average value under ordinary conditions for metal bearings is from to

1

We can now at once determine the work wasted in turning through any angle; generally we reckon per revolution. Thus

Work wasted per revolution=friction moment x angle turned, =f'Srx 2 =πf'Sd,

where

d=diameter of bearing.

Graphic Representation. -Graphic construction enables us to represent very simply quantities of energy or work. Let

Peffort in lbs.

s=space moved in feet.

On a scale of x inches to I foot make AB to represent s ft. AB is then sx inches.

C

r may be a whole number

D

- 12"

or a fraction.

On a scale of y inches to 1 lb. make AC

to represent P lbs. AC is then Py inches.

Complete the rectangle AD. Then

Area of ADAC × AB,

Thus the area AD

xy sq. ins. to 1 ft.-lb.

represents Ps ft.-lbs. on a scale of AD then represents the energy

exerted by P, on a scale compounded of the scales of

length and force originally chosen. For example:

a piston, s the

Let P be 2000 lbs. pressure on stroke being 3 ft.

Then the scales would be say:

in. to I ft. Thus AB=3=11⁄2 in.

and

2000 in. to 1 lb. Thus AC=2000 × = 1 in. These are the dimensions of Fig. 54. pound scale is then

Then

× sq. in. to I ft.-lb.

The com

Area AD=1 sq. in.,

which is the energy exerted.

=6000 ft.-lbs.

In this method the path of s ft. may be either straight or curved. If it be straight AB is an actual representation of it, but if curved then AB represents it straightened

out.

We can in this manner represent the work done in a turning pair. For let

R=resistance,

r=radius at which R acts.

Then for one revolution, we make AB to represent 2π ft., and AC to represent R lbs.

But now the work done is also given by the moment Rr overcome through 27, and we shall see that exactly

the same diagram will also represent the energy or work

when reckoned in this way. For let

R=80 lbs., r=3 ft.

Then draw AB on a scale of I in. to 4 ft. or 1/4′′ in. to I ft. to represent 2πr, i.e. 6π ft.

... Work done = × 4π × 100=480π ft.-lbs.

Next Method.-Resisting moment

Rr=240 lbs.

ft., and AC above will represent this on a scale of in. to 240 lbs.-ft., ie. in. to 1 lb.-ft.

Angle turned=2π,

Also

and AB will represent this on a scale of 1 in. to 2π, i.e. 3/4 in. to unit angle. And then

as before.

Work done = Rrx 2π,

=AC × AB=area AD,

And for the scale we have in. x 3/4 in. to I lb.ft. x unit angle, i.e. 1/400 sq. in. to 1 ft.-lb. of work, which is exactly the scale on which we measured AD before; which of course should be, as the result must be

the same.

The same diagram then which we use for a sliding pair represents work done in a turning pair; only we use a scale of moment and a scale of angle, instead of a scale of force and a scale of distance. The compounded scale gives in each case the scale of energy, or of work done.

There is no gain in such simple cases as we have just taken in graphic representation, but the method is

best understood in simple cases, and the gain will be seen when we come to more complicated ones.

Power-Horse Power. The definitions of Energy and of Work done contain no reference to the speed at which the operations considered have to be carried on; thus to lift one lb. through one ft. requires the same amount of energy to be exerted whether it be done in one second, or one minute, or one hour.

[The speed of a certain operation has an effect generally on the energy required, because it affects the values of the useless resistances, i.e. those which waste energy; but this does not affect what is stated above where we take the resistance as given.]

But to compare the value of different working agents, we must consider the energy they can respectively exert in a given time. This quality of an engine, or other energy exerting agent, we call its Power. The Power of an engine, for example, then means the energy it can exert in a given time.

To measure power we require some Unit Power, and this we take as the capacity for exerting 33,000 ft.-lbs. in one minute. The unit is called a Horse Power, and is due to Watt; being introduced by him, so that purchasers of engines should be able to compare their values with those of horses. The unit is much above the power of any horse in continuous work; but the extra margin was allowed to prevent any possibility of a mistake in the other direction.

The Horse Power then of any agent is equal to the number of ft.-lbs. of energy it can exert in a minute divided by 33,000.

If, for example, a steam engine work with constant steam pressure p lb. per sq. in., piston area A sq. ins., stroke s ft., revolutions per minute N. The calculation of the Horse Power proceeds thus

Effort=p× A=pA lbs.

Distance moved in one revolution = 25 ft.

Energy exerted in one revolution=pA x 2s ft.-lbs.

1. A man weighing 150 lbs. carries loads of 144 lbs. to a height of 30 ft. Find the work done in each journey. If the man exerted the same amount of energy in lifting the weights by means of a winch, which wasted one-third of the energy · applied, how much more useful work could he do?

Ans. 8820 ft.-lbs.; .3 times. 2. Iron pigs six inches square lying originally on the ground, are built into a stack 6 ft. by 5 ft. by 4 ft. high. Find the work done. Ans. 90720 ft. -lbs.

3. A bicyclist and machine weigh 180 lbs. he exerts when riding at 20 miles per hour, on ance of which is I per cent of the weight.

a

Find the H. P. track the resistAns. .096.

4. A colliery engine raises loads weighing 21 cwt. from a depth of 1200 ft. in 45 seconds. Find the H. P. required.

Ans. 114.

5. Compute the nett II. P. required to pump out a basin with vertical sides in 48 hours, the area of the water surface being 50,000 sq. yds., and depth of water 20 ft., the water being delivered at a height of 26 ft. above the bottom of the basin. (35 c. ft. of sea water weigh I ton.) Ans. 97.

The

6. A boiler 12 ft. diameter, 10 ft. long, full of water, is to be pumped out, the water being delivered at the sea level. ship's draught is 21 ft., and the bottom of the boiler is ft. above the keel. Find how long two men each capable of exerting 2500 ft.-lbs. per minute would take, using a hand pump in which of the energy is wasted. Ans. 3 hrs. 28 min.

7. A pumping engine 9 ins. diameter, 6 ins. stroke, making 80 revolutions per minute, is fitted in the above ship. Assuming a constant steam pressure (effective) of 40 lbs. per sq. in.; find how long the engine would take in pumping out the boiler, assuming that engine and pump together waste one half the energy. Ans. 8 minutes.