Chapter 7 PSYCHROMETRY

PSYCHROMETRYDR.K.ARUMUGAM

1

PSYCHROMETRY

The study of the properties of air and vapourpertaining to air conditioning problems is calledpsychrometry.Dry air: it is the mixture of nitrogen, oxygen and smallpercentage of other gases. Air contains 79% nitrogen,21% oxygen by volume and has a molecular weight of29.Moist air: it is the mixture of dry air and water vapour.The amount of water vapour varies according to thetemperature of air and reaches saturation at one point.Saturated air mixture: saturated air mixture is amixture of dry air and water vapour in which the partialpressure of the vapour is equal to the saturationpressure of water at the temperature of the mixture.

2

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

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

Dew point temperature: when the unsaturated air iscooled at constant pressure, the mixture reachessaturation temperature corresponding to the partialpressure of water vapour. This temperature at whichcondensation of the vapour begins resulting information of liquid droplets or dew when themixture is cooled is called dew point temperature. 3

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’.

4

Dalton’s law of partial pressures: the totalpressure of the mixture of gases is the sum of thepartial pressures exerted by each gas when itoccupies the same volume of the mixture at thesame temperature of the mixture.

Dry bulb temperature: it is the equilibriumtemperature of the mixture indicated by anordinary thermometer denoted by Tdb.

Wet bulb temperature: it is the temperatureindicated by a wet bulb thermometer which hasits temperature sensitive element (bulb) coveredwith a wick soaked in water. It is denoted by Twb.

5

6

Relative humidity: it is defined as the ratio ofpartial pressure of water vapour in a mixture tothe saturation pressure of water at dry bulbtemperature.

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

Hence

Hence relative humidity is also defined as theratio of the mass of water vapour in a certainvolume of moist air at a given temperature tothe mass of water vapour in the same volume ofsaturated air at the same temperature.

7

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

8

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

9

Carrier equation: When DBT and WBT are given, forcalculating the partial pressure of water vapour inair many equations have been proposed of whichDr. 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 air

tdb = dry bulb temperature, 0C

twb = wet bulb temperature, 0C10

Psychrometric chart:As the calculations of various properties ofmoist air are tedious and time consuming, allthe essential data for complete thermodynamicand psychrometric analysis of air conditioningprocesses can be summarized and can bepresented in the form of a chart.

Such a chart which makes it possible to obtainthe necessary information readily forengineering calculations related to moist airknown as a psychrometric chart.

A psychrometric chart is constructed for a givenmixture pressure, using dry bulb temperatureand the specific humidity as co-ordinates.

11

At a given mixture pressure, the vapour pressure pv

is a function of specific humidity only and hencethere is only one value of pv for each value of ω.

12

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

Relative humidity lines: these are the curves withhumidity ranging from maximum 100% tominimum 0%.

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

DBT lines: these are vertical lines and are equallyspaced. DBT increases as we move from left toright.

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

13

Sensible heating:

Heating of air withoutaddition or subtraction ofmoisture is called sensibleheating.

This can be achieved bypassing the air over a heatingcoil.

The heat added increases theDBT of air.

This is useful in winter airconditioning.

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

14

Sensible cooling:Cooling of air withoutaddition or subtraction ofmoisture is called sensiblecooling.

This can be achieved bypassing the air over a coolingcoil.

This is useful in summer airconditioning.

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

15

Cooling and dehumidification:

Water vapour may be removed from air by cooling itbelow its dew point temperature.

As a result of the cooling process a portion of the vapourin the air is condensed. Dehumidification will take placealong with cooling when the saturation temperature ofthe cooling coil is below the dew point temperature ofthe cooling air.

16

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

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

17

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

In this case it is given by

18

Adiabatic humidification: if humidification iscarried out adiabatically, the energy required forthe evaporation of the added moisture mustcome from the entering air.

19

As the dry bulb temperature of air decreases duringadiabatic humidification process, it is also known asevaporative cooling process or cooling withadiabatic 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.

20

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

21

Making energy balance,

ma1(ha1 + ω1hv1) + ma2(ha2 + ωhv2) = ma3(ha3 + ω3hv3)

Making mass balance

ma1 + ma2 = ma3

Making moisture balance

ma1 ω1 + ma2 ω2 = ma3 ω3

Combining we get,

22

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 theoccupants.

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

4. To maintain the purity of air by removing odourand dust.

23

1. Oxygen supply: human body takes in oxygenthrough the system and gives out CO2. Normallyeach person requires nearly 0.65m3 of O2 per hourand produces 0.2m3 of CO2.

The percentage of CO2 in atmosphere is about0.6% and it is necessary to maintain this to ensurecomfort for easy breathing. Thus the quantity of airsupply to an air conditioned space should beregulated properly to see that percentage of CO2

should not be exceeded.

2. Heat removal: it is a well known fact that humanbeings dissipate good amount of heat to theatmosphere during breathing etc.

24

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%.25

4. Air movement:

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

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

Air movement without proper air-distribution ispermissible for local cooling sensation known asdraft.

A velocity of about 8m/min associated withtemperature differential of 10C do not result innoticeable draft.

Velocities greater than this producesuncomfortable drafting conditions.

26

5. Purity of air:

It is important to maintain quality of air in any air-conditioned space.

Odour, dust, toxic gases and bacteria areconsidered for defining the purity of air.

The various factors which makes air impure are:

i. The evaporation on the surface of the body addsodour to the air.

ii. The smoke from the surroundings which has a badeffect on nose, eyes and heart.

iii. The toxic gases are objectionable as they causeirritation.

27

Summer air conditioning systems(for Hot and dry outdoor conditions): this system isused when outdoor conditions are hot and dry.That means atmospheric temperature is higher than

the comfort temperature with less moisture content(lesser RH).

28

Atmospheric air at condition ‘1’ (higher DBT and lowerRH) enters the air dampers and passes over the coolingcoil via an air filter.

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

Air now enters an adiabatic humidifier and is passedover water eliminators.

29

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

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

30

Summer air conditioning systems:Hot and humid outdoor conditions: this system is usedwhen the out door condition is hot and humid, like incoastal areas. That means the atmospheric temperature ishigher but at the same time contains large moisturecontent. This condition eliminates the need for anadiabatic humidifier as shown in fig.

31

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

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

32

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

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

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

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

33

Winter air conditioning system:During winter air in the atmosphere is at lowertemperature than the required conditions.

Also the relative humidity may be more or lesser than theactual relative humidity required for human comfort. Figshows an arrangement for such a system where dryconditions prevail in the atmosphere.

34

Cold air from the atmosphere is filtered in an air filterbefore 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 heatcoil is ‘2’. Condition line 1-2-3 represents the changes inthe DBT of air while it passes over the heating coil.

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

35

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

As the humidity and temperature of air have still notreached comfort conditions air is heated again up to point‘7’ by another heating coil.

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

36

Problem 1. Moist air at 30°C,1.01325 bar has a relativehumidity of 80%. Determine without using thepsychrometry chart Partial pressures of water vapourand air , Specific humidity , Specific Volume and Dewpoint temperature (V.T.U. July2004)

37

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

Problem 2: Atmospheric air at 101.325 kPa ha 30°C DBT and15°C DPT. Without using the psychometric chart, using theproperty values from the table, Calculate Partial pressureof air and water vapour, Specific humidity , Relativehumidity,Vapour density and Enthalpy of moist air

38bar 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

Problem 3:

Air at 30°C DBT and 25°C WBT is heated to 40°C. ifthe air is 300 m3/min, find the amount of heatadded/min and RH and WBT of air. Take air pressureto be 1 bar

Solution:

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

40

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

4) One stream of air at 5.5m3/min at 15°C and 60% RHflows into another stream of air at 35m3/min at 25°C and70%RH, calculate for the mixture 1) Dry bulb temperature,2) Wet bulb temperature 3) Specific Humidity and4) Enthalpy

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

43airdry of /006343.0

)01023.001325.1(

01023.0622.0

)p-(p

622.0

min/672.6

288287.0

5.510)01023.001325.1()p-(pair of

01023.0017051.06.0

017051.0p

1

1

2

kgkg

xp

kgm

x

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

Problem 5:

An air conditioning system is designed under the following conditions

Outdoor conditions: 30°C DBT, 75% RH

Required indoor conditions: 22°C DBT,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

The capacity of the cooling coil in tons of refrigeration

Capacity of the heating coil in kW

The amount of water vapour removed in kg/hr46

Solution:

47

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

3

sa

cd

ba

18.92kW48)-3.78(53

)(mcoil heating ofCapacity

ionrefrigerat of 84.375.3

)4883(78.3

5.3

)(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

Problem 6:A summer air conditioning system for hot and humidweather (DBT=32°Cand 70% RH)Consists in passing the atmosphere air over a coolingcoil where the air is cooled and dehumidified. The airleaving the cooling coil is saturated at the coiltemperature. It is then sensibly heated to the requiredcomfort condition of 24°C and 50%RH by passing it overan electric heater then delivered to the room.Sketch the flow diagram of the arrangement andrepresent the process undergone by the air on askeleton 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 50

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

Problem 7

It is required to design an air conditioning plant for anoffice 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 andthen by adiabatic humidifying. Determine thefollowing.

Heating capacity of the coil in kW and the surfacetemperature required, if the by pass factor of thecoil is 0.4 Capacity of the humidifier.

53

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

3

sa

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

dd

a

b

Problem 8An 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% RHIndoor condition = 220C DBT and 60% RHVolume of air required = 0.4m3/min/personThe 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 0.25.57

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/min58

Solution:

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

59

Problem 9An air conditioned system is to be designed for acinema hall of 1000 seating capacity when thefollowing conditions are given:Outdoor condition = 110C DBT and 70% RHRequired indoor condition = 200C DBT and 60% RHAmount of air required = 0.3m3/min/personThe required condition is achieved first by heating,then by humidifying and finally by heating. Thecondition of air coming out of the humidifier is 75%RH.Find the following .Heating capacity of the first heater in kW and conditionof the air coming out of the first heater in kW andcondition of the air

60

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/min61

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

62