78

MINE VENTILATION.

§ 30.

if only one continuous current were used. Let us, as an illustration, assume a mine divided into four equally developed districts of the same area; and also the same mine arranged so as to form one continuous airway, whose length will be four times greater than that of one of the districts. Let a = area, p = perimeter, 1 = length, and Q = quantity of air circulating; then we have, as the value of the friction of the air,

multiplied by a co-efficient to be determined by experiment. In the first case, the airways may be considered as a single airway, with a length 1, a section 4a, and a perimeter 4p: the equation then becomes

Secondly, the area and perimeter being a and p, the length will be 4l, whose "drag" or resistances due to

On dividing equation (a) by (b) we find (b) is

of equation (a), i.e., the friction of (a) is that of

(b); or we may say "that the resistance to the motion of the air in a mine requires in a single current a motive-force sixty-four times greater than when the air is divided into four currents of the same sectional area and aggregate length.

The same is true for bends and contractions that require a motive-force equal to the square of the volume of air circulating in each district. Thus, in the Q first case, the volume of air was the motive-force 4'

required to overcome this resistance

m X

Q2 16'

m being

a numerical co-efficient depending upon the number of bends or strictures.

In the second case, the volume must be multiplied by a co-efficient, which, owing to the quadruple length of the airway, must be equal to the sum of the four co-efficients of each district, or 4mQ2, a value sixty-four times the first.

31. By splitting the air-current, or leading it off from the downcast into the different districts to be ventilated, we have two decided advantages over the old method. In the first place, we obtain purer air for each district; for each split takes its fresh air directly from the downcast.

80

MINE VENTILATION.

§ 31.

In the next place, we obtain more air by decreasing the velocity. As the friction increases as the square of the velocity, we lessen the friction by lessening the velocity, and using the same power: we also decrease the cost of ventilation more than by any known means as yet. Were it not for the resistances of the shafts, the results of splitting the air would be easily calculated.

Problem. - To show the effect of splitting, without considering the shaft resistances: suppose we have a mine circulating 18,000 cubic feet per minute, and that the area of the gallery is 25 feet, the length 1,000 feet. What amount of air will circulate when the current is split into two, three, four, and five equal divisions, the pressure remaining constant?

The effect of splitting into two, three, four, and five equal air-courses will be to double, treble, etc., the areas without altering the rubbing-surface, because the area after splitting is two, three, etc., times that of the original airway, although the rubbing-surface remains the same; and as the quantity may be found by the formula

we may substitute the values of the symbols, and find the quantities of the various splits under consideration.

Since, however, in the above formula, p, k, and s remain constant, and a varies, we may simplify the work by eliminating them, and use √a × a = q, which corresponds to one of our former laws; viz., "that the relative quantities will be according to the square root of the area multiplied by the area, and then multiplied by the original quantity flowing through the mine." The pressure may be found thus:

ksv2 0.0217 × 20000 × (0.72)2

p=

a

Using now formula (x

q =

q =

q=

q =

25

8.999 pounds.

× 50 = 50910

× 75 = 93525

× 100 = 144000

× 125 = 200070.

The question may be worked, without reference to the actual dimensions of the areas and rubbing-surfaces, thus:

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MINE VENTILATION.

√ĩ x 1:√2 x 2 :: 18000 : 50910 √1 x 1 : √3 x 3:: 18000:93528 √1 x 1 : √4 x 4 :: 18000: 144000 √1 x 1 : √5 x 5 :: 18000 : 200070.

§ 31.

The difference between the two calculations is owing to the treatment of the decimals, and assuming the pressure more than it really is.

(b) There is a difference between splitting the air, and adding an air-course of the same length and area. In the above example the area was doubled while the length remained unchanged; but, if we were to add an air-course of the same length and area, we would double the rubbing-surface and the area. In fact, we cannot call such an arrangement a split; for, if both received an equal volume of air, they would both have the same pressure; but that pressure would only be a small part of what it was in the above example, first case, on account of the increased area: therefore, instead of increasing the quantity of air, we diminish it.

To illustrate this let a = 25 feet, s = 24,000 feet in an airway passing 15,000 cubic feet per minute: add an airway of the same length and area, what quantity will flow through each? Ans. 22,900 cubic feet.

The units of power necessary to circulate the above quantity may be found by multiplying the pressure by the quantity of air circulating per minute.