1

Gas-Vapor Mixtures and Air-Conditioning

Content:� Dry and Atmospheric Air� Dew-point Temperature� Properties in the Psychrometric Chart� Air Conditioning Processes

Dry and Atmospheric AirDry and Atmospheric Air

hhdrydry airair = = CCpp(T(T--TTrefref) = ) = CCppTT (14.1a)(14.1a)hhdrydry airair = C= Cpp(T(T22 ––TT11) = C) = Cpp T (14.1b)T (14.1b)

Dry airDry air is a mixture of N2, O2, and small amount of other gases. Normally, atmospheric airatmospheric air contain moisture (or water vapor).

In air-conditioning applications ranges from about –10 to about 50oC, the atmospheric air can be the atmospheric air can be treated as an ideal gastreated as an ideal gas with constant specific heats. Taking 0oC as the reference temperature then

Atmospheric air = dry air + water vapor

2

Total Atmospheric AirTotal Atmospheric Air PressurePressureWhen the mixture temperature is below 50oC, the water vapor is an ideal gaswater vapor is an ideal gas with negligible error (<0.2%). Then, the atmospheric airatmospheric air (Dry air + Water Vapor) can be treated as ideal gas mixture.ideal gas mixture.

The total atmospheric air pressure (PP) is sum of partial pressure of dry air (PPaa) and that of water pressure (PPvv):

P = PP = Paa + + PPvv (14.2)

Enthalpy of Water VaporEnthalpy of Water Vapor

hhg g (T) = (T) = hhgg(T(T=0=0ooC) + Cp(TC) + Cp(T22 ––TT00) ) hhg g (T) = 2,500.9 + 1.82(T) (14.4)(T) = 2,500.9 + 1.82(T) (14.4)

For Air conditioning temperature (-10oC to 50oC), the enthalpy of water vaporenthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at the same enthalpy of saturated vapor at the same temperaturetemperature.

and can be expressed as

hhvv(T(T, low P) ~ , low P) ~ hhgg(T(T) (14.3)) (14.3)

3

Specific HumiditySpecific Humidity ((ZZ))Absolute humidity or specific humidityspecific humidity (sometimes called humidity ratio, humidity ratio, ZZ) is the ratio of mass of water vapor to mass of dry air.

(14.8) )(

622.0

(14.7) 0.622

(14.6) humidity; specific

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Relative humidity ( ) is the ratio of mass of water vapor at the specified condition to mass of water mass of water vapor if vapor if saturated conditionsaturated condition at the same temperature.

(14.11b) 622.0

and

(14.11a) )622.0(

Then;

(14.10) where

(14.9)

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Relative HumidityRelative Humidity ( )

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eTemperatur Bulb-Dry

(14.12) ; since

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Total Enthalpy of Air (Water Vapor Total Enthalpy of Air (Water Vapor Mixture)Mixture)

For an amount of moist air, the mass of dry air is not change but water vapor may condense causing mass of vapor change. Then enthalpy of moist air is expressed per unit mass of dry airper unit mass of dry airinstead of per unit mass of air-water vapor mixture.

Example 14.1 A 5-m x 5-m x 3-m room contains air at 25oC and 100 kPa at a relative humidity of 75 percent. Determine (a) the partial pressure of dry air, (b) the specific humidity (c) the enthalpy per unit mass of dry air, and (d) the masses of dry airand water vapor in the room.

Solution

5

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air)dry (kJ/kg hhh

PPP

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or

answer kg 1.30 ............

answer kg 85.61 ............ )(

answerair dry kJ/kg 8.63

(c)answer air dry water/kgkg 0.0152

.......... )(

622.0 (b)

answer kPa 97.62 .......

and (a) @

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Dew Point TemperatureDew Point Temperature

Dew-point temperature (Tdp) is the temperature at temperature at which condensation beginswhich condensation begins when the air is cooled when the air is cooled at constant pressureat constant pressure or Tdp is the saturation temperature of water corresponding to the vapor pressure.

Tdp = Tsat@Pv (14.13)

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Example 14.2 In clod weather, condensation frequently occurs on the inner surfaces of the windows due to the lower air temperatures near the window surface. consider a house shown, that contains air at 20oC and 75 percent relative humidity. At what window temperature will the moisture in the air start condensation on the inner surfaces of the windows?

Solution

Tdp = Tsat@Pv

Pv = Pg@20C

Tdp = 15.4oC

Adiabatic Saturation TemperatureAdiabatic Saturation TemperatureRelative humidity, and Specific humidity, Ä�o�n°¥¤µ�Â�nª´�¥µ�����( = Pv/Pg, = mv/ma = 0.622 Pv/Pa) Â�n�oµ¦¼o�dew point temp. �È�³®µ�Pv�Å�o�(Tdp = Tsat@Pv)�¨³�ε�ª�®µ� �¨³� �Å�oÃ�¥Å¤n¥µ�

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(14.15) 622.0

, 1.0 where2, statefor (14.11b) from

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yield,which

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Consider an Adiabatic Saturation Process

So, just measure TSo, just measure T11, T, T22and and PPtotaltotal, we can get , we can get 11

WetWet--Bulb TemperatureBulb Temperature (experiment)To make exit stream saturated, it need a very long channel or spray mechanism. At normal atmospheric WetWet--Bulb TemperatureBulb Temperature and Adiabatic Adiabatic Saturation TemperatureSaturation Temperature of air and water vapor mixture are approximately equal.

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Example 14.3 The Specific and Relative Humidity of Air

The dry-bulb and wet-bulb temperatures of atmospheric air at 1 atm (101.325 kPa) pressure are measured with a sling psychrometer and determined to be 25 and 15oC, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air.

Solution Tdb=T1 ; Twb=T2 ; Patm = P2

airdry kJ/kg 41.8 C (c)

33.2%or 0.332 )622.0(

:a)eqn.(14.11 (b)

air dry water/kgkg 0.00653 )( C

(a)

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Properties in the Properties in the PsychrometricPsychrometric ChartChart

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1. What is a dry bulb temperature?� It is measured with an ordinary thermometer.� It is independent of moisture.� It is located on the “X” axis of a psychometric

chart.

0oC dry bulb temperature 50oC

2. What is absolute humidity (humidity ratio)?

� It is the ratio of the weight of moisture contained in 1 kg of dry air.

� It is the kg water / kg of dry air.� It is located on the “Y” axis of the chart.� It is also called Humidity Ratio or Specific Humidity

.000

.001

.002

.003

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3. What is the saturation curve?� It includes the wet bulb and dew point temperatures.� It completes the psychrometric chart outline.

“X” Axis

“Y”Axis

Saturation Curve

4. What is the wet bulb temperature?� The temperature at which an equilibrium exists

between an air-vapor mixture and water.� It is dependent on moisture in the air.� Values are on the saturation curve.� The lines slope downward to the “X” axis.

“ X” Axis

“ Y”Axis

50100

250300

350

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5. What is dew point?� It is the temperature at which condensation occurs

as heat is removed from an air-vapor mixture.� The answer is read on the saturation curve

horizontally to the left of the point where the dry bulb andwet bulb meet.

80q

65ï

50ï

How to Find Dew Point Temp. ?How to Find Dew Point Temp. ?

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6. What is relative humidity?

� It is the ratio of actual pressure of water vapor in the air to the pressure if the air were saturated and with a constant temperature.

100%

60%30%

20%

7. What are enthalpy lines?� Enthalpy is a thermal (heat) property.� It is the heat in an air vapor mixture.� Lines are parallel to the wet bulb temp. lines.� Values are in kJ per kg dry air.

kJ per kg dry air

4050

60

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8. What is specific volume?� The volume occupied by 1 kg of dry air.� It represents the m3 / kg of dry air.� The values are read below the dry bulb readings.

0.80 0.84 0.88 0.92Specific Volume = m3 / kg of dry air

PsychrometricPsychrometric ChartChartThe psychrometric chart has seven lines.

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Example 14.4 The Use of t he Psychr omet r ic Char t

Consider a r oom t hat cont ains air at 1 at m (101.325 kPa) 35oC, and 40% r elat ive humidit y. Using t he psychr omet r icchar t , det er mine (a) t he specif ic humidit y, (b) t he ent halpy (c) of wet -bulb t emper at ur e, (d) t he dew-point t emper at ur e, and (e) t he specif ic volume of t he air

Solut ion

= 0.0142 kg water/kg.da.h = 71.5 kJ/kg.da.Twb = 24 oCTdp = 19.4 oCv = 0.893 m3/kg.da.

Human ComfortHuman Comfort� Human wants:

Not hot, not cold, not humid, not dry Îjust COMFORTABLE Î Air-conditioning.

� Typical comfort conditions:22 22 –– 2727ooC and 40 C and 40 –– 60 %RH60 %RH

�� Air motionAir motion also effect human feeling (body heat loss).

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AirAir--Conditioning ProcessesConditioning ProcessesMaintaining space desired temperature and humidity requires “ airair--conditioning processesconditioning processes””including simple heating, cooling, humidifying and dehumidifying.

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The processes are normally considered as SSSF process;The processes are normally considered as SSSF process;* mass of dry air remains the same* mass of dry air remains the same* water vapor added/removed* water vapor added/removed* 1* 1stst law treat dry air and water vapor separately.law treat dry air and water vapor separately.

AirAir--Conditioning ProcessesConditioning Processes

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Example 14.5 Heating and Humidification of Air

An air-conditioning is to take in outdoor air at 10oC, and 30% relative humidity at a steady rate of 45 m3/min and to condition it to 22oC, and 60% relative humidity. The outdoor air is first heated to 22oC in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming the entire process takes place at a pressure of 100 kPa, determine (a) the rate of heat supply in the heating section, and (b) the mass flow rate of the steam required in the humidifying section.

Solution

(a) = 673 kJ/min; (b) = 0.539 kg/min

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Example 14.6 Cooling and Dehumidif icat ion of Air

Air ent er s a window t ype air condit ioner at 1 at m, 30oC, and 80% r elat ive humidit y at a r at e of 10 m3/ min and leaves as sat ur at ed air at 14oC det er mine t he r at e of heat and moist ur e r emoval f r om t he air .

Solut ion

(a) = 511 kJ/min; (b) = 0.131kg/min

Evaporative Cooling Evaporative Cooling •�°µ��Êε–Á¥È�­�µ¥����µÂ°¨�°±°¦r–Á¥È�ªµ�•��ÊεÄ�ð��·�Á�µ�®¦º°Á®¥º°��·�Á�µ�Á¥È��ªnµ���·•�¡´�¨¤�Êε�Ä®o¨¤�¸ÉÁ¥È��ªnµ¡´�¨¤�¦¦¤�µ• Evapor at ive cooling pr ocess: As wat er evapor at es, t he lat en heat of vapor isat ion is absor bed f r om t he wat er body and t he sur r ounding air . As a r esult , bot h of t hem ar e cooled.• Evapor at ive cooling pr ocess ~ Adiabat ic sat ur at ion pr ocess:

TTwbwb ~ const , and h ~ const .~ const , and h ~ const .

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Example 14.7 Evapor at ive cooling

Air ent er s an evapor at ive cooler at 1 at m, 35oC, and 20% r elat ive humidit y and it exit s at 80% r elat ive humidit y det er mine (a) t he exit t emper at ur e of air , and (b) t he lowest t emper at ur e t o which t he air can becooled by t his evapor at ive cooler .

Solut ion(a) 21.3oC (b) 18.9oC

Adiabatic Mixing of Airstreams Adiabatic Mixing of Airstreams •Many air-conditioning applications require the mixing of two airstream. •ie. Fresh air + Return air. •Normally Q very small, no work, and KE and PE =0

(14.24) - -

- -

: Elminating

(14.23) SSSF :BalanceEnergy

(14.22) :balance massWater

(14.21) :balance massair

13

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20

Example 14.8 Mixing of Conditioned Air with Outdoor Air

Satureated air leaving the cooling section of an air conditioning system at 14oC with a rate of 50 m3/min is mixed adiabatically with the outside air at 32oC, and 60% relative humidity at a rate of 20 m3/min. Assuming that the mixing occure at 1 atm, determine, the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture.

Solution

3 = 0.0122 kg/kg da, 3 = 89%;

Tbd3 = 19.0oC ; V3 = 70.1 m3/min

Wet Cooling Tower Wet Cooling Tower •Power plant, large A-C system, some industrial process need to reject a large amount of waste heat.•A wet cooling tower ~ semi-closed evaporative cooling cooler.•Type of cooling tower: Mechanical draft (Induced, Forced), Natural draft•Other cooling method: Spray pond, Cooling pond

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Example 14.9 Cooling of a Power Plant by a Cooling TowerCooling water leaves the condenser of a power plant and enters a wet cooling tower at 35oC at a rate of 100 kg/s. Water is cooled to 22oC in the cooling tower by air that enters the tower at 1 atm, 20oC, and 60% relative humidity and leaves satureted at 30oC. Neglecting power input to the fan, determine (a) the volume flow rateof the air into the cooling tower, (b) the mass flow rate of the required makeup water.

Solution (a) Va = 81.6 m3/s, (b) mf = 1.80 kg/s

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