Where q = quantity of air discharged in a given time, a = area of cross section of flue, v = velocity of current.
By means of this formula, the area of chimney required to discharge a given volume of air at a given average velocity can be ascertained. Thus—
a = q ∕ v.
The application of the preceding principles and formulæ will be rendered clearer by the following examples.
How much inlet and outlet area per head will be required to give 10 persons in a room of 5,000 cubic feet capacity, 2,000 cubic feet of air per head per hour, supposing that the outside temperature is 40°, while the internal temperature is 60°, and the height of the heated column of air 20 feet?
First ascertain the velocity of entrance and exit of air.
v = 8·2√(h(t - t1)/492)
= 8·2√(20(60 - 40)/492) = 8·2 × ·902.
= 7·3964 = velocity in feet per second.
If we allow one-fourth for friction, then there remains a velocity of 5·5473 feet per second.
5·5473 feet per second = 19700·8 feet per hour.
Now, a = q ∕ v
= 2,000 ∕ 19700·8 = ·1015 square feet.
= 14·6 square inches.
Thus the size of the outlet required per head is 14·6 square inches. The size of the room and the number occupying it do not enter into the question, except for a short time at the beginning. (See page [135].)