Unit V – Psychrometry

Psychrometrics
• The science which investigates the thermal properties of moist air, considers the measurement
and control of the moisture content of air, and studies the effect of atmospheric moisture on
material and human comfort may properly be termed “psychrometrics’’.

• Dry air. The international joint committee on Psychrometric Data has adopted the following
exact composition of air expressed in mole fractions (Volumetric) Oxygen 0.2095, Nitrogen
0.7809, Argon 0.0093, Carbon dioxide 0.0003. Traces of rare gases are neglected. Molecular
weight of air for all air conditioning calculations will be taken as 28.97. Hence the gas constant,

Dry air is never found in practice. Air always contains some moisture. Hence the common
designation “air” usually means moist air. The term ‘dry air’ is used to indicate the water free
contents of air having any degree of moisture.

23 October 2020 Dr. R. Shanthi 2

• Saturated air. For a given temperature, a given quantity of air can be saturated with a fixed
quantity of moisture. At higher temperatures, it requires a larger quantity of moisture to
saturate it. At saturation, vapour pressure of moisture in air corresponds to the saturation
pressure given in steam tables corresponding to the given temperature of air.

• Dry-bulb temperature (DBT). It is the temperature of air as registered by an ordinary

thermometer (Td).

• Wet-bulb temperature (WBT). It is the temperature registered by a thermometer when the

bulb is covered by a wetted wick and is exposed to a current of rapidly moving air (Tw).

• Adiabatic saturation temperature. It is the temperature at which the water or ice can
saturate air by evaporating adiabatically into it. It is numerically equivalent to the measured
wet bulb temperature (Tw).

• Wet bulb depression. It is the difference between dry-bulb and wet bulb temperatures (Td –
Tw).
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• Dew point temperature (DPT). It is the temperature to which air must be cooled at constant
pressure in order to cause condensation of any of its water vapour. It is equal to steam table
saturation temperature corresponding to the actual partial pressure of water vapour in the air
(Tdp).

• Dew point depression. It is the difference between the dry bulb and dew point temperatures
(Td – Tdp).

• Specific humidity (Humidity ratio). It is the ratio of the mass of water vapour per unit
mass of dry air in the mixture of vapour and air, it is generally expressed as grams of water per
kg of dry air. For a given barometric pressure it is a function of dew point temperature alone.

• Relative humidity (RH), (φ). It is the ratio of the partial pressure of water vapour in the
mixture to the saturated partial pressure at the dry bulb temperature, expressed as
percentage.

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• Sensible heat. It is the heat that changes the temperature of a substance when added to or
abstracted from it.

• Latent heat. It is the heat that does not affect the temperature but changes the state of
substance when added to or abstracted from it.

• Enthalpy. It is the combination energy which represents the sum of internal and flow energy
in a steady flow process and is expressed as kJ per kg of dry air (h).

• Note. When air is saturated DBT, WBT, DPT are equal.

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PSYCHROMETRIC RELATIONS
• Pressure: Dalton’s law of partial pressure is employed to determine the pressure of a mixture

of gases. This law states that the total pressure of a mixture of gases is equal to the sum of

partial pressures.

• For calculating partial pressure of water vapour in the air many equations have been proposed,

probably Dr. Carrier’s equation is most widely used.

where pv = Partial pressure of water vapour,

pvs = Partial pressure of water vapour when air is fully saturated,
pt = Total pressure of moist air,
tdb = Dry bulb temperature (ºC), and
twb = Wet bulb temperature (ºC).

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Specific humidity W :

Ra = 0.287 kJ/kg K
Rv = 0.462 kJ/kg K

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 The masses of air and water vapour in terms of specific volumes are given by expression as

where va = Specific volume of dry air, and vv = Specific volume of water vapour.

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Degree of saturation (μ):
It is the ratio of actual specific humidity and saturated specific humidity, both at the same
temperature, T

where, Ws = Specific humidity of air when air is fully saturated

where pvs = Partial pressure of water vapour when air is fully saturated (pvs can be calculated
from steam tables corresponding to the dry bulb temperature of the air)

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Relative humidity (RH), φ : It is the ratio of the partial pressure of water vapour in
the mixture to the saturated partial pressure at the same dry bulb temperature,
expressed as percentage.

𝑝𝑣
∅=
𝑝𝑣𝑠
𝑝𝑣 𝑉 = 𝑚𝑣 𝑅𝑇
𝑝𝑣𝑠 𝑉 = 𝑚𝑣𝑠 𝑅𝑇

𝑚𝑣 𝑅𝑇
∅=
𝑚𝑣𝑠 𝑅𝑇

𝑚𝑣 𝑝𝑣
∅= =
𝑚𝑣𝑠 𝑝𝑣𝑠

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Enthalpy of moist air: It is the sum of enthalpy of dry air and enthalpy of water
vapour associated with dry air. It is expressed in kJ/kg of dry air

𝒉 = 𝑪𝒑 𝑻𝒂 + 𝑾 𝒉𝒈 + 𝟏. 𝟖𝟖 𝑻𝒅 − 𝑻𝒅𝒑

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PSYCHROMETRIC CHARTS
1. The dry bulb temperature (ºC)
2. The mass of water vapour in kg (or
grams) per kg of dry air
3. Pressure of water vapour in mm of
mercury
4. Dew point temperatures
5. Constant relative humidity lines
6. Enthalpy or total heat at
saturation temperature in kJ/kg of
dry air
7. Wet bulb temperatures
8. The volume of air vapour mixture
per kg of dry air (specific volume)

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 1. The dry bulb temperature (ºC)
2. The mass of water vapour in kg
(or grams) per kg of dry air
3. Pressure of water vapour in mm
of mercury
4. Dew point temperatures
5. Constant relative humidity
lines
6. Enthalpy or total heat at
saturation temperature in kJ/kg
of dry air
7. Wet bulb temperatures
8. The volume of air vapour
mixture per kg of dry air
(specific volume)

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Problem 1: Atmospheric air at 1.0132 bar has a DBT of 32°C and a WBT of 26°C. compute (a) the
partial pressure of watervapour, (b) the specific humidity, (c) the dew point temperature, (d) the
relative humidity, (e) the degree of saturation, (f) the density of the air in the mixture, (g) the density
of the vapour in the mixture and (h) the enthalpy of the mixture.

Given:
p = 1.0132 bar
h= 80 kJ/kg
DBT, Td = 32°C
WBT, Tw = 26°C (0.0182)

Solutuion:
(b) the specific humidity
From Psychrometric chart,
W1 = 0.0182 kg vap/ kg of dry air

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(a) the partial pressure of watervapour,

where pv = Partial pressure of water vapour,

pvs = Partial pressure of water vapour when air is fully saturated,
pt = Total pressure of moist air,
tdb = Dry bulb temperature (ºC), and
twb = Wet bulb temperature (ºC).

From steam tables, Tw = 26°C

(pvs)wb = 0.0336 bar
(1.0132 − 0.0336) × (32 − 26)
𝑝𝑣 = 0.0336 −
1572.4 − (1.3 × 26)
𝒑𝒗 = 𝟎. 𝟎𝟑 𝒃𝒂𝒓

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(b) the specific humidity
(d) the relative humidity,
𝒑𝒗
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒉𝒖𝒎𝒊𝒅𝒊𝒕𝒚, 𝑾 = 𝟎. 𝟔𝟐𝟐 𝑝𝑣
𝒑 − 𝒑𝒗 ∅=
𝑝𝑣𝑠
0.03
𝑊 = 0.622 From Stream tables for Td = 32°C, p = 0,048 bar
1.0132 − 0.03
𝑝𝑣𝑠 = 0.048 𝑏𝑎𝑟
𝑾 =?
𝑾 = 𝟎. 𝟎𝟏𝟖𝟗kg vap/ kg of dry air 0.03
∅=
0.048
(c) the dew point temperature ∅=?

Saturation temperature at pv will be the dew point ∅ = 𝟎. 𝟔𝟐𝟓 = 𝟔𝟐. 𝟓%

temperature. (e) the degree of saturation

From steam tables for pv = 0.03 bar, T = 24.1°C 𝑝𝑣 𝑝𝑡 − 𝑝𝑣𝑠 0.03 1.0132 − 0.048
𝜇= =
𝑝𝑣𝑠 𝑝𝑡 − 𝑝𝑣 0.048 1.0132 − 0.03
Dew point temperature, Tdp = 24.1°C
𝝁 =?
𝝁 = 𝟎. 𝟔𝟏𝟒
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(f) the density of the air in the mixture (h) the enthalpy of the mixture
We Know that p= pa + pv From chart, h= 80 kJ/kg
Partial pressure of dry air in the mixture, pa= p – pv (or)
pa= 1.0132 – 0.03 = 0.9832 bar 𝒉 = 𝒉𝒂 + 𝑾𝒉𝒗
pa Va = Ra Td
𝒉 = 𝑪𝒑 𝑻𝒂 + 𝑾 𝒉𝒈 + 𝟏. 𝟖𝟖 𝑻𝒅 − 𝑻𝒅𝒑
1 𝑝𝑎 0.9832 × 105 From steam tables, for Td = 32°C, hg = 2559.9 kJ/kg
𝜌𝑎 = = =
𝑉𝑎 𝑅𝑎 𝑇𝑎 0.287 × 103 × (32 + 273) ℎ = 1.005 × 32 + 0.0189 × 2559.9 + 1.88 32 − 24.1
𝝆𝒂 =? 𝒉 =?
𝝆𝒂 = 𝟏. 𝟏𝟐 𝒌𝒈/𝒎𝟑 𝒅𝒓𝒚𝒂𝒊𝒓 𝒉 = 𝟖𝟎. 𝟖𝟐 𝒌𝑱/𝒌𝒈 (approx)
(g) the density of the vapour in the mixture (or)
𝝆𝒗 = 𝑾 × 𝝆𝒂 𝒉 = 𝑪𝒑 𝑻𝒂 + 𝑾 𝒉𝒈 + 𝟏. 𝟖𝟖 𝑻𝒅 − 𝑻𝒅𝒑
𝑘𝑔 𝑣𝑎𝑝 𝑘𝑔 𝑑𝑟𝑦 𝑎𝑖𝑟 From steam tables, for Tdp = 24.1°C, hg = 2545.5 kJ/kg
𝜌𝑣 = 0.0189 3 air × 1.12 3
𝑚 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑚 𝑑𝑟𝑦𝑎𝑖𝑟
ℎ = 1.005 × 32 + 0.0189 × 2545.5 + 1.88 32 − 24.1
𝝆𝒗 =? 𝒉 =?
𝝆𝒗 = 𝟎. 𝟎𝟐𝟏𝒌𝒈 𝒗𝒂𝒑/𝒎𝟑 𝒅𝒓𝒚𝒂𝒊𝒓 𝒉 = 𝟖𝟎. 𝟓𝟓 𝒌𝑱/𝒌𝒈
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Problem 2: The atmospheric conditions are ; 20°C and specific humidity of 0.0095 kg/kg of dry air.
Calculate the following : (i) Partial pressure of vapour (ii) Relative humidity (iii) Dew point temperature.
Given: From Steam tables for Td = 20°C, p = 0.0234 bar
DBT, Td = 20°C 𝑝𝑣𝑠 = 0.0234 𝑏𝑎𝑟
Specific humidity, W = 0.0095 kg/kg of dry air
0.001524
∅=
Solution: 0.0234
∅=?
(i) Partial pressure of vapour, pv :
∅ = 𝟎. 𝟔𝟓 = 𝟔𝟓%
𝑝𝑣
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, 𝑊 = 0.622 (iii) Dew point temperature, Tdp :
𝑝 − 𝑝𝑣
𝑝𝑣 The dew point temperature is the saturation
0.0095 = 0.622
1.0132 − 𝑝𝑣
temperature of water vapour at a pressure of 0.01524
𝒑𝒗 =?
bar, Tdp [from steam tables by interpolation]
𝒑𝒗 = 0.01524 bar
Tdp= 13.24°C
(ii) Relative humidity:
𝑝𝑣
∅=
𝑝𝑣𝑠
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Problem 3: 0.004 kg of water vapour per kg of atmospheric air is removed and temperature of air after
removing the water vapour becomes 20°C. Determine : (i) Relative humidity (ii) Dew point temperature.
Assume that condition of atmospheric air is 30°C and 55% R.H. and pressure is 1.0132 bar.

Solution: The specific humidity after removing 0.004 kg of water vapour

Corresponding to 30ºC, from steam tables, becomes,
pvs = 0.0425 bar 0.01468 – 0.004 = 0.01068 kg/kg of dry air
𝑝𝑣
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅ = and the temperature Td = 20ºC (given).
𝑝𝑣𝑠
𝑝𝑣
0.55 = The partial pressure of water vapour, pv, at this condition can
0.0425
be calculated as follows :
𝒑𝒗 =?
𝑝𝑣
𝒑𝒗 = 0.02337 bar 𝑊 = 0.622
𝑝𝑣 𝑝 − 𝑝𝑣
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, 𝑊 = 0.622
𝑝 − 𝑝𝑣 𝑝𝑣
0.02337 0.01068 = 0.622
𝑊 = 0.622 1.0132 − 𝑝𝑣
1.0132 − 0.02337
𝒑𝒗 =?
𝑾 =?
𝒑𝒗 = 0.0171 bar
𝑾 = 0.01468 kg/kg of dry air
Corresponding to 20ºC, from steam tables, pvs = 0.0234 bar.
23 October 2020 Dr. R. Shanthi 19
 (i) Relative humidity

0.0171
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅ =
0.0234
∅ =?
∅ = 0.73 or 73%.
(ii) Dew point temperature, Tdp :
The dew point temperature is the saturation temperature of water vapour at a pressure
of 0.0171 bar, Tdp
Tdp= 15°C

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 PSYCHROMETRIC PROCESSES
• The processes affecting the psychrometric properties of air are called psychrometric
processes.

• The important psychrometric processes are :

1. Sensible heating

2. Sensible cooling

3. Cooling and dehumidification

4. Cooling and humidification

5. Heating and dehumidification

6. Heating and humidification.

7. Mixing of air streams

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 Sensible Heating
• When air passes over a dry surface which is at a temperature greater
than its (air) dry bulb temperature, it undergoes sensible heating.

• Thus the heating can be achieved by passing the air over heating coil
like electric resistance heating coils or steam coils.

• During such a process, the specific humidity remains constant

but the dry bulb temperature rises and approaches that of the
surface.

• The by-pass factor (BF) for the process is defined as the ratio of the
difference between the mean surface temperature of the coil and
leaving air temperature to the difference between the mean surface
temperature and the entering air temperature.

23 October 2020 Dr. R. Shanthi 22

 Sensible Heating
• air at temperature Td1, passes over a heating coil with an average
surface temperature Td3 and leaves at temperature Td2

𝑻𝒅𝟑 − 𝑻𝒅𝟐
𝑩𝑭 =
𝑻𝒅𝟑 − 𝑻𝒅𝟏
𝑾𝟏 = 𝑾𝟐

Mass balance for dry air, 𝒎𝒂𝟏 = 𝒎𝒂𝟐 = 𝒎𝒂

Mass balance for water vapour, 𝒎𝒗𝟏 = 𝒎𝒗𝟐

𝑾𝟏 × 𝒎𝒂𝟏 = 𝑾𝟐 × 𝒎𝒂𝟐

Energy balance,

Energy in = energy out

𝑚𝑎1 × ℎ1 + 𝑄 = 𝑚𝑎2 × ℎ2
𝑸 = 𝒎𝒂 × 𝒉𝟐 − 𝒉𝟏 Td3 Td2 Td1

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Problem 4: 90 m3 of air per minute at 20°C and 75% R.H. is heated until its temperature becomes
30°C. Calculate : (i) R.H. of the heated air. (ii) Heat added to air per minute.
Given:
Td1 = 20°C
∅1 = 75 %
Td2 = 30°C
h2 = 58.5 kJ/kg
Solution:
From Chart,
h1 = 47.5 kJ/kg
h1 = 47.5kJ/kg
∅ 2 = 42.5%
W = 0.0175 kg/kg of dry air W = 0.0175
kg/kg of d
h2 = 58.5kJ/kg
∅𝟐 = 𝟒𝟐. 𝟓 %

(i) R.H. of the heated air

∅𝟐 = 𝟒𝟐. 𝟓 %

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(ii) Heat added to air per minute
Corresponding to 20ºC, from steam tables, 1.0132 − 0,01755 × 105 × 90
𝑚𝑎 =
0.287 × 103 (20 + 273)
pvs = 0.0234 bar

𝑝𝑣 𝒎𝒂 =?
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅ =
𝑝𝑣𝑠
𝒎𝒂 = 𝟏𝟎𝟔. 𝟓 𝒌𝒈/𝒎𝒊𝒏
𝑝𝑣
0.75 =
0.0234 Amount of heat added per minute = 𝒎𝒂 (h2 –h1)
𝒑𝒗 =?
= 106.5 (58.5 – 47.5) = ?
𝒑𝒗 = 0.01755 bar
Mass of dry air in 90 m3 of air supplied = 1171.5 kJ/kg
𝑝𝑎 𝑉 = 𝑚𝑎 𝑅𝑎 𝑇

𝑝𝑎 𝑉 𝑝 − 𝑝𝑣 𝑉
𝑚𝑎 = =
𝑚𝑎 𝑅𝑎 𝑇 𝑅𝑎 𝑇

23 October 2020 Dr. R. Shanthi 25

 Sensible Cooling
• Air undergoes sensible cooling whenever it passes over a surface
that is at a temperature less than the dry bulb temperature of the
air but greater than the dew point temperature.

• Thus sensible cooling can be achieved by passing the air over cooling
coil like evaporating coil of the refrigeration cycle or secondary brine
coil.

• During the process, the specific humidity remains constant and dry
bulb temperature decreases.

• The by-pass factor (BF) for the process is defined as the ratio of the
difference between the mean surface temperature of the coil and
leaving air temperature to the difference between the mean surface
temperature and the entering air temperature.
Td3 Td2 Td1

23 October 2020 Dr. R. Shanthi 26

 Sensible Cooling
• air at temperature Td1, passes over a cooling coil with an average
surface temperature Td3 and leaves at temperature Td2

𝑻𝒅𝟐 − 𝑻𝒅𝟑
𝑩𝑭 =
𝑻𝒅𝟏 − 𝑻𝒅𝟑
𝑾𝟏 = 𝑾𝟐

Mass balance for dry air, 𝒎𝒂𝟏 = 𝒎𝒂𝟐 = 𝒎𝒂

Mass balance for water vapour, 𝒎𝒗𝟏 = 𝒎𝒗𝟐

𝑾𝟏 × 𝒎𝒂𝟏 = 𝑾𝟐 × 𝒎𝒂𝟐

Energy balance,

Energy in = energy out

𝑚𝑎1 × ℎ1 = 𝑄 + 𝑚𝑎2 × ℎ2
𝑸 = 𝒎𝒂 × 𝒉𝟏 − 𝒉𝟐 Td3 Td2 Td1

23 October 2020 Dr. R. Shanthi 27

Problem 5: 40 m3 of air at 35°C DBT and 50% R.H. is cooled to 25°C DBT maintaining its
specific humidity constant. Determine : (i) Relative humidity (R.H.) of cooled air ; (ii) Heat
removed from air.
Given: W = 0.0175
kg/kg of dry air

Td1 = 35°C h1 = 80 kJ/kg

∅1 = 50 % h2 = 71 kJ/kg

Td2 = 25°C
∅ 2 = 82.5%
Solution:

From Chart,
h1 = 80 kJ/kg
W = 0.0175 kg/kg of dry air
h2 = 71 J/kg
∅𝟐 = 𝟖𝟐. 𝟓 %

(i) R.H. of the heated air

∅𝟐 = 𝟖𝟐. 𝟓 %
23 October 2020 Dr. R. Shanthi 28
(ii) Heat removed from air.
Corresponding to 25ºC, from steam tables, 1.0132 − 0.0262 × 105 × 40
𝑚𝑎 =
0.287 × 103 (25 + 273)
pvs = 0.0262 bar

𝑝𝑣 𝒎𝒂 =?
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅ =
𝑝𝑣𝑠
𝒎𝒂 = 𝟒𝟕 𝒌𝒈
𝑝𝑣
0.825 =
0.0317 Amount of heat added per minute = 𝒎𝒂 (h1 –h2)
𝒑𝒗 =?
= 47 (80 – 71) = ?
𝒑𝒗 = 0.0262 bar
Mass of dry air in 90 m3 of air supplied = 423 kJ/kg
𝑝𝑎 𝑉 = 𝑚𝑎 𝑅𝑎 𝑇

𝑝𝑎 𝑉 𝑝 − 𝑝𝑣 𝑉
𝑚𝑎 = =
𝑚𝑎 𝑅𝑎 𝑇 𝑅𝑎 𝑇

23 October 2020 Dr. R. Shanthi 29

Mixing of Air Streams
• Mixing of several air streams is the process which is
very frequently used in air conditioning.

• This mixing normally takes place without the addition

or rejection of either heat or moisture, i.e.,
adiabatically and at constant total moisture content. 𝒎𝟏 𝑾𝟑 − 𝑾𝟐 𝒉 𝟑 − 𝒉𝟐
= =
Thus we can write the following equations : 𝒎𝟐 𝑾𝟏 − 𝑾𝟑 𝒉 𝟏 − 𝒉𝟑

𝒎𝟏 + 𝒎𝟐 = 𝒎𝟑

𝒎𝟏 𝑾𝟏 + 𝒎𝟐 𝑾𝟐 = 𝒎𝟑 𝑾𝟑 -------(1)

𝒎𝟏 𝒉𝟏 + 𝒎𝟐 𝒉𝟐 = 𝒎𝟑 𝒉𝟑 --------(2)

(1) and (2) Can be written as

𝑚1 𝑊1 + 𝑚2 𝑊2 = 𝑚1 + 𝑚2 𝑊3
𝑚1 𝑊1 − 𝑊3 = 𝑚2 𝑊3 − 𝑊2
𝑚1 ℎ1 − ℎ3 = 𝑚2 ℎ3 − ℎ2
23 October 2020 Dr. R. Shanthi 30
 Problem 6: One kg of air at 35°C DBT and 60% R.H. is mixed with 2 kg of air at 20°C DBT and
13°C dew point temperature. Calculate the specific humidity, enthalpy, relative humidity and
DBT of the mixture W1 = 0.0216
h1 = 80 kJ/kg kg/kg of dry air
Given:
Td1 = 35°C, ∅1 = 60 %
𝒎1 = 1 𝑘𝑔
h3 = 56 kJ/kg
Td2 = 20°C, Tdp2 = 13°C
𝒎2 = 2 𝑘𝑔
h2 = 44 kJ/kg W3 = 0.013
Solution: kg/kg of dry air

From chart,
h1 = 80 kJ/kg 𝑻d2 = 13°C

W2 = 0.0086
W1 = 0.0216 kg/kg of dry air kg/kg of dry air

h2 = 44 kJ/kg
W2 = 0.0086 kg/kg of dry air
Td2 = 20°C Td3 = 22°C
Td1 = 35°C
23 October 2020 Dr. R. Shanthi 31
 𝒎𝟏 + 𝒎𝟐 = 𝒎𝟑
From Chart,
𝑚3 = 2 + 1 = 3 𝑘𝑔
For 𝒉𝟑 = 𝟓𝟔 𝒌𝑱/𝒌𝒈 and 𝑾𝟑 = 𝟎. 𝟎𝟏𝟐𝟗 𝒌𝒈/𝒌𝒈𝒐𝒇𝒅𝒓𝒚𝒂𝒊𝒓
𝒎𝟏 𝑾𝟏 + 𝒎𝟐 𝑾𝟐 = 𝒎𝟑 𝑾𝟑
DBT of the mixture, Td3 = 22°C
1 × 0.0216 + 2 × 0.0086 = 3 × 𝑊3 RH of the mixture, ∅3 = 80%
𝑾𝟑 =?
𝑾𝟑 = 𝟎. 𝟎𝟏𝟐𝟗 𝒌𝒈/𝒌𝒈𝒐𝒇𝒅𝒓𝒚𝒂𝒊𝒓
𝒎𝟏 𝒉𝟏 + 𝒎𝟐 𝒉𝟐 = 𝒎𝟑 𝒉𝟑
1 × 80 + 2 × 44 = 3 × ℎ3
𝒉𝟑 =?
𝒉𝟑 = 𝟓𝟔 𝒌𝑱/𝒌𝒈

23 October 2020 Dr. R. Shanthi 32

 Problem 7: Saturated air leaving the cooling section of an air conditioning system at 14°C at a rate of
50m3/min is mixed adiabatically with the outside air at 32°C and 60% R.H at rate of 20m3/min. assuming
that the mixing process occurs at a pressure of 1 atm., . Calculate the specific humidity and enthalpy of
the mixture
Given:
Td1 = 14°C, ∅1 = 100 %
h2 = 79 kJ/kg
𝑽1 = 50m3/min
Td2 = 32°C, ∅2 = 60 % W2 = 0.018
kg/kg of dry air
𝑽2 = 20m3/min
Solution:
From chart,
h1 = 39 kJ/kg
h1 = 39 kJ/kg
𝑻w1= 14°C
W1 = 0.0099 kg/kg of dry air W1 = 0.0099
kg/kg of dry
h2 = 79 kJ/kg
W2 = 0.018 kg/kg of dry air
v1=0.83 m3/kg
v2=0.89 m3/kg
Td1 = 14°C v1=0.83 m3/kgTd2 = 32°C
v2=0.89 m3/kg

23 October 2020 Dr. R. Shanthi 33

 𝑉1 50
𝑚1 = = =?
𝑣1 0.83
𝒎𝟏 = 𝟔𝟎. 𝟐𝟒 𝒌𝒈/𝒎𝒊𝒏

𝑉2 20
𝑚2 = = =?
𝑣2 0.89
𝒎𝟐 = 𝟐𝟐. 𝟒𝟕 𝒌𝒈/𝒎𝒊𝒏
𝒎𝟏 𝑾𝟏 + 𝒎𝟐 𝑾𝟐 = 𝒎𝟑 𝑾𝟑
60.24 × 0.0099 + 22.47 × 0.018 = (60.24 + 22.47) × 𝑊3
𝑾𝟑 =?
𝑾𝟑 = 𝟎. 𝟎𝟏𝟐𝟏 𝒌𝒈/𝒌𝒈𝒐𝒇𝒅𝒓𝒚𝒂𝒊𝒓
𝒎𝟏 𝒉𝟏 + 𝒎𝟐 𝒉𝟐 = 𝒎𝟑 𝒉𝟑
60.24 × 39 + 22.47 × 79 = (60.24 + 22.47) × ℎ3
𝒉𝟑 =?
𝒉𝟑 = 𝟒𝟗. 𝟖𝟕 𝒌𝑱/𝒌𝒈

23 October 2020 Dr. R. Shanthi 34

Humidification Process: Addition of moisture
Problem 8: 150 m3 of air per minute is passed through the adiabatic humidifier. The
condition of air at inlet is 35°C DBT and 20 per cent relative humidity and the outlet condition is
20°C DBT and 15°C WBT. Determine the following :
(i) Dew point temperature
(ii) Relative humidity of the exit air
(iii) Amount of water vapour added to
the air per minute.
Given:
Td1 = 35°C, ∅1 = 20 %
h1 = 55 kJ/kg
Td2 = 20°C, Tw2 = 15°C
Solution: From chart,
Tw2= 15°C
(i) Dew point temperature
Tdp2= 11°C ∅ 2 = 60% W2
Tdp1= 9°C Tdp1= 9°C W1
Tdp1= 11°C
(ii)Relative humidity of the exit air,
∅ 2 = 60% v1=0.885 m3/kg
Td2= 25°C Td1 = 35°C
23 October 2020 Dr. R. Shanthi 35
(iii) Amount of water vapour added to the air per minute.
From chart, v1=0.885 m3/kg of dry air

𝑉 150
𝑚𝑎 = =
𝑣1 0.885
𝒎𝒂 = ?
𝒎𝒂 = 𝟏𝟔𝟗. 𝟒𝟗 𝒌𝒈 𝒐𝒇 𝒅𝒓𝒚 𝒂𝒊𝒓/𝒎𝒊𝒏
Amount of water vapour added to the air per minute, 𝑚𝑣 = 𝑚𝑎 𝑊2 − 𝑊1
From chart, W1= 0.0065 kg/kg of dry air and W2= 0.0084 kg/kg of dry air
𝑚𝑣 = 169.49 0.0084 − 0.0065
𝒎𝒗 =?
𝒎𝒗 = 𝟎. 𝟑𝟐𝟐 𝒌𝒈/𝒎𝒊𝒏

23 October 2020 Dr. R. Shanthi 36

Problem 9: An air conditioning system is to be taken in outdoor air at 10ºC and 30% RH at a steady
rate of 45 m3/min and to condition it to 25ºC and 60 % RH. The out door air is first heated to 22ºC in the
heating section and then humidified by the injection of the hot steam in the humidified section.
Assuming the entire process takes place at a pressure of 1 atm, determine
a) The rate of heat supply in the heating section,
b) The mass flow rate of the steam requires in the humidifying section.

Given:
Td1 = 10°C,
∅1 = 30 %
Td2 = 22°C 3

Td3 = 25°C
∅3 = 60 %
1 2

10ºC 22ºC 25ºC

DBT

23 October 2020 Dr. R. Shanthi 37

Given: W3 = 0.012 kg/kg of dry a
Td1 = 10°C,
∅1 = 30 %
Td2 = 22°C
Td3 = 25°C
h3 = 55 kJ/kg
∅3 = 60 %
Solution: From Chart,
h1= 15 kJ/kg
h2= 27 kJ/kg 3
h2 = 27 kJ/kg
h3= 55 kJ/kg h1 = 15 kJ/kg

W1= W2= 0.002 kg/kg of dry air

W3 = 0.012 kg/kg of dry air
v1= 0.805 m3/kg

1 2
Td2 = 22°C Td3 = 25°C
v1=0.805 m3/kg W1=W2= 0.002kg/kg of dry
T = 10°C
23 October 2020 Dr. R. d1
Shanthi 38
(a) The rate of heat supply in the heating section,
Mass balance -(2-3 process)
1 - 2 process:
𝒎𝒂 𝑾𝟐 + 𝒎𝒗𝟐 = 𝒎𝒂 𝑾𝟑
The rate of heat supply in the heating section = 𝒎𝒂 (h2 –h1)
𝑚𝑣2 = 𝑚𝑎 (𝑊3 −𝑊2 )
From chart, v1=0.805 m3/kg of dry air
= 55.9 × (0.012 − 0.002)
𝑉1 45
𝑚𝑎 = = 𝒎𝒗𝟐 = 𝟎. 𝟓𝟓𝟗 𝒌𝒈/𝒎𝒊𝒏
𝑣1 0.805
𝒎𝒂 = ?
𝒎𝒂 = 𝟓𝟓. 𝟗 𝒌𝒈 𝒐𝒇 𝒅𝒓𝒚 𝒂𝒊𝒓/𝒎𝒊𝒏
The rate of heat supply in the heating section = 55.9 ×(27 –15)
𝐐 = 𝟔𝟕𝟎. 𝟖 𝒌𝑱/𝒌𝒈
(b) The mass flow rate of the steam requires in the
humidifying section.
Let 𝒎𝒗𝟐 be the amount of steam sprayed in the humidifying
section.

23 October 2020 Dr. R. Shanthi 39

Problem 10: Saturated air at 3ºC is required to be supplied to a room where the temperature must be
held at 22ºC with a relative humidity of 55%. The air is heated and then water at 10ºC is sprayed
to give the required humidity. Determine :
(i) The mass of spray water required per m3 of air at room conditions.
(ii) The temperature to which the air must be heated.
Neglect the fan power. Assume that the total pressure is constant at 1.0132 bar

Given:
Td1 = 3°C,
∅1 = 100 %
Td3 = 22°C,
∅3 = 55 %

23 October 2020 Dr. R. Shanthi 40

Solution: Corresponding to Td1 = 3°C, from steam tables,
(i) Mass of spray water required pvs1 = 0.0076 bar
Corresponding to Td3 = 22°C, from steam tables, 𝑝𝑣1
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅1 =
pvs3 = 0.0264 bar 𝑝𝑣𝑠1
𝑝𝑣1
𝑝𝑣3 1=
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, ∅3 = 0.0076
𝑝𝑣𝑠3
𝒑𝒗𝟏 =?
𝑝𝑣3
0.55 = 𝒑𝒗𝟏 = 0.0076 bar
0.0264
𝑝𝑣1
𝒑𝒗𝟑 =? 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, 𝑊1 = 0.622
𝑝 − 𝑝𝑣1
𝒑𝒗𝟑 = 0.01452 bar 0.0076
𝑝𝑣3 𝑊1 = 0.622
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦, 𝑊3 = 0.622 1.0132 − 0.0076
𝑝 − 𝑝𝑣3 𝑾𝟏 =?
0.01452
𝑊3 = 0.622 𝑾𝟏 = 0.0047 kg/kg of dry air
1.0132 − 0.01452
𝑾𝟑 =? 𝑊3 − 𝑊1 = 0.00904 − 0.0047 =?
𝑾𝟑 = 0.00904 kg/kg of dry air 𝑾𝟑 − 𝑾𝟏 = 0.00434 kg/kg of dry air

23 October 2020 Dr. R. Shanthi 41

pa3 va3 = Ra Td3

𝑅𝑎 𝑇𝑑3 𝑅𝑎 × 𝑇𝑑3
𝑣𝑎3 = =
𝑝𝑎2 (𝑝 − 𝑝𝑣3 )

0.287 × 103 × (22 + 273)

=
(1.0132 − 0.01452) × 105
𝒗𝒂𝟑 =?
𝑣𝑎3 = 0.843 m3/kg of dry air

𝑊3 − 𝑊1 0.00434
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑝𝑟𝑎𝑦 𝑤𝑎𝑡𝑒𝑟 = = =?
𝑣𝑎3 0.843
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒑𝒓𝒂𝒚 𝒘𝒂𝒕𝒆𝒓 = 0.005124 kg moisture/m3

23 October 2020 Dr. R. Shanthi 42

Td1 = 3°C,
∅1 = 100 %
Td3 = 22°C,
∅3 = 55 %

(ii) The temperature to which the air must

be heated.
From Chart,
Td3 = 35°C,
∅𝟑 = 𝟏𝟒 % 3

1 2

Td3 = 22°C
23 October 2020 Td1 = 3°C
Dr. R. Shanthi 43
 Cooling and dehumidification
• The removal of water vapour from air is termed as dehumidification. Td3

• Dehumidification is possible only if the air is cooled below the dew point
temperature of the air.
• It is necessary to maintain the coil surface temperature below the dew
point temperature for effective dehumidification. This temperature is
called ‘apparatus dew point’ (ADP).
• Td1 – Temp entering the coil
• Td3 – Coil surface temperature
• Td2 – dew point temperature
• Under ideal conditions, the air leaving the coil will at condition ‘3’. No
cooling 100 % efficient. SO the condition of the air coming out of the coil
will be represented by point ‘4’ somewhere between ‘1’ and ‘3’ depending
upon the coil effectiveness
Td3 Td4 Td1

23 October 2020 Dr. R. Shanthi 44

Problem 11: It is required to design an air-conditioning system for an industrial process for the
following hot and wet summer conditions :
Outdoor conditions ...... 32ºC DBT and 65% R.H.
Required air inlet conditions ...... 25ºC DBT and 60% R.H.
Amount of free air circulated ...... 250 m3/min.
Coil dew temperature ...... 13ºC.
The required condition is achieved by first cooling and dehumidifying and then by heating.
Calculate the following :
(i) The cooling capacity of the cooling coil and its by-pass factor.
(ii) Heating capacity of the heating coil in kW and surface temperature of the heating coil if the
by-pass factor is 0.3.
(iii) The mass of water vapour removed per hour.
Solve this problem with the use of psychrometric chart.
Given:
Td1 = 32°C, ∅1 = 65 %
Td2 = 25°C, ∅2 = 60 %
ADP = 13°C
Amount of free air circulated Va= 250 m3/min
Solution:
From Chart,
23 October 2020 Dr. R. Shanthi 45
 Given:
Td1 = 32°C, ∅1 = 65 % h1 = 82 kJ/kg
W1 = 0.0197
Td2 = 25°C, ∅2 = 60 % kg/kg of dry air

ADP = 13°C 1

Solution:
h2 = 56 kJ/kg
Fromh1 Chart,
= 82 kJ/kg
h3 = 48 kJ/kg

h4 = 37 kJ/kg
h3 = 48 kJ/kg
3
2 W2 = 0.012kg/kg
h2 = 56 kJ/kg of dry air
𝑨𝑫𝑷= 13°C
h4 = 37 kJ/kg
4

3 2
4

Td4= 13°C
Td3= 17°C 0.892 m3/kg

23 October 2020 Dr. R. Shanthi 46

h1 = 82 kJ/kg
= 280.26 (82 – 48) = ?
W1 = 0.0197 kg/kg of dry air
= 9528.84 kJ/min
h2 = 56 kJ/kg
1 TR = 210 kJ/min
h4 = 37 kJ/kg
h3 = 48 kJ/kg 9528.84
The capacity of the cooling coil = =?
210
Td3 = 17°C
The capacity of the cooling coil = 𝟒𝟓. 𝟕𝟖 TR
W2 = W3 = 0.012 kg/kg of dry air
v1= 0.892 kg/m3 𝑻𝒅𝟑 − 𝑻𝒅𝟒 17 − 13
𝑩𝑭 = = =?
𝑉 𝑻𝒅𝟏 − 𝑻𝒅𝟒 32 − 13
The mass of air supplied per minute, 𝑚𝑎 = 𝑣𝑎
1
𝑩𝒚𝒑𝒂𝒔𝒔 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒄𝒐𝒐𝒍𝒊𝒏𝒈 𝒄𝒐𝒊𝒍, 𝑩𝑭 = 𝟎. 𝟐𝟏𝟏
250 (ii) Heating capacity of the heating coil in kW and
𝑚𝑎 = =? surface temperature of the heating coil if the by-
0.892
pass factor is 0.3.
𝒎𝒂 = 𝟐𝟖𝟎. 𝟐𝟔 𝒌𝒈/𝒎𝒊𝒏
𝑯𝒆𝒂𝒕𝒊𝒏𝒈 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝒎𝒂 × 𝒉𝟐 − 𝒉𝟑
(i) The cooling capacity of the cooling coil and its by-pass
factor = 280.26 × 56 − 48 = ?
The capacity of the cooling coil= 𝒎𝒂 (h1 –h3)
= 𝟐𝟐𝟒𝟐. 𝟎𝟖 𝒌𝑱/𝒎𝒊𝒏= 37.37 kW

23 October 2020 Dr. R. Shanthi 47

 𝑻𝒄 − 𝟐𝟓
𝑩𝒚𝒑𝒂𝒔𝒔 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒉𝒆𝒂𝒕𝒊𝒏𝒈 𝒄𝒐𝒊𝒍, 𝑩𝑭 =
𝑻𝒄 − 𝟏𝟕
𝑇𝑐 − 25
0.3 = =?
𝑇𝑐 − 17
𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒉𝒆𝒂𝒕𝒊𝒏𝒈 𝒄𝒐𝒊𝒍, 𝑻𝒄 = 𝟐𝟖. 𝟒𝟑°C
(iii) The mass of water vapour removed per hour.
𝑚𝑣 = 𝑚𝑎 𝑊1 − 𝑊2
= 280.26 0.0197 − 0.012
𝒎𝒗 =?
𝒎𝒗 = 𝟐. 𝟏𝟔 𝒌𝒈/ 𝐦𝐢𝐧 = 𝟏𝟐𝟗. 𝟒𝟖 𝒌𝒈/𝒉𝒓

23 October 2020 Dr. R. Shanthi 48