1.005x[266

Cair of effect/kg Re

53.216]3.205300[85.0300'

' '

12p

41p

4

433443

43

TTC

TTnfrigeratio

kxT

TTTTTT

TTTE

71

kgkJ

TTT

/06.729.8096.152

WWair W of Input/kgNet work

kJ/kg9.84216.53]-1.005x[300

'Cair W of work/kgurbine

TCnet

43pT

73.006.72

73.46..

Work

REPOC

C.O.P

capacityion Refrigerat

ionrefergerat of per tons required

Power

72

kWxxmassofairWPower

skg

net 42.5075.006.72sec/

/075.047.50

5.3

RE

capacityion Refrigeratair of Mass

3.5kJ/s ton1capacityion Refrigerat

73

Problem5: 20 tons of ice is produced from water at 200C to

ice at -60C in a day of 24 hours, when the temperature range in the compressor is from -150C to 250C. The condition of the vapour is dry at the end of compression.

Assuming relative C.O.P as 80%, calculate the power required to drive the compressor.

Take Cpice=2.1kJ/kg, Latent heat of ice=335k/kg

74

Temp ºC

Liquid

Vapour

Enthalpy hf Entropy Sf Enthalpy hg Entropy Sg

25 100.04 0.347 1319.2 4.4852

-15 -54.55 -2.1338 1304.99 5.0585

75

KT

TLn

T

TLnC

To

b

b

b

bp

15.312

7019.0303

56.06853.0ss

aat Entropy bat Entropy

b'.'point at vapour ofcondition thefind

ga'

gb'

kgkJHork

kgkJH

TTChH

kgkJh

TTChH

a

c

ccPLfcc

ga

bbpgbb

/55.2119.18374.204Hw

/61.123575.5919.183HR.E

kJ/kg 59.57525)-1.003(30-64.59

)(

/19.183H

/kg204.74kJ303)-50.56(312.1100.62

)('

b

a

''

'a

'

76

431.346 x 2.133520 x 4.187

)]6(0[Cheat

)020(Cice of kgextracted/Heat

7.130.8x8.913C.O.P ctual

0.8C.O.P Re

913.898.122

18.1096..

pice

pw

Latent

A

lativework

REPOC

kWPowerActual

Actual

Actual

skgx

Mass

147.13

99.84 work/sec

work/sec

ecextrated/sheat ActualC.O.P Actual

99.84kJ/s 31431.34x0.2secextracted/heat

/231.0360024

20x1000ecproduced/s ice of

77

skJ

capacityf

nfrigeratiowork

REPOC

/424.0123.61

52.5

RE

.Referon of Mass

52.5kJ/kg15x3.5

15tonscapacity Re

73.555.21

61.123..

9.152kW421.55x0.42

freon/s of xMasswork/kgrequired

Power

78

Problem6: A food storage locker requires a refrigeration

system of 12 tons capacity at an evaporator temperature of -80C and a condenser temperature of 300C. The refrigerant freon-12 is sub cooled to 250C before entering the expansion valve and the vapour is superheated to -20C before entering the compressor. The compression of the refrigerant is reversible adiabatic. A double action compressor with stroke equal to 1.5 times the bore is to be used operating at 900 rpm.

DetermineCOPTheoretical piston displacement/minMass of refrigerant to be circulated/minTheoretical bore and stroke of the compressor.Take liquid specific heat of refrigerant as 1.23 kJ/kg K and the

specific heat of vapour refrigerant is 0.732 kJ/kg K.

79

Temp ºC

EnthalpyEntropy

hf hg Sf Sg

30 64.59 199.62 0.24 0.6853

-8 25.75 184.2 0.1142 0.7002

Solution:From tables the properties of Freon 12 are

80

265

271Ln 733.07002.0

303

TLn 732.06853.0

SS

aat entropy bat

/235.1,/732.0

b

'ga'

'gb'

a

ap

b

bp

PLp

T

TLnC

T

TLnC

Entropy

kgKkJCkgKkJC

KTb 22.317

kJ/kg02.210

)30322.318(732.062.991

)(

kJ/kg59.188

)265271(732.02.841

)(

''

''

bbpgbb

aapgaa

TTChH

TTChH

81

kgkJHork

kgkJH

TTChH

a

c

ccPLgcc

/43.2159.18802.210Hw

/181.13041.5859.188HR.E

kJ/kg41.85298)-1.235(303-64.59

)(

b

a

''

07.643.21

18.130..

work

REPOC

a

a

a

ga

T

PV

T

PV

kgm

skg

capacityf

'

'

3ga' /0441995.0V C,8-at tablesFrom

n19.35kg/mi0.322x60

/322.0130.18

12x3.5

RE

.Retrefrigeran of Mass

82

0.0203m

9004

5.1287462.0

)5.1( 4

2

min/0.87462m5219.35x0.04

xVmass Vnt displacemepiston

0452.00441995.0265

271

2

2

3

a

'

'

xdxxd

dLLNdV

lTheoretica

xT

xVTV

a

gaaa

11.08cm

1.5x7.381.5dL

7.38cm

0.0738m d

83

Temp Pressure Vf Vg hf hg Sf Sg

°C bar m3/kg   kJ/kg   kJ/kgK

-10 226 0.7x10-3 0.08 190 345 0.95 1.5

30 7.5 0.77x10-3 0.02 220 220 1.10 1.45

Problem7: A vapour compression refrigeration system of 5kW cooling

capacity operates between -10ºC and 30ºC. The enthalpy of refrigerant vapour after compression is 370kJ/kg. Find the COP, refrigerating effect, mass flow rate of the refrigerant and the compressor power. The extract of the refrigerant property table is given below

84

kgkJork

kgkJ

givenkgkJH

kgkJh

kgkJhH

b

ga

fcc

/25345370HHw

/125220345HHR.E

)(/370

/345H

/220

ab

ca

'a

Solution:Assume the condition before compression as dry saturated vapour

525

125..

work

REPOC

85

1kW25x0.04-

trefrigeran of mass x .

/04.0125

5

RE

.Retrefrigeran of Mass

kJ/sor 5kW capacity ion Refrigerat

kgworkworkCompressor

skg

capacityf

86

Problem8: A vapour compression refrigerator uses methyl chloride and works in the pressure rang of 1.19 bar and 5.67 bar. At the beginning of compression, the refrigerant is 0.96 dry and at the end of isentropic compression, its temperature is 55ºC. The refrigerant liquid leaving the condenser is saturated. If the mass flow of refrigerant is 1.8kg/min,DetermineCOPThe rise in temperature of cooling water if the water flow rate is16 kg/min. the properties of methyl chloride is given below

87

Temp ºC

Pressurebar

EnthalpyEntropy

hf Hfg hg Sf Sg

30 1.19 64.59 135.03 199.62 0.24 0.6853

-10 5.67 26.87 156.31 183.19 0.108 0.7019

kgkJ

hhxhhxhH

CT

x

Solution

Take

fagaafafgaafaa

b

a

/196.43830.1)-0.96(455.2430.1

)()(

55

96.0

K 0.75kJ/kg

as chloride methylheat super ofheat specific

88

kgkJHork

kgkJH

kgkJh

TTChH

a

c

fc

bbpgbb

/8.60196.438499Hw

/669.3375.100196.438HR.E

/5.100H

kJ/kg49925)-(5575.05.764

)(

b

a

c

''

55.58.60

669.337..

work

REPOC

watercoolingby gain heat

condenser in the vapour by thelost

Heat

CriseeTemperatur

riseetemperaturx

riseetemperaturxCmhmTTCm pwfgbrbbpr

7.10

x 187.4 16

100.5)-1.8(476.525)-0.75(55 x 8.1

)( ''

89

90