The loss by friction in two similar tubes of equal sectional area varies (1) directly with the square of the velocity of the air currents; and (2) directly with the length of the outlet channel. In two similar tubes of unequal size the loss by friction is (3) inversely as the diameter of the cross-section in each. When two tubes are of different shapes, the loss by friction is inversely as the square roots of the sectional areas. Owing to the variable value of the co-efficient of friction (called _c_ in the first formula given), it is usually preferable to measure the actual rate of progress of air through a given flue by means of an anemometer (wind measure). Then the velocity of the current of air and the area of the cross section of the flue being given, the volume of air discharged in a given time is represented by the product of these two and the time which has elapsed. Thus, q = a × v. 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 - t^1)/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.) The amount of inlet required will also be 14·6 square inches per head. Theoretically it ought to be slightly less than that required for outlet, as the outgoing air is more expanded than that entering the room; but practically no allowance need be made for this fact. The total amount of inlet and outlet required per head = +29·2 square inches.+ _If the mean temperature of a room is 61°, the external temperature 45°, while the heated column of air is 50 feet, and the required delivery of air 2,000 cubic feet per hour, find the size of inlet and outlet._ v = 8·2√(h(t - t^1)/492) = 8·2√(50(61 - 45)/492) = _10·55 feet per second._ = _37,980 feet per hour._ If we make no allowance for friction, then