Chapter 10

PSYCHROMETRY

Atmospheric air is moist in nature, consisting of both air and water vapour. The amount of water vapour in the atmospheric air plays an important role in human comfort. Study of such air - water vapour mixture is known as psychrometry. For simplicity the analysis is carried out with the following assumptions:

(i) The gas mixture of interest is assumed to be composed of dry air and water vapour, each of which are considered to behave as ideal gases.

(ii) The specific heats of air are assumed to be constant.(iii) Enthalpy of water vapour depends only on temperatureThese assumptions are justified as the engineering applications of

psychrometry fall generally in the temperature range of 0 to 50oC and the total pressure of the mixture is generally about one standard atmosphere (101.32 kPa). Before considering analysis, definitions and terminology related to psychrometry are first presented.10.1 Dry bulb, Dew point and Wet bulb temperatures

Temperature of the air-water vapour mixture that would be measured by an ordinary thermometer with a dry sensing element is known as dry bulb temperature.

According to Dalton’s law of partial pressure the total pressure of a mixture of dry air and water vapour is the sum of the component pressures.

Thereforep pa + pv

...(10.1)where p is the total pressure

pa is the partial pressure of dry airpv is the partial pressure of water vapour

1

In general, the state of water vapour in the atmospheric air is superheated. It can be represented in a T-s diagram as indicated by state 1 in Figure 10.1. When cooled at constant pressure, temperature of a sample of such air will decrease until it reached the saturation temperature of steam corresponding to the partial pressure Rr. This temperature is known as dew point temperature. Air becomes saturated at this state so that further cooling will result in condensation of water vapour.

Let a stream of unsaturated air flows over the bulb of a thermometer covered with thoroughly wetted wick as shown in Figure 10.2. When air passes over the wick, evaporation of water in the wick is taking place. This vapour is taken along by the unsaturated air.

The driving potential for this evaporation is the difference in partial pressure of water vapour in the unsaturated air and that of water in the wick. Thermodynamic states of water vapour in air and water in the wick are represented respectively as 1

2

and l in a T-s diagram as shown in Figure 10.3. Heat required for this evaporation is partially taken from water in the wick and the remaining from air. Hence, temperature of water and air decreases. Partial pressure of water vapour in the air decreases partial pressure pf water vapour in the air increases and that of water in the wick decreases until there exists an equilibrium between these two pressures. For further evaporation heat is supplied by the air alone. Therefore there will not be any further decrease in temperature of water in the wick. This temperature is known as wet bulb temperature. Thus wet bulb temperature is the temperature measured by the wick-covered bulb of a thermometer in which the wick is thoroughly wetted with water. The difference between drybulb and wet bulb temperatures is known as wet bulb depression. That is

WBD DBT WBT ...(10.2)

10.2 Specific humidity (or) Absolute humidity (or) Humidity ratio

The specific humidity (or) absolute humidity (or) humidity ratio (or) moisture content is defined as the ratio of mass of water vapour to the mass of dry air in a given volume of the moist air. It is usually denoted as . Thus for a sample of moist air of volume Vm3 and temperature T K :

ω=mvma

Assuming ideal gas behaviour for both air and water vapour

...(10.3)

Where pV is the partial pressure of water vapour in the mixturepa is the partial pressure of dry air in the mixtureRv is the characteristic gas constant of water vapour

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p V R Tp V R T

v v

a ap Rp R

v a

a v

Ra is the characteristic gas constant of dry air.

As defined earlier, R

Thus Equation 10.3 can be modified as :

w=pvpa

×(RuM a)(RuM v)

w=pvpa

×(M v

M a)

...(10.4)Where Mv is the molecular weight of water vapour

Ma is the molecular weight of dry airThe molecular weight of water vapour can be easily obtained knowing its

chemical formula H2O consisting of two hydrogen atoms and one oxygen atom. That is

18.02

The average molecular weight of dry air can be computed as follows: Table 10.1 gives the composition of dry air

Table 10.1 composition of dry air

Components Molecular Part byweight (Mi) volume (yi)

N2 28.02 0.7803O2 32.00 0.2099Ar 39.91 0.0094CO2 44.00 0.0003H2 2.02 0.0001

Therefore the average molecular weight of dry air Ma

0.7803 28.02

4

RM

u

+ 0.2099 32.00 + 0.0094 39.91 + 0.0003 44.0 + 0.0001 2.02

28.96 Substituting the numerical values of Ma and Mv

w=pvpa

×(M v

M a)

=0 .622pvpa

=0 .622pvp−pv

kg of w .v .kg of d .a .

…(10.5) 10.3 Relative Humidity and Degree of saturation

Let us consider a certain volume V m3 of unsaturated air at temperature T K. The state of water vapour in the mixture is represented by ‘1’ as given in Figure 10.4. If the state of water vapour is changed to state ‘s’ at the same temperature T, then the partial pressure will be ps. This pressure is nothing but the saturation pressure of steam corresponding to the mixture temperature T. The ratio of p v to ps is known as Relative humidity denoted usually by (or) RH. That is

φ=pvps

...(10.6)

Where pv is the partial pressure of water vapour in the mixture.ps is the partial pressure of water vapour in the same mixture if it is

saturated at the same temperature.Since ideal gas behavior is being assumed for both dry air and water vapour.

p m RT

Vvv

Therefore (or) RH

5

p m RTVss

mm

v

s

Thus, the relative humidity is the ratio of mass of water vapour in a certain volume of unsaturated moist air at a given temperature to the mass of water vapour in the same volume of saturated air at the same temperature. For dry air relative humidity is zero ( 0) and one for saturated air ( 100%).

The ratio of specific humidity of the given unsaturated air to that of saturated air at the same temperature is known as degree of saturation. That is

μ= wws ...

(10.7)

Fig 10.4 Illustration for relative humidity and degree of saturation

Where is the actual humidity ratio of given unsaturated air.

s is the humidity ratio of the same volume of air if it is saturated at the same temperature.

10.4 Specific volume of moist airAs the entire volume of the mixture is occupied by air, as well as water vapour,

the specific volume can be expressed in terms of per kg of dry air, per kg of water vapour and per kg of the mixture. Therefore,

v V/ma ...(10.8)

v V/mv ...(10.9)

v V/(ma+mv) ...(10.10)

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10.5 Enthalpy of moist airOn the basis of unit mass of moist air, the specific enthalpy of moist air can be

expressed as follows : h ha+ hv

...(10.11)Whereha Specific enthalpy of dry air

Humidity ratiohv Specific enthalpy of water vapour

Assuming ideal gas behaviour, the specific enthalpy of dry air is given as ha 1.005 t

...(10.12)The datum for enthalpy is taken as : h 0 at t 0oC the temperature t should

be substituted in Celsius.For water vapour the common empirical relation recommended for specific

enthalpy is given below : hv 2500 + 1.88t

...(10.13)Where t is the temperature in CelsiusTherefore specific enthalpy of moist. air is given as

h 1.005 t + (2500 + 1.88t) ...(10.14)

10.6 Adiabatic saturation and Thermodynamic wet bulb temperature

Let us consider a pool of water kept inside an insulated container as shown in Figure 10.5. When unsaturated air is blown over the surface of water in the pool, evaporation of water occurs. The moisture content of air increases. Heat required for evaporation is supplied partially by air and the remaining from water. Therefore, temperature of both air and water decreases. The process of evaporation continues until the energy supplied by the air to the water is equal to the energy required to vaporize the water. When this point is reached, thermal equilibrium exists among water, air and water vapour and consequently air is saturated. The corresponding temperature is known as adiabatic saturation temperature or thermodynamic wet bulb temperature. Figure 10.6 shows the process on T-s coordinates.

The thermodynamic wet bulb temperature is almost equal to the wet bulb temperature. For all the practical purposes, the adiabatic saturation temperature (or) thermodynamic wet bulb temperature and wet bulb temperature are considered to be equal.

7

This steady flow process can be analyzed on the basis of mass and energy to develop a relationship between the inlet-air conditions and the adiabatic-saturation temperature.

Applying mass balance of air and water vapour we get ...(10.15)

...(10.16)

Where is the mass flow rate of dry in the inlet streamis the mass flow rate of dry in the exit streamis the mass flow rate of water vapour in the exit streamis the mass flow rate of make up water

Equation (10.16) can be modified as

Substituting from Equation (10.15) and rearranging, we get

8

m m ma1 a2 a

m m m1 f 2v v

1 a1 f 2 a2m m m

m ( )mf 2 1 a ...(10.17)Applying energy balance,

..(10.18)Substituting from Equation (10.15) and (10.17), Equation (10.18) becomes

...(10.19)Where h1 and h2 are specific enthalpy’s of moist per kg of dry air. It can be

expressed as follows:h1 Cpa (t1) + 1 hv1 ...(10.20)h2 Cpa (t2) + 2 hv2 ...(10.21)

Substituting for h1 and h2 in equation 10.19, we haveCpa (t2 t1) + 2hv2 1hv1 (w2 w1) wf

Therefore, 1

...(10.22)Examination of the above equation reveals that all the quantities on the right

hand side can be determined if the dry bulb temperature t1 at the inlet and the adiabatic saturation temperature t2 at exit can be measured. As mentioned earlier, the adiabatic saturation temperature is almost equal to the wet bulb temperature and hence by knowing dry bulb and wet bulb temperature, the moisture content of humid air can be computed using the above expression. 10.7 Psychrometric chart

The chart that presents all the psychrometric properties is known as psychrometric chart. It is generally constructed for standard atmospheric pressure of 760 mm Hg. Dry bulb temperature is taken on x-axis and moisture content on y-axis. Lines corresponding to constant relative humidity wet bulb temperature, specific volume and specific enthalpy are plotted on the chart. A typical psychrometric chart is shown in Figure 10.7.

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m h m h m ha2 2 a1 1 f f

m (h h ) m ( )ha 2 1 a 2 1 f

C (t t ) w (h h )h h

pa 2 1 2 v2 f

v1 f

10.8 Psychrometric ProcessesTemperature and moisture content of the moist is to be controlled for human

comfort and many process applications. Principles of psychrometry play an important role in establishing the necessary heat and mass interactions. Some of the basic processes are presented in this section.10.8.1 Sensible Heating

Moist air is heated without altering the moisture content of the air. The following changes are observed due to sensible heating :

1) Dry bulb temperature - Increases2) Wet bulb temperature - Increases3) Specific enthalpy - Increases4) Specific volume - Increases5) Relative humidity - DecreasesFigure 10.8 shows the schematic arrangements of the process. Figure 10.9

shows the process on a psychrometric chart.

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Amount of heated added can be expressed as

Q m (h h )a 2 1 ...(10.23)

10.8.2 Sensible coolingIt is just the reverse of sensible heating, that is, moist air is cooled such that

the moisture content remains unaltered. The following changes are observed in a sensible cooling process.

1. Dry bulb temperature - Decreases2. Wet bulb temperature - Decreases3. Specific enthalpy - Decreases4. Specific volume - Decreases5. Relative humidity - IncreasesFigure 10.10 shows the schematic arrangement of a sensible cooling process.

Figure 10.11 shows the process on a psychrometric chart.

Quantity of heat to be removed is given as

... (10.24)

11

Q m (h h )a 1 2

10.8.3 Dehumidification by coolingWhen moist air is cooled below its dew point temperature water vapour is

removed from air as it starts condensing below the dew point temperature. Air is in saturated state at the end of cooling as shown in Figure 10.12.

Heat removal rate is given as Q m (h h )+m ( )ha 1 2 a 1 2 f

...(10.25)where hf is the specific enthalpy of the condensate.The changes observed are listed below :1. Drybulb temperature - Decreases2. Wet bulb temperature - Decreases3. Specific enthalpy - Decreases4. Moisture content - Decreases5. Relative humidity - Increases

10.8.4 Cooling and DehumidificationLet us consider conditioning of air existing at state 1 to state 2 as shown in

Figure 10.13. To remove the necessary quantity of moisture, air is cooled to a state 2'. From this state it is heated sensibly to state 2. Schematic arrangement of such a process is shown in Figure 10.14.

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Heat load on the cooling coil is given as

...(10.26)Heat load on the heater is given as

...(10.27)10.8.5 Adiabatic Humidification

Increase in moisture content of air is known as humidification. If humidification is carried out adiabatically, the energy required for evaporation must be supplied by the entering air. Consequently, the dry bulb temperature of air must decrease. Figure 10.16 shows the schematic arrangement. Figure 10.17 shows the process on psychrometric chart.

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Q m (h h )+m ( )hcooling a 1 2 a 2 1 f

Q m (h h )heating a 2 2

On the basis of unit mass of dry air, the energy equation is given ash1 + (2 1) hf h2 ...

(10.28)In comparison with the magnitude of h1 and h2, (w2 w1)hf is negligible.

Therefore, it can be assumed that h-specific enthalpy of air-water vapor mixture remains constant during adiabatic humidification process.10.8.6 Heating and humidification

Moisture content and temperature are considerably less in winter. Air is to be heated and humidified as shown in Figure 10.18. The process can be represented on the psychrometric chart as shown in Figure 10.19.

The energy interaction at the heater is given as Q m (h h )a 2 1 ...(10.29)

Mass to be added by the sprayer

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m m ( )a 3 2 ...(10.30)10.8.7 Adiabatic Mixing

In many practical applications, two or more streams are to be mixed adiabatically. Let us consider two humid air streams 1 and 2 are mixed adiabatically in a chamber. The mixture formed is at state 3.

Mass balance of air gives m +m ma1 a2 a3

...(10.31)Mass balance of water vapour gives

15

mm 1 1a

1a2

1

3

2

3

m +m m1 2 3v v v ...(10.32)

m +m ma1 1 a2 2 a3 3

Energy balance of the entire control volume gives m h +m h m ha1 1 a2 2 a3 3 ...(10.33)

Where h1 ha1 + 1hv1

h2 ha2 + 2 hv2

h3 ha3 + 3hv3

Consider equation (10.31) and divide all the terms by .ma2 It gives

mm 1 m

ma1

a2

a3

a2

...(10.34)

Dividing equation (10.32) by , we have

mm

mm

a1

a21 2

a3

a23

mm

mm

a3

a2

a1

a2

1

3

2

3

...(10.35)

Substituting equation (10.35) in (10.34) we get

mm 1 m

ma1

a2

a1

a2

1

3

2

3

...(10.36)

Similarly it can be demonstrated that

16

mm

a1

a2

3 2

3

3

1 3

mm

a1

a2

3 2

1 3

...(10.37)

Solved ProblemsProb : 10.1 An air-conditioning unit receives an air-water vapour mixture

at 101 kPa, 35oC and 80% relative humidity.Determine

a) the dew pointb) the humidity ratioc) the partial pressure of aird) the mass fraction of water vapour

System : Air-water vapour mixtureKnown :

1) DBT 35oC2) p 101 kPa3) 80%

Diagram :

To find : 1) Dew point temperature2) Humidity ratio3) Partial pressure of air4) Mass fraction of water vapour

Analysis : 1) Dew point temperatureIt is the saturation temperature of steam corresponding to the

partial pressure pv. To find pv consider relative humidity. 0.8

Where ps is the saturated pressure corresponding to 35oC as shown in the T-s diagram. From steam table (appendix) ps 5.63 kPa. Therefore

pv 0.8 ps 0.8 5.63 4.5 kPa

Thus the dew point is the saturation temperature at 4.5 kPa. From steam table, it is obtained that

DPT 30.93oC.

17

mm

h hh h

a1

a2

3 2

1 3

2) Humidity ratio () 0.622 pp

v

s

=0.622

0.029

3) Partial pressure of air p pv

101 4.5 96.5 kPa

4) Mass fraction of water vapour

Mass associated with each kg of dry air as indicated by humidity ratio is 0.029 kg. ThereforeMass fraction of water vapour 0.02818

Result : 1) Dew point temperature 30.93oC2) Humidity ratio 0.029

3) Partial pressure of air 96.5 kPa4) Mass fraction of water vapour 0.02818

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4.5(101 4.5)

kg of w.vkg of d.a

mm m

v

v a

kg of w.vkg of d.a

Prob : 10.2 Given for an air-water vapour mixture that Tmix 70oC and pmix 200 kPa and Pair = 180 kPa, find the dew point, humidity ratio and relative humidity

System : Air-water vapour mixtureKnown :

1) Tmix 70oC2) pmix 200 kPa3)Pair 180 kPa

To find : 1) Dew point temperature (DPT)2) Humidity ratio ()3) Relative humidity ()

Diagram :

Analysis : 1) Dew point temperatureIt is the saturation temperature of steam corresponding to the

partial pressure (pv) of the water vapour in the mixture. Form Dalton’s law of partial pressure.

p pa + pv

200 180 + pv

pv 20 kPaFrom steam table for 20 kPa, saturation temperature is 60.06oC. Therefore

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DPT 60.06oC 2) Humidity ratio () 0.622

0.0691 kg of w.vkg of d.a

3) Relative Humidity ()

Where ps is the saturation pressure corresponding to the mixture temperature 70oC. From steam table at 70oC, saturation pressure is 31.19 kPa.Therefore 0.641 (64.1 %)

Result : 1) Dew point temperature 60.06oC2) Humidity ratio 0.06913) Relative Humidity () 64.1 %

Exercises1. State Dalton’s law of Partial pressure2. What does Psychometry mean ?3. Define the following

a) Dew point temperatureb) Specific humidityc) Relative humidityd) Degree of saturatione) Wet bulb depression

4. For saturated air wet bulb depression is zero (T/F)[Ans : True]

5. In psychrometric chart ___________ is taken on x-axis[Ans : DBT]

6. In psychrometric chart ___________ is taken on y-axis[Ans : Humidity ratio]

7. What properties are represented by straight lines on psychrometric chart[Ans : h ; WBT ; v]

8. Which one of the following property can not be read from the psychrometric chart.

a) Relative humidity

20

pp

v

s

pp

v

a

b) Absolute humidityc) Wet bulb temperatured) Degree of saturation

[Ans : d]

9. Dew point temperature is same as the thermodynamic wet bulb temperature (T/F)

[Ans : False]10. During sensible cooling wet bulb temperature increases (T/F)

[Ans : False]11. During sensible heating dew point temperature increases (T/F)

[Ans : False]12. For all the practical purposes thermodynamic wet bulb temperature is taken

as adiabatic saturation temperature (T/F) [Ans : True]13. Wet bulb depression can never be negative (T/F)

[Ans : True]14. During evaporative cooling of moist air ___________ remains constant

[Ans : h]15. During sensible heating ___________ remains constant

[Ans : Humidity ratio]16. By use of psychrometric chart, estimate the missing information is table

given below. The system is a mixture of water vapour and dry air, and the total pressure is 1 atm.

17. Use the formulas and the steam tables to find the missing property of , and DBT, total pressure is 100 kPa; repeat the answers using the psychrometric chart.

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a) 60%, 0.010 b) DPT 10oC, 50 %c) DTP 25oC, WBT 20oC

[Ans : a) DBT = 23oC b) DBT = 21oC ; = 0.0076 c) = 0.0128 ; = 0.64]

18 A rigid of volume vessel 300 litre contains air water vapour mixture at 150 kPa, 40oC, and 10% relative humidity. Calculate the following

a) partial pressure of air and water vapourb) Moisture content in kg per kg of dry airc) Dew point temperatured) degree of saturation

[ Hint : Since the pressure 150 kPa, standard psychrometric charts should not be referred] [Ans : a) pv= 6.645 kPa ; pa=143.35 kPa b) w = 0.0288

c) DPT = 37.62oC d) m = 0.894 ]19. 10m3/min of moist air at 10oC, 90% relative humidity is passed over a heater

until the temperature reaches 20oC. Determine the quantity of heat added per minute.

[Ans : 117.3 kJ/min]20. 100 kg/minute of air at drybulb temperature 40oC and wet bulb temperature

passes over the cooling of a vapour compression refrigeration plant of overall COP 3. If the preferred exit temperature is 25oC, what is the power required for the plant ?

[Ans : 8.61 kW]21. Moist air in a piston/cylinder is at 35oC, 100 kPa, and a relative humidity of

20%. It is to be compressed isothermally until the pressure increases to 200 kPa. Find the final volume and final relative and specific humidity.

[Ans : 0.5 m3; 40 %; 7.07 g.w.v/kg of air]22. A vessel of volume 1m3 containing steam at 20 MPa, 400oC is kept inside a

room of volume 100 m3 containing dry air at 100 kPa, and 27oC. As the wall of the vessel ruptures, steam fills the entire volume of the room. What will be the final room pressure.

23. Atmospheric air at 35oC, relative humidity 10%, is too warm and also too dry. An air conditioner should deliver it at 20oC and 60% relative humidity in the amount of 3600 m3/min. Sketch a setup to accomplish this, find any amount of liquid (at 20oC) that is needed or discarded and any heat transfer.

[Ans : 18.43 kg/min to be added]24. In an evaporative cooling process 600 m3/hr of at 39oC, 20% relative humidity

is cooled to 25oC. Calculate.a) Final relative humidityb) final humidity ratioc) required water quantity in kg/hr

22

[Ans : 83 %; 0.012 kg w.v/kg of d.a; 3.75 kg/hr]

25. 100m3/hr of atmospheric air at 40oC, 50% relative humidity is cooled until it becomes saturated at 15oC. Determine :

a) Moisture content at the exitb) Heat removal rate in kWc) The rate of condensate formed in kg/min [Ans : 0.0105 kg w.v/kg d.a; 1.755

kw; 0.0225 kg/mco]26. In an air-conditioning process 100 m3/min of atmospheric air at 5oC, 50%

relative humidity is to be conditioned to 25oC, 60% relative humidity. The processing is done is two stages : Sensible heating followed by evaporative cooling. Determine the condition of air at the end of sensible heating and the required heating load. Also compute the rate of moisture addition in kg/min.

27. Two streams of moist air are mixed adiabatically as shown in the Figure p.28. Determine the condition of the air after mixing.

[Ans : 27oC DBT; 60% RH]28. A common method for dehumidifying air employs silica gel to absorb some of

the moisture in the air. Suppose that 2 kg/s of moist air with DBT 30oC and RH 80% enters the dehumidifier and leaves with a relative humidity of 30%.Determine :

a) DBT at the exit of the humidifierb) Humidity Ratio at the exit of the humidifierc) Mass of water vapour removed per hour

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[Ans : 43oC; 0.016 kg w.v/ kg of d.a; 43.2 kg/hr]

29. Atmospheric air with a dry bulb temperature 30oC and a relative humidity 70% is cooled to 12oC dry bulb temperature in a steady flow process. Determine the heat transfer from the air per unit mass of dry air and the amount of water vapour condensed per unit mass of dry air.

[Ans : 44 kJ/kg; 0.01 kg w.v/kg of d.a]30. A combination of cooling and reheat process is used to deliver air at a dry

bulb temperature of 20oC and a relative humidity of 40 %. The air enters at a dry bulb temperature of 30oC with a relative humidity of 70% and a volumetric flow rate of 45 m3/min. Determine :

a) the heat transfer rate in the cooling sectionb) the heat transfer rate in the heating section and c) the mass flow rate from the cooling section

[Ans : 48.2 kW; 12.52 kW; 0.97 kg/min]

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