**Introduction:**Atmospheric air contains mostly a mixture of dry air and water vapour with some amount of other gases. Psychrometry deals with determination of thermodynamic properties of moist air and utilisation of these properties in analysis of conditions and processes including moist air.**Properties of Moist Air:**Almost all air contains some moisture but the actual amount of water vapour which the air can hold is governed by mixture temperature. Moisture content required to saturate the air is very little at low temperature but it is very high once the temperature is high. Some general terminologies used for moist air are defined below:2.1. **DRY BULB TEMPERATURE (DB): .**Temperature of air registered by an ordinary thermometer.2.2. **WET BULB TEMPERATURE (WB):**Temperature recorded by a thermometer whose bulb is covered with wet wick and placed in the rapidly moving air stream. It is a measure of temperature at which water evaporates into the air to bring it to saturation adiabatically at exactly the same temperature and pressure.The amount of water that can evaporate from the wet wick into the air depends upon the difference between the amount of water vapour which it can hold when fully saturated. If the air flowing past the wet bulb were already saturated with moisture, no water could evaporate from the bulb into the air and there would be no cooling of the wet bulb thermometer. The drier the air flowing past the wick of the thermometer, the larger will be the moisture that will evaporate into the air steam, the greater will be the latent heat absorbed, the lower will be the reading of the wet bulb thermometer. So wet bulb temperature will always be lower or equal to dry bulb temperature.

2.3. **HUMIDITY RATIO :**Ratio of weight of water vapour to the weight of dry air contained in the sample.w = m

_{w}/ m_{a}2.4. **DEGREE OF SATURATION:**It is the ratio of air humidity ratio to the humidity ratio of saturated air at same temperature and pressure.h = w / w

_{s}2.5. **SPECIFIC HUMIDITY :**Ratio of weight of water vapour to the total weight of the moist air, i.e.q

_{s}= m_{w}/ (m_{w}+ m_{a})2.6. **ABSOLUTE HUMIDITY :**Ratio of weight of water vaopur to the total volume i.e.q

_{a}= m_{w}/ v2.7. **RELATIVE HUMIDITY :**Ratio of the actual water vapour pressure of air to the saturated water vapour pressure of the air at same temperature.2.8. **SPECIFIC VOLUME :**Volume of air per unit weight of dry air. ( defined as v / m_{a})2.9. **ENTHALPHY :**Total quantity of heat i.e. sum of sensible heat and latent heat above an arbitrary datum. The datum for dry air is 00 F and, for the moisture content 320 F water.Enthalpy of moist air = Enthalpy of air + Enthalpy of water vapour i.e. h = h

_{a}+ m_{w}x h_{g}where h

_{a}is the enthalpy of dry air at temperature t (In Kilo Joules per Kg) h_{g}is specific enthalpy of saturated water vapour at the temperature of the nixture and = 2501 + 1.805 t KJ/KgTotal enthalpy, H, represents, the enthaphy of m kilograms & specific enthalpy, h, is enthalpy of one kilogram.

2.10. **SUPER HEAT:**Actual temperature – Saturated temperature at the presure of gas.2.11. **DEW POINT TEMPERATURE :**The temperature at which condensation of moisture begins when the air is cooled. The dew point temperature is very important in the design of AC system:- Temperature od supply air should be less than dew point (DP) of outdoor otherwise condensation starts at the air surface of the duct and to avoid this we may have to insulate duct and pipe.
- Apparatus Dew Point (ADP): As temperature of cooling coil is equal to DP, so if condensation is there, heat exchange will not be as good compared to when condensation is not there.

The dry bulb, Wet bulb, and dew point temperatures and the relative humidity are so connected that given any two properties all other can be determined. Also when air is saturated, D.B.T., W.B.T. and dew point temperatures are equal.

**Psychometric Chart:**A Graphical representation of various properties of air, in which various properties of moist air are so related that if two properties are known all other properties may then be determined easily.Steps to plot psychometric chart:

- Choose X-axis to represent D.B. temperature and Y-axis for moisture contents.
- Plot saturated points of air with various temperatures. We thus plot saturation lines.
- Draw R.H. lines by dividing vertical coordinates into various parts since R.H. is the ratio of actual water vapour pressure of air to the saturated water vapour pressure of the air at same temperature.
- Dew points will also represent W.B. and thus we have one point on W.B. line. now let us dehumidify the air chemically which is an adiabatic process, and measure state of air and plot it. It will indicate a constt. W.B. line. Since the process is adiabatic and humidity ratio decreases, the temperature of air must increase.
- The total heat of the air is a function of W.B.T. remains constant the enthalpy does not change. This is used for drawing constant enthalpy lines.
- Specific volume lines: As Pv = RT, so v = R/p × T; so for a given volume, T and P can be found out and by above relation specific volume can be determined.

Psychometric chart can indicate following properties :

- Dry bulb temperature.
- Wet bulb temperature.
- Relative humidity.
- Total heat of air.
- Sensible heat of air.
- Latent heat of air.
- Dew point of air.
- Specific volume of air.
- Moisture content per unit weight of air, etc.

An example of this is given in table below. Here figures in ittalics are calculated from the chart:

DB (°F) WB (°F) RH (%) Sp. Vol of air (ft³/lb _{a})Moisture content (grains/lb _{a})Enthalpy (Btu/lb _{a})*80*66.6 *50*13.8 80 31.5 *85**74*60 14.1 110 37.7 *95*75 40 14.3 *100*38.6 *100*71 24 14.3 67 *34.9**90*64 23 *14*48 29.3 *75**75*100 13.8 132 38.6 94 *70**30*14.2 72 34.1 65 *65*100 *13.5*92 30.0 83 *70*54 14.0 *90*34.1 83 72.2 *60**14*102 36.0 86 69 44 *14**80*33.4 84 70 *50*14.0 88 *34.18***Processes on Psychometric Chart:**From any state on the psychometric chart, a process of simple heating (at constant humidity) would be represented by horizontal movement to the right and thus an increase in the sensible heat at constt. latent heat. The reverse movement is thus a simple cooling process. Simple humdification (at constant temperature) would be represented by vertical movement upward and thus an increase in latent heat with no change in sensible heat. Accordingly, vertical movement downward gives rise to decrease in latent heat but no change in sensible heat and the process is called simple dehumidification.#### Note

Total heat of air = Sensible heat of air + Sensible heat of water vapour + Latent heat of water vapour + Super heat of water vapour

**Sensible Heating of Moist Air:**A case of horizontal movement towards right on the psychrometric chart. For steady flow conditions, the required rate of heat addition is -q = w

_{a}(h_{2}- h_{1})**Cooling an Dehumidification of Moist Air:**When moist air is cooled to a temperature below its initial dew point separation of moisture takes place. Though water may get separated at various temperatures ranging from the initial dew point to the final saturation temperature, yet it is assumed that condensed water is cooled to the final air temperature before it is drained from the system.Entry balance equation is as given below :

m

_{a}h_{1}= m_{a}h_{2}+ q + m_{w}hw_{2}m

_{a}w_{1}= m_{a}w_{2}+ m_{w}Thus m

_{w}= m_{a}(w_{1}- w_{2})q = m

_{a}[(h_{1}- h_{2}) - (w_{1}- w_{2}) hw_{2}]**Adiabatic Mixing of two streams of Moist Air :**A most common process involved in the A/C system for which energy balance equation is as given below :ma

_{1}h_{1}+ ma_{2}h_{2}= ma_{3}h_{3}ma

_{1}+ ma_{2}= ma_{3}ma

_{1}w_{1}+ ma_{2}w_{2}= ma_{3}w_{3}On eliminating ma

_{3}we get(h

_{2}- h_{3})/(h_{3}- h_{1}) = (w_{2}- w_{3})/(w_{3}- w_{1}) = ma_{1}/ma_{2}

**Example:**A stream of 5000 cfm of outdoor air at 40 °F D.B.T. and 35 °F W.B.T. is adiabatically mixed with 15000 cfm of recirculating air at 75 °F and 50 % R.H. Find the D.B.T. and W.B.T. of the resulting mixture.**Solution:**V_{1}= 12.65 cft/1ba, V_{2}= 13.68 cft/1bama

_{1}= 5000/12.65 = 395 lb/minma

_{2}= 15000/13.68 = 1096 lb/minNow, ma

_{3}= ma_{2}+ ma_{1}= 1096 + 395 = 1491ma

_{2}/ ma_{3}= 1096 / 1491 = 0.735Thus the length of the segment is 0.735 times the entire length. Values of D.B.T. and W.B.T. and W.B.T. can be read on the chart as 65.92 °F and 55.21 °F respectively.

[DBT = (395 × 40 + 1096 × 75)/(1096 + 395) = 65.728 °F]

[WBT = (395 × 35 + 1096 × 625)/(1096 + 395) = 55.21 °F]

**Example:**Moist air saturated at 35 °F enters a heating coil at a rate of 20 000 cfm. Air leaves the coil at 100 °F. Find the required rate of heat addition in Btu/hr and capacity of heater in kW.**Solution:**For moist air at 35 °F, h_{1}= 13.11 Btu/lb dry airV

_{1}= 12.55 ft³/lb dry air, W_{1}= 0.00428 1b/1b airAt 100 °F (move horizontally from 35 °F W.B.T. and find the coordinate of 35 °F W.B.T. and 100 °F D.B.T.). Read h2.

Thus h

_{2}= 28.77 Btu/lbq = (20 000 × 60)/12.55 × (28.77 - 13.01) = 1 507 000 Btu/h

Heater capacity = 1 507 000/3410 = 44 193 kW

**Example:**Moist air at 85 °F D.B.T. and 50 % R.H. enters a cooling coil at 10 000 cfn and is processed to a final condition of saturation at 50 °F. Find the tonnes of refrigeration required.**Solution:**h_{1}= 34.62 Btu/lb, w_{1}= 0.01292 1bw / 1ba, V=14.01 cft/1bah

_{2}= 20.30 Btu/1b, w_{2}= 0.00766 1bw/1baH

_{w}= 8.4 Btu/1bwm

_{a}= 10000/14.01 = 7138 lb dry air / minuteq = 713.8 [(34.62 - 20.30) - (0.01292 - 0.00766) 8.4] = 10190 x 60 Btu/h = 50.95 T.R.

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