so will it increase or decrease according to the boundinglines or perimeter of the airway. From this it becomes evident that the form which has the least perimeter will have the least rubbing-surface. Above, it was perimeter than a square airway of the same diameter; and hence, if the two airways have the same length, the circular will have less rubbing-surface.

shown that a circular airway had less

6

17. In large airways the friction will be less than in smaller airways the sum of whose areas are equal to the area of the large airway. Let us suppose an airway twelve feet square in section: its area will be 144 square feet, and its perimeter 12 + 12 +12+12 48'.

=

12

Let us now take three smaller airways, two 6' x 6' in section, and one 6' x 12' in section. The aggregate areas of the three airways will be 36 +36 +72144 square feet: the sum of the perimeters will be 24+ 24 +36

84. Hence we see, that, while the large airway has the same area as the aggregated smaller airways, its perimeter is much smaller than the aggregate perimeters of the lesser airways.

18. The force used to overcome the resistances

offered to the passage of air in mines is estimated in pounds to the square foot, and may be expressed as so much head of air, motive-column, or water-gauge. The motive-column has already been treated of, so our attention may be given to the water-gauge. The watergauge is an instrument used to measure the dynamic force of a current of air. It consists of a U-shaped tube of equal area throughout. The arms are about six inches long, provided with a scale divided into inches and fractional parts of an inch, so that the difference between the height of the water in one arm of the tube and that of the other may be measured. One arm is placed in connection with the air passing in the mine, while the other is open to the air away from the mine. The difference in water-level will indicate the drag, or the resistance to the air in the mine. In some gauges, oil is substituted for water. They are made in different shapes; but the principle is the same in all. The weight of one cubic foot of water at 62° F. and 30" barometrical pressure is 62.32102 pounds avoirdupois: 62.321021728 0.036 pound is the weight of one cubic inch of water. When the gauge measures one inch, the pressure is 0.036 × 144 = 5.184, or 5.2 pounds (nearly) to the square foot.

Example.-Suppose a water-gauge read 0.4", what pressure would it indicate?

0.036 × 0.4 × 144 2.0734 pounds to the square foot.

This gauge may be used to show the force of a current produced by a fan or by a furnace, and hence is very useful as a check to the furnace-man. As it tells the amount of resistance to the air in the air-courses, their state or condition may be inferred. If the pressure per square foot exerted by the motive-column be known, the height of the motive-column may be determined.

Problem. Suppose the temperature of the motivecolumn be 62° F., and the water-gauge reads 0.4", what is the length of such motive-column?

Solution.-100 cubic feet of air at 62°, barometer 30", weigh 7.629 pounds. 1 cubic foot of air at 62°, barometer 30", weighs 0.07629 pounds. The pressure per square foot as indicated by 0.4" water-gauge is 0.036 0.4 × 144 2.0736 pounds. Dividing the × pressure per square foot by the weight of a cubic foot

• 2.0736 of air gives 0.076

column in feet.

27.28′ as the length of the motive

When the height of the motive-column is known, we may find the velocity of the air in feet per second which the motive-column will produce.

Problem. Suppose the motive-column be 27.25' in height, what velocity per second will it produce? A body falling, acted upon by gravity, would, according to Eq. 2, attain a velocity represented by 8.02√h; or

46

MINE VENTILATION.

§ 18.

substituting the value of h, which in this case is 27.25', we have 8.02/27.25 = 42′ per second.

Table IV. gives the comparative height of the watergauge and air-column at a temperature of 62° F., with pressure in pounds per square foot, and the theoretical velocity of air due to this pressure. This table, it must be remembered, is only theoretically true, on account of the enormous power required to overcome friction to the passage of air in an airway: hence, in practice, from ten to twenty times this amount of motivecolumn is required in order to produce the theoretical velocity.

19. "Co-efficient of friction" is a term used to represent the constant resistance met with by air during its journey through the mine. This resistance must be overcome at each point, before the air can pass that point. It varies, of course, under different conditions; but, the smoother the rubbing-surface, the less will be this resistance. This co-efficient cannot be determined with any degree of certainty except by actual experiment; and even then experimenters differ in the exact amount, because of the different conditions under which the experiments were made. Sir John Atkinson, after comparing the results of a number of experimenters, took an average between their results, and used, as the co-efficient, 0.26881 feet of air-column of the same density as the flowing air. He appears to remain in