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19.18 SECTION NINETEEN

FIGURE 19.9 Humidity chart (air-water at 1 atm). ( From McCabe and SmithUnit Operations of Chemical Engineering .)

Calculation Procedure

1. Find the humidity. In Fig. 19.9, the humidity is the ordinate (along the right side of the graph)of the point on the saturation line (the 100 percent humidity line) that corresponds to the dew point,the latter being read from the abscissa along the bottom. In the present case, the humidity is found tobe 0.011 lb water per pound of dry air (0.011 kg water per kilogram of dry air).

2. Find the percentage humidity. Find thedry-bulb temperature, that is, 130 F, along the abscissa,erect a perpendicular to intersect the 0.011-lb humidity line (at point A in Fig. 19.9), and nd thepercentage-humidity line (interpolating a line if necessary) that passes through that intersection. Inthis case, the 10 percent line passes through, so the percentage humidity is 10 percent.

3. Find the adiabatic-saturation temperature. Find the adiabatic-cooling line (these are thestraight lines having negative slope) that passes through point A, interpolating a line if necessary,and read the abscissa of the point (point B) where this line intersects the 100 percent humidity line.This abscissa is the adiabatic-saturation temperature. In the present case, it is 80 F (300 K).

4. Find the humidity at adiabatic saturation. The humidity at adiabatic saturation is the ordinate,along the right side of the graph, of point B. Its value is 0.022 lb water per pound of dry air (0.022 kg

water per kilogram of dry air).

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OTHER CHEMICAL ENGINEERING CALCULATIONS 19.19

5. Find the humid heat. Find the intersection (point C ) of the 0.011-lb-humidity line with thehumid-heat-versus-humidity line, and read the humid heat as the abscissa of point C along the top of the graph. This abscissa is 0.245 Btu/( F)(lb dry air) [or 1024 J/(K)(kg dry air)].

6. Find the humid volume. Erect a perpendicular through the abscissa (along the bottom of thegraph) that corresponds to 130 F, thedrybulb temperature. Label the intersection of this perpendicularwith the saturated-volume-versus-temperature lineas point D, andthe intersectionof theperpendicularwith the specic-volume-dry-air-versus-temperature line as point E . Then, along line ED , nd pointF by moving upward from point E by a distance equal to

E D [(percentage humidity) / 100]

or, in the present case ( E D )(10 / 100),where E D is the length of line segment E D . The humid volumeis the ordinate of point F as read along the left side of the graph. In this case, the humid volume is15.1 ft 3 /lb dry air (0.943 m 3 /kg dry air).

Related Calculations. Do not confuse percentage humidity with relative humidity. Relative hu-midity is the ratio of the partial pressure of the water vapor to the vapor pressure of water at thetemperature of the air, this ratio usually being expressed as a percent. Percentage humidity is theratio of the actual humidity to the saturation humidity that corresponds to the gas temperature, whichis also usually expressed as a percent. At all humidities other than 0 or 100 percent, the percentagehumidity is less than the relative humidity.

19.11 BLOWDOWN AND MAKEUP REQUIREMENTS FOR COOLING TOWERS

A cooling tower handles 1000 gal/min (0.063 m 3 /s) of circulating water that is cooled from 110 to80 F (316 to 300 K). How much blowdown and makeup are required if the concentration of dissolvedsolids is allowed to reach three times the concentration in the makeup?

Calculation Procedure

1. Set out material-balance equations for the cooling tower. When the system is at equilibrium,the makeup must equal the losses, so, by denition,

M = E + B + W (19.1)

where M is makeup, E is evaporation loss, B is blowdown, and W is windage loss, all being expressedas percent of circulation.Since the evaporation water will be essentially free of dissolved solids, all solids introduced with

the makeup water must be removed by the blowdown plus windage loss, or

Mpm = ( B + W ) pc

where pm is concentration of solids in the makeup and pc is concentration of solids in the circulatingwater, both in parts per million.

For cooling towers, the concentration in the recirculating water is arbitrarily dened as cycles of concentration C , namely, C = (concentration in cooling water) / (concentration in makeup water).

Adapted from Chemical Engineering , June 21, 1976.