CHAPTER XIV

WORK, POWER, AND ENERGY; HORSEPOWER OF BELTING

97. Work. Whenever a force causes a body to move, work is done. Unless the body is moved, no work is accomplished. A man may push against a heavy casting for hours, but unless he moves it, he does no work, no matter how tired he may feel at the end of the time. It is evident that there are two factors to be considered in measuring work-force and distance. In the study of levers, tackle blocks, and inclined planes we dealt with the problem of work. In any of these machines the work delivered or accomplished in lifting a weight is measured by the product of the weight times the height it is raised. The work supplied, or put into the machine, is the product of the force exerted times the distance through which this force must move. It is a known fact that if we neglect the work lost in friction, the work supplied or put into a machine is equal to the work delivered or accomplished by it. The actual difference between the work supplied to the machine and the work delivered is the amount that is lost in friction. The following expressions may make these relations clearer.

98. Unit of Work.-The unit by which work is measured is called the foot-pound. This is the work done in overcoming a resistance of 1 lb. through a distance of 1 ft.; that is, if a weight of 1 lb. is lifted 1 ft., the work done is equal to 1 ft.-lb. All work is measured by this standard. If a weight of 2 lb. is lifted through a distance of 1 ft., the work done is 2 X 12 ft.-lb.; or if a weight of 1 lb. is lifted a distance of 2 ft., the work done is 1 X 2 = 2 ft.-lb. The work in foot-pounds is the product of the force in pounds times the distance in feet through which the force

moves.

In lifting a weight vertically, the resistance, and, hence, the force that must be exerted, are equal to the weight itself in pounds. The work done is the product of the weight times the vertical distance that it is raised. If a weight of 80 lb. is lifted a distance of 4 ft., the work done is 80 X 4, or 320 ft.-lb. It would require the same amount of work to lift 40 lb. 8 ft., or to lift 20 lb. 16 ft.

When a body is moved horizontally, the only resistance to be overcome is the friction. When a team of horses pulls a loaded wagon, the only resistances which it must overcome are the friction between the wheels and the axles and the resistance on the tires caused by the unevenness of the road. The actual work done by the horses is equal to the work required by friction. However, if the wagon is being drawn up a hill, in addition to the work done in overcoming friction, the horses will have to perform an additional amount of work equal to that of raising the wagon the height of the hill. Considering the wagon as a machine, the work supplied or "put in," is the total work done by the horses, that is, the work of friction plus the work of raising the wagon. The work done in overcoming friction is practically wasted; therefore, the work delivered or "got out" is only that required to raise the wagon the height of the hill.

The work necessary to pump a certain amount of water is the weight of the water times the height through which it is lifted or pumped, plus the work lost in friction between the water and the sides of the pipe. The work necessary to hoist a casting is the weight of the casting times the height it is raised, plus the work required to overcome friction in the hoist. The actual work accomplished is merely the weight of the casting times the height, since the work of friction is wasted. The work done by a belt is the effective pull of the belt times the distance in feet which the belt travels. The work done in hoisting an elevator is the weight of the cage and of the load it carries times the height of the lift. Numerous other illustrations of work will suggest themselves.

99. Power. Power is the rate of doing work; that is, in calculating power the time required to do a certain number of footpounds of work is considered. If 10,000 lb. are lifted 7 ft., the work done is 70,000 ft.-lb., regardless of how long it takes. But, if one of two machines can do this in one-half the time that the other machine requires, then the first machine has twice the power of the second.

The engineer's unit of power is the horsepower, which may be defined as the ability to do 33,000 ft.-lb. of work per minute. Therefore, if a machine can perform 33,000 ft.-lb. of work in 1 min., its power is 1 hp.; if it can perform 66,000 ft.-lb. in 1 min., its power is 2 hp., etc. The horsepower required to perform a certain amount of work is found by dividing the foot-pounds done per minute by 33,000. If an engine can do 1,980,000 ft.-lb. in a minute, its horsepower will be 1,980,000 ÷ 33,000 = 60. An engine that can raise 66,000 lb. to a height of 10 ft. in 1 min. will do 66,000 lb. × 10 ft. 660,000 ft.-lb. per

=

20 hp. If another engine

takes 4 min. to do this same amount of work, it is only one

fourth as powerful; the work done per minute will be

165,000 ft.-lb. per minute; and its horsepower is

165,000
33,000

=

5 hp.

Example:

An electric crane lifts a casting weighing 3 tons to a height of 20

ft. from the floor in 30 sec.; what is the horsepower used?

In calculating horsepower, it should be remembered that the work done should be reduced to foot-pounds per minute before diriding by 33,000.

100. Horsepower of Belting.-A belt is used for the transmission of power from one shaft to another. The driving pulley exerts a certain pull in the belt, and this pull is transmitted by the belt and exerted on the rim of the driven pulley.

The power transmitted by any belt depends upon two things— the effective pull of the belt tending to turn the wheel, and the speed with which the belt travels. It is easily seen that these include the three items necessary to measure power. The pull of the belt is the force. The speed, given in feet per minute, includes both distance and time. Force, distance, and time, therefore, are the three items necessary for the measurement of power.

The total pull that a belt will stand depends upon its width and thickness. It should be wide enough and heavy enough to stand for a reasonable time the greatest tension put upon it. This is, of course, the tension on the driving side. This tension, however, does not represent the force tending to turn the pulley. The force tending to turn the pulley (or the Effective Pull, as it is called) is the difference in tension between the tight and the slack sides of the belt.

The effective pull that can be allowed in a belt depends primarily upon the width, thickness, and strength of the leather, or whatever material the belt is made of. In addition, every time a belt causes trouble from breaking or becoming loose, it means a considerable loss in time of the machine, of the men who are using it, and of the men required to make the repairs. Therefore, it should not be loaded so heavily as might otherwise be allowed. Leather belts are called "single," "double," "triple," or "quadruple," according to whether they are made of one, two, three, or four thicknesses of leather. Good practice allows an effective pull of 35 lb. in a single leather belt per inch of width. In a double belt a pull of 70 lb. per inch of width may be allowed. The allowable pull per inch of width times the width of the belt in inches gives the total effective pull or force which can be safely transmitted by the belt.

The total effective pull in the belt times the velocity in feet per minute will give the foot-pounds transmitted by it in 1 min. One horsepower is a rate of 33,000 ft.-lb. per minute; hence, the horsepower of a belt is obtained by dividing the foot-pounds transmitted by it per minute by 33,000. The velocity of the belt is calculated from the diameter and revolutions per minute of either one of the pulleys over which the belt travels. From these considerations, the formula for the horsepower that a belt will transmit may be written

W

=

V

width of belt in inches

velocity of belt in feet per minute.

Stated in words, this formula would read as follows: "The horsepower that may be transmitted by a belt is found by multiplying

together the allowable pull per inch of width of the belt, the width of the belt in inches, and the velocity of the belt in feet per minute, and then dividing this product by 33,000."

Example:

Find the horsepower that should be carried by a 12-in. double leather belt if one of the pulleys is 14 in. in diameter and runs 1100 r.p.m.

Explanation: To get the horsepower, we must first find the values of P, W, and V. We will take P as 70 lb. since this is a double belt. W is given as 12 in. V, the velocity, is obtained by multiplying the circumference of the pulley by the r.p.m. which gives 4032. Multiplying these three together gives 3,387,000 ft.-lb. per minute, and dividing by 33,000 gives 102+ as the horsepower that this belt might be required to carry.

101. Widths of Belts. It is possible, also, to develop a formula with which to calculate the width of belt required to transmit a certain horsepower at a given velocity.

One horsepower is 33,000 ft.-lb. per minute. Then the given number of horsepower multiplied by 33,000 gives the number of foot-pounds to be transmitted per minute.

Foot-pounds per minute = 33,000 X H

If we know the velocity in feet per minute, we can divide the foot-pounds per minute by the velocity; the quotient will be the force or the effective pull in the belt.

Now the force can be divided by the allowable pull per inch of width of belt. The result will be the necessary width.

Stated in words, this formula would read: "To obtain the width of belt necessary for a certain horsepower multiply the horsepower by 33,000 and divide by the product of the allowable pull per inch of width of belt times the velocity of the belt in feet per minute."