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Trains of gears are used for hoisting, for clocks, and when a change in velocity ratio is necessary.

Each pair has its driver and follower. If the shafts are parallel, the usual arrangement for obtaining a change in velocity is to key two wheels of unequal size on every shaft except the first and last.

The velocity ratio for a train of gears is found in the same way as for a single

Substituting the values r=80, t=12 pair. The rule to remember in calculatand T-32, we have

R

rt -- - or T

80X12

30.

32

Then the driver will make 30 revolutions per minute.

The ratio of the speed of one gear to that of the other which meshes with it, is called the velocity ratio. In the case of the above gears, A and B, the velocity ratio is 80:30-2 2-3. That is, A (the That is, A (the larger wheel) makes one revolution

ing for a train of gears is as follows:

The product of the number of teeth on all the drivers, or the result obtained by multiplying them together, divided by the product of the number of teeth on all of the followers, is equal to the velocity ratio.

Suppose we have a train of gears composed of six wheels, three drivers, A, B and C, and three followers, L, M and N. A has 14 teeth and drives its follower which has 70 teeth. The pinion B, keyed to the same shaft with L, has 13

L,

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on the ceiling of the first floor. We shall still further suppose switch 2 in the position shown, and the person has just entered the house and turned switch I in the position shown. This will cause the light to burn. If, now, the person passes upstairs and turns switch 2, the light will be put out, but will be again lighted if switch I is turned. In this way the light is under perfect control of either switch.

Comparative Weight of Motors (2) Question: Why is a slow-speed motor heavier than a high-speed motor? P. R.

Answer: It will be observed from an inspection of the formula for motor out2 TnT put that for a given amount of 33,000 horse-power, there are two variables, speed and torque. If the horse-power is to remain constant at a reduced speed, the torque must be increased. Torque depends on the number of lines cut per unit time by the revolving conductor. If the speed is decreased, the number of lines evidently must be increased. In order to obtain this, without working the iron at an uneconomical density, the cross-section of all magnetic paths, yoke, poles, and armature core, must be increased. The increase of size of these paths is the main reason for the large size of low-speed motors.

Again, looking at this from the side of counter-E.M.F., any motor must generate a counter-E. M. F. approximately equal to the impressed E.M.F. If it is

to do this at a slower speed, it is evident that either the number of armature conductors must be increased, or the number of lines of force must be increased, since, in order to obtain any given voltage, a definite number of lines of force must be cut per unit time. In designing slowspeed motors, the number of armature conductors ordinarily is not materially increased, because of the fact that this increases armature reaction, which is a very undesirable feature; and hence, the other factor, the number of lines of force, must be increased, and, as stated above, this requires a larger magnetic path, hence a larger, heavier motor.

Transferring Heat to Boilers

(3) Question: What is the method. of transferring heat from heating furnaces to boilers, and what is the temperature of the heat when it reaches the boilers? W. McP.

Answer: The boilers should be set over the furnaces, in which are heated large billets or blooms of iron or steel, in such a manner that the grate or front end of the boiler is placed over and above the stack end of the heating furnace. P this arrangement, waste heat of the heating furnace to the boilers is discharged, the heat striking the boilers at the same place that they would get their heat if fired direct.

The temperature of the heat at the time it reaches the boilers is from 2,500 to 2,800 degrees.

Where coal is cheap there is not much saving by placing the boilers over furnaces. The boilers keep the stack ends of the furnaces much colder than the same would be without a boiler over them. Any furnace with a cold stack works slower than a furnace with a hot stack, and a slow furnace will not turn out the tonnage that a hot furnace will, and tonnage counts for a great deal in a rolling mill.

Effect of Cold Rolling

(4) Question: What is the effect of cold rolling iron and steel? S. A.

Answer: Cold rolling iron and steel increases the elastic limit and ultimate

Size of Pipe for Condenser

(5) Question: We have had a great deal of trouble with our jet condenser. Our engine is a 600-H. P. tandem compound-condensing engine, and is running under full load. We get our condensing water, the maximum temperature of which is 70° F, from a bay, through 800 feet of 6-inch pipe. There are 8 right-angled elbows. The entrance of the water into the condenser is 16 feet above the level of the water in the bay. We wish to have a 24-inch vacuum. The water is drawn into the condenser by means of the vacuum.

The trouble is that when working under full load, the vacuum often breaks and we are compelled to take off some of the load until we get the vacuum again. I think this difficulty would be overcome if our water supply could be increased. I wish you would figure out the size of the pipe to give the condenser a good flow of water. W. A. L.

Answer: With the data you have given, the problem can be solved as follows:

Since the vacuum desired is 24-inch, which corresponds to a pressure of 3 lbs. per square inch, the heat in the liquid is 109.8, and the total heat in the steam is 1,125, its temperature being 141° F according to the steam tables given in Steam Engines, Part II. I should suggest, if you have difficulty in following this work, that you refer to page 38 of this instruction paper.

From the above, the pounds of water necessary to condense one pound of steam if the temperature of the condensing water is 70° F, will be

1,125-11O 141-70

14.5 lbs. This will be the minimum amount of water that will condense the steam, because in the above the most favorable conditions are assumed. If the engine is assumed to use 22 lbs. of steam per indicated horse-power, then, since there are 600 horse-power, the steam used will be 22 x 600 13,200 lbs. per hour. From the above, 14.5 lbs. of water are necessary per pound of steam. Therefore, 13,200x14.5 (=191,500) lbs. of water are

÷ =

water weighs 621⁄2 lbs., we shall have 191,500 62% 3,060 cubic feet per hour. Dividing this by 3,600, we get .851 cubic feet per second. The area of a 6-inch pipe is about .196 sq. foot; therefore, .851.1964.34 feet per second, the velocity necessary to supply enough water to condense the steam.

A 24-inch vacuum corresponds to a head of 271⁄2 feet. This can be found by multiplying 24 inches or 2 feet by 13.6 because the density of mercury is 13.6. Since the height of the entrance of the pipe into the condenser is 16 feet above the level of the condensing water in the bay, the effective suction head in the condenser will be 271⁄2-16=111⁄2 feet.

According to Carpenter, the resistance of one elbow is equivalent to 520 diameters of straight pipe. Therefore, since there are 8 elbows which are 1⁄2 foot in

=

diameter, the equivalent length of pipe will be 8 x 1⁄2 x 520 = 2,080 feet. The total equivalent length of pipe will be 2,080 + 800 2,880, since there are 800 feet of straight pipe. The velocity of the flow can be found. h D 1+48 D' in which is the velocity in feet per second; h is the head in feet of water; D is the diameter of the pipe in feet; and L is the equivalent length. Therefore, for a 6-inch pipe the velocity will be = 48 By

by means of the formula v =

I I

v=

=2.14 per second.

2,880+48 ×1⁄2 referring to the above calculated velocity necessary to condense the steam, you will see that this pipe is too small.

If the 90° elbows were replaced by long-sweep bends, the resistance to flow would be very much diminished, as one of these large bends will give very little more friction than an ordinary straight pipe. If this is done, the equivalent length of pipe will be but 800 feet. Calculating the velocity by means of the above formula for an 800-foot pipe, we will get 4.0+ feet per second, which, although a little smaller than that which was found to be necessary, would probably do, because in solving for it the most unfavorable condition of temperature was assumed. Therefore, if it were possible to straighten out the pipe, using no 90° elbows, or if these elbows

v=

were replaced by long-sweep curves, a 6-inch pipe would barely satisfy the conditions.

If an 8-inch pipe is used whose diameter is about .66+ foot, and whose area is .35 of a square foot, the velocity necessary in the pipe would be .851.35= 2.43 feet per second. Substituting in the above formula again to find the velocity which would be obtained under a head of 111⁄2 feet and a pipe .66 foot in diamII 1⁄2 X.66 =2.46 eter, we get v=48, 2,880+48X.66 feet per second, which is more than necessary. Therefore, if an 8-inch pipe were used, all the water necessary would be supplied.

Compound-Wound Generator Connections

(6) Question: Please send sketch for connections of two compound-wound generators in parallel. T. B. & F. W.

Answer: We enclose a sketch to make this description clearer. The corresponding leads from each machine are connected to the bus bar of the same polarity. The equalizer is connected between the two machines, starting from a point on each between the brush and the connection to the series coil. A rheostat is shown in the shunt winding for regulating the voltage; also an ammeter for each machine, which should be connected on the opposite side to the series

coil.

This arrangement shown represents the simplest possible connection, and would have to be modified to suit the special requirements of the generator and switchboard under consideration.

In this case the equalizer runs directly from machine to machine and does not go to the switchboard, and the switch for closing this circuit is generally located on a pedestal near one of the machines. This arrangement allows the shortest possible equalizer, hence the lowest resistance, and secures the most sensitive regulation. This equalizer may run from each machine to the switchboard if it is not too far away, and terminate on the two poles of the switch which is placed on the face of the switchboard. Or, if more than two machines are used, an equalizer bus bar may be mounted on the back of the board, and the equalizer

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remove the difficulty, we should advise using a heater or adding a condenser if conditions are favorable.

We should be pleased to hear from you again regarding this matter.

Relations of Compression and Energy

(8) Question: Please tell me how to find the foot-pounds of mechanical energy given up in reducing the pressure of a cubic foot of compressed air from 100 pounds per square inch to 30 pounds per square inch. L. J. S.

W= 144 x 100 x log

The

Answer: If I cubic foot of air is expanded from 100 pounds to any other pressure in an engine cylinder, the work done will be represented by the formula. final volume original volume The 100 in this formula is the original pressure; and by substituting any other pressure for the 100, a formula will be obtained by which the work for any difference in pressure can be found. loge should be taken from the table of Naperian or natural logarithms. The initial volume in this case is I cubic foot, and the final volume would be found by use of the formula PV=C, where C equals the product of the initial volume and the initial pressure. P is the final pressure in pounds per square inch. Substituting in the case of reduction to 30 pounds, W = 144 x 100 x loge 3 1-3, giving 17,325 approximate foot-pounds. The loge of 3 1-3 is 1.203.

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