1

Chapter 7

PSYCHROMETRY

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PSYCHROMETRY

The study of the properties of air and vapour pertaining to air conditioning problems is called psychrometry.

Dry air: it is the mixture of nitrogen, oxygen and small percentage of other gases. Air contains 79% nitrogen, 21% oxygen by volume and has a molecular weight of 29.

Moist air: it is the mixture of dry air and water vapour. The amount of water vapour varies according to the

temperature of air and reaches saturation at one point. Saturated air mixture: saturated air mixture is a mixture of

dry air and water vapour in which the partial pressure of the vapour is equal to the saturation pressure of water at the temperature of the mixture.

3

Unsaturated air mixture: it is a mixture of dry air and super heated water vapour. The partial pressure of vapour being less than the saturation pressure of water at the temperature of the mixture.

Super saturated air mixture: it is a mixture of dry air and water vapour in which the partial pressure of the water vapour is greater than the saturation pressure of water at the temperature of the mixture.

Dew point temperature: when the unsaturated air is cooled at constant pressure, the mixture reaches saturation temperature corresponding to the partial pressure of water vapour. This temperature at which condensation of the vapour begins resulting in formation of liquid droplets or dew when the mixture is cooled is called dew point temperature.

4

Absolute humidity or specific humidity or humidity ratio:

The ratio of mass of vapour to the mass of dry air is called absolute humidity. Denoted by ‘ω’.

5

Dalton’s law of partial pressures: the total pressure of the mixture of gases is the sum of the partial pressures exerted by each gas when it occupies the same volume of the mixture at the same temperature of the mixture.

Dry bulb temperature: it is the equilibrium temperature of the mixture indicated by an ordinary thermometer denoted by Tdb.

Wet bulb temperature: it is the temperature indicated by a wet bulb thermometer which has its temperature sensitive element (bulb) covered with a wick soaked in water. It is denoted by Twb.

6

7

Relative humidity: it is defined as the ratio of partial pressure of water vapour in a mixture to the saturation pressure of water at dry bulb temperature.

using perfect gas relation, pv.vv = pg.vg

Hence Hence relative humidity is also defined as the

ratio of the mass of water vapour in a certain volume of moist air at a given temperature to the mass of water vapour in the same volume of saturated air at the same temperature.

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Degree of saturation or saturation ratio: it is defined as the ratio of specific humidity of actual air (ω) to the specific humidity of saturated air (ωg) at the same temperature.

9

Enthalpy of moist air: Enthalpy of moist air = enthalpy of dry air + enthalpy of

water vapour associated.h = hair + ω hvapour = Cp tdb + ω hvapour

But, hair = Cpa tdb = 1.005 tdb

Where Cpa = is the specific heat of air = 1.005 kJ/kg/K

hvapour = hg + Cps (tdb – tdp)

= 2500 + 1.88tdb

Cps = Specific heat of water vapour = 1.88kJ/kg/K & tdp = 0

Therefore, h = 1.005 tdb + ω(2500 + 1.88tdb) kJ/kg of dry air

10

Carrier equation: When DBT and WBT are given, for calculating the partial pressure of water vapour in air many equations have been proposed of which Dr. Carriers equation is most widely used.

Where,(pg)wb = saturation pressure at wet bulb temperature.

pv = partial pressure of water vapour

pg = partial pressure of saturated vapour

p = total pressure of moist airtdb = dry bulb temperature, 0C

twb = wet bulb temperature, 0C

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Psychrometric chart: As the calculations of various properties of

moist air are tedious and time consuming, all the essential data for complete thermodynamic and psychrometric analysis of air conditioning processes can be summarized and can be presented in the form of a chart.

Such a chart which makes it possible to obtain the necessary information readily for engineering calculations related to moist air known as a psychrometric chart.

A psychrometric chart is constructed for a given mixture pressure, using dry bulb temperature and the specific humidity as co-ordinates.

12

At a given mixture pressure, the vapour pressure pv is a function of specific humidity only and hence there is only one value of pv for each value of ω.

13

Saturation curve: the saturation line represents the states of saturated air at different temperatures. Wet bulb and dew point temperatures fall on this curve. Relative humidity on the curve is 100%.

Relative humidity lines: these are the curves with humidity ranging from maximum 100% to minimum 0%.

Constant specific volume lines: these are inclined lines and are equally spaced. The WBT lines are much flatter than constant enthalpy lines.

DBT lines: these are vertical lines and are equally spaced. DBT increases as we move from left to right.

Specific humidity lines: these are horizontal lines and value of ω increases from bottom towards top of chart.

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Sensible heating: Heating of air without

addition or subtraction of moisture is called sensible heating.

This can be achieved by passing the air over a heating coil.

The heat added increases the DBT of air.

This is useful in winter air conditioning.

The heat added is given by, Qs = ma (h2 – h1)

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Sensible cooling: Cooling of air without

addition or subtraction of moisture is called sensible cooling.

This can be achieved by passing the air over a cooling coil.

This is useful in summer air conditioning.

The heat removed is given by, Qr = ma (h1 – h2)

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Cooling and dehumidification: Water vapour may be removed from air by cooling it below

its dew point temperature. As a result of the cooling process a portion of the vapour in

the air is condensed. Dehumidification will take place along with cooling when the saturation temperature of the cooling coil is below the dew point temperature of the cooling air.

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In the above process warm air t1 enters the cooling coil maintained at temperature t2.

The surface temperature of the cooling coil t2 is lesser than the dew point temperature of the incoming air, t4. Under ideal conditions air leaves at t2, but due to inefficient cooling it leaves at a higher temperature t3.

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The temperature t2 corresponding to point 2 on the saturation curve is known as apparatus dew point, ADP. The ratio of actual heating/cooling to the ideal heating/cooling is known as by-pass factor.

In this case it is given by

19

Adiabatic humidification: if humidification is carried out adiabatically, the energy required for the evaporation of the added moisture must come from the entering air.

20

As the dry bulb temperature of air decreases during adiabatic humidification process, it is also known as evaporative cooling process or cooling with adiabatic humidification of air.

When warm air is passed through a spray chamber, part of water is vaporized and carried away with air. This results in humidification of air.

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Mixing process: In this process two or more streams are mixed to produce a stream with desirable state of temperature and relative humidity.

22

Making energy balance,ma1(ha1 + ω1hv1) + ma2(ha2 + ωhv2) = ma3(ha3 + ω3hv3)

Making mass balancema1 + ma2 = ma3

Making moisture balancema1 ω1 + ma2 ω2 = ma3 ω3

Combining we get,

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Requirement of Comfort air conditioning:The following 5 factors determine the comfort

feeling of the people in an air conditioned space.

1. Supply of O2 and removal of CO2.

2. Removal of body moisture dissipated by the occupants.

3. To provide sufficient air movement and air distribution in the occupied space.

4. To maintain the purity of air by removing odour and dust.

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1. Oxygen supply: human body takes in oxygen through the system and gives out CO2. Normally each person requires nearly 0.65m3 of O2 per hour and produces 0.2m3 of CO2.

The percentage of CO2 in atmosphere is about 0.6% and it is necessary to maintain this to ensure comfort for easy breathing. Thus the quantity of air supply to an air conditioned space should be regulated properly to see that percentage of CO2 should not be exceeded.

2. Heat removal: it is a well known fact that human beings dissipate good amount of heat to the atmosphere during breathing etc.

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The atmosphere should be capable of absorbing the heat dissipated by persons otherwise discomfort exist.

Thus sufficient circulation of air should be provided through proper ventilation system to avoid rise in temperature of air in the air conditioned space.

3. Moisture removal: A moisture loss of up to 50% from human body is commonly observed.

This should be properly taken into account while designing a air conditioning unit.

The ventilation system must be capable of maintaining an RH of below 70%.

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4. Air movement:

In addition to providing air motion, proper air distribution is very important.

Air distribution is defined as a uniform supply of air to an air conditioned system.

Air movement without proper air-distribution is permissible for local cooling sensation known as draft.

A velocity of about 8m/min associated with temperature differential of 10C do not result in noticeable draft.

Velocities greater than this produces uncomfortable drafting conditions.

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5. Purity of air: It is important to maintain quality of air in any air-

conditioned space. Odour, dust, toxic gases and bacteria are

considered for defining the purity of air. The various factors which makes air impure are:i. The evaporation on the surface of the body adds

odour to the air.ii. The smoke from the surroundings which has a

bad effect on nose, eyes and heart.iii. The toxic gases are objectionable as they cause

irritation.

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Summer air conditioning systems (for Hot and dry outdoor conditions): this system is used

when outdoor conditions are hot and dry. That means atmospheric temperature is higher than the

comfort temperature with less moisture content (lesser RH).

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Atmospheric air at condition ‘1’ (higher DBT and lower RH) enters the air dampers and passes over the cooling coil via an air filter.

Temperature of the air is reduced to condition ‘2’ in the cooling coil.

Air now enters an adiabatic humidifier and is passed over water eliminators.

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Air enters the conditioned space at condition ‘3’. Point ‘4’ represents the condition of air after passing over the cooling coil if efficiency of the coil were 100%.

Condition line 1-2-4 represents the changes in DBT of air when passes over the cooling coil and the condition line 2-3-5 represents the changes in DBT of cooled air in the adiabatic humidifier.

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Summer air conditioning systems: Hot and humid outdoor conditions: this system is used when

the out door condition is hot and humid, like in coastal areas. That means the atmospheric temperature is higher but at the same time contains large moisture content. This condition eliminates the need for an adiabatic humidifier as shown in fig.

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Atmospheric air at condition ‘1’ enter the air filter via air dampers. Filtered air is cooled when it passes over the cooling coil at condition ‘3’. Condition line 1-2-3 represents the changes in DBT of air.

While passing over the cooling coil with an efficient coil the air would have been cooled to 2.

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As the air temperature is still below the required comfort condition, it is now passed over a heating coil.

The air coming out of the coil is at condition ‘5’ which is delivered to the space.

Line 3-5-4 represents changes in DBT of air while it passes over the heating coil.

Point ‘4’ represents the maximum temperature that could have been attained by using an efficient heating coil.

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Winter air conditioning system: During winter air in the atmosphere is at lower temperature

than the required conditions. Also the relative humidity may be more or lesser than the

actual relative humidity required for human comfort. Fig shows an arrangement for such a system where dry conditions prevail in the atmosphere.

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Cold air from the atmosphere is filtered in an air filter before it passes over the heating coil.

Air at condition ‘1’ is heated to ‘3’ in the heating coil. However due to losses, condition of air leaving the heat coil is ‘2’. Condition line 1-2-3 represents the changes in the DBT of air while it passes over the heating coil.

Hot air at ‘2’ now enters the humidifier and it is cooled to ‘4’.

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Any water particles suspended in air is removed by the water eliminator.

As the humidity and temperature of air have still not reached comfort conditions air is heated again up to point ‘7’ by another heating coil.

Due to heat losses the final temperature of air entering the conditioned space is at point ‘6’.

37

Problem 1. Moist air at 30°C,1.01325 bar has a relative humidity of 80%. Determine without using the psychrometry chart Partial pressures of water vapour and air , Specific humidity , Specific Volume and Dew point temperature (V.T.U. July2004)

air.dry of kg/kg 0.213

397.3325.101

397.3622.0

622.0

397.32461.48.0

2461.4p tablefrom C30At : s

xpp

p

kPaxpp

p

kPaSolution

s

Corresponding to Pv =3.397 kPa from tables, we get dew point temperature = 28.9°C

38

Problem 2: Atmospheric air at 101.325 kPa ha 30°C DBT and 15°C DPT. Without using the psychometric chart, using the property values from the table, Calculate Partial pressure of air and water vapour, Specific humidity , Relative humidity,Vapour density and Enthalpy of moist air

bar 0.984274

017051.001325.1p-pair of pressure

017051.0p have weC,51DPT to

042461.0p have weC,30DBT to

C15 DPT ,30

01325.1325.101

:

s

Partial

baringCorrespond

baringCorrespond

tableFrom

CDBT

barkpap

Solution

39

airdry of 57.69kJ/kg

1.88x30)5000.010775(21.005x30

)10882500( t1.005Enthalphy

40.15%

4015.0042461.0

017051.0 Re

airdry of kg0.01077kJ/

984274.0

017051.0622.0622.0

db

db

s

a

t

p

phumiditylative

x

p

phumiditySpecific

3w

3

a

/12.0847.0

010775.0density

/874.010098425.0

0.2872x303

air,dry of volume

mkgVapour

kgmx

P

RTSpecific

a

40

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 barSolution:At 25°C WBT from tables page no 14 Pvs(wbt)=0.03166 bar

41

airdry of /0179.0

0284.01

0284.0622.0

622.0

bar 0.0284 2544.11547

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

44.11547

))(()(p

1

kgkJ

pp

p

x

t

ttppP

wb

wbdbswbtwbtVS

42

airdry of 86.29kJ/kg

)4088.12500(0179.040005.1H

38.5%0.385

07375.0

0284.0

0284.0

constant p and

07375.0

40

2

xx

p

pRH

barp

remainheatingsensibleDuring

barP

DBTCAt

s

VS

C27.2chart WBT From

3449kJ/min76)-29335.18(86.added/min

min/18.335303287.0

0x100.0284)x30-(1

)(min/300m of

2

3

Heat

kgx

RT

VppofairWeight

43

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) EnthalpySolution: For air at 15°C and 60%RH, V=5.5m3/min

airdry of /006343.0

)01023.001325.1(

01023.0622.0

)p-(p

622.0

min/672.6288287.0

5.510)01023.001325.1()p-(pair of

01023.0017051.06.0

017051.0p

1

1

2

kgkg

xp

kgmx

xx

RT

VMass

barxpp

pRH

bar

s

s

44

barxpp

pRH

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

6299.6006343.01

672.6

1m

airmoist of massair/Unit dry of Mass

airdry of /59.60

)2588.12500(01392.0)25005.1(H

airdry of /01392.0)02218.001325.1(

02218.0622.0

min.55.40m 298287.0

)350.02218x10-(1.01325air of

1

1a1

2

2

2

2

2

m

kgkJH

xx

kgkgx

kgx

xMass

45

airdry of kJ/kg 55.96

)55.40672.6

56)39.993(60.12)6.6299(34.

)()(H

air, mixed theofenthalpy

993.3901392.01

55.40

1m Since

21

2211mix

2

2a2

mm

HmHm

Then

m

aa

Cttxt

tt

mm

mm

dbdbdb

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

67%RH , C19chart WBT From

C24.42mixture theof

DBT

46

Problem 5:An air conditioning system is designed under the following conditionsOutdoor conditions: 30°C DBT, 75% RHRequired indoor conditions: 22°C DBT,70% RHAmount of Free air circulated 3.33 m3/sCoil dew point temperature DPT=14°The required condition is achieved first by cooling and dehumidification and then by heating. Estimate 

The capacity of the cooling coil in tons of refrigerationCapacity of the heating coil in kWThe amount of water vapour removed in kg/hr

47

Solution:

heatingcd

cationdehumidifi 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

Locate

Locate

Locate

48

kgmkgkgWW

kgkgW

kgkJkgkJ

kgkJkgkJ

From

dc

a

/88.0V ,air dry of /0118.0

airdry of /0202.0

,air of /48H ,air of /53H

air of /40H ,air of /83H

chart

3sa

cd

ba

18.92kW48)-3.78(53

)(mcoil heating ofCapacity

ionrefrigerat of 84.375.3

)4883(78.35.3

)(mcoil cooling ofCapacity

/78.388.0

33.3air of

a

a

cd

ca

a

HH

tons

HH

skgV

VMass

49

114.3kg/hr

00.0118)360-23.78(0.020

3600)(m removedour water vapof a

daAmount

50

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 coil2. 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 and4. The heat added per kg of dry air in the heating coil

51

airdry of /0092.0

airdry of /021.0

air of /5.48H

air of /38H

air of /86H

chart

c

b

a

kgkg

kgkg

kgkJ

kgkJ

kgkJ

From

b

a

52

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

Amount

CThe

heatingb

cationdehumidifi 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

At

Locate

Locate

53

Problem 7 It is required to design an air conditioning plant for an

office room with the following conditions.Outdoor conditions: 14°CDBT, 10°CWBTRequired conditions: 20°CDBT,60% RHAmount of air circulated 0.3m3/min/personStarting 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.

54

tionhumidifica b

heatingab

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

At

Locate

Locate

55

sec/3.060

0.3x60Vsuppliedair of

/8175.0V volomeSpecific

airdry of kg/kg 00875.0

airdry of /006.0

air of /43HH ,air of /30H

chart

3

3sa

cba

mVolume

kgm

kgkg

kgkJkgkJ

From

c

ba

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

Let

C

HH

V

VWeight

ab

a

56

r3.63kg/hou

0.006)3600-08750.3669(0.0

3600)(m humidifier theof

8.34 4.1T

5.26T4.0

266.54.0T

Tfactor passing

a

d

d

d

d

xCapacity

CT

kJTTT

TBy

bc

d

dda

b

57

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 .1. DBT of the air leaving the dehumidifier.2. The quantity of water vapour removed in the

dehumidifier per hour.3. The capacity of cooling coil in tons of refrigeration.4. Surface temperature of the coil if the by pass factor of the

coil is 0.25.

58

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 line to cut the previous line at point b.DBT of air leaving the dehumidifier Tb = 40.50C

From chart Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air

Hc = 45 kJ/kg, b = 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

Solution:

59

Quantity of water vapour removed/hour = Wa(a-b)60

= 91.42(0.13-0.009)60 = 21.94 kg/hr

Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5)

= 91.42(65-45)/(60 X 3.5)

= 8.7 tons

By pass factor = (Tc-Td)/( Tb-Td) = 0.25

Td = Temperature of cooling coil = 15.830C

60

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 humidifying and finally by heating. The condition of air coming out of the humidifier is 75% RH.

Find the following . 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

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 line to cut the previous line at point b.DBT of air leaving the dehumidifier Tb = 40.50C

From chart Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air

Hc = 45 kJ/kg, b = 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

61

62

Quantity of water vapour removed/hour

= Wa(a-b)60

= 91.42(0.13-0.009)60 = 21.94 kg/hr

Capacity of cooling coil

= Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5)

= 8.7 tons

By pass factor = (Tc-Td)/( Tb-Td) = 0.25

Td = Temperature of cooling coil = 15.830C