Processes and Properties of Dry Air
Interpreting the Air Conditioning Charts  Example 1: Reading Properties of Air  Example 2: Reading Properties of Air  Example 3: Heating Process  Example 4: Cooling and Dehumidifying Process  Example 5: Mixture of Air
Drybulb, wetbulb, dewpoint temperatures, relative humiditythese terms are so closely related that if two properties are known, all others shown in the figure below may be read from the chart. When air is saturated, drybulb, wetbulb, and dewpoint temperatures are identical. (See Example 2.)
Enthalpy of air for any given condition is the enthalpy at saturation corrected by the enthalpy deviation due to the air not being in a saturated state. The enthalpy (h) in Btu per pound of dry air is the enthalpy at the saturation h_{wb} plus the enthalpy deviation h_{d}. See Example 2.
h = h_{wb} + h_{d}
If the air's moisture content increases or decreases in a psychrometric process, the heat added (q) or removed (q) is the difference between the enthalpy of the final or leaving air h_{la} and the initial or entering air h_{ea} minus the enthalpy of the moisture (water in liquid or ice state) added h_{w} or rejected h_{w}.
q = h_{la} + h_{ea}  h_{w}
See Examples 4 and 5.
The enthalpy of added or rejected moisture is shown in the small graphs at the top of the chart.
Enthalpy of added or rejected moisture and enthalpy deviation are usually omitted in applications not requiring precise results  for example, comfort air conditioning. Errors due to omissions for wetbulb temperatures below 32°F are much larger than for omissions above 32°F.
Sensible heat factor. This is part of certain calculations for installing air conditioning equipment. A scale along the right side of the figure in Example 4 below used with an origin at 80° drybulb temperature and 50% RH provide a reasonable heat factor value. See Example 4.
Barometric pressures. In comfort air conditioning, a mercury reading of one inch or less either above or below the standard 29.92 inches of mercury is considered a standard reading.
When drybulb and dewpoint temperatures are known for air at nonstandard barometric pressures, values of percent RH and grains of moisture per cubic foot are correct on a standard chart. But for given drybulb and wetbulb readings at nonstandard barometric pressures, all properties must be corrected.
Interpreting the Air Conditioning Charts
Generally, in graphic presentations, humidifying is shown by an upward line and dehumidifying is shown by a downward line.
Heating and cooling air without changes in moisture content involve only a change in sensible heat and appear as a horizontal line, to right or left respectively. Changes occur in drybulb, wetbulb, RH, and enthalpy. Specific humidity and dewpoint temperature remain constant.
In heating and humidifying, both sensible heat and specific humidity increase  shown as a line sloping upward and to the right. Changes occur in drybulb, wetbulb, dew point temperatures, and enthalpy. A difference in RH depends on the slope of the line.

For cooling and dehumidifying, both sensible heat and specific humidity decrease, so the line slopes downward and to the left. Drybulb, wetbulb, dewpoint temperatures, and enthalpy all change. Changes in RH are dependent on the slope of the line.
Evaporative cooling refers to air brought in contact with spray water at a temperature equal to the wetbulb temperature of the air. The process takes place upward along the wetbulb line. As sensible heat of the initial air vaporizes the water, the air's drybulb temperature falls. The sensible heat used to vaporize the water enters the air as latent heat in added vapor; thus no heat is added or removed. Wetbulb temperature remains constant. Dewpoint temperature, RH, specific humidity, and enthalpy increase. (In most evaporative cooling installations, heat may be added or removed during the process due to outside sources, this amount is usually negligible.) 
In chemical dehydration, the air that contacts the chemical either adsorbs or absorbs moisture from the air. Thus in this energy constant process, heat is liberated and added to the airand this amount is basically equally to the latent heat of vaporization of the moisture removed. Indicated by a downward sloping line approximating the wetbulb line, the slope of the chemical dehydration line may be either slightly greater or less than the wetbulb line, depending on if heat is stored, liberated, or absorbed.
AIR CONDITIONING PROCCESSES such as heating, cooling, humidifying and dehumidifying may be shown graphically on the chart. See Figure 1.
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Example 1. Reading Properties Of Air
Given:{ 
DB = 70°F WB = 60°F 
Find:{ 
% RH DP Volume GR of moisture per lb dry air GR of moisture per cu ft 
Locate point of intersection on the chart of vertical line representing 700DB and oblique line representing 600WB. All values are read from this point of intersection.
Interpolate between relative humidity lines on 700DB line, read RH = 56%.
Follow horizontal line left to saturation curve, read DP = 53.6°F.
Interpolate between lines representing cubic feet per pound of dry air, read v = 13.53 cu ft.
Follow horizontal line to right, read grains of moisture per pound of dry air, W  61.4 gr.
Grains of moisture per pound of dry air (61.4) divided by cubic feet per pound of dry air (13.53) = 4.54 gr per cu ft
Example 2. Reading Properties of Air
Given:{ 
RH = 50% WB = 60°F 
Find:{ 
DB DP Gr. of moisture per lb. of dry air Enthalpy 

Locate point of intersection on the chart of 50% RH line and oblique line representing 600WB. All values are read from this point.
Follow vertical line downward to drybulb temperature scale, read DB = 71.9°F
Follow horizontal line left to saturation curve, read DP = 52.3°F.
Follow horizontal line to right, read grains of moisture per pound of dry air, W = 58.4 gr.
Follow wetbulb line to "Enthalpy at saturation" scale and read Hwb = 26.46 Btu. Read enthalpy deviation for point of intersection hd = .08 Btu.
Enthalpy of air at given condition
h = hwb + hd = 26.46 + (.08) = 26.38 Btu per Ib of dry air. 
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Example 3. Heating Process
Given:{ 
Initial Air:{ 
DB = 30°F RH = 80% 
Find:{ 
Final Air:{ 
%RH WB DP 

Air heated to 75° DB 
Heat Added 

Locate the condition of initial air on the chart. Follow horizontal line to 75°DB. Read: RH = 15%: WB = 51.5°F: DB = 25.2°F.
Exact Solution  Heat added:
Read enthalpy at saturation initial air h_{cwb} = 10.10 Btu Read enthalpy deviation initial air h_{cd}= .06 Btu Enthalpy of initial air h_{ca} = h_{cwb} + h_{cd} = 10.10 + .06 = 10.16 Btu Read enthalpy at saturation of final air h_{lwb} = 21.14Btu Read enthalpy deviation of final air h_{id} = 0.10 Btu Enthalpy of final air hla = hlwb + hld = 21 .14 + (0.10)= 21.04 Btu Heat added q = hla  hca = 21.04  10.16 = 10.88 Btu per Ib of dry air
Approximate Solution  Heat added:
q = h_{lwb}  h_{cwb} = 21.14 10.10 = 11.04 Btu per Ib of dry air.
The approximate solution is 1.5% higher than exact solution.
Example 4. Cooling and Dehumidifying Process
(a) Moisture rejected as water condensate
Given:{ 
Initial Air:{ 
DB = 83°F WB = 69°F 




Find:{ 
Heat removed Sensible Heat Factor 
Final Air:{ 
DB = 56°F WB = 55°F 







Condensate rejected at 55°F 
Locate initial and final conditions of air on chart.
Read: 
h_{cwb} = 33.25 Btu h_{lwb} = 23.22 Btu H_{cd} = 0.12 Btu h_{ld} = 0.01 Btu H_{ca} = 33.25+ (0.12) = 33.13 Btu h_{la} = 23.22 +(0.01)= 23.21 Btu 
Read grains of moisture in initial air W_{ca} = 84 Read grains of moisture in final air W_{Ia} = 63 w = W_{la}  W_{ca} = 63  84 = 21 (moisture rejected)
Read enthalpy of rejected moisture (hw) from diagrams at top of chart for 21 gr grains and 55 F = 0.08 Btu. Exact Solution  Heat removed: q = h_{la}  h_{ca}  h_{w} = 23.21  33.13  (0.08) = 9.84 Btu per Ib dry air.
Approximate Solution  Heat removed:q = h_{lwb}  h_{cwb} = 23.22  33.25 = 10.03 Btu per Ib dry air.
Approximate solution is 1.9% higher than exact solution.
To determine Sensible Heat Factor, draw a line between initial and final conditions. Draw a line parallel to this line from reference point (80 DB, 50 RH) to Sensible Heat Factor scale, read SHF = 0.68.
Example 5. Mixture Of Air
Given:{ 
Inside Air: 3 parts by weight: 
{ 
DB = 75°F WB = 62°F 
Find:{ 
Properties of Mixture 

Enter Air: 1 part by weight: 
{ 
DB = 95°F WB = 75°F 

Locate on chart conditions of inside and entering air. Draw line connecting two points. Measure off distance equal to 1/4 of line, starting from inside air condition. Point thus established represents condition of mixture of inside and entering air.

Read properties of mixture:
DB = 80°F, WB = 65.6°F, h = 30.50 + (0.11) = 30.39 Btu Moisture content (W) = 71.3 gr per Ib of dry air.
When air quantities being mixed are at widely different temperatures, the above method is slightly in error. For exact solution calculate properties of mixture on basis of specific humidity and enthalpy. 
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