Chapter 20 Air-Conditioning and Psychrometrics – Thermal Engineering

Chapter 20

Air-Conditioning and Psychrometrics


The air-conditioning is that branch of engineering which deals with the study of supplying and maintaining desirable internal room atmospheric conditions of air for human comfort. The four important factors for human comfort are: temperature, humidity, purity and motion of air.

The working substance for air-conditioning is moist air, which a mixture of two gases. One of these is dry air, which itself is a mixture of a number of gases and the other is water vapour which may exist in a saturated or superheated state. Moist air is not a pure substance in any process in which condensation or evaporation of moisture occurs. Both dry air and water vapour can be considered as perfect gases since both exist in the atmosphere at low pressures. Hence, perfect gas laws to both and Dalton’s law of partial pressures for non-reactive mixture of gases can be applied to dry air part only.


The psychrometry is that branch of engineering which deals with the study of moist air i.e. dry air mixed with water vapour.

Dry air: The dry air consists of O2 = 0.2099, N2 = 0.7803, Ar = 0.0094, CO2 = 0.0003 and H2 = 0.0001 parts by volume.

Molecular weight of air = 28.966

Gas constant, Ra = 0.287 kJ/kg K

Density of air = 1.293 kg/m3 at 1.01325 bar and 0°C

Specific heat of air at constant pressure = 1.005 kJ/kg K

Molecular weight of water (vapour) = 18.02

Gas constant for water vapour, Rv = 0.461 kJ/kg K

Barometric pressure = 1.01325 bar or 760 mm Hg

Dalton’s law of partial pressures: The Dalton’s law of partial pressures states that for a mixture of ideal gases, the total pressure is equal to the sum of the partial pressures, which each constituent would exert if it occupied the whole space alone. Thus,

Barometric pressure of mixture of dry air and water vapour is given by

where pa = partial pressure of dry air

pv = partial pressure of water vapour

Moist air: It is a mixture of dry air and water vapour.

Saturated air: It is a mixture of dry air and water vapour when air has absorbed maximum amount of water vapour. At saturation pv = ps, the saturation pressure.

Degree of saturation (µ):

Specific humidity or humidity ratio w: It is the mass of water vapour present in one kg of dry air. It is expressed in terms of kg per kg of dry air (kg/kg d.a)

Absolute humidity: It is the mass of water vapour present in 1m3 of dry air. It is expressed in terms of kg per cubic metre of dry air (kg/m3 d.a.)

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

Dry Bulb Temperature (DBT, td): It is the temperature of air recorded by a thermometer, when it is not affected by the moisture present in the air.

Wet Bulb Temperature (WBT, tw): It is the temperature of air recorded by a thermometer, when its bulb is surrounded by a wet cloth exposed to the air.

The DBT and WBT are measured by a Sling Psychrometer, as shown in Fig. 20.1.

Wet Bulb Depression, WBD = DBT − WBT

Dew Point Temperature (DPT, tdp): It is the temperature of air recorded by a thermometer, when the water vapour present in it begins to condense. Thus, DPT is the saturation temperature (tsat) corresponding to the partial pressure of water vapour pv

Dew Point Depression, DPD = DBT − DPT

Figure 20.1 Sling psychrometer

  1. Specific (or absolute) humidity, humidity ratio or moisture content, w: It is defined as the ratio of the mass of water vapour to the mass of dry air in a given volume of the mixture.

    Let p, V, T, m = pressure, volume, absolute temperature and mass respectively

    R, M, v, = gas constant, molecular mass, specific volume, and universal gas constant respectively

    Subscripts a, v = air and vapour respectively.

    For air,


    For water vapour,

    Substituting for mv and ma from these equations in Eq. (20.2), we get

    Now actual total pressure, pb = pa + pv

    For ma = 1 kg, mv = w kg and total mass of moist air

    For saturated air, pv = ps and maximum humidity ratio

    where ps = partial pressure of air corresponding to saturation temperature i.e. td.

  2. Degree of saturation or percentage humidity, µ: The ratio of actual specific humidity w to the specific humidity ws of saturated air at the same temperature is termed as the degree of saturation.

    The degree of saturation is a measure of the capacity of air to absorb moisture.

  3. Relative Humidity (RH or ϕ): It is defined as the ratio of mass of water vapour mv in a certain volume of moist air at a given temperature to the mass of water vapour mvs in the same volume of saturated air at the same temperature.

    For a perfect gas pvvv = psvs


  4. Pressure of water vapour:
    1. The partial pressure of water vapour, according to Carrier’s equation, is given by

      where pw = saturation pressure corresponding to WBT

      pb = barometric pressure

      td = DBT

      tw = WBT

    2. Modified Apjohn equation:
    3. Modified Ferrel equation:
  5. Vapour density or Absolute humidity:

    Let pv = density of water vapour corresponding to its partial pressure and DBT, td.

    pa = density of dry air.

    Then mv = Vvρv and ma = Vaρa

    Since Va = Vv, therefore,

    Or ρv = a

    Now pa Va = maRaTd

    For ma = 1 kg,


The temperature-entropy diagram is shown in Fig. 20.2.

Enthalpy of moist air = Enthalpy of dry air + Enthalpy of water vapour associated with dry air

h = ha + whv per kg of dry air

where ha = cpa td = 1.005 td kJ/kg

cpa = 1.005 kJ/kg. K
td = DBT of air in °C.
hv = cpw tdp +[(hfg)dp + cpv (tdtdp)]kJ/kg
= 4.1868tdp + (hfg)dp +1.88(tdtdp)

where cpw = 4.1868 kJ/kg K is the specific heat of liquid water

(hfg)dp = latent heat of vaporization at DPT
cpv = 1.88 kJ/kg.K is the specific heat of superheated water vapour

Figure 20.2 Temperature-entropy diagram

tdp = DPT, °C
w = specific humidity
h = 1.005 td + w [4.1868 tdp + (hfg)dp + 1.88 (td + tdp)]

Eq. (20.16) can be written as,

h= (cpa + wcpv)td + w(hfg)0°C

is termed as the humid specific heat. It is the specific heal of moist air (1 + w) kg per kg of dry air.

At low temperatures of air-conditioning the value of w is very small. An approximate value of cp of 1.0216 kJ/(kg d.a.).K may be taken for nil practical purposes

Example 20.1

A mixture of dry air and water vapour is at a temperature of 20°C under a total pressure of 735 mm Hg. The dew point temperature is 15°C. Calculate the following:

  1. Partial pressure of water vapour,
  2. Relative humidity,
  3. Specific humidity,
  4. Specific enthalpy of water vapour
  5. Enthalpy of air per kg of dry air, and
  6. Specific volume of air per kg of dry air.


  1. From steam tables, the partial pressure of water vapour at tdp = 15°C
    pv =12.79mm Hg

    Now  760mm Hg = 1.01325×105 N/m2

    1mm Hg = 133.32 N/m2

    ∴   pv =12.79×133.32=1705.16N/m2

  2. Saturation pressure of water vapour at 20°C DBT
    ps =17.54mm Hg = 17.54 × 133.32 = 2338.4 N/m2

    Relative humidity,

  3. Specific humidity, w =
  4. Latent heat of vaporization of water at td = 20°C, (hfg)20°C = 2454.1kJ/kg

    Latent heat of vaporization of water at tdp =15°C, (hfg)15°C = 2465.9 kJ/kg

    Specific enthalpy of water vapour,

    hv =[4.1868tdp + (hfg)dp +1.88(tdtdp)]

    =[4.1868×15+ 2465.9+1.88(20 −15)]

    = 2538.1 kJ/kg w.v

  5. Enthalpy of air per kg d.a., h = 1.005 td + whv
    = 1.005 × 20 + 0.011 × 2538.1
    = 48.02 kJ/kg d.a.
  6. Specific volume of air per kg of d.a.
    = 0.873 m3/kg d.a

Example 20.2

A sling psychrometer gave the following readings:

DBT = 30°C, WBT = 20°C, Barometer reading = 740 mm of Hg.

Determine: (a) dew point temperature, (b) relative humidity, (c) specific humidity (d) degree of saturation, (e) vapour density, and (f) enthalpy of mixture per kg of dry air


Given: td = 30°C, tw = 20°C, pb = 740 mm Hg

  1. Corresponding to tw = 20°C, from steam tables, saturation pressure, pw = 0.023385 bar

    Barometric pressure, pb = 740 × 133.32 × 10−5 = 0.986568 bar

    Partial pressure of water vapour

    Corresponding to pv = 0.01703 bar pressure, from steam tables, dew point temperature

    tdp = 15°C
  2. Corresponding to td = 30°C, from steam tables, saturation pressure of water vapour,
    ps = 0.04246 bar
  3. Specific humidity,
    = 0.010925 kg/kg d.a
  4. Specific humidity of saturated air,

    Degree of saturation,

  5. Vapour density,
    = 0.01218 kg/m3 of d.a.
  6. From steam tables, (hfg)dp = 15°C = 2465.9 kJ/kg

    Enthalpy of mixture per kg d.a.,

    h = 1.0216 td + w[(hfg)dp + 2.3068 tdp]
    = 1.0216 × 30 + 0.010925 [2465.9 + 2.3068 × 15] = 57.966 kJ/kg d.a.

Example 20.3

A room 5m × 4m × 3m contains moist air at 40°C. The barometric pressure is 1 bar and relative humidity is 70%. Determine: (a) humidity ratio, (b) dew point temperature, (c) mass of dry air, and (d) mass of water vapour.

If the moist air is further cooled at constant pressure until the temperature is 15°C, find the amount of water vapour condensed.


Given: V = 5 × 4 × 3 = 60 m3, td = 40°C, pb = 1 bar, ϕ = 0.70.

  1. Saturation pressure of vapour corresponding to td = 40°C, from steam tables is,
    ps = 7.384 kPa or 0.07384 bar
    pv = 0.7× 0.07384 = 0.05169 bar

    Humidity ratio,

    = 0.0339 kg/kg d.a or 33.9 g/kg d.a
  2. DPT, tdp = saturation temperature corresponding to pv = 0.05169 bar or 5.169 kPa
    = 33.46°C from steam tables
  3. pa = pbpv = 1.0 − 0.05169 = 0.94831 bar

    Mass of dry air,

  4. Now w =

    Mass of water vapour, mv = 0.0339 × 63.34 = 2.147 kg

    Saturation pressure corresponding to 15°C, from steam tables is,

    ps = pv = 1.707 kPa or 0.01707 bar

    Now  pa = pbpv = 1 − 0.01707 = 0.98293 bar or 98.293 kPa

    Mass of dry air,

    Mass of water vapour, mv = wma = 0.0108 × 71.35 = 0.77 kg

    Amount of water vapour condensed = 2.147 − 0.77 = 1.377 kg


The thermodynamic wet bulb temperature or AST is the temperature at which the air can be brought to saturation state adiabatically by the evaporation of water in the flowing air. A schematic representation of this process, called the adiabatic saturation process, is shown in Fig. 20.3. It consists of an adiabatic enclosure containing adequate quantity of water. There is also an arrangement for feed water from the top.

The unsaturated air enters the enclosure at section 1. As the air flows through the enclosure water evaporates which is carried by the flowing stream of air. Thus the specific humidity of air increases. The water level in the enclosure is maintained by the feed water. Both the air and water are cooled as the evaporation takes place. The process continues until the energy transferred from the air to water is equal to the energy required to vaporise water. When steady conditions are reached, the air flowing at section 2 becomes saturated with water vapour. The temperature of the saturated air at section 2 is known as thermodynamic wet bulb temperature or adiabatic saturation temperature. The adiabatic saturation process is represented on T-s diagram as shown by the curve 1-2 in Fig. 20.4. The adiabatic saturation temperature is taken equal to the wet bulb temperature.

Let   td1 = DBT of initial unsaturated air

td2 = DBT of saturated air

     = adiabatic saturation temperature, tw

h1, h2 = enthalpy of unsaturated and saturated air respectively

w1, w2 = specific humidity of air at sections 1 and 2 respectively

hfw = sensible heat of water at adiabatic saturation temperature

For energy balance of air at inlet and outlet, we have

h1 + (w2w1)hfw = h2

or   h1 + whfw1 = h2 + whfw1

Now   h1 = ha1 + w1 hs1

and  h2 = ha2 + w2 hs2

where  ha1 = enthalpy of 1 kg dry air at DBT = td1 = cpa td1

hs1 = enthalpy of superheated vapour at td1 per kg of vapour

ha2 = enthalpy of 1 kg dry air at WBT = tw = cpa td2

hs2 = enthalpy of saturated vapour at WBT = tw per kg of vapour

Thus, we get

(ha1 + w1hs1) − w1hfw = (ha2 + w2hs2) − w2hfw
w1(hs1 + hfw) = w2 (hs2 + hfw) + (ha2ha1)

Also  h1 = cpatd1 + w1hs1

h2 = cpatd2 + w2hs2

Figure 20.3 Adiabatic saturator

Figure 20.4 T-s diagram for adiabatic saturation process

where  hs2 = hf2 + hfg2

hs1 = hs2 + cpv (td1td2) = hf2 + hfg2 + cpv (td1td2)

From the energy balance, we get

cpatd1 + w1[hf2 + hfg2 + cpv (td1td2)]+ (w2w1) hf2
= cpatd2 + w2 (hf2 + hfg2)

or  (cpa + w1 cpv)td1 − (cpa + w2 cpv)td2 = hfg2 (w2w1)

where cp1 and cp2 are humid specific heats

If cp1 = cp2 = cp, then

cp (td1td2) = hfg2 (w2w1)

Example 20.4

Atmospheric air at 1 bar enters the adiabatic saturator. The dry bulb temperature is 30°C and wet bulb temperature 20°C during adiabatic saturation process. Calculate the following:

  1. Humidity ratio of entering air,
  2. Vapour pressure and relative humidity at 30°C, and
  3. Dew point temperature.


Given: pb = 1 bar, td = 30°C, tw = 20°C

  1. Saturation pressure of vapour at 20°C, from steam tables is,

    pv2 = 2.34 kPa or 0.0234 bar

    Enthalpy of saturated vapour at 20°C, hs2 = hg2 = 2538.2 kJ/kg

    Sensible heat of water at 20°C, hfw = 83.9 kJ/kg

    Enthalpy of saturated vapour at 30°C, hs1 = hg1 = 2556.4 kJ/kg

    Enthalpy of 1 kg unsaturated air at td = 30°C,

    ha1 =mcpatd =1×1.005×30 = 30.15kJ/kg

    Enthalpy of 1 kg saturated air at tw = 20°C,

    ha2 =mcpatw =1×1.005× 20 = 20.10 kJ/kg
  2. Now

    0.0107 − 0.0107 pv1 = 0.622 pv1

    pv1 = 0.0169 bar

    Saturation pressure corresponding to 30°C from steam tables is,

    ps = 4.246 kPa or 0.04246 bar

    Relative humidity,

  3. Dew point temperature is the saturation temperature corresponding to the partial pressure of water vapour, pv1 = 0.0169 bar or 1.69 kPa

    From steam tables, we have

    tdp = 14.85°C

The psychrometric chart is a graphical representation of the complete thermodynamic properties of moist air and psychrometric analysis of air-conditioning processes. The chart most commonly used is the w-td chart, which specifies specific humidity (ω) or water vapour pressure (pv) along the ordinate and the dry bulb temperature (td) along the abscissa. The chart is normally constructed for a standard atmospheric pressure of 760 mm Hg or 1.01325 bar corresponding to the pressure at the mean sea level. A typical layout of this chart is shown in Fig. 20.5.

Some of the important lines on the psychrometic chart are explained below:

DBT lines: These lines are vertical, i.e. parallel to the ordinate and are uniformly, spaced. These lines are drawn at an interval of 5°C and up to the saturation line.

Specific humidity or moisture content lines: These lines are horizontal i.e parallel to the abscissa and are uniformly spaced.

DPT lines: These lines are horizontal, i.e. parallel to the abscissa and are non-uniformly spaced. At any point on the saturation line, the DBT and DPT are equal.

WBT lines: These are inclined straight lines and are non-uniformly spaced. At any point on the saturation line DBT and WBT are equal.

Enthalpy line: These are inclined straight lines and uniformly spaced. These lines are parallel to the WBT lines. The enthalpy values are given above the saturation line. Enthalpy deviation
(ws w)hw is read form the deviation lines which are curved non-equally spaced with +ve and −ve correction.

Specific volume lines: These are obliquely inclined straight lines and are uniformly spaced.

Vapour pressure lines: These lines are horizontal and uniformly spaced. These lines are not drawn on the chart. A scale showing vapour pressure in mm of Hg is given on the extreme left or right side of the chart.

Relative humidity lines: These are curved lines and follow the saturated line. The saturation line represents 100% RH.

The psychrometric chart is shown in Fig. 20.6.

Figure 20.5 Layout of psychrometric chart

Figure 20.6 Psychrometric chart


The psychrometric processes by which the state of moist air can be altered are shown in Fig. 20.7. These processes are:

  1. Sensible heating-process OA
  2. Sensible cooling-process OB
  3. Humidifying-process OC
  4. Dehumidifying-process OD
  5. Heating and humidifying-process OE
  6. Cooling and dehumidifying-process OF
  7. Cooling and humidifying-process OG
  8. Heating and dehumidifying-process OH

Figure 20.7 Basic psychrometric processes

20.8.1 Sensible Heating or Cooling Process

The heating/cooling of air without any change in its specific humidity is known as sensible heating/cooling. The process of sensible heating (line 1-2) or cooling (line 2-1) is shown by a horizontal line in Fig. 20.8. In this process, the specific humidity of air remains constant, i.e. w1 = w2. The heat has to be transferred which goes to change the temperature of air. The sensible heat transfer rate is given by,

s = a (h2h1)
= a [cpa (t2t1)+ wcpv (t2t1)]
= a cp (t2t1)

where   cp = cpa + wcpv = humid specific heat

a = mass flow rate of dry air, kg d.a./s

Figure 20.8 Sensible heating process

= Volume of air in m3/min × density of air ρ in kg/m3

For the purpose of calculations, standard air is taken at 20°C and 50% RH. The density of standard air is approximately 1.2 kg/m3 d.a. The value of humid specific heat is taken as 1.0216 kJ/(kg d.a.)(K)

The sensible heating of moist air can be done to any desired temperature. But the sensible cooling can be done only up to the dew point temperature tdp, as shown in Fig. 20.8.

20.8.2 Humidification or Dehumidification Process

Humidification is the addition of moisture to air without change in its DBT. Similarly dehumidification is the removal of moisture without change in DBT. When the state of air is altered along the constant DBT line, such as 1-2 in Fig. 20.9, moisture in the form of vapour has to be transferred to change the humidity ratio of air. This transfer of moisture is given by,

Because of this change in humidity ratio, there is also a change in the specific enthalpy of air, as shown in Fig. 20.9. This change in enthalpy due in the change in humidity ratio is considered to cause a latent heat transfer, given by

L = a(h2h1) = a[(cptd2+hfgw2) − (cptd1 + hfgw1)]

Since td1 = td2

Figure 20.9 Humidification or dehumidification

where   hfg = latent heat of vaporisation at td1

20.8.3 Heating and Humidification

This process involves both a change in temperature as well as in the humidity ratio as shown in Fig. 20.10. The change in temperature causes a sensible heat load, given by,

Qs = a (hAh1) = acp (td2td1)

The change in humidity ratio causes a moisture transfer, given by

G = ma (w2wA) = ma (w2w1)

Latent heat load  QL =ma (h2hA) = mahfg (w2w1)

Figure 20.10 Heating and humidification

20.8.4 Sensible Heat Factor-SHF

The ratio of the sensible heat transfer to the total heat transfer is termed as the sensible heat factor.


The process line 1-2 is called the sensible heat factor line (or process or condition line). The point A divides the total enthalpy change in the ratio of SHF and (1 − SHF). Qs is proportional to SHF and QL to (1 − SHF).

20.8.5 Cooling and Dehumidification

In this process, the DBT as well as specific humidity of air decreases. The final relative humidity of the air is generally higher than that of the entering air. This is achieved by passing the air over a cooling coil or through a cold water spray. The effective surface temperature of the coil is known as apparatus dew point (ADP). The cooling and dehumidification process is shown in Fig. 20.11.

Qs = hAh2
QL = h1hA
Q = Qs + QL = h1h2

Figure 20.11 Cooling and dehumidification

20.8.6 Air Washer

The schematic representation of an air washer is shown in Fig. 20.12. It involves the flow of air through a spray of water. During the course of flow, the air may be cooled or heated, humidified or dehumidified, or simply adiabatically saturated, depending on the mean surface temperature of water. The water is, accordingly, externally cooled or heated or simply circulated by a pump. Make-up water is added for any loss in the case of humidification of air. Eliminator plates are provided to minimise the loss of water droplets.

The thermodynamic changes of state of air along path 1-2 in an air washer, depending on the mean surface temperature of water droplets ts (equal to tu), is shown in Fig. 20.13. The following processes are possible:

  1. Process 1-2a: Heating and humidification (ts > td1). The water is externally heated.
  2. Process 1-2b: Humidification (ts = td1). The enthalpy of air increases. The water is externally heated.
  3. Process 1-2c: Cooling and humidification (tw1 < ts < td1). The mean surface temperature of water is greater than the wet bulb temperature of air, tw1. The enthalpy of air increases as a result of humidification. The water has to be externally heated.
  4. Process 1-2d: Adiabatic saturation (tw1 = ts). The water is simply recirculated without any external heating or cooling.
  5. Process 1-2e: Cooling and humidification (tdp < ts < tw1). The enthalpy of air decreases. The water is required to be externally cooled.
  6. Process 1-2f: Cooling (ts = tdp). Water is required to be cooled.
  7. Process 1-2g: Cooling and humidification (ts < tdp). The condition line drawn from the initial state 1 is tangential to the saturation line.

The mass balance of an air washer is shown in Fig. 20.14.

Figure 20.12 Air-Washer

Figure 20.13 Range of psychrometric processes with an air washer

Figure 20.14 Mass balance of an air washer

Let  a = mass flow rate of dry air

w = mass flow rate of water

Energy balance gives,

a (h2h1) = wcpwtw3 − [wa (w2w1)]cpwtw4

or   a (h2h1) = wcpw(tw3tw4)tw3 + a(w2w1)cpwtw4

Neglecting the effect of temperature of water in the last term, we have

For adiabatic saturation case, dh = 0

dtw = 0 and tw3 = tw4

20.8.7 Cooling with Adiabatic Humidification

When the air is passed through an insulated chamber, as shown in Fig. 20.15(a), having water sprays maintained at a temperature (tf) higher than the DPT of entering air (tdp1), but lower than its DBT (td1) of entering air or equal to the WBT of entering air (tw1), then the air, is said to be cooled and humidified. Since the chamber is insulated and same water is circulated again and again, therefore, a condition of a adiabatic saturation is reached. Ultimately tf = tw1. This process is shown by line 1-3 in Fig. 20.15(b). This line follows the path along constant WBT line or constant enthalpy line.

Figure 20.15 Cooling with adiabatic humidification

For perfect humidification, the final condition of air will be at point 3, where td3 = tw1. In actual practice the humidification is not perfect, therefore the final condition of air at outlet will be represented by point 2 on line 1-3.

Effectiveness or humidifying efficiency of spray chamber, for tf = tw1, is given by

When tf < tw1, then the process follows the path 1-2′-3′. In that case

When tf > tw1, then the process follows the line 1-2′′-3′′. In that case

20.8.8 Cooling and Humidification by Water Injection
(Evaporative Cooling)

Let liquid water at temperature tf be injected and sprayed into air stream with the help at nozzles. The condition of air will change depending on the amount of water that evaporates. The enthalpy of evaporation will come from the enthalpy of air. The process is shown in Fig. 20.16.

Let a = mass flow state of air

v = mass flow rate of water evaporated

= mass flow rate of water supplied.

hf = enthalpy of liquid water

Then mass and enthalpy balances given

The term (w2w1) hf is extremely small as compared to h1 and h2.

Hence h1 = h2. Thus the water injection process is a constant enthalpy process, irrespective of the temperature of injected water.

Figure 20.16 Cooling and humidification by water injection

Figure 20.17 Heating and humidification by steam injection

20.8.9 Heating and Humidification by Steam Injection

Steam is normally injected into fresh outdoor air which is then supplied for the conditioning of textile mills where high humidity needs to be maintained. The process is shown in Fig. 20.17. If ms is the mass of steam supplied with enthalpy hs and ma the mass of air, then from the mass balance,

and for the enthalpy balance,

20.8.10 Heating and Adiabatic Chemical Dehumidification

This process is mainly used in industrial air-conditioning. It can also be used for some comfort air-conditioning requiring low relative humidity or low dew point temperature in the room. In this process, the air is passed over chemicals which have infinity for moisture. As the air comes in contact with these chemicals, the moisture gets condensed out of air and gives up its latent heat. Due to condensation, the specific humidity decreases and the heat of condensation supplies sensible heat for heating of air. As a result of this the DBT increases. The process is shown in Fig. 20.18 by line 1-2. The path followed during the process is along the constant WBT line or constant enthalpy line.

The chemicals used are hygroscopic solutions or brines of calcium chloride, lithium chloride, lithium premie and ethyl glycol. These are called absorbents which undergo a chemical or physical or both changes during taking moisture. The other type of chemicals are called adsorbents which are in the solid state to take up moisture but do not undergo changes chemically or physically. These include silica gel and activated alumina. The effectiveness or efficiency of the dehumidifier,

Figure 20.18 Heating and adiabatic chemical dehumidification


Consider two air streams 1 and 2 mixing adiabatically, as shown in Fig. 20.19(a)

Let   ma1, h1, w1 = mass, enthalpy and specific humidity of entering air at 1.

ma2, h2, w2 = corresponding values of entering air at 2.

ma3, h3, w3 = corresponding values of leaving air at 3.

The mixing process on psychrometic chart is shown in Fig. 20.19(b).

For the mass balance,

ma1 + ma2 = ma3

For the energy balance,

ma1h1 + ma2 h2 = ma3 h3

For the mass balance of water vapour,

ma1w1 + ma2 w2 = ma3 w3

Eq. (20.42) can be written as,

Simplifying, we get

Figure 20.19 (a) Adiabatic mixing of air streams, (b) Mixing process on the psychrometric chart

The second term in the above expression being negligible, we get

Example 20.5

The atmospheric air at 760 mm of Hg, DBT = 20°C and WBT = 10°C enters a heating coil whose temperature is 40°C. Assuming bypass factor of heating coil as 0.5, determine for the air leaving the coil: (a) DBT, (b) WBT, (c) relative humidity, and (d) sensible heat added to the air per kg of dry air.


Given: pb = 760 mm Hg, td1 = 20°C, tw1 = 10°C, td3 = 40°C, BPF = 0.5

  1. Let td2 = DBT of air leaving the coil

    The psychrometric process is shown in Fig. 20.20.

    (b) and (c) from the psychrometric chart we find that tw2 = 14.4°C

    and RH, ϕ2 = 15%

  1. d. From the psychrometric chart,
    h1 = 30 kJ/kg. d.a.

    and h2= 40 kJ/kg. d.a.

    Figure 20.20

    Sensible heat added to air = h2h1

    = 40 − 30
    = 10 kJ/kg.d.a.

Example 20.6

Moist air enters a refrigeration coil at the rate of 120 kg d.a/min at 40°C and 50% RH. The apparatus dew point of coil is 10°C and by-pass factor is 0.20. Determine the outlet state of moist air and capacity of coil in TR.


Given: ma =120 kg d.a/min, td1 = 40°C, ϕ1 = 50%, ADP = 10°C, BPF = 0.20

  1. Let td2 and ϕ2 be the DBT and RH of air leaving the cooling coil respectively

    The psychrometric process is shown in Fig. 20.21.

    Figure 20.21

    Locate point 1 (td1 = 40°C, ϕ1 = 50%) on the psychrometric chart. Join point 1 with ADP = 10°C point on the saturation line.

  2. Draw a vertical line at td2 = 16°C to intersect line 1-3 at point 2. Then ϕ2= 93%
  3. From the psychrometric chart, h1 = 101 kJ/kg d.a, and h2 = 43 kJ/kg d.a.

    Cooling capacity of coil, Q = ma (h1h2)

Example 20.7

The atmospheric air at 30°C DBT and 70% RH enters a cooling coil at the rate of 220 m3/min. The coil DPT = 15°C and its BPF = 0.15. Determine

  1. the temperature of air leaving the cooling coil,
  2. the capacity of the cooling coil in TR,
  3. the amount of water vapour removed per minute, and
  4. sensitive heat factor for the process.


Given: td1 = 30°C, ϕ1 = 70%, V1 = 220 m3/min (cmm). ADP = td3 = 15°C, BPF = 0.15

  1. Let td2 = temperature of air leaving the cooling coil

    The psychromatric process is shown in Fig. 20.22.

    At td1 = 30°C and ϕ1 = 70%, tdp1 = 24°C from psychrometric chart.

    Figure 20.22

    From psychrometric chart, we have

    w1 = 18.8 g/kg.d.a.
    w2 = 11.8 g/kg.d.a.
    v1 = 0.885 m3/kg
    h1 = 78.5 kJ/kg.d.a.
    h2 = 47.8 kJ/kg.d.a.
    hA = 61.8 kJ/kg.d.a.

    Mass of air flowing through cooling coil,

  2. Capacity of cooling coil = a (h1h2) = 248.6(78.5 − 47.8) = 7632 kJ/min
  3. Amount of water vapour removed per minute,
    v = a(w1w2)
    = 248.6 (18.8 − 11.8) × 10−3 = 1.74 kg/min
  4. Sensible heat factor, SHF =

Example 20.8

In a certain environment, the DBT of air is 25°C and RH is 40%. Determine the specific humidity, dew point and wet bulb temperature of air. This air is cooled in an air washer using recirculated spray water and having a humidifying efficiency of 0.85. Find the DBT and DPT of air leaving the air washer.


Given: td1 = 25°C, ϕ1 = 40%, ηH = 0.85

The psychrometric process is shown in Fig. 20.23

At point 1 (td1 = 25°C, ϕ1 = 40%), w1 = 7.6 g/kg d.a., tdp1 = 10.4°C

and  tw1 = 16°C = td3

Let  td2 = DBT of air leaving the air washer

td2 = 17.35°C

Then  tdp2 = 15.2°C

Figure 20.23

Example 20.9

The atmospheric air at 25°C DBT and 15°C WBT is flowing at the rate of 120 cmm through a duct. The dry saturated steam at 100°C is injected into the air stream at the rate of 75 kg/h. Calculate the specific humidity, enthalpy, DBT, WBT, and RH of leaving air.


Given: td1=25°C, tw1 =15°C, V1 =120 cmm, ts = 100°C, ms = 75 kg/h or 1.25 kg/min.

The psychrometric process is shown in Fig. 20.24

At state 1 (td1 = 15°C, tw1 =15°C), w1 = 6.6 g/kg d.a., v1 = 0.853 m3/kg, h1 = 42.2 kJ/kg d.a.

Mass flow rate of air,

Figure 20.24

Enthalpy of saturated steam at 100°C from steam tables, hs = 2676 kJ/kg

Locate point 2 (w2 = 0.0155 kg/kg d.a., h2 = 65.97 kJ/kg) on the psychrometric chart

Then td2 = 260°C, WBT, tw2 = 22.4°C, ϕ2 = 73%.

Example 20.10

Saturated air leaving the cooling section of an air-conditioning system at 15°C at the rate of 60 m3/min is mixed adiabatically with outside air at 30°C and 60% RH at the rate of 20 m3/min. Find the specific humidity, relative humidity, DBT and the volume flow rate of the mixture.


Given: td1 = 30°C, ϕ1 = 60% V1 =20 m3/min, td2 = 15°C, V2 =60 m3/min

The psychrometric process for mixing is shown in Fig. 20.25.

Locate point 1 (td1 = 30°C, ϕ1 = 60%) and point 2 (td2 = 15°C on the saturation curve)

From the psychrometric chart, we have


h1 = 71.6 kJ/kg.d.a., h2 = 42.1 kJ/kg.d.a.
v1 = 0.88 m3/kg.d.a., v2 = 0.83 m3/kg.d.a.

Figure 20.25


0.3144(71.6 − h3) = h3 − 42.1
h3 = 49.16kg/kg.d.a.
w3 = 0.012 kg/kg.d.a.,   ϕ3 = 88% and td3 = 18.8°C,
v3 = 0.843 m3/kg.d.a.
3 = (ma1 + ma2)v3 = (22.73 + 72.29) × 0.843 = 80.1 m3/min

The rate at which body produces heat is called the metabolic rate. The heat produced by a normal healthy person while sleeping is called the basal metabolic rate, which is of the order of 60 W. The maximum value may be 10 times as much as this for a person engaged in sustained hard work. Human comfort is influenced by physiological factors determined by the rate of heat generation within the body and the rate of heat dissipation to the environment.

The body loses heat to the environment mainly by convection (Qconv), radiation (Qrad) and evaporation of moisture (Qevap). In addition, there is heat loss by respiration having sensible component (QSr) and latent component (QLr) the total heat loss from the body is thus,

Q = Qconv + Qrad + Qevap + (QSr + QLr)
= (Qconv + Qrad + QSr) + (Qevap + QLr)

where   QST = total sensible heat loss

QLT = total latent heat loss.

The total sensible heat component depends on the temperature difference between the surface of the body and the surroundings. The latent heat component depends in the difference in the water vapour pressures.

In summer, the temperature difference available for sensible heat transfer is less. Thus, the convective and radiative heat losses are reduced. To maintain thermal equilibrium, the body starts perspiring to increase the evaporative loss. On the other hand in winter the sensible heat transfer is increased so that the evaporative losses tend towards zero.

The heat exchange between human body and environment can be expressed by the following energy balance equation:

where  QM = metabolic heat produced within the body

W= useful work done by human being

QMW = heat to be dissipated to environment

Q = Qconv + Qrad + Qevap+ (QSr+ QLr)

Qs = heat stored in the body

  1. The metabolic heat produced depends upon the rate of food energy consumption in the body.
  2. The heat loss by evaporation is always positive. It depends upon the vapour pressure difference between the skin surface and the surrounding air.

    where Cd = diffusion coefficient, kg of water evaporated per unit surface area and pressure difference per hour.

    A = skin surface area = 1.8 m2 for a normal human being

    ps = saturation vapour pressure corresponding to skin temperature

    pv = vapour pressure of surrounding air

    hfg = latent heat of vaporisation = 2450 kJ/kg

    Cc = factor which accounts for clothing worn.

  3. The heat loss or gain by radiation from the body to the surroundings depends upon the mean radiant temperature. It is the average surface temperature of the surrounding objects. Qrad is positive when the mean radiant temperature is lower than the DBT of room air, i.e. the human body will undergo a radiant heat loss. When Qrad is negative, the body will undergo a radiant heat gain.
  4. The heat loss by convection from the body to the surroundings is given by

    where   U = body film coefficient of heat transfer

    A = body surface area = 1.8 m2 for normal human being

    tb = body temperature

    ts = surroundings temperature.

    The heat will be gained by body when ts > tb and Qconv is negative

  5. QS is negative when Q > (QMW), i.e. body temperature falls down. QS is positive when

    Q < (QMW), i.e. a human being gets fever.

    A human body feels comfortable when there is no change in the body temperature, i.e. when QS = 0. Any variation in the body temperature acts as a stress on the brain which results in either perspiration or shivering.

20.10.1 Factors Affecting Human Comfort

The following factors affect human comfort:

  1. Effective temperature.
  2. Heat production and regulation in human body.
  3. Heat and moisture losses from the human body.
  4. Moisture content of air.
  5. Quality and quantity of air.
  6. Air velocity.
  7. Hot and cold surfaces.
  8. Air stratification.

20.10.2 Physiological Hazards Resulting from Heat

The physiological hazards resulting from rise in body temperature are:

  1. Heat exhaustion: It is due to the failure of normal blood circulation. The symptoms of heat exhaustion include fatigue, headache, dizziness, vomiting and abnormal mental reactions such as irritation. Severe heat exhaustion may cause fainting. It does not cause permanent injury to the body and recovery is usually rapid when the person is shifted to a cool place.
  2. Heat Cramp: It results from loss of salt due to an excessive rate of body perspiration. It causes severe pain in high muscles. This may be avoided by using salt tablets.
  3. Heat stroke: It occurs when a person is exposed to excessive heat and work. If the body temperature rises rapidly to 40.5°C or 105°F, sweating ceases and a person may enter a coma, with imminent death. It may lead to permanent damage to brain. Heat stroke can be a avoided by drinking sufficient water at frequent intervals.

Effective temperature (ET) is defined as the temperature of saturated air (RH = 100%) at which the human beings would experience the same feeling of comfort as experienced in the actual unsaturated environment. It may also be defined as that index which correlates the combined effects of air temperature, relative humidity and air velocity on the human body.

Correspondingly, another effective temperature (ET*) can also be defined as the temperature at 50% RH at which the human body would experience exactly same feeling of comfort as at ET at 100% RH, and as experienced in the actual environment.

The effective temperature is defined to evaluate the combined effect of DBT, RH, and air velocity. The numerical value of effective temperature is made equal to the temperature of still (i.e. 5 to 8 m/min air velocity) saturated air, which produces the same sensation of warmth or coolness as produced under the given conditions.

20.11.1 Comfort Chart

The comfort chart is defined as that index which correlates the combined effects of air temperature, relative humidity and air velocity on the human body. The numerical value of effective temperature is made equal to the temperature of still (i.e. 5 to 8 m/min air velocity) saturated air, which produces the same sensation of warmth or coolness as produced under the given conditions.

The comfort chart, shown in Fig. 20.26, represents the concept of effective temperature in the comfort chart, DBT is taken as the abscissa and WBT as ordinates. The RH lines are replotted from the psychrometric chart. The statistically prepared graphs corresponding to summer and winter season are also superimposed. These graphs have effective temperature scale as abscissa and percentage of people feeling comfortable as ordinate.

There is a correlation between comfort level, temperature, humidity sex, length of exposure, etc. According to ASHRAE (American Society for Heating, Refrigeration, and Air conditioning Engineering) thermal sensation scale for exposure for a period of one hour, the value of index y for comfort for men, women and both sexes separately are expressed as follows

where td = DBT, °C

pv = vapour pressure, kPa

The values for feeling of comfort of y are as follows:

The comfort chart shown in Fig. 20.26 represent the range for both summer and winter conditions within which a condition of comfort exists for most people. The general practice is to recommend the following optimum inside design conditions for comfort:

Figure 20.26 Comfort chart for still air (air velocities from 5 to 8 m/min)

Summer air-conditioning: Winter air-conditioning:
ET = 22°C ET = 19°C
DBT = 25° ± 1°C DBT = 21°C
RH = 50 ± 5% RH = 50%
Room air velocity = 0.4 m/s Room air velocity = 0.15 − 0.20 m/s

The comfort chart does not take into account the variations in comfort conditions when there are wide variations in the mean radiant temperature. The comfort chart shown in Fig. 20.27 has become obsolete due to its shortcomings of over exaggeration of humidity at lower temperature and under estimation of humidity at heat tolerance level. The modified comfort chart shown in Fig. 20.27 is commonly used now.

Figure 20.27 Modified comfort chart

20.11.2 Factors Affecting Optimum Effective Temperature

The factors affecting optimum effective temperature are:

  1. Climate and seasonal differences: The ET changes with season. During summer it is 22°C and during winter 19°C.
  2. Clothing: Persons with light clothings need lower optimum ET than a person with heavy clothings.
  3. Age and sex: Women require higher ET (about 0.5°C) than men. Children also need higher ET than adults.
  4. Duration of stay: For shorter stay, higher ET is required than that needed for longer stay.
  5. Kind of activity: Low ET is required for people doing heavy activity, like factory workers, dancers, etc.
  6. Density of occupants: Lower ET is needed when density of persons in a place is high.

20.12.1 Selection of Inside Design Conditions

The inside design conditions depend on the particular air-conditioning system. The important air-conditioning systems are:

1. Cold storage, 2. Industrial air-conditioning, 3. Comfort air conditioning


Table 20.1 Storage conditions and properties of food products

  1. Cold Storage: In a cold storage air-conditioning system, the room air is cooled to much lower temperature over a cooling coil and supplied back to the storage space. The condition-maintained inside the storage space depend on the nature of the product stored. In cold storages, strict control of both temperature and relative humidity is required. The required storage conditions for a number of products are given in Table 20.1.
  2. Industrial Air-conditioning: The inside design conditions for industrial air-conditioning depend on the category of application. One category comprises those where constancy of temperature is the prime consideration, such as metrology laboratories, precision machine tools, computer centres, etc. In these applications a variation in relative humidity of 10 to 20 percent will not have much effect. The other category may comprise paper and textile mills where the relative humidity is to be maintained constant at a high value of 70 to 75 percent. The temperature requirements of such spaces are not severe. In another category of applications, strict control of both temperature and relative humidity is required, such as chemical and biological process industries.
  3. Comfort air-conditioning: This has been dealt with in art 20.11.

20.12.2 Selection of Outside Design Conditions

For the outside design conditions in summer, it is recommended to use the mean monthly maximum dry bulb temperature and its corresponding wet bulb temperature.

During winter it has been observed that the fuel consumption for heating of buildings varies also directly as the difference between the outside temperature and a reasonably comfortable inside temperature of 18.5°C. Thus the power consumption would be practically nil if the outside temperature is 18.5°C. On the other hand, the power consumption would double if the outside temperature drops from 18.5°C to 8.5°C. A degree-day is obtained for every degree when the mean outside temperature is below 18.5°C during the 24 hour period. If in a given locality the outside temperature average of 30 days is 10°C, then the degree days (D) for the period would be

D = (18.5 − 10) × 30 = 255

The outside design temperature may be calculated as follows:

where tdo = outside design temperature.


The components of a cooling load are:

  1. Sensible heat gain: This is the direct addition of heat to the space to be conditioned. It may occur due to any one or all of the following sources of heat transfer:
    1. The heat flowing into the building by conduction through exterior walls, floors, ceilings, doors and windows due to the temperature difference on their two sides.
    2. The heat received from solar radiation. It consists of the following:
      1. The heat transmitted directly through glass of windows, ventilators, and doors.
      2. The heat absorbed by walls and roofs exposed to solar radiation and later on transferred to the space to be conditioned by conduction.
    3. The heat conducted from adjoining un-conditioned rooms.
    4. The heat generated by fans, lights, machinery, cooking operations, industrial processes, etc.
    5. Occupancy load.
    6. Air infiltration load through doors, windows, and their frequent opening.
    7. Heat gain from walls of ducts, etc.
  2. Latent heat gain: The latent heat load occurs due to the presence of water vapour in the air of conditioned space. This load occurs due to the following sources:
    1. Moisture in the air entering by infiltration
    2. Condensation of moisture from occupants, food cooking process.
    3. Moisture entering directly into the conditioned space through permeable walls, partitions, etc.

    Total cooling load = Sensible heat load + Latent heat load.

20.13.1 Heat Transfer Through Walls and Roofs

The heat transferred through a plane wall by conduction is given by Fourier law. For a single wall shown in Fig. 20.28(a), we have

where   k = thermal conductivity of wall, W/(m. °C)

Ac = area of wall cross-section, m2

Δt = temperature difference on two sides of wall, °C

Δx = wall thickness, m

The heat transfer by convection from a plane wall is given by Newton’s law

Qconv = hAs Δt

where   h = heat transfer coefficient, W/(m2 °C)

As = surface area of wall, m2

Δt = temperature difference between wall and surroundings, °C

Heat gained through a wall by the combined effect of conduction and convention becomes,

Figure 20.28 Heat transfer by conduction through a plane wall: (a) Single wall, (b) Composite wall, (c) Composite wall with air space

where subscripts o, i refer to outside and inside wall conditions respectively

and   U = overall coefficient of heat transfer of wall

for a single wall (Fig. 20.28a)

for a composite wall (Fig. 20.28b)

for a composite wall with air gap(Fig. 20.28c)

where  ka = thermal conductance of air space

20.13.2 Heat Gain from Solar Radiation

The heat from solar radiation is received by building surfaces by direct radiation and diffuse radiation. The direct radiation is the impingement of the Sun’s rays upon the surface. The diffuse radiation is received from moisture and dust particles in atmosphere which absorbs part of the energy of the Sun’s rays. The diffuse radiation is received by surfaces which do not face the sun.

The heat gain through outside walls and roofs is given by,

where  Δte = equivalent temperature differential.

20.13.3 Sol Air Temperature

It is a hypothetical temperature used to calculate the heat received by the outside surface of a building wall by the combined effect of convection and radiation.

Total heat received by the outside surface of the wall per unit area.


qos = qconv + qrad = ho (totos) + I α

where = sol air temperature.

t o = temperature of outside air

tos = temperature of outside surface of wall

ho = outside film coefficient

I = total radiation intensity

α = absorptivity of the surface

20.13.4 Solar Heat Gain Through Glass Areas

20.13.5 Heat Gain Due to Infiltration

There are two methods of estimating the infiltrated air

(i) Crack length method, and (ii) Air change method.

The crack length method is usually used where greater accuracy is required. In most cases the air change method is used for calculating the quantity of infiltrated air. According to this method, the quantity of infiltrated air through windows and walls is,

where L, W, H = length, width and height of room respectively, m

Ac = air changes per hour

Factor is used because infiltration takes place on the windward side of building only.

20.13.6 Heat Gain from Products

This heat gain is very important in case of cold storages.

  1. Chilling load above freezing,

    where m = mass of product

    cpm = mean specific heat of product

    T1, T2 = initial and final temperature of product respectively

    tch = chilling time.

  2. where hfg = latent heat of freezing

    tf = freezing time

  3. Cooling load below freezing,

    where c' pm = mean specific heat of freezing product

    T1′ T2′ = actual storage and freezing temperatures of product

    tc = cooling time

20.13.7 Heat Gain from Lights

Heat gain from lights, Qt = total wattage of lights × use factor × allowance factor

= 0.5 for industrial workshops

Allowance factor = 1.25 for flourescent tubes.

20.13.8 Heat Gain from Power Equipments

where ηem = efficiency of electric motor.

20.13.9 Heat Gain Through Ducts

where U = overall heat transfer coefficient

AD = surface area of duct

ta, ts = temperatures of ambient and supply airs respectively

20.13.10 Empirical Methods to Evaluate Heat Transfer Through Walls and Roofs

There are two approaches to calculate empirically heat transfer through walls and roofs.

They are:

  1. The decrement factor and time lag method.
  2. The equivalent temperature differential method.

Both the methods use analytical-experimental results for their formulations. The equivalent temperature differential method is more commonly used by the air-conditioning engineers as it is also applicable to sunlit walls and roofs.

  1. Using Decrement Factor and Time Lag: If the thermal capacity of the wall is ignored, then the instantaneous rate of heat transfer through the wall at any time τ is given by
    Qτ = UA(teti)

    where te = sol-air temperature.

    and on an average basis, the mean heat flow is given by

    Qm = UA(tem – ti)

    where tem = mean sol-air temperature.

    For the sake of simplicity, the dot above Q has been dropped to indicate the rate.

    But most building materials have a finite thermal capacity which is expressed as

    mc = ρcV = ρc(AΔx)

    where m = Mass of wall, V = volume of wall

    ρ, c = Density and specific heat of wall material

    A = Cross-sectional area of wall

    Δx = Wall thickness.

    It has been seen that there is a two-fold effect of thermal capacity on heat transfer

    1. There is a time lag between the heat transfer at the outside surface qo and the heat transfer at the inside surface qi, defined by ϕ.
    2. There is a decrement in the heat transfer due to the absorption of heat by the wall and subsequent transfer of a part of this heat back to the outside air when its temperature is lower, defined by λ.

    The rigorous analytical method to determine the time lag ϕ and decrement factor λ is quite complicated. In the limiting case, when the wall thickness approaches zero λ → 1 and ϕ → 0

    Then Qτ = UA(teti)

    i.e. the heat transfer through the wall is equal to its instantaneous value.

    Considering the effect of thermal capacity, the actual heat transfer at any time τ is,

    Qτ = UA(temti) + UA λ(te τϕtem)

    where te τϕ = sol-air temperature at time τϕ. i.e., ϕ hours before the heat transfer is to be calculated.

    In a locality where the daily range of variation of the outside air temperature is small it is immaterial what thickness of wall is provided. But in a locality when the daily range of temperature is large, it is desirable to have thick walls so as to cut the cooling load in summer and the heating load in winter. Such walls also do not allow the inside temperature to rise very much during the day and drop at night.

  2. Equivalent temperature differential (ETD) method: The actual heat transfer can be written in terms of an equivalent temperature differential ΔtE defined by the equation.
    Q = UAΔtE

    where ΔtE = (temti) + λ(te τϕtem)

    ΔtE depends on the following:

    1. l and f, which in turn depend on the thermo-physical properties of construction.
    2. Outside air temperature t0 and solar radiation intensity I.
    3. Room temperature ti.

    Thus, the equivalent temperature differential approach takes care of the exposure of the wall or roof to the sun.


An estimate of the heating load is made on the basis of maximum probable heat loss of the conditioned heated space. The heating loads are:

  1. Transmission heat loss: The transmission heat loss from walls roofs, etc., is given by,

    where ti, to = inside and outside design temperature respectively

  2. Solar radiation: The solar radiation heat gain is neglected as peak heating load occurs in the early hours of morning.
  3. Internal heat gains: The internal heat gains from occupants, lights, appliances, etc., decrease the heating load. These negative loads should be accounted for in applications, such as theatres, assembly halls, stores, office buildings, etc.,

It is defined as the ratio of room sensible heat to the total heat (Fig. 20.29).

where RSH, RLH, RTH = room sensible, latent and total heats respectively.

Figure 20.29 Room and supply air conditions and ADP

Figure 20.30 Grand sensible heat factor

20.15.1 Estimation of Supply Air Conditions

When the supply air conditions are not known, then the RSHF line may be drawn from the RSHF calculated value, as described below (Fig. 20.30).

  1. Mark point ‘a’ on the sensible heat factor scale drawn on the right hand corner of psychrometric chart. The point ‘a’ represents the calculated value of RSHF.
  2. Joint point ‘a’ with the alignment circle or reference point b (td − 26°C, ϕ = 50%). The line ab is called the base line.
  3. Mark point R on the psychrometric chart to represent the room design conditions.
  4. Draw a line RR’ through point R parallel to the base line ab. This line is required RSHF line.

It is defined as the ratio of total sensible heat to the grand total heat which the cooling coil or conditioning apparatus is required to handle (Fig. 20.31).

where TSH = total sensible heat = RSH + OASH

TLH = total latent heat = RLH + OALH

GTH = grand total heat = TSH + TLH = (RSH + OASH) + (RLH + OALH)

Let 1 = volume rare of outside air m3/min

td1, td2 = DBT of outside and room air respectively, °C

w1, w2 = specific humidity of outside and room air respectively kg/kg.d.a

h1, h2 = enthalpy of outside and room air respectively, kJ/kg d.a

Outside air total heat, OATH = OASH + OALH

Figure 20.31 RSHF line estimation


It is defined as the ratio of the effective room sensible heat to the effective room total heat

where ERSH = Effective room sensible heat


= RSH + 0.0204 1(td1td2) × BPT

Figure 20.32 Effective room sensible heat factor

ERLH = Effective room latent heat


= RLH + 50 1(w1w2) × BPF

ERTH = Effective room total heat


BPF = By-pass factor

The line joining the point 2 and point 6, i.e. ADP as shown in Fig. 20.32, gives the effective room sensible heat factor line (ERSHF line). From point 4, draw line 4-4′ parallel to line 3-2. From similar triangles 6-4-4′ and 6-3-2, we have


20.18.1 Summer Air-conditioning System with Ventilation
Air and Zero By-pass Factor

The schematic arrangement of summer air-conditioning system with ventilation without by-pass factor is shown in Fig. 20.33. The outside air flows through the damper and mixes with the recirculated air. The mixed air then passes through the filter to remove dirt and dust and then through cooling coil whose temperature is kept much below the DBT of air in the conditioned space. The cooled air then passes through a perforated membrane where it losses its moisture in the condensed form. The moisture is collected in a sump. After that, the air is made to pass through a heating coil which heats the air slightly as per the requirement of designed air DBT and RH. Now the conditioned air is supplied to the room by a supply air fan. From the room a part of the used air is exhausted to the atmosphere by ventilation. The remaining part of the used air (recirculated air) is mixed with fresh air for conditioning.

Figure 20.33 Schematic diagram of system with ventilation air without bypass factor

Figure 20.34 Summer air conditioning process with ventilation air without BPF

Figure 20.34 shows the air-conditioning processes.

o, i = outside and inside air states respectively

1 = state of air after mixing recirculated room air with ventilation air

ai, a0 = mass flow rate of recirculated room air and ventilation air respectively

2 = supply air state for a minimum rate of supply air

line i-2 = ADP, tdp2
= RSHF line
line 1-2 = GSHF line
as = mass flow rate of mixture air
= as(hih2) + (ai hi + ao ho)
(asao)hi + ao hoas h2
  1. Ventilation load

    Let (cmm)o = outside ventilation air volume flow rate = as × vo

  2. Room load

    Let (cmm)1 = mixture air volume flow rate

    a1 = ai + ao = as

    RTH = RSH + RLH
  3. Air-conditioning equipment load
    TSH = RSH + OASH
    TLH = RLH + OALH
    GTH = TSH + TLH

20.18.2 Summer Air-conditioning System with Ventilation Air
and By-pass Factor

The summer air-conditioning processes with ventilation air and BPF is shown Fig. 20.35.

Figure 20.35 Summer air-conditioning processes with ventilation air and BPF

20.18.3 Winter Air-conditioning System

The schematic arrangement of winter air-conditioning system is shown in Fig. 20.36. The outside air flows through a damper and mixes up with the recirculated air. The mixed air passes through a filter to remove dirt, dust and other impurities. The air now passes through a preheating coil to prevent the possible freezing of water and to control the evaporation of water in the humidifier. Then the air is passed through a reheating coil to bring the air to the designed DBT. This conditioned air is supplied to the room by means of a supply an fan. From the room, a part of the used air is exhausted to the atmosphere and the remaining part is recirculated again.

Figure 20.36 Winter air-conditioning system

The air-conditioning processes are shown in Fig. 20.37.

Process 1-2: Preheating

Process 2-3: Adiabatic saturation.

Process 1-3: Heating and humidifying air in air washer with pumped recirculation and external heating of water.

Process 3-s: Reheating

Figure 20.37 Winter air-conditioning processes

Example 20.11

Air at 10°C DBT and 90% RH is to be brought to 35°C DBT and 22.5°C WBT with the help of winter air-conditioner. If the humidified air comes out of the humidifier at 90% RH, determine: (a) the temperature to which the air should be preheated, and (b) the efficiency of the air washer.


Given: td1 =10°C, ϕ1 = 90%, tds = 35°C, tws = 22.5°C

The psychrometric processes are shown in Fig. 20.38

Locate point l(tdl, ϕ1) and point s(tds, tws).

Draw horizontal lines from points 1 and s. From point 3 draw WBT = const. line to locate point 2.

From psychrometric chart, the temperature to which air should be preheated

td2 = 31.2°C
td3 = 18.5°C and td3 = 17.5°C

Efficiency of air-washer =

Figure 20.38

20.18.4 Comfort Air-conditioning System

In comfort air-conditioning, the air is brought to the required DBT and RH for the human health, comfort and efficiency. The comfort conditions have been discussed in detail in art. 20.11.1. Normally it is assumed to be 21°C DBT and 50% RH. The SHF is generally kept between 0.7 to 0.9. The comfort air-conditioning may be adopted for homes, offices, shops, restaurants, theatres, hospitals, schools, etc.

20.18.5 Industrial Air-conditioning System

In an industry the inside DBT and RH of air have to be kept constant for proper working of the machines and manufacturing processes. Special type of air-conditioning systems have to be provided for textile mills, paper mills, machine-parts manufacturing plants, tool rooms, CNC machining centres, photo-processing plants, etc.

Example 20.12

Atmospheric air at a pressure of 1-0132 bar has a dry bulb temperature of 30°C and Relative Humidity of 70%. Calculate:

  1. partial pressure of Water Vapour and air in moist air,
  2. the specific humidity,
  3. the degree of saturation, and
  4. the Dew Point temperature.


Given: pb = 1.0132 bar, td = 30°C, ϕ = 70%

  1. Corresponding to td = 30°C, saturation pressure of water vapour, from steam tables, is
    ps = 0.04246 bar,

    Now pb = pa + pv

    Partial pressure of air, pa = 1.0132 − 0.02972 = 0.98348 bar
  2. Specific humidity,
  3. Specific humidity of started air:

    Degree of saturation,

  4. DPT = saturation temperature corresponding to pv = 0.02972 bar
    From steam tables, corresponding to pv = 0.02972 bar

Example 20.13

In a laboratory test on a particular day, a psychrometer recorded the dry bulb and wet bulb temperatures for atmospheric air as 35°C and 25°C respectively. The atmospheric pressure is 1.013 bar 1. Calculate:

  1. the relative humidity,
  2. the specific humidity,
  3. the dew point temperature, and
  4. the degree of saturation.

The partial pressure of water vapour (pv) can be calculated with the help of Carrier’s Equation, given as;

where pwb = Saturation pressure corresponding to wet bulb temperature,

p = Barometric pressure, and

tdb and twb = are the dry bulb and wet bulb temperatures of air (in 0°C).

Gas constant of air, Ra = 0.287 kJ/kg.K.


Given: td = 35°C, tw = 25°C, pb = 1.013 bar,

  1. Corresponding to td = 35°C, ps = 0.5622 bar, and for tw = 25°C, pw = 0.03166 bar

    Partial pressure of water vapour,

    Relative humidity,

  2. Specific humidity,
  3. Corresponding to pv = 0.02516 bar,
  4. Specific humidity of saturated,

    Degree of saturation,

Review Questions

  1. Explain the following terms:
    1. Degree of saturation
    2. Specific humidity
    3. Relative humidity
    4. Dew point temperature
    5. Thermodynamic wet bulb temperature
  2. Represent the following processes on psychrometric chart:
    1. Sensible heating and cooling
    2. Adiabatic heating and dehumidification
    3. Adiabatic cooling and humidification
    4. Cooling and dehumidification
  3. What is the difference between WBT and thermodynamic WBT?
  4. Define SHF and LHF.
  5. What is by-pass factor?
  6. What are the fa ctors that determine human comfort?
  7. What is effective temperature? List the factors that affect effective temperature.
  8. Differentiate between heat stroke and heat camp.
  9. What is the difference between summer and winter air conditioning?
  10. What is alignment circle?


20.1. An air conditioned space is to be supplied with 300 cmm of air at 20°C dry bulb and 40% relative humidity. Unconditioned air from atmosphere at 32°C dry bulb and 28°C dew point is supplied to the dehumidifier coils, from where the same passes on to a heating coil before entering the conditioned space. Determine the cooling required. Solve the problem both from first principles and with Psychrometric chart.

20.2. Room air at 26°C dry bulb and 50% relative humidity is mixed with outdoor air at 43°C dry bulb and 40% relative humidity in the ratio of 4 : 1. The mixture is passed through a cooling coil maintained at 5°C with a by-pass factor of 0.15. The air from the cooling coil is mixed with room air in the ratio of 4 : 1. The mixture is then reheated to 20°C dry bulb and supplied to the conditioned space.

  1. Show the flow diagram schematically.
  2. Show the different process on a Psychrometric chart.
  3. For 450 kg of supply air per minute determine the quantity of fresh air needed, the refrigeration load and the heat supplied to the reheater coils.


Given: Summer Design = 43°C DBT, 24°C WBT
Inside Design = 25°C DBT, 50% RH
Room Sensible Heat = 300000 kJ/hr
Room Latent Heat = Nil
Ventilation = 30 cmm


  1. effective sensible heat factor
  2. dehumidified air-quantity
  3. supply air temperature

    Assume by-pass factor = 0.05.

20.4. In an Industrial Application:

Summer Design = 40°C DBT; 24°C WBT
Inside Room Design = 60% RH
Room Sensible Heat = 200000 kJ/hr (55.556 kW)
Room Sensible Heat Factor = 0.9.

Use all outdoor air and evaporative cooling process with washer efficiency 80%. Determine

  1. Room DBT maintained
  2. Supply quantity in cmm.

20.5. The following data refer to a laboratory

Summer Design = 40°C DBT, 24°C WBT
Inside Design = 25°C DBT, 50% RH
Room Sensible heat = 60000 kJ/kg
Room Latent Heat = 12000 kJ/hr
Ventilation = 40 cmm.

Also all out side fresh air to be supplied through coil, and no return air. Determine

  1. effective sensible heat factor
  2. ADP
  3. actual quantity of outdoor air to be supplied to take care of the loads
  4. dehumidified air quantity supplied.

    Assume by-pass factor = 0.05.

20.6. Air 35°C dry bulb temperature and 25°C wet bulb temperature is passed through a cooling coil at the rate of 280 m3/min. The air leaves the cooling coil at 26.5°C dry bulb temperature and 50% relative humidity. Find

  1. Capacity of the cooling coil in tonnes of refrigeration;
  2. Wet bulb temperature of the leaving air;
  3. Water vapour removed per minute: and
  4. Sensible heat factor.

[Ans. 31.98 TR; 19.2°C; 1.56 kg/kg of dry air:0.39]

20.7. The outside air at 30°C dry bulb temperature and 18°C wet bulb temperature enters a cooling coil at the rate of 40 m3/min. The effective surface temperature of the cooling coil is 4.5°C and its cooling capacity is 12.5 kW of refrigeration. Find:

  1. dry bulb and wet bulb temperatures of the air leaving the coil,
  2. enthalpy of air leaving the coil, and
  3. by-pass factor of the coil.

[Ans. 18.8°C; 12.7°C; 35.6 kJ/kg of dry air; 0.52]

20.8. Air at 40°C dry bulb temperature and 15% relative humidity is passed through the adiabatic humidifier at the rate 200 m3/min. The outlet conditions of air are 25°C dry bulb temperature and 20°C wet bulb temperature. Find (a) dew point temperature, (b) relative humidity of exit air; and (c) amount of water vapour added to the air per minute.

[Ans. 17.8°C; 65%; 1 26 kg/min]

20.9. Air at 0°C and 95% relative humidity has to be heated and humidified to 25°C and 40% relative humidity by the following three processes:

  1. preheating
  2. adiabatic saturation in a recirculated water air washer and
  3. reheating to final state.


  1. The heating required in two heaters
  2. The makeup water required in washer and temperature of washer

    Assume effectiveness of washer as 80 per cent.

[Ans. 27.3 kJ/kg of dry air; 9 kJ/kg of dry air; 4.1 g/kg of dry air 12.6°C]

20.10. The following data relates to a cold storage for storing 450 tonnes of vegetables

Inside design conditions 19°C DBT 60% RH
Outside design conditions 36°C DBT 28°C WBT
Infiltrated air 180 m3/hr
Fresh air supply 4500 m3/hr
No. of persons working in cold store 20
Sensible heat gain through glass, walls, ceiling etc., 16.3 kW
Water content of vegetables 74%
Loss of water content 0.01% per hour
Heat from equipment and reaction heat of vegetables 3.1 kW

If the air conditioning is achieved by first cooling and dehumidifying and then heating and the temperature of air entering the room is not to exceed 16°C, determine

  1. amount of recirculated air, if the recirculated air is mixed with fresh air before entering the cooling coil.
  2. capacity of the cooling coil in tonnes of refrigeration and its by-pass factor if the dew point temperature of the coil is 6°C and
  3. capacity of heating coil

[Ans. 3.384 kg/s; 46.75 TR; 34 82 kW]

20.11. The following data relates to a conference room having a seating capacity of 80 persons.

Inside design conditions 22°C DBT 55% RH
Outside design conditions 38°C DBT 28°C WBT
Sensible and latent heat loads per person 75W and 45W respectively
Lights and fan loads 12000 W
Sensible heat gain through glass, walls, ceiling etc. 12000 W
Air infiltration 18 m3/min
Fresh air supply 80 m3/min
By pass factors of the coils 0.1

If two third of recirculated room air and one third of fresh air are mixed before entering the cooling coils determine

  1. apparatus Dew point
  2. grand total heat load
  3. effective room sensible heat factor.

20.12. A sling psychrometer reads 40°C DBT and 28°C WBT. Calculate the following:

  1. Specific humidity
  2. Relative humidity
  3. Dew point, temperature
  4. vapour density
  5. Enthalpy of mixture per kg of dry air.

    Assume atmospheric pressure = 1.03 bar

[Ans. 0.0189 kg/kg d.a., 40.4%, 24°C, 0.0338 kg/m3, 88.8 k/kg d.a]

20.13. The atmospheric conditions are 25°C DBT and specific humidity of 10 g/kg of d.a. Determine the partial pressure of vapour and relative humidity.

Take atmospheric pressure = 1 bar.

[Ans. 0.0158 bar 49%]

20.14. Air enters an evaporative cooler at a pressure of 1 atm. DBT of 38°C and 25% RH. The air leaves the cooler at 1 atm and 18°C DBT. Assuming adiabatic process, determine the relative humidity of cool air and quality of water that must be supplied to cooler for each kg of dry air.

[Ans. 63%, 0.00808 kg/kg. d.a]