in which w weight in pounds; and = = velocity in feet per second; R radius in feet; g = = = acceleration due to gravity, or 32.16; f force in pounds pulling the body towards the center of its revolution. This is a case like that in which a pound weight might be made to revolve on the end of a string, and should the string at any moment be cut, the weight would simply move off in a line tangent to the curve. In the second case, the centrifugal force, the velocity, and the radius are constantly increased, and correspond to the centrifugal force developed by a ventilating fan, which develops a velocity outwards from the center of revolution. To explain the difference, take a case in which a pound weight is made to revolve within a tube instead of being attached to the end of a string. Now, if the tube moves with the same speed as the string, the pound weight will move outwards along the inside of the tube. The velocity thus acquired will cause the weight to move in a path situated between the tangent to the curve and the prolongation of the radius at the point of disengagement, instead of in a path tangential to the curve of the outer circle. The body revolving on the end of the fixed radius has, at the moment of disengagement, acquired only sufficient centrifugal force to make it describe a path tangential to its circle; whereas, the body moving through the tube has, in addition to the centrifugal force of the former body, acquired a force due to an increased outward acceleration. The force acquired in the second case is calculated by the formula in which the factors are the same as in formula 70, except that the constant 3.1416 is substituted for R. 1034. In (a), Fig. 161, both of the above cases are illustrated. The radius of the circle described by the first body revolving is o e, or o h, and if this body is by any means disengaged, as, for example, by the breaking of a string, at the instant it is passing the point e, the body will reach the point g at the moment it should have arrived at h, having moved along the line e g, which is tangent to the curve e h. If, however, the same body is made to revolve in a tube and also to commence its outward journey at the center o, by the time it reaches e it will have acquired a high outward velocity that the fixed body can not possess. Therefore, at the moment of disengagement it will move off in the path e ƒ, and will arrive at ƒ in as short a time as the first body will require to reach g. The result is that the velocity developed by a continual deflection and acceleration is much greater in amount than that due to a body revolving with a uniform velocity on the outer extremity of a radius of constant length. DEFINITIONS OF TERMS. 1035. The centrifugal force developed by a fan is calculated by the following formula: Before giving examples that involve the use of formula 72, the meaning and use of its factors must be explained. For example, the velocity is obtained from the length of the radius of gyration, and this is obtained by adding to the radius of the port of entry a fraction of the radial length of the blades, the latter being found by multiplying the radial length by .6. The radial length of the blades is found by the formula D-d (73.) in which D = diameter of the fan; and diameter of the port of entry; EXAMPLE. The diameter of a fan is 30 feet, and the diameter of the port of entry is 12 feet. What is the radial length of the blade? SOLUTION. By formula 73, the radial length of the blades is 30 - 12 -9 ft. Ans. 2 1036. In (b), Fig. 161, is given a graphic illustration. of the terms in question. For example, taking o as the center of a fan, c is situated at the center of gyration, and oc is the radius of gyration. As oh is the radius of the port of entry, oh hc (or, what is the same thing, oh + abx.6) is equal to the radius of gyration. This is expressed by the formula in which is the radius of gyration. When air is flowing along the blades of a fan, all the air on the blades, from the periphery of the port of entry to the outer periphery of the fan itself, is subject to centrifugal force, and, as the blades may be 9 or 10 feet long, to find the weight of air subject to centrifugal force, the weight of a cubic foot of air is multiplied by 7, the length of the blades; this is the meaning of the expression in formula 72, where w occurs as two of the factors. Now, it must be clear that the velocity of the moving air at b is much less than it is at a; and, therefore, the mean of the squares of the velocities, multiplied by the weights, occurs at the center of gyration c, for a stream of air lies on every blade, as that shown at de gf. An example will make this clear: EXAMPLE.-A ventilating fan is 30 feet in diameter, the radial length of the blades is 9 feet, and the length of the radius of gyration is 11.4 feet; (a) what is the mean velocity generating centrifugal force when the fan is making 50 revolutions per minute? (b) What is the total pressure produced? (c) What is the quantity of air passed per minute? = = = SOLUTION. The velocity generating centrifugal force in feet per 11.4 X 2 X 3.1416 × 50 second will be equal to 59.69 feet per second, 60 the required velocity. From this velocity, the total pressure in pounds per square foot to produce the two depressions already noticed and the compression for blowing out can be found by formula 72. In the case for which the velocity has been calculated, the length of the blade is 9 feet; if the average weight of a cubic foot of air is taken at .076 .076 × 9 × 59.692 pound, formula 72 gives ƒ = 24.11 pounds per 3.1416 × 32.16 square foot. Next, suppose that the mine resistance in a case like this is equal to 3 inches of water-gauge, or 15.6 pounds per square foot, and the depression within the fan is 4 pounds per square foot below that in the fan drift, and that the pressure per square foot for blowing out is 4.51 pounds above the atmosphere. These figures yield the factors for calculating the quantity of air that this fan is exhausting_out of the mine in cubic feet per minute. By formula 64, v= = 18 p 18 X 4 = 36 feet per second, the mean velocity of the air entering the fan. Next, the port of entry, which is circular, is 30 — 2 7, or 30 — 18 = 12 feet in diameter, and its area is 122 × .7854 = 113.0976 square feet. 36 is the mean velocity in feet per second, and, therefore, 36 × 60 = 2,160, the velocity in feet per minute. If the area found be multiplied by the mean velocity in feet per minute, the result will be the quantity of air exhausted by the fan in cubic feet per minute, as 113.0976 × 2,160 244,290.8 cubic feet of air per minute. Ans. = = EXAMPLE.-A ventilating fan is 28 feet in diameter, and the diameter of the port of entry is 10 feet; what is the radial length of the blades? EXAMPLE. A ventilating fan is 28 feet in diameter, and the orifice of entry is 10 feet in diameter; what is the length of the radius of gyration? SOLUTION. By formula 74, or, r = 1o + (.6 × 9) =10.4 ft., the radius of gyration. Ans. EXAMPLE. The radius of gyration for a ventilating fan is 9.5 feet, and the length of blade is 7.5 feet; the diameter of the orifice of entry is 10 feet; the angular velocity is 50 revolutions per minute; what is the total range of the fan's ventilating pressure? SOLUTION. The velocity per second is equal to 49.742 feet; then, by formula 72, f= 9.5×2×3.1416×50 = the total ventilating pressure. 60 1037. The Center of Gyration.—The calculations thus far given are based on the assumption that the blades of the fan lie longitudinally in radial lines, but in many cases this does not occur. Therefore, it is necessary to be able to make the expressions adaptable for fans in which blades make different angles with the radii. Now, when the blade makes an angle with the radius, its efficient length is practically shortened in the proportion of the cosine of the angle. For example, suppose a blade makes an angle of 45° with the radius projected from the circumference of the port of entry; then the efficient length of the blade is only .7 of its actual length, as shown in Fig. 162, in which AB is the radius, CD the actual length of the blade, and CB the efficient length of the blade. Again, when the blades make an angle with the radii, the efficient angular velocity, that is, the number of revolu-. tions, is practically reduced in the proportion of the cosine of the angle which the blades make with the radii. For, sup FIG. 162. B pose a case in which the fan is making 50 revolutions per minute. The 50 would be reduced by the cosine of the angle, because where the blades decline from the radii, the relative efficiency of the velocity is only for an angle of 45°, which is equal to 50 × .7 = 35 revolutions per minute. To make this allowance, the expressions may be simplified by multiplying the number of revolutions by the square of the cosine; formula 72 thus becomes |