## Rule 6

Ascertain the weight of moisture in the air before it is heated, and compute the weight of moisture required to produce the desired degree of humidity in the same weight of air at the temperature at which it is to be used; the difference between the quantities of moisture thus found will be the amount of moisture to be supplied.

## Example

A certain room is supplied with air having a temperature at the registers of 120°, at the rate of 300 cubic feet per minute. The temperature of the room is to be maintained at 70°, and the humidity at 70 per cent. The temperature of the air before entering the heater is 45°, and its humidity is also 70 per cent. What weight of moisture must be supplied to the air-current to secure the desired humidity in the room?

## Solution

It is necessary to know the volumes of the air at the time it is used and before it enters the heater, and these must be computed from the volume and temperature at the register as given. The original volume is 300 cubic feet, and the original temperature is 120° (580° absolute). Applying rule 4, to find the volume at 70° (530° absolute),

 V1 = VT1 = 300 X 530 = 274.1 cu. ft.; T 580

and at 45° (505° absolute),

 V1 = VT1 = 300 X 505 = 261.2 cu. ft. T 580

Thus, at the beginning we have 261.2 cubic feet of cold air at 70 per cent. humidity. The weight of that volume of steam at 45° is, from Table 8, 261.2 X .00048 = .1254 pound, and 70 per cent. of this equals .1254 X .70 = .08778 pound, which is the weight of moisture originally contained in the air.

The air when used will have a volume of 274.1 cubic feet and a temperature of 70°. The weight of an equal volume of steam at 70° is 274.1 X .00115 = .3152 pound. The humidity is required to be 70 per cent.; therefore, the total moisture required will be .3152 X.70 = .22064 pound.

Then, .22064 - .08778 = .13286 pound is the amount of moisture, per minute, that must be added to the air-current. Ans.