Relative humidity = 15 percentWet-bulb temperature = 51.5F

Dew point = 25.2F

The enthalpy of the inlet air is obtained from Fig. 12-2 as h1 = h1 + D1 =10.1 + 0.06 = 10.16 Btu/lb dry air; at the exit, h2 = h2 +D2 = 21.1 0.1 = 21 Btu/lbdry air. The heat added equals the enthalpy difference, or

qa = h = h2 h1 = 21 10.16 = 10.84 Btu/lb dry airIf the enthalpy deviation is ignored, the heat added qa is h = 21.1 10.1 = 11Btu/lb dry air, or the result is 1.5 percent high. Figure 12-7 shows the heatingpath on the psychrometric chart.

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

Solution. Figure 12-8 shows the path on a psychrometric chart. The leavingdry-bulb temperature is obtained directly from Fig. 12-2 as 72.2F. Since thespray water enters at the wet-bulb temperature of 70F and there is no heatadded to or removed from it, this is by definition an adiabatic process and therewill be no change in wet-bulb temperature. The only change in enthalpy is thatfrom the heat content of the makeup water. This can be demonstrated as fol-lows:

Inlet moisture H1 = 70 gr/lb dry airExit moisture H2 = 107 gr/lb dry air

H = 37 gr/lb dry airInlet enthalpy h1 = h1 + D1 = 34.1 0.22

= 33.88 Btu/lb dry airExit enthalpy h2 = h2 + D2 = 34.1 0.02

= 34.08 Btu/lb dry air

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

Then qa = h2 h1 + hw= 34.08 33.88 + 0.2 = 0

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

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

Inlet enthalpy h1 = 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 airExit moisture H2 = 19 gr/lb dry air

Moisture rejected H = 59 gr/lb dry airEnthalpy 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 loadwould be about 5 percent low.

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

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

PSYCHROMETRY 12-5

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

12-6 PSYCHROMETRY, EVAPORATIVE COOLING, AND SOLIDS DRYING

FIG. 12-3 Psychrometric charthigh temperatures. Barometric pressure, 29.92 inHg. To convert British thermal units per pound tojoules 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.

PSYCHROMETRY 12-7

FIG. 12-4 Humidity chart for air-water vapor mixtures. To convert British thermal units per pound to joules per kilo-gram, 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. 12-5 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).]

12-28 PSYCHROMETRY, EVAPORATIVE COOLING, AND SOLIDS DRYING

FIG. 12-36 Psychrometric chart: properties of air and water-vapor mixtures from 20 to 120C. (Carrier Corp.)

where hc = heat-transfer coefficient by convection, J/(m2sK) [Btu/(hft2F)]; t = air temperature, K; tw = wet-bulb temperature of air, K;kg = mass-transfer coefficient, kg/(sm2) (kg/kg) [lb/(hft2)(lb/lb)]; =latent heat of evaporation at tw, J/kg (Btu/lb); Hw = saturated humidityat tw = kg/kg of dry air; and Ha = humidity of the surrounding air, kg/kgof dry air.

For airwater-vapor mixtures, it so happens that hc /kg = Cs approx-imately, although there is no theoretical reason for this. Hence, sincethe ratio (Hw Ha)/(tw t) equals hc /kg /, which represents the slopeof the wet-bulb-temperature lines, it is also equal to Cs /, the slope ofthe adiabatic-saturation lines as shown previously.

A given humidity chart is precise only at the pressure for which it is evaluated. Most airwater-vapor charts are based on a pressure of 1 atm. Humidities read from these charts for given values of wet- anddry-bulb temperature apply only at an atmospheric pressure of 760mmHg. If the total pressure is different from 760 mmHg, the humid-ity at a given wet-bulb and dry-bulb temperature must be correctedaccording to the following relationship.

Ha = Ho + 0.622pw (12-23)where Ha = humidity of air at pressure P, kg/kg of dry air; Ho = humid-ity of air as read from a humidity chart based on 760-mm pressure atthe observed wet- and dry-bulb temperatures, kg/kg dry air, pw =vapor pressure of water at the observed wet-bulb temperature,mmHg; and P = the pressure at which the wet- and dry-bulb readingswere taken. Similar corrections can be derived to correct specific vol-ume, the saturation-humidity curve, and the relative-humidity curves.

HUMIDITY CHARTS FOR SOLVENT VAPORS

Humidity charts for other solvent vapors may be prepared in an anal-ogous manner. There is one important difference involved, however,in that the wet-bulb temperature differs considerably from the adia-batic-saturation temperatures for vapors other than water.

1760 pw

1P pw

Figures 12-37 to 12-39 show humidity charts for carbon tetrachlo-ride, benzene, and toluene. The lines on these charts have been cal-culated in the manner outlined for airwater vapor except for thewet-bulb-temperature lines. The determination of these linesdepends on data for the psychrometric ratio hc /kg, as indicated by Eq.(12-22). For the charts shown, the wet-bulb-temperature lines arebased on the following equation:

Hw H = (hc /w kg)(t tw) (12-24)where = radiation correction factor, a value of 1.06 having been usedfor these charts. Values of hc /kg, obtained from values of hc /kgCs aspresented by Walker, Lewis, McAdams, and Gilliland (Principles ofChemical Engineering, 3d ed., McGraw-Hill, New York, 1937), whereCs = humid heat of air with respect to the vapor involved, are as fol-lows:

CarbonMaterial tetrachloride Benzene Toluene

hc /kgCs 0.51 0.54 0.47

A discussion of the theory of the relationship between hc and kg maybe found in the psychrometry part of this section. Because both theo-retical and experimental values of hc /kg apply only to dilute gas mix-tures, the wet-bulb lines at high concentrations have been omitted.For a discussion of the precautions to be taken in making psychro-metric determinations of solvent vapors at low solvent wet-bulb tem-peratures in the presence of water vapor, see the paper by Sherwoodand Comings [Trans. Am. Inst. Chem. Eng., 28, 88 (1932)].

GENERAL CONDITIONS FOR DRYING

Solids drying encompasses two fundamental and simultaneousprocesses: (1) heat is transferred to evaporate liquid, and (2) mass is

SOLIDS-DRYING FUNDAMENTALS 12-29

FIG. 12-37 Humidity chart for air-carbon tetrachloride vapor mixture. To convert British thermal units per pound tojoules per kilogram, multiply by 2326; to convert British thermal units per pound dry air-degree Fahrenheit to joules perkilogram-kelvin, multiply by 4186.8; and to convert cubic feet per pound to cubic meters per kilogram, multiply by 0.0624.

FIG. 12-38 Humidity chart for air-benzene-vapor mixture. To convert British thermal units per pound to joules per kilo-gram, 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. 12-39 Humidity chart for air-toluene-vapor mixture. To convert British thermal units per pound to joules per kilo-gram, 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.

12-30

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Prepared by a staff of specialists under the editorial direction of

Late EditorRobert H. Perry

EditorDon W. Green

Deane E. Ackers Professor of Chemicaland Petroleum Engineering,

University of Kansas

Associate EditorJames O. Maloney

Professor Emeritus of Chemical Engineering,University of Kansas

PERRYSCHEMICAL

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Perrys chemical engineers handbook. 7th ed. / prepared by a staffof specialists under the editorial direction of late editor Robert H. Perry : editor, Don W. Green : associate editor, James OHara Maloney.

p. cm.Includes index.ISBN 0-07-049841-5 (alk. paper)1. Chemical engineeringHandbooks, manuals, etc. I. Perry,

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Figure 12-4: Figure 12-4Figure 12-5: Figure 12-5Figure 12-37: Figure 12-37Figure 12-38: Figure 12-38Figure 12-39: Figure 12-39