Refrigeration

Definition Refrigeration is the process of providing and maintaining temperature of the system below that of the surrounding atmosphere. Carnot Cycle The reversed carnot cycle can be considered in refrigeration system.

Unit of Refrigeration The common unit used in the field of refrigeration is known as Ton of refrigeration. 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 0ºC Consider a refrigerator of T tons capacity, Refrigeration capacity = 3.5 kJ/s Heat removed from refrigerator = Refrigeration effect =R.E. kJ/s Power of the compressor =work/kg of refrigerant x mass flow rate Air Refrigeration system working on Bell-coleman cycle In air refrigeration system, air is used as the refrigerant which always remains in the gaseous phase. The heat removed consists only of sensible heat and as a result, the coefficient of performance (C.O.P) is low. The various processes are:

1212

2 where.. TTTT

TPOC <−

=

kJ/s 5.3360024

3351000ionrefrigerat of ==x

xTon

Process 1-2: The air leaving the evaporator enters a compressor. Where it is compressed isentropically to higher pressure and temperature. Process 2-3: This high pressure, high temperature air, then enters a cooler where it is cooled at constant pressure to a low temperature. Process 3-4: This high pressure, low temperature air is then expanded in an expander to lower pressure and temperature in a isentropic manner.At point 4, the temperature of the air will be lowest. Process 4-1: This low temperature air is then passed through the heater coils where it absorbs heat from the space to be cooled namely the refrigerator and the air gets heated back to the initial temperature, but in the process, it cools the refrigerator. And the cycle repeats. Air refrigeration system

Expression C.O.P when compression and expansion are Isentropic Refrigeration Effect = Heat removed from the refrigerator

( ) kgkJTTCp /41 −=

−−

−

−−

=−=11

Winput 44331122C γ

γγ

γ VPVPVPVPWWork E

( ) ( )[ ]

( ) ( )[ ]

( ) ( )[ ]4312p

p

4312

4312C

C1

C

1

1 Winput

TTTTW

RBut

TTTTRW

TTRTTRWWork

net

net

E

−−−=−

=

−−−

−

=

−−−

−

=−=

γγ

γγ

γγ

( )( ) ( )[ ]

( )( ) ( )[ ] )1(

1

1..

CC

..

41

324312

41

4312p

41p

−−−

−−

=−−−

−=

−−−

−==

TTTTTTTT

TTPOC

TTTTTT

WorkREPOC

4

3

1

2

1

1

2

1

4

3

4

3

1

1

2

1

2

TT

TT

(3) and (2) From

-(3)--- PP

PP

TT

isentropic is 4-3 Pr

-(2)--- PP

TT

isentropic is 2-1 Pr

=

=

=

=

−−

−

γγ

γγ

γγ

ocess

ocess

(4)----- TT

TTTT

TTT

TTT

1TT1

TT

TT

TT

4

3

41

32

4

41

3

32

4

1

3

2

4

1

3

2

=−−

−=

−

−=−

=

43

4

4

3

T-TTC.O.P

1TT

1C.O.P

(4) and (1)

=

−=

From

Advantages of air refrigeration system

1. Air is cheap, easily available. 2. It is not flammable. 3. For a given capacity, weight of air refrigeration system is less compared to other

system and hence it is widely used for aircraft cooling. Disadvantages

1. Since heat removed by air consists only of sensible heat, weight of air required is high.

2. C.O.P of the system is low compared to other systems. Problem 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)

( ) ( )[ ]

( )

−

−−

=

−−−

−

−=

γγ

γγ

11

11

W

Net workprocess Polytropic

43

4

4312net

nnTT

TCOP

TTTTCn

n

For

p

7.6682308

268

TTT ideal C.O.P

cycle.carnot on based C.O.Pbut nothing isplant ideal theof C.O.P

2685T 30835T-:Solution

21

2

1 2

=−

=

−=∴

=°==°= kCkC

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) Problem 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 I 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:-

(i) Co-efficient of performance. (ii) Quantity of air circulated per minute.

(iii) Piston displacement of compressor and expander (iv) Bore of compression and expansion cylinder when the unit runs at 240 rpm and is

double acting with stroke length =200 mm (v) Power required to drive the unit

233.27.631

..31 ..

==

=

x

POidealCPOCActual

kW 12.98required 233.229

.. W

..

29

2

2

2

=

==

=

=

PowerPOCActual

QWQPOCActual

kWQ

bar2.5P 41T

1bar P 18T-:Solution

23

11

=°=

=°=

CC

( ) ( )[ ]( ) ( )[ ] kgkJ

TTTTWork

/57196314291466005.1

Cinput 4312p

=−−−=

−−−=

21kJ/s6x3.5 tons6capacity Re

67.157

95.42

inputWork Re..

===

==

=

nfrigeratio

effectngriferatioPOC

13.2kg/min0.22x60air/min of Mass12.54kW0.22 x 57

air/sec of Massair x of /kgrequired

/22.095.42

21

R.Eapacity Reair/sec of

====

=

==

=

workdonePower

skg

cngriferatioMass

min/42.7Vexpander ofnt displaceme

min/42.7101

196287.02.13min/11V compressor ofnt displaceme

min/11101

291287.02.13

34

32

4

44

31

32

1

11

mPiston

mx

xxP

mRTV

mPiston

mx

xxP

mRTV

=

===

=

===

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

π

π

π

π

Problem3 An air refrigerator system operating on Bell Coleman 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 20ºC. Air expands in expander where the index of expansion is 1.35. Calculate:

i) C.O.P of the system ii) Quantity of air circulated per minute for production of 1500 kg of ice per day at

0°C from water at 20ºC. iii) Capacity of the plant.

( )

K8.376

5.5268 25.1125.1

1

1

212

=

=

=

−−γγ

PPTT

Solution

K83.1925.5

130035.1

135.11

3

434 =

=

=

−−γγ

PPTT

( )

( ) kgkJ

TTCn

np

/2.1562688.376005.14.1

14.1125.1

25.1

11

W 12C

=−

−

−=

−

−

−=

γγ

( )

( ) kgkJ

TTCn

np

/69.11883.192300005.14.1

14.1135.1

35.1

11

W 43E

=−

−

−=

−

−

−=

γγ

25.3754.75..

/54.75)83.192268(005.1)(./5.3769.1182.156

41

===

=−=−==−=−=

workREPOC

skJTTCERkgkJWWNetwork

p

EC

Problem 4 An air refrigeration system is to be designed according to the following specifications Pressure of air at compressor inlet=101kPa Pressure of work at compressor outlet=404kPa Pressure loss in the inter cooler=12kPa Pressure loss in the cold chamber=3kPa Temperature of air at compressor inlet=7° Temperature of air at turbine inlet=27° Isentropic efficiency of compressor =85% Isentropic efficiency of turbine =85% Determine

i) C.O.P of cycle ii) Power required to produce 1 ton of refrigeration

iii) Mass flow rate of air required for 1 ton of refrigeration

skg

LHeat

/0173.024x3600

1500ecproduced/s ice of Mass

g418.74kJ/k3354.187(20)

)020(Cice of kgextracted/ pw

==

=+=

+−=

skg

Mass

tons

heatActual

/096.0 54.7526.7

efection RefrigeratCapacityion Refrigeratrate flow

02.25.3

7.267.26kJ/scapacityion Refrigerator

173418.74x0.0secextracted/

=

==

===

=

85.0;85.0 27T

101kPaP 7T-:Solution

CT3

11

==°=

=°−=

ηηCC

K4.395101404266

'Hence ,isentropic is 2-1 Pr

4.114.1

1

1

212

=

=

=

−

−γγ

PPTTocess

kPaxPPPPPkPaxPPPPP

kT

TTTTTT

C

39240497.097.0 03.010410103.103.1 03.0

2.418'88.0

2664.395'or '

23232

14114

2

1212

12

===∴=−===∴=−

=

−=−

−−

=η

K3.202392104300

,isentropic is 4-3 Pr

4.114.1

1

3

434

=

=

=∴

−

−γγ

PPTTocess

( )

( )

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

'Cair of work/kgompressor kJ/kg47.05216.53]-1.005x[266

Cair of effect/kg Re53.216]3.205300[85.0300'

' '

12p

41p

4

433443

43

==

−===

−==−−=

−−=∴−−

=

TTC

TTnfrigeratiokxT

TTTTTTTT

TE ηη

( )

kgkJ

TTT

/06.729.8096.152WWair W of Input/kgNet work

kJ/kg9.84216.53]-1.005x[300

'Cair W of work/kgurbine

TCnet

43pT

=−=−===

−=

73.006.7273.46.. ===

WorkREPOC

C.O.Pcapacityion Refrigerat

ionrefergerat of per tons required

=

Power

kWxxmassofairWPower

skg

net 42.5075.006.72sec/

/075.047.505.3

REcapacityion Refrigeratair of Mass

3.5kJ/s ton1capacityion Refrigerat

===

==

=

==

Problem 1 Moist air at 30°C,1.01325 bar has a relative humidity of 80%. Determine without using the psychrometry chart

1) Partial pressures of water vapour and air 2) Specific humidity 3) Specific Volume and 4) Dew point temperature (V.T.U. July2004)

Corresponding to Pv =3.397 kPa from tables, we get dew point temperature = 28.9°C Problem 2: Atmospheric air at 101.325 kPa ha 30°C DBT and 15°C DPT. Without using the pschrometric chart, using the property values from the table, Calculate

1. Partial pressure of air and water vapour 2. Specific humidity 3. Relative humidity 4. Vapour density and 5. Enthalpy of moist air

air.dry of kg/kg 0.213 397.3325.101

397.3622.0622.0397.32461.48.0

2461.4p tablefrom C30At : s

=−

=−

=

==

=

=°

xppp

kPaxppp

kPaSolution

s

υ

υ

υ

υ

υ

υ

ω

φ

bar 0.984274 017051.001325.1p-pair of pressure

017051.0p have weC,51DPT to042461.0p have weC,30DBT to

C15 DPT

,3001325.1325.101

:

s

=−===°==°=

°=°=

==

υ

υ

υ

PartialbaringCorrespondbaringCorrespond

tableFrom

CDBTbarkpap

Solution

Pscychrometrics and Air-conditioning Systems

Problem 3: Air at 30°C DBT and 25°C WBT is heated to 40°C. if the air is 300 m3/min, find the amount of heat added/min and RH and WBT of air. Take air pressure to be 1 bar Solution: At 25°C WBT from tables pwswbt=0.03166 bar

airdry of 57.69kj/kg 1.88x30)5000.010775(21.005x30

)10882500(1.005tEnthalphy40.15%

4015.0042461.0017051.0 Re

airdry of kg0.01077kJ/ 984274.0

017051.0622.0622.0

db

=++=

++==

===

=

==

db

s

a

t

pphumiditylative

xpphumiditySpecific

ω

υ

υ

υ

3w

3

a

/12.0847.0

010775.0density

/874.010098425.0

0.2872x303

air,dry of volume

mkgVapour

kgmx

PRTSpecific

a===

==

=

υωρ

υ

airdry of /0179.00284.01

0284.0622.0

622.0

bar 0.0284 2544.11547

25)-00.03166)(3-(1-0.03166

44.11547))(()(p

1

kgkJ

ppp

x

tttppP

wb

wbdbswbtwbtVS

=

−

=

−=

=−

=

−−−

−=∴

υ

υ

υυ

ω

Problem 4: One stream of air at 5.5m3/min at 15°C and 60% RH flows into another stream of air at 35m3/min at 25°C and 70%RH, calculate for the mixture 1) Dry bulb temperature, 2) Wet bulb temperature 3) Specific Humidity and 4) Enthalpy Solution: For air at 15°C and 60%RH, V=5.5m3/min

airdry of 86.29kJ/kg )4088.12500(0179.040005.1H

38.5%0.385 07375.00284.0

0284.0constant p and

07375.0 40

2

=++=

==

===

=

=°

xx

ppRH

barpremainheatingsensibleDuring

barPDBTCAt

s

VS

υ

υ

υ

υ

φ

ω

C27.2chart WBT From3449kJ/min76)-29335.18(86.added/min

min/18.335303287.0

0x100.0284)x30-(1

)(min/300m of

2

3

°===∴

==

−=

Heat

kgx

RTVppofairWeight υ

airdry of /006343.0 )01023.001325.1(

01023.0622.0)p-(p

622.0min/672.6

288287.05.510)01023.001325.1()p-(pair of

01023.0017051.06.0

017051.0p

1

1

2

kgkg

xpkgm

xxx

RTVMass

barxpppRH

bar

s

s

=−

==

=

−==

==∴

==

=∴

υ

υ

υ

υ

υ

υ

υ

ω

φ

barxpppRH

barP

ttH

s

s

dbdb

02218.07.003169.0

03169.0min/35m V RH, 70% and C25at air For

airdry of 34.12J/kg 1.88x15)5000.006343(21.008x18 )88.12500(005.1

3

11

==

==

==°

=++=

++=

υ

υ

υ

υ

φ

ω

barxpppRH

barP

ttH

s

s

dbdb

02218.07.003169.0

03169.0min/35m V RH, 70% and C25at air For

airdry of 34.12J/kg 1.88x15)5000.006343(21.008x18 )88.12500(005.1

3

11

==

==

==°

=++=

++=

υ

υ

υ

υ

φ

ω

6299.6006343.01672.6

1m

airmoist of massair/Unit dry of Massairdry of /59.60

)2588.12500(01392.0)25005.1(H

airdry of /01392.0)02218.001325.1(

02218.0622.0min.55.40m

298287.0)350.02218x10-(1.01325air of

1

1a1

2

2

2

2

2

=+

=+

=

=++=

=−

=

=

=

ω

ω

m

kgkJHxx

kgkgxkg

xxMass

airdry of kJ/kg 55.96 )55.40672.6

56)39.993(60.12)6.6299(34.

)()(H

air, mixed theofenthalpy

993.3901392.0155.40

1m Since

21

2211mix

2

2a2

=++

=

++

=

=+

=+

=

mmHmHm

Then

m

aa

ω

Problem 5: An air conditioning system is designed under the following conditions Outdoor conditions: 30°CDBT, 75% RH Required indoor conditions: 22°CDBT,70% RH Amount of Free air circulated 3.33 m3/s Coil dew point temperature DPT=14° The required condition is achieved first by cooling and dehumidification and then by heating. Estimate

1) The capacity of the cooling coil in tons of refrigeration 2) Capacity of the heating coil in kW 3) The amount of water vapour removed in kg/hr

Solution:

Cttxt

tt

mmmm

db

dbdb

dbmixdb

aa

°=++=

++==

++

=

++

=

42.24)88.12500(01234.0005.196.55

)88.12500(005.1HBut airdry of kg0.01268kg/ 40.556.672

01932)(39.993x0.006343)(6.6299x0.

)()(air, mixed theofHumidity Specific

mix

21

2211mix

ω

ωωω

67%RHC19chart WBT From

C24.42mixture theof

=°=

°=DBT

heatingcdcationdehumidifi and coolingac

c.point at ab line cut the toline horizontal a draw d,at

abJoin re temperatusurface coil DPT, C14 b''point

condition required 70%RH DBT, C22 d''point conditiondoor out 75%RH DBT, C30 a''point

→→

°°°

LocateLocateLocate

Problem 6: A summer air conditioning system for hot and humid weather (DBT=32°Cand 70% RH) Consists in passing the atmosphere air over a cooling coil where the air is cooled and dehumidified. The air leaving the cooling coil is saturated at the coil temperature. It is then sensibly heated to the required comfort condition of 24°C and 50%RH by passing it over an electric heater then delivered to the room. Sketch the flow diagram of the arrangement and represent the process undergone by the air on a skeleton psychometric chart and determine

1) The temperature of the cooling coil 2) The amount of moisture removed per kg of dry air in the cooling coil. 3) The heat removed per kg of dry air in the cooling coil and 4) The heat added per kg of dry air in the heating coil

kgm

kgkgWWkgkgW

kgkJkgkJkgkJkgkJ

From

dc

a

/88.0V

airdry of /0118.0airdry of /0202.0

air of /48Hair of /53Hair of /40Hair of /83H

chart

3sa

c

d

b

a

=

=======

18.92kW48)-3.78(53 )(mcoil heating ofCapacity

ionrefrigerat of 84.375.3

)4883(78.35.3

)(mcoil cooling ofCapacity

/78.388.033.3air of

a

a

==−=

=−

=

−=

===

cd

ca

a

HH

tons

HH

skgVVMass

114.3kg/hr 00.0118)360-23.78(0.020

3600)(m removedour water vapof a

==

−= daAmount ωω

airdry of /0092.0airdry of /021.0

air of /5.48Hair of /38Hair of /86H

chart

c

b

a

kgkgkgkg

kgkJkgkJkgkJ

From

b

a

=====

ωω

Problem 7 It is required to design an air conditioning plant for an office room with the following conditions. Outdoor conditions: 14°CDBT, 10°CWBT Required conditions: 20°CDBT,60% RH Amount of air circulated 0.3m3/min/person Starting capacity of the office= 60 The required condition is achieved first by heating and then by adiabatic humidifying. Determine the following. Heating capacity of the coil in kW and the surface temperature required, if the by pass factor of the coil is 0.4 Capacity of the humidifier.

heatingbcationdehumidifi and coolingab

abJoin b''point at line

saturation cut the toline horizontal a draw c condition required 50%RH DBT, C24 c''point

conditiondoor out 70%RH , C32 a''point

→→

°°

c

AtLocateLocate

airdry of kJ/kg 10.5 385.48addedHeat airdry of kJ/kg 48 3886removedHeat

airdry of g0.0108kg/k0.0092-0.021 removed moisture of

13Tcoil cooling theof re temperatu

ba

b

=−=−=

=−=−===

−=°==

bc

ba

HH

HH

AmountC

The

ωω

tionhumidifica bheatingab

abJoin b''point at line horozontal cut the tolineenthalpy constant a draw c''At

line horizontal a draw a condition required 60%RH DBT, C20 c''point

condition)door (out CWBT10 and , C14 a''point

adiabaticc

AtLocateLocate

→→

°°°

Problem 8 An air conditioned system is to be designed for a hall of 200 seating capacity when the following conditions are given: Atmospheric condition = 300C DBT and 50% RH Indoor condition = 220C DBT and 60% RH Volume of air required = 0.4m3/min/person The required condition is achieved first by chemical dehumidification and after that by sensible cooling. Find the following .

a) DBT of the air leaving the dehumidifier. b) The quantity of water vapour removed in the duhumidifier per hour. c) The capacity of cooling coil in tons of refrigeration. d) Surface temperature of the coil if the by pass factor of the coil is 0.25.

sec/3.060

0.3x60Vsuppliedair of

/8175.0V volomeSpecific

airdry of kg/kg 00875.0airdry of /006.0

air of /43HHair of /30H

chart

3

3sa

cb

a

mVolume

kgm

kgkgkgkJ

kgkJFrom

c

ba

===

=

=====

=

ωωω

d

b

a

a

T be re temperatusurface coil 5.26Tchart From

4.77kW30)-0.3669(43 )(mcoil heating theofCapacity

ec0.3669kg/s 8175.0

3.0m suppliedair of

LetC

HH

VVWeight

ab

a

°===−=

=

==

r3.63kg/hou 0.006)3600-08750.3669(0.0

3600)(m humidifier theof 8.34

4.1T5.26T4.0

266.54.0TTfactor passing

a

d

d

d

d

==

−=°=−−

=

−=−=−−

=

xCapacityCT

kJTTTTBy

bc

d

dda

b

ωω

Solution: a ω

c d b 220C 300C

DBT Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition. Locate point ‘c’, 220C DBT, 60% RH, the required indoor condition. “Since chemical dehumidification process follows constant enthalpy line” at a draw a line parallel to constant enthalpy line. At ‘c’ draw a constant ω line to cut the previous line at point b.

a) DBT of air leaving the dehumidifier Tb = 40.50C From chart Hb = Ha = 65kJ/kg, ωa = 0.013 kg/kg of dry air Hc = 45 kJ/kg, ωb = 0.009 kg/kg of dry air Vsa = 0.875 m3/min Volume of air = 200 X 0.4 = 80 m3/min Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min

b) Quantity of water vapour removed/hour = Wa(ωa-ωb)60 = 91.42(0.13-0.009)60 = 21.94 kg/hr

c) Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5) = 8.7 tons

d) By pass factor = (Tc-Td)/( Tb-Td) = 0.25 Td = Temperature of cooling coil = 15.830C

Problem 9 An air conditioned system is to be designed for a cinema hall of 1000 seating capacity when the following conditions are given: Outdoor condition = 110C DBT and 70% RH Required indoor condition = 200C DBT and 60% RH

Amount of air required = 0.3m3/min/person The required condition is achieved first by heating, then by humidifuing and finaly by heating. The condition of air coming out of the humidifier is 75% RH. Find the following .

a) Heating capacity of the first heater in kW and condition of the air coming out of the first heater in kW and condition of the air

Solution: a ω

c d b 220C 300C

DBT Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition. Locate point ‘c’, 220C DBT, 60% RH, the required indoor condition. “Since chemical dehumidification process follows constant enthalpy line” at a draw a line parallel to constant enthalpy line. At ‘c’ draw a constant ω line to cut the previous line at point b.

e) DBT of air leaving the dehumidifier Tb = 40.50C From chart Hb = Ha = 65kJ/kg, ωa = 0.013 kg/kg of dry air Hc = 45 kJ/kg, ωb = 0.009 kg/kg of dry air Vsa = 0.875 m3/min Volume of air = 200 X 0.4 = 80 m3/min Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min

f) Quantity of water vapour removed/hour = Wa(ωa-ωb)60 = 91.42(0.13-0.009)60 = 21.94 kg/hr

g) Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5) = 8.7 tons

h) By pass factor = (Tc-Td)/( Tb-Td) = 0.25 Td = Temperature of cooling coil = 15.830C