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Perfect gas or Ideal gas: It is defined as a state of substance, whose evaporation from

its liquid state is complete. It strictly obeys all the gas laws under all conditions of temperature and pressure.

Vapour: If its evaporation is partial, the substance is called vapour. For example carbon dioxide, sulphur dioxide, and ammonia are regarded as vapours.

Note: vapour becomes dry, when it is completely evaporated. If further heated it will become superheated vapour, this vapour behavior is similar to perfect gas.

Steam: A vapour, therefore, contains some particles of liquid in suspension. It is thus obvious that steam.

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Real or Actual gas:In actual practice, there is no real or actual gas which strictly

obeys the gas laws over the entire range of temperature and pressure.

Real gases which are ordinarily difficult to liquefy (Cryogenic liquids) such as Oxygen, Nitrogen, Hydrogen and air, Within certain temperature and pressure limits, may be regarded as perfect gases.

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Air or dry air: Dry air is a theoretical sample of air that has no water vapor. Its composition.

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water vapor: Water in its gaseous state, especially in the

atmosphere and at a temperature below the boiling point. Water vapor in the atmosphere serves as the raw material for cloud and rain formation.

It also helps regulate the Earth's temperature by reflecting and scattering radiation from the Sun and by absorbing the Earth's infrared radiation.

Moist air: Moist air is a mixture of dry air and any amount of

water vapor.

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The degree of moisture has a strong effect on

1. heating, cooling, and comfort2. insulation, roofing, stability and deformation of building materials3. odor levels, ventilation4. industry and agriculture

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Saturated air: It is a mixture of dry air & water vapor, when the air

has diffused the maximum amount of water vapor into it.The water vapor usually occurs in the form of

superheated steam as an invisible gas.when the saturated air is cooled, the water vapor in

the air starts condensing, and the same may be visible in the form of moist, fog or condensation on cold surfaces.

Humidity or Specific humidity or Humidity ratio: It is the mass of water vapor present in 1 kg of dry

air. It is expressed generally gram/kg of dry air.

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Humidity ratio, W:

w

w

w

w

da

w

PPP

PPP

mmW

622.0

5.4619.286

wsPPwsP

sW

622.0

Saturated Humidity ratio:

Saturation is when the air contains the maximum amount of water vapor at its current temperature. The saturation pressure is taken from the steam tables at the moist air temperature.

sW

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Absolute Humidity:Mass of water vapour present in 1 m3 of dry air. It is

generally expressed in terms of g/m3 of dry air or grains/m3

of dry air, (1 kg of water vapour is equal to 15,430 grains)

Relative humidity( , RH): It is the ratio of actual mass of water vapour in a given volume of moist air to the actual mass of water vapour in the same volume of saturated air at the same temperature and pressure.

wsPwP

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Degree of saturation:

Degree of saturation is the ratio of the amount of water contained in the moist air to that which would be contained if the air were saturated

sWW

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whWdahah

Enthalpy: kJ/kg of dry air

=1.0 T +W(1.805 T +2501 )

wPPTdaR

av

Specific volume: cu. m /kg of dry air

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Dry-bulb Temperature & Wet-bulb Temperature

►In engineering applications involving moist air, two readily-measured temperatures are commonly used: the dry-bulb and wet-bulb temperatures.

►The dry-bulb temperature, Tdb, is simply the temperature measured by an ordinary thermometer placed in contact with the moist air.►The wet-bulb temperature, Twb, is the temperature measured by a thermometer whose bulb is enclosed by a wick moistened with water.

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Dry-bulb Temperature & Wet-bulb Temperature

►The figure shows wet-bulb and dry-bulb thermometers mounted on an instrument called a psychrometer. Flow of moist air over the two thermometers is induced by a battery-operated fan.

►Owing to evaporation from the wet wick to the moist air, the wet-bulb temperature reading is less than the dry-bulb temperature: Twb < Tdb.►Each temperature is easily read from its respective thermometer.

MoistAir in

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Wet bulb depression:It is the difference between dry bulb temperature and

wet bulb temperature.Dew Point Temperature:

When moist air is cooled, partial condensation of the water vapor initially present can occur. This is observed in condensation of vapor on window panes, pipes carrying cold water, and formation of dew on grass.

An important special case is cooling of moist air at constant mixture pressure, p.Dew point depression:

It is the difference between dry bulb temperature and dew point temperature.

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Using the Psychrometric ChartIf any two properties of air are known then the

other four can be found from the psychrometric chart.

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Humidification

►The aim of a humidifier is to increase the amount of water vapor in the moist air passing through the unit.►This is achieved by injecting steam or liquid water.

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HumidificationThe figure shows a control volume enclosing a humidifier operating at steady state.

►Moist air enters at state 1.►Steam or liquid water is injected.►Moist air exits at state 2 with greater humidity ratio, w2 > w1.

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙

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Humidification►For adiabatic operation, the accompanying psychrometric charts show states 1 and 2 for each case.

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙

►With relatively high-temperature steam injection, the temperature of the moist air increases.►With liquid injection the temperature of the moist air may decrease because the liquid is vaporized by the moist air into which it is injected.

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Humidification

►For the control volume, let us evaluate►The humidity ratio, w2, and►The temperature, T2.

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙Wcv = 0, Qcv = 0∙∙

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Humidification►Mass rate balances. At steady state, mass rate balances for the dry air and water are, respectively

2v31v

2a1ammm

mm

(dry air)

(water)

Then, since mv1 = w1ma and mv2 = w2ma, where ma denotes the common mass flow rate of the dry air, we get

a

312 m

m

ww (1)

∙∙ ∙ ∙∙

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙

ma1∙

3

m3∙h3,

Wcv = 0, Qcv = 0∙∙Wcv = 0, Qcv = 0∙∙

Since w1, ma, and m3 are specified, the humidity ratio w2 can be calculated from Eq. 1

∙ ∙

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Humidification►Energy rate balance. With no significant kinetic and potential energy changes, the energy rate balance for the control volume reduces to

With mv1 = w1ma and mv2 = w2ma, Eq. 2 becomes ∙ ∙ ∙∙

)()(0 v22va2a33v11va1acvcv hmhmhmhmhmWQ

Solving Eq. 3

(3)

Since Wcv and Qcv are each zero in this case∙∙

)()(0 v2v2a2a33v1v1a1a hmhmhmhmhm (2)

)()()(0 v22a23a

3v11a1 hhh

mm

hh ww

3a

3v11a1v22a2 )()()( h

mm

hhhh

ww (4)

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Humidification

►Options for determining T2 from Eq. (4) include

►Use the psychrometric chart:

3a

3v11a1v22a2 )()()( h

mm

hhhh

ww (4)

w11

(ha1 + w1hv1)

w2

T1 T2

(ha2 + w2hv2)

2

w

Spec

ific enth

alpy o

f moist

air,

in kJ/kg

(dry

air)•The first term on the right side

of Eq. (4) can be read from the chart using T1 and w1 to fix the state.•Since the second term on the right is known, the value of (ha2 + w2hv2) can be calculated. •This value together with w2 fixes the exit state, which allows T2 to be determined by inspection.

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Humidification

►Options for determining T2 from Eq. (4) include

►An iterative solution using data from Table A-22: ha(T) for the dry air and Table A-2: hv = hg(T) for the water vapor:

3a

3v11a1v22a2 )()()( h

mm

hhhh

ww (4)

• The value of the right side of Eq. (4) is known because the data are either known or can be obtained from the indicated tables using T1.• On the left side of Eq. (4), w2 is known from the mass rate balance.• Accordingly, the only unknown is T2, which can be found iteratively:

For each assumed value of T2, Table A-22 gives ha2 and Table A-2 gives hv2. This allows the left side to be calculated. Iteration with T2 continues until the calculated value on the left agrees with the known value on the right.

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Dehumidification

►The aim of a dehumidifier is to remove some of the water vapor in the moist air passing through the unit.►This is achieved by allowing the moist air to flow across a cooling coil carrying a refrigerant at a temperature low enough that some water vapor condenses.

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Dehumidification►The figure shows a control volume enclosing a dehumidifier operating at steady state.

►Moist air enters at state 1.►As the moist air flows over the cooling coil, some water vapor condenses.►Saturated moist air exits at state 2 (T2 < T1).►Condensate exits as saturated liquid at state 3. Here, we take T3 = T2.

ma, T1, w1

T3 = T2

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3

2 = 100%, T2 < T1, w2 < w1

∙

mw∙

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Dehumidification

►For the control volume, let us evaluate►The amount of condensate exiting per unit mass of dry air: mw/ma and►The rate of heat transfer between the moist air and cooling coil, per unit mass of dry air: Qcv/ma.

∙ ∙

∙∙

T3 = T2

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3

mw∙

ma, T1, w1∙ 2 = 100%,

T2 < T1, w2 < w1

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Dehumidification►Mass rate balances. At steady state, mass rate balances for the dry air and water are, respectively

2vw1v

2a1ammm

mm

(dry air)

(water)

Solving for the mass flow rate of the condensate

2v1vw mmm

Then, with mv1 = w1ma and mv2 = w2ma, where ma denotes the common mass flow rate of the dry air, we get the following expression for the amount of water condensed per unit mass of dry air

21a

w ww mm

(1)

∙∙ ∙ ∙∙

T3 = T2

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3

mw∙

ma, T1, w1∙ 2 = 100%,

T2 < T1, w2 < w1

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Dehumidification►Energy rate balance. With Wcv = 0 and no significant kinetic and potential energy changes, the energy rate balance for the control volume reduces at steady state to

With mv1 = w1ma, mv2 = w2ma, and Eq. (1), Eq. (2) becomes ∙ ∙ ∙∙

(2))()(0 v22va2awwv11va1acv hmhmhmhmhmQ

∙

Since heat transfer occurs from the moist air to the cooling coil, Qcv/ma will be negative in value.∙ ∙

w211va2vaa

cv )()()( hhhhhmQ

wwww

(3)

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Dehumidification

►Options for evaluating the underlined terms of Eq. (3) include

w211va2vaa

cv )()()( hhhhhmQ

wwww

(3)

►w1 and w2 are known. Since T1 and T2 are also known, ha1 and ha2 can be obtained from ideal gas table Table A-22, while hv1 and hv2 can be obtained from steam table Table A-2 using hv = hg.

►For the condensate, hw = hf (T2), where hf is obtained from Table A-2.

►Alternatively, using the respective temperature and humidity ratio values to fix the states, (ha + whv) at states 1 and 2 can be read from a psychrometric chart.

(ha + whv)2

(ha + whv)1

T1T2

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Adiabatic Mixing of Two Moist Air Streams

►In air-conditioning systems, a frequent component is one that mixes moist air streams as shown in the figure:

►For the case of adiabatic mixing, let us consider how the following quantities at the exit of the control volume, ma3, w3, and T3, can be evaluated knowing the respective quantities at the inlets.

∙

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Adiabatic Mixing of Two Moist Air Streams►Mass rate balances. At steady state, mass rate balances for the dry air and water vapor are, respectively

3v2v1v

3a2a1ammmmmm

(dry air)

(water vapor)

)( a2a13a22a11 mmmm www

With mv = wma, these equations combine to give∙∙

Alternatively

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23

a2

a1wwww

mm

(1)

These equations can be solved for w3 using known values of w1, w2, ma1, and ma2.∙ ∙

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Adiabatic Mixing of Two Moist Air Streams►Energy rate balance. Ignoring the effects of kinetic and potential energy, the energy rate balance for the control volume reduces at steady state to

)()()(0 v33va3a3v22va2a2v11va1a1cvcv hmhmhmhmhmhmWQ

Since Wcv and Qcv are each zero in this case∙∙

)()()( v33a3a3v22a2a2v11a1a1 hhmhhmhhm www Eq. 2

(2)

The enthalpies of the water vapor are evaluated using hv = hg. With ma3 = ma1 + ma2, Eq. 2 can be solved to give an expression with the same form as Eq. 1

)()(

)()(

g33a3g11a1

g22a2g33a3

a2

a1hhhh

hhhh

mm

ww

ww

Using known data, this equation can be solved for (ha + whg)3, from which T3 can be evaluated.

∙ ∙ ∙

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Adiabatic Mixing of Two Moist Air Streams►From study of Eqs. (1) and (2) we conclude that on a psychrometric chart state 3 lies on a straight line connecting states 1 and 2, as shown in the figure

(2))()(

)()(

g33a3g11a1

g22a2g33a3

a2

a1hhhh

hhhh

mm

ww

ww

31

23

a2

a1wwww

mm (1)

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Example: For adiabatic mixing of two moist air streams with the data provided in the table below, use the psychrometric chart to determine(a) w3, in kg (vapor)/kg (dry air), and (b) T3 in oC.

State

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T(oC)245

w(kg (dry air)/kg (vapor))

0.00940.002

ma

(kg (dry air)/min)497180

(ha + whg)*

(kJ/kg (dry air))4810

*The values of (ha + whg) are read from Fig. A-9 using the respective temperature and humidity ratio values.

∙

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(a) Inserting known valuesSolution:

3

30094.0

002.0180497

ww

we get w3 = 0.0074 kg (vapor)/kg (dry air).

T3 = 19oC

(b)

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(a) Inserting known valuesSolution:

3

30094.0

002.0180497

ww

we get w3 = 0.0074 kg (vapor)/kg (dry air).

T3 = 19oC

(b) Alternatively

(ha + whg)3 = 38 kJ/kg (dry air).

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