48 MINE VENTILATION. 19. doubt whether the mere change of temperature does or does not affect the co-eficient of resistance. As it is now taken for granted that it is not influenced by change of temperature (although it probably is), we may, for a velocity of 1.000 cubic feet per minute, consider the friction equal to an air-column 0.26881 feet in height, of the same density as the flowing air. This air-column is equal to a pressure of 0.0217 pound per square foot of area of section with air at 32°. 1.32529 x 30 × 0.26881 = 0.0217 pound per square foot. 459 +32 When we consider the air-column, we may use the co-efficient 0.26881; afterwards, if we desire, we may reduce the height of the air-column thus found to pounds per square foot, as above. We may shorten this work, however, by using 0.0217 pound per square foot of area for every square foot of rubbing-surface exposed to the air-current, at a velocity of 1,000 feet per minute, or 0.0000000217 pound for a velocity of 1 foot per minute. We must also bear in mind, that, while we use h in the following formulas for the air-column which by its weight will produce pressure, we use 0.0217 to find the pressure, P, direct, considering all-air in passing to have the same co-efficient, viz., 0.0217. 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 Ρ = motive-column in feet. 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. rubbing-surface. v = the velocity of the air in thousandths of feet per minute, 1000 feet per minute being taken as the unit of velocity. 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. α 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 corre spond 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 crea tion of velocity: hence they will be more correct for long than for short airways. formulas are so connected with each other, that, when The symbols in these 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 × 7 = 56 square feet. k = 0.26681. S= 8 +8 +7 +7 v2 = 15 x 60 = = 30 × 2000 = 60000 square feet. 900' per minute, or 0.9 of 1000 feet per minute, which squared is equal 0.81. 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. 0.0763 x 233.288 = 3.42" water-gauge. 52 MINE VENTILATION. § 21. 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 x 33000. The HP in the last formula stands for horse-power ventilation. (11) Qp HP = 33000 33000 of |