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unit of surface. This is of great importance in ventilation, as it allows of a greater quantity of air passing with the expenditure of less power than any other means known. This principle is the one upon which the splitting of air is reasoned. In the case of the eight-foot and four-foot passages in (6), while the former required double the amount of pressure for its doublesized perimeter, that pressure would propel four times the quantity: otherwise the same expenditure of power on air in a four-foot passage would propel double the quantity it would force through a two-foot passage, or an equal quantity would be propelled by half the power. The entire mass of moving air in an airway may be considered as a column of water passing through a pipe, exposing a certain amount of surface to resistance, and hence requiring a fixed amount of pressure acting upon its sectional area to overcome such resistance when the velocity is constant: therefore, the greater the area of section, the less the amount of pressure requisite for each individual unit of such area in order to make up the gross amount of such pressure required. The steam-engine piston will require less pressure per square inch to produce a given force as the area of the piston is greater, and vice versa. Taking the areas given in (6), we have the following propor




8. The pressure required to overcome the frictional resistances encountered by air in passing through a confined passage has been found to vary with the nature of the material composing the inner surface of the airway to which the moving air is exposed in its route, as well as the mechanical state of its surface: in other words, the smoother the rubbing-surface, the less the resistance.



23. 1. IN airways of the same sectional area, and which only differ in length, "the volume and velocity of air-currents are inversely proportional to the square roots of the lengths."

Problem. When an airway 4' x 5' area, and length of 4,000 feet, passes 20,000 cubic feet of air per minute, what will another airway 1,000 feet in length, with the same area and pressure, pass?

V1000: √4000 :: 20000 : x

√1: √4 :: 20000 : 40000 cubic feet. Ans.



Again: the velocity in the second airway will be

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1:2::1000:2000′ per minute. Ans.


That is, the volume and velocity in the 1,000-foot airway are twice the volume and velocity in the 4,000-foot airway.

2. The volume passing through airways of similar form but unequal size will be greater as the area of section is greater, other data being the same; or, the pressure and other data remaining constant, the quantity will be directly proportional to the areas:

a: A::v: V.

3. The volume passing, when areas and other data are equal, is inversely proportional to the perimeter of the sections of the airways. The circle passes the greater quantity of two airways, one of which is a square with a side equal to the diameter of the circle. (See § 16.)

4. The quantity or volume of air passing through an airway varies as the square root of the rubbing-surface.

5. Friction diminishes in proportion to the square root of the velocity, and increases according to the square of the velocity; i.e., friction varies as the square of the velocity.

6. Pressures are proportional to the squares, and the powers are proportional to the cubes, of the quantities of air passing through airways.

7. The quantity of air passing through airways of different areas, other things being equal, is according to the square root of the area multiplied by the area.


24. (a) If 20,000 cubic feet of air can be produced in an airway of 60 feet area with a certain pressure, how much air will the same pressure produce in an airway of 30 feet area?

Assume the perimeters to be 32 feet for 60-foot airway, and 22 feet for 30-foot airway. Had the relationship which existed between the perimeter and the area of the larger airway been maintained in the smaller airway, then the quantity of air flowing through the latter would be directly proportional to its area (2), pressure, etc., remaining constant. That is to say, the ratio of the area to the perimeter of the larger airway is as 60 to 32. In the smaller airway we have one-half the area; and, had the smaller perimeter been one-half




the larger, then the ratio would have been unaltered, and the quantity of air would have been one-half also (3). But, instead of the smaller perimeter being 16 feet, it is 22 feet. Now, the problem resolves itself simply into finding a fourth proportional; and, bearing in mind (4) that the quantity varies inversely as the square root of the rubbing-surface, we have,

√22:√16 :: 10000 cubic feet: x

... X= 8528 cubic feet per minute. Ans.

In the above calculation the length of the airway is taken as unity; because any given length will be a factor common to both terms of the first ratio, and hence may be eliminated.

To prove this method of reasoning, we may work it out according to Atkinson's method:

Taking the length of both airways as 1,000 feet, we have, for the larger,

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