## Use Of Psychrometric Charts

Three charts for the air-water vapor system are given as Figs. 12-1 to 12-3 for low-, medium-, and high-temperature ranges. Figure 12-4 shows a modified Grosvenor chart, which is more familiar to the chemical engineer. These charts are for an absolute pressure of 1 atm. The corrections required at pressures different from atmospheric are given in Table 12-2. Figure 12-5 shows a psychrometric chart for combustion products in air. The thermodynamic properties of moist air are given in Table 12-1.

Examples Illustrating Use of Psychrometric Charts In these examples the following nomenclature is used:

t = dry-bulb temperatures, °F tw = wet-bulb temperature, °F td = dew-point temperature, °F H = moisture content, lb waterflb dry air

AH = moisture added to or rejected from the air stream, lb waterflb dry air h' = enthalpy at saturation, Btuflb dry air D = enthalpy deviation, Btuflb dry air h = h' + D = true enthalpy, Btuflb dry air hw = enthalpy of water added to or rejected from the system,

Btuflb dry air qa = heat added to the system, Btuflb dry air qr = heat removed from system, Btuflb dry air

Subscripts 1, 2, 3, etc., indicate entering and subsequent states.

Example 2: Determination of Moist Air Properties Find the properties of moist air when the dry-bulb temperature is 80°F and the wet-bulb temperature is 67°F.

Solution. Read directly from Fig. 12-2 (Fig. 12-6 shows the solution dia-grammatically).

Moisture content H = 78 grflb dry air

= 0.011 lb waterflb dry air Enthalpy at saturation h' = 31.6 Btuflb dry air Enthalpy deviation D = -0.1 Btuflb dry air True enthalpy h = 31.5 Btuflb dry air Specific volume v = 13.8 ft3flb dry air Relative humidity = 51 percent Dew point td = 60.3°F

Example 3: Air Heating Air is heated by a steam coil from 30°F dry-bulb temperature and 80 percent relative humidity to 75°F dry-bulb temperature. Find the relative humidity, wet-bulb temperature, and dew point of the heated air. Determine the quantity of heat added per pound of dry air. Solution. Reading directly from the psychrometric chart (Fig. 12-2),

Volume in Cubic Feet Per Pound of Dry Air

FIG. 12-1 Psychrometric chart—low temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogram-kelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

### Volume in Cubic Feet Per Pound of Dry Air

FIG. 12-1 Psychrometric chart—low temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogram-kelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-2 Psychrometric chart—medium temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogram-kelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

Relative humidity = 15 percent Wet-bulb temperature = 51.5°F Dew point = 25.2°F

The enthalpy of the inlet air is obtained from Fig. 12-2 as h = h1 + Di = 10.1 + 0.06 = 10.16 Btu/lb dry air; at the exit, h = h2 + D2 = 21.1 - 0.1 = 21 Btu/lb dry air. The heat added equals the enthalpy difference, or qa = Ah = h2 - h = 21 - 10.16 = 10.84 Btu/lb dry air

If the enthalpy deviation is ignored, the heat added qa is Ah = 21.1 - 10.1 = 11 Btu/lb dry air, or the result is 1.5 percent high. Figure 12-7 shows the heating path on the psychrometric chart.

Example 4: Evaporative Cooling Air at 95°F dry-bulb temperature and 70°F wet-bulb temperature contacts a water spray, where its relative humidity is increased to 90 percent. The spray water is recirculated; makeup water enters at 70°F. Determine exit dry-bulb temperature, wet-bulb temperature, change in enthalpy of the air, and quantity of moisture added per pound of dry air.

Solution. Figure 12-8 shows the path on a psychrometric chart. The leaving dry-bulb temperature is obtained directly from Fig. 12-2 as 72.2°F. Since the spray water enters at the wet-bulb temperature of 70°F and there is no heat added to or removed from it, this is by definition an adiabatic process and there will be no change in wet-bulb temperature. The only change in enthalpy is that from the heat content of the makeup water. This can be demonstrated as follows:

Inlet moisture H\ = 70 gr/lb dry air Exit moisture H2 = 107 gr/lb dry air AH = 37 gr/lb dry air Inlet enthalpy h = h1 + D1 = 34.1 - 0.22

= 33.88 Btu/lb dry air Exit enthalpy h = h'2 + D2 = 34.1 - 0.02 = 34.08 Btu/lb dry air

Then

Enthalpy of added water hw = 0.2 Btu/lb dry air (from small diagram, 37 gr at 70°F)

Example 5: Cooling and Dehumidification Find the cooling load per pound of dry air resulting from infiltration of room air at 80°F dry-bulb temperature and 67°F wet-bulb temperature into a cooler maintained at 30°F dry-bulb and 28°F wet-bulb temperature, where moisture freezes on the coil, which is maintained at 20°F.

Solution. The path followed on a psychrometric chart is shown in Fig. 12-9.

Inlet enthalpy h = h1 + D1 = 31.62 - 0.1 = 31.52 Btu/lb dry air Exit enthalpy h2 = h2 + D2 = 10.1 + 0.06 = 10.16 Btu/lb dry air Inlet moisture H1 = 78 gr/lb dry air Exit moisture H2 = 19 gr/lb dry air Moisture rejected AH = 59 gr/lb dry air Enthalpy of rejected moisture = -1.26 Btu/lb dry air (from small diagram of Fig. 12-2) Cooling load qr = 31.52 - 10.16 + 1.26 = 22.62 Btu/lb dry air

Note that if the enthalpy deviations were ignored, the calculated cooling load would be about 5 percent low.

Example 6: Cooling Tower Determine water consumption and amount of heat dissipated per 1000 ft3/min of entering air at 90°F dry-bulb temperature and 70°F wet-bulb temperature when the air leaves saturated at 110°F and the makeup water is at 75°F.

Solution. The path followed is shown in Fig. 12-10.

FIG. 12-3 Psychrometric chart—high temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogramkelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-3 Psychrometric chart—high temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogramkelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-4 Humidity chart for air-water vapor mixtures. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogramkelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-4 Humidity chart for air-water vapor mixtures. To convert British thermal units per pound to joules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules per kilogramkelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-5 Revised form of high-temperature psychrometric chart for air and combustion products, based on pound-moles of water vapor and dry gases. [Hatta, Chem. Metall. Eng., 37, 64 (1930).]

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### Responses

• Neftalem
How to master the use of the psychrometric chart?
7 years ago