Mr. Thomas and several other writers consider this co-efficient as much too large: be that as it may, we can see at present no reason why it should be rejected, until the experiments already referred to approach more nearly to each other in their results. The co efficients in practice can never be the same in different mines; but they may approach each other by making the airways as smooth, and free from obstructions, as possible. PRESSURE. 20. Some of the formulas relating to the friction of air in mines, adopted by Mr. Atkinson in his book on "Mine Ventilation," will be retained in this treatise, so that any one studying this book will better understand his method of reasoning. Let h = motive-column in feet. p = pressure per square foot due to weight of h. 0.0217 pounds per square foot of area of section for a velocity of 1000 feet per minute with air at 32°. It is to be taken in the same terms or unit as p is taken in. 8 rubbing-surface. 8= the velocity of the air in thousandths of feet per minute, 1000 feet per minute being taken as the unit of velocity. The total pressure is found by the formula (6) pa = ksv2 which, expressed in words, gives us the following rule: To find the total pressure due to the friction of air passing through an airway, multiply the co-efficient of friction by the rubbing-surface, and the product by the square of the velocity; or, pressure being known, multiply the pressure per square foot by the area of the airway. divide the total pressure by the area of the airway. By the clearing of fractions, division, and other algebraical operations, we may find formulas which correspond to rubbing-surface and velocity, and also find the co-efficient of friction. These formulas embrace only pressure due to friction, and not that due to the creation of velocity: hence they will be more correct for long than for short airways. The symbols in these formulas are so connected with each other, that, when a sufficient number of them are known, those unknown may be found. To show the application of these for mulas, suppose we have an airway 8' x 7', 2,000 feet travelling at the rate of 15 feet per What is the resistance due to fric long, with the air second through it. tion, or the motive-column required to overcome friction in the airway? a = 8 x 7 = 56 square feet. k = 0.26681. S = 8 +8 +7 +7 = 30 × 2000 = 60000 square feet. = 15 × 60 = 900′ per minute, or 0.9 of 1000 feet per minute, which squared is equal 0.81. Substituting the numerical values of these symbols •in (7), h = 0.26881 × 60000 × 0.81 56 =233.288 feet of air-column as the pressure required to overcome the friction, and produce circulation. Taking the air at 62° F., one cubic foot, with the barometer at 30 inches, would weigh 0.0763 of a pound. If we now multiply this motive-column by the weight of a cubic foot of the air it is composed of, and divide the product by 5.2 pounds, -the pressure per square foot when the water-gauge is one inch, we may find the water-gauge due to this pressure, and also the pressure on each square foot. The pressure per square foot is 17.7 pounds. From this we may obtain a formula for water-gauge. The same result may be obtained by multiplying the length of the motive-column by the weight of a cubic foot of air of the same density, thus obtaining the pressure direct without first finding the water-gauge. The quantity of air passing may be found by multiplying the velocity in feet per minute by the area of airway in square feet, or The u in the last equation represents units of work, foot-pounds applied to circulate the air. (10) u = Q× p = vpa = HP × 33000. The HP in the last formula stands for horse-power of ventilation. When we represent the length of an airway by l, the rubbing-surface by s, the perimeter by o, we have THE LAWS AFFECTING AIR IN MINES. 22. THE laws affecting the circulation of air through mines or confined passages, such as gangways, etc., have been ascertained principally by such eminent men as Magnus, Regnault, Gay-Lussac, Daubisson, Peclet, and others, and are as follows: 1. The volume assumed by a given weight of air is inversely proportional to the pressure on each unit of surface under which it exists, so long as the temperature remains unaltered. Consequently, if we take a cubic foot of air under a pressure of five pounds, it will only be one-half a cubic foot under a pressure of ten |