Unit - 3
Properties of dry and wet air
Q1) What is dry air? What are various important properties of dry air?
A1) Dry Air:
Air which does not contain any moisture is called Dry Air.
Properties like
- Specific heats
- Ratio of specific heats
- Dynamic viscosity
- Thermal conductivity
- Prandtl number
- Density
- Kinematic viscosity
- Thermal diffusivity
Of dry air vary at different temperatures and pressures.
Q2) What is wet air? Explain any three of its properties in detail.
A2) Wet Air or Moist Air:
Air along with water vapour content, is called as Wet or Moist Air.
Properties of Wet Air:
Dry-Bulb Temperature:
Dry bulb temperature is measured with a normal thermometer.
Dew Point:
Dew point is the temperature at which water vapor begins to condense out of the air.
Wet-Bulb Temperature:
Wet bulb temperature is relatively easy to measure, but requires special equipment.
A normal thermometer is fixed to a rotating sling.
The bulb of the thermometer is covered with a wet wick of cotton.
The sling is rotated till the temperature reading is stabilized.
The air passes over the wet wick and cools down before reaching the bulb of the thermometer.
This temperature is called as wet bulb temperature.
Wet bulb temperature will never be higher than dry bulb temperature.
Wet bulb temperature is a critical parameter for sizing, and measuring the performance of evaporative-cooled cooling water systems.
Relative Humidity:
Relative humidity is the ratio of the water vapor actually present in the air, to the maximum amount that the air can hold at that particular temperature and pressure.
Density:
Density is the total mass of moist air per unit volume.
Q3) Explain the various lines present on the psychrometric chart.
A3) All the values represented in the chart are calculated at standard atmospheric pressure.
The horizontal line at the bottom represents the scale of Dry bulb temperature.
The vertical lines in the chart show the various Dry bulb temperatures.
Horizontal lines emanating from the extreme right vertical line represent the constant moisture values.
The diagonal lines emerging from the curve on the left side represent the Wet Bulb temperature lines.
Q4) Explain the process of Sensible heating.
A4) Sensible heating:
Sensible heating refers to addition of heat at constant humidity, that is without any change in moisture content of air.
Q = m Cp ΔΤ
The process shown in the above figure shows the increase in temperature at constant humidity ratio (HR). There is increase in enthalpy of the system due to heat addition. The RH of the system increases.
Q5) Explain the process of Sensible cooling.
A5) Sensible Cooling:
Sensible cooling refers to removal of heat at constant humidity, that is without any change in moisture content of air.
The process on the psychometric chart would be exactly opposite to the first one, that is, it will follow the path 2-1. The enthalpy of the system would decrease at constant HR but decrease in RH.
Q6) Explain the process of Cooling and dehumidification.
A6) Cooling and Dehumidification:
This process is removal of heat as well as moisture from the air. It involves removal of sensible as well as latent heat from the air.
The above figure shows the dehumidifying cooling process.
Ideally the process should be combination of two stages.
In the first stage, the sensible heat is removed from the air till dew point temperature is reached.
In the second stage, the latent heat is removed from the air till the desired temperature is achieved.
But practically, the process takes place in a single process as shown in the graph.
Q7) Explain the process of Heating and Humidification.
A7) Heating and Humidification:
This process is addition of heat as well as moisture to the air. It involves addition of sensible as well as latent heat to the air.
The above figure shows the humidifying heating process.
Ideally the process should be combination of two stages.
In the first stage, the latent heat is added to the air till the desired temperature is achieved.
In the second stage, the sensible heat is added to the air till dew point temperature is reached.
But practically, the process takes place in a single process as shown in the graph.
Q8) What is Dew point?
A8) Dew point temperature (DPT) or simply dew point is the temperature at which the water vapour present in the air starts to condense to form water droplets.
DPT varies with pressure.
It can be computed easily with the help of psychrometric chart.
The left most curve represents the Dew point temperature.
If any other two properties are known for the air, DPT can be easily found from the chart.
Q9) Atmospheric air at 1.0132 bar has a DBT of 30°C and a WBT of 25°C. Compute:
- The partial pressure of water vapour,
- The specific humidity,
- The dew point temperature,
- The relative humidity,
- The degree of saturation,
- The density of air in the mixture,
- The density of vapour in the mixture and
- The enthalpy of the mixture. Use thermodynamic table.
A9)
Given data:
Dry bulb temperature, (td1) = 30oC
Wet bulb temperature, (td2) = 25oC
Atmospheric pressure, (pb) = 1.0132 bar
- Humidity ratio, = 0.622 pv/(pb – pv)
Partial pressure vapour, pv = psw – 
Where,
psw – saturation pressure corresponding to WBT
pb – barometric pressure = 1 bar
td – Dry bulb temperature
tw – Wet bulb temperature
From steam table, for 25oC WBT, corresponding pressure is 0.03166 bar
i.e. psw = 0.03166 bar
pv = 0.0336 – 
Partial pressure of vapour,
pv = 0.0296 bar
Substituting pv and pb value in equation ,
= 0.622 0.0296/(1 – 0.0296)
Specific humidity, = 0.0189 kJ/kg of air
2. Dew point te,perature (tdp)
From steam table, we find that partial vapour pressure, pv = 0.03 bar,
Corresponding temperature is 24.1oC
So, the dew point temperature tdp = 24.1oC
3. Relative humidity
∅ = pv/ps
Where, ps – Saturation pressure corresponding DBT
From steam table,
For 30oC DBT, corresponding pressure is 0.04242 bar
i.e. ps = 0.04242 bar
∅ = 0.0296/0.04242
∅ = 0.6977 = 69.77 %
4. Degree of saturation, μ = pv/ps (pb- ps)/(pb – pv)
μ = 
μ = 0.6886
5. Specific volume (va) of air:
From gas law,
pava = RaTa
Where,
Gas constant, R = 0.287 kJ/kg
Ta = td + 273 = 30 + 273 = 303 K
pa = pb – pv = 1.013 – 0.0296
pa = 0.9834 bar
va = RaTa/pa = 
= 88.428 m3/kg
Density of air,
a = 1/va = 1/88.428 = 0.011kg/m3
6. Density of water vapour (v)
From steam table, corresponding to 30oC DBT, specific volume (vg) is
32.929 m3/kg
v = 1/vg = 1/32.929 = 0.0303 kg/m3
Vapour density of relative humidity,
∅ = 0.6977
Vapour density, v = 0.0303 0.6977
v = 0.0211 kg/m3
7. Total enthalpy
Enthalpy, h = Cptd + hg
Where, Cp – Specific heat = 1.005 kJ/kg K
hg – Specific enthalpy of air corresponding DBT
From steam table,
For 30oC DBT, corresponding specific enthalpy is 2430.7 kJ/kg
h = (1.005 30) + (0.0189 2430.7)
h = 76.09 kJ/kg
Q10) Calculate the sensible heat gain of a ventilation flow rate of 10,000 cfm when the temperature (to) of outside air is 88oF and inside air temperature (ti) is 78oF.
A10)
Qsensible = 1.10 x cfm x (to – ti)
Qsensible = 1.10 x 10000 x (88 – 78)
Qsensible = 110,000 Btu/h
Hence, the sensible heat gain is 110,000 Btu/h.
