Modern Machine-Shop Practice Part 244

If we take the first pair of wheels B and C, we have that the velocity will vary in the same ratio or degree as their diameters vary, notwithstanding that their revolutions are equal.

Radius. Diameter. Circ.u.mference.

B = 5-1/8 inches. 10-1/4 inches. 32.2 inches.

C = 10-1/4 " 20-1/2 " 64.4 "

D = 7-5/8 " 15-1/4 " 47.9 "

E = 15-1/4 " 30-1/2 " 95.8 "

The velocity is the s.p.a.ce moved through in a unit of time, and as it is the circ.u.mference of the pulley that is considered, the velocity of the circ.u.mference is that taken; thus, if we make a mark on the circ.u.mferences of the two pulleys, B and C, Fig. 3354, the velocity of that on C will be twice that upon B, or in the same proportion as the diameters.

Let there be suspended from the circ.u.mference of B 10 lbs. weight, and let us see the degree to which this power will be distended by this arrangement of pulleys, supposing the weight to rotate B, and making no allowance for the friction of the shaft.

Suppose the weight to have fallen 32.2 inches, and we have 10 lbs.

moving through 32.2 inches, this power it will have transmitted to pulley B.

To find what this becomes at the perimeter of C, we must reduce the number of lbs. in the same proportion that the perimeter of C moves faster than does that of B; hence we divide the circ.u.mference of one into the other, and with the sum so obtained divide the amount of the weight; thus, 64.4 (circ.u.mference of C) 32.2 (circ.u.mference of B), = 2; and 10 lbs. 2 = 5 lbs., which, as the circ.u.mference of C is twice that of B, will move twice as fast as the 10 lbs. at B, hence for C we have 5 lbs. moving through 64.4 inches.

Now C communicates this to D by means of the belt H, hence we have at D the same 5 lbs. moving through 64.4 inches.

Now E moves twice as fast as D, because its circ.u.mference is twice as great, and both are fast upon the same shaft, hence the 5 lbs. at D becomes 2-1/2 lbs. at E, but moves through a distance equal to twice 64.4, which is 128.8 inches. To recapitulate, then, we have as follows:

The weight gives 10 lbs. moving through 32.2 inches.

Pulley B " 10 " " " 32.2 "

" C " 5 " " " 64.4 "

" D " 5 " " " 64.4 "

" E " 2-1/2 " " " 128.8 "

That the amount of power is equal in each case, may be shown as follows:

For C, 5 lbs. moving through 64.4 inches is an equal amount of power to 10 lbs. moving through 32.2 inches, because if we suppose the first pair of pulleys to be revolving levers, whose fulcrum is the centre of the shaft, it will be plain that one end of the lever being twice as long as the other, its motion will be twice as great, and the 5 at 10-1/4 inches just balances 10, at 5-1/8 inches from the fulcrum, as in the common lever.

In the case of D we have the same figures both for weight and motion as we have at C, because D simply receives the weight or force and the motion of C. In the case of E, we have the motion of the weight multiplied four times; for the distance E moves is 128.8 inches, which, divided by 4, gives 32.2 inches, which is the amount of motion of the weight, hence the 10 lbs. of the weight is decreased four times, thus 10 lbs. 4 = 2-1/2 lbs., hence the 2-1/2 lbs. moving through 128.8 inches is the same amount of power as 10 lbs. moving 32.2 inches, and we may concentrate or convert the one into the other, by dividing 128.8 by 4, and multiplying the 2-1/2 lbs. by 4, giving 10 lbs. moving 32.2 inches.

If, therefore, we make no allowance for friction, nothing has been lost and nothing gained.

Thus far, we have taken no account of the time in which the work was done, more than as one wheel is caused to move by the other, and all of them by the motion of the weight, they must all have begun and also have to move at the same time. Suppose, then, that the time occupied by the weight in falling the 32.2 inches was one minute, and the amount of power obtained may be found by multiplying the lbs. of the weight by the distance it moved through in the minute, thus 10 lbs. moving 32.2 inches in a minute gives 32.2 inch lbs. per minute, being the amount of power developed by the 10 lb. weight in falling the 32.2 inches.

We may now convert the power at each pulley perimeter or circ.u.mference into inch pounds by multiplying the respective lbs. by the distance moved through in inches, as per the following table:

Distance moved.

Lbs. Inches. Inch lbs. of power.

Weight at B 10 32.2 = 322 " " C 5 64.4 = 322 " " D 5 64.4 = 322 " " E 2-1/2 128.8 = 322

If we require to find the power in foot lbs. per minute, we divide by 12 (because there are 12 inches in a foot), thus 322 inch lbs. 12 = 26.83 foot lbs. per minute.

Now suppose that B was moved by a belt, with a pull of 10 lbs. at its perimeter, and made 100 revolutions in a minute instead of one, then the pull at the perimeters of C, D, and E would remain the same, but the motion would be 100 times as great, and the work done would therefore be increased one hundred fold. It will be apparent, then, that the time is as important an element as the weight.

The velocity and power of gear wheels are calculated at the pitch circle.

Now suppose the gear A in Fig. 3355 has 30, gear B 60, gear C 10 and gear E 80 teeth, and that 5 lbs. be applied at the pitch circle of A; to find what this 5 lbs. would become at the pitch circle of E, we multiply it by the number of teeth in B and divide it by the number of teeth in C, thus:

Lbs.

At pitch of circle A 5 Number of teeth in B 60 --- Number of teeth in C 10 ) 300 --- 30 --

Answer, 30 lbs. at the pitch circle of E.

Now suppose that on the shaft of A there is a pulley 20 inches in diameter, and that on this pulley there is a belt exerting a pull of 5 lbs., while on the shaft of E there is a pulley 16 inches in diameter, and to find how much this latter pulley would pull its belt, we proceed as follows:

2 ) 20 = Diameter of pulley on A.

-- 10 = Radius of pulley on A.

5 = Pull on pulley A.

-- Number of teeth in A = 30 ) 50 = Pull at centre of shaft ----- of A.

1.666 60 = Number of teeth on B.

------ Number of teeth on C = 10 ) 99.960 = Pull at axis of shaft ------ of B.

9.996 80 = Number of teeth on E.

------- Radius of pulley on shaft of E 8 ) 799.680 Pull at axis of shaft E.

------- 99.96 Pull at perimeter of last pulley.

We have in this case treated each pulley as a lever whose length equalled the radius of the pulley, while in the case of the wheel we have multiplied by the number of teeth when the power was transmitted from the circ.u.mference to the shaft, and divided by the number of teeth (the number of teeth representing the circ.u.mference) when the power was transmitted from the shaft to the teeth.

We thus find that power is composed of three things, first, the amount of impelling force; second, the distance that force moves through; and third, the time it takes to move that distance.

If we take a number of pulleys, say four, and arrange them one after another so that they drive by the friction of their circ.u.mferences, then the amount of power transmitted by each will be equal and the velocities will be equal, whereas, if we arrange them as in Fig. 3354, the power will be equal for each, but the velocities or s.p.a.ce moved through in a given time will vary.

What is known as the unit of power is the foot lb., being the amount of power exerted in raising or lifting one lb. one foot, and from what has already been said, it will be perceived that this is the same amount of power as 12 lbs. moving a distance of one inch.

Watt determined that the power of a horse was equal to that necessary to raise 33,000 lbs. one foot high in a minute, and this is accepted, in English speaking countries, as being a horse power.

An engine or machine has as much horse power as it has capacity to lift 33,000 lbs. a foot high in a minute.

CALCULATING THE HORSE POWER OF AN ENGINE.

The horse-power of an engine may be calculated as follows:

_Rule._--Multiply the area of the piston by the average steam pressure upon the piston throughout the stroke, and by the length of the stroke in inches, which gives the number of inch pounds received by the piston from the steam during one stroke.

As there are two piston strokes to one revolution of the engine, we multiply by two, and thus get the number of inch pounds received by the piston in one revolution.

By multiplying this by the number of revolutions the engine makes in a minute, we get the number of inch pounds of power received by the piston in a minute.

By dividing this by 12, we get the number of foot pounds the piston receives per minute, and dividing this by 33,000 lbs. we get the horse-power of the engine.

It has already been stated that Watt determined that a horse was capable of exerting a power equal to the raising of 33,000 lbs. one foot high in a minute, hence, having foot pounds of the engine per minute, dividing them by 33,000 gives the horse power.

This gives the amount of power received by the piston, but it is evident that the engine cannot exert so much power, because part of it is expended in overcoming the friction of the moving parts of the engine.

The amount of the piston power expended in overcoming the friction depends upon the fit of the parts, upon the lubrication and the amount of the load.