Modern Machine-Shop Practice Part 245

Thus, the friction of the cross head guides, of the cross head pin, of the crank pin and of the crank shaft bearings will increase with the amount of resistance offered to the piston motion.

The average pressure on the piston is a difficult thing to find, however, for several reasons.

First, because the pressure in the cylinder may, during the live steam period, vary from that in the steam chest because of the ports being too small or from the pa.s.sages being choked from a defective casting.

Second, because the steam is wire drawn during the time that the slide valve is closing the port to effect the cut off.

Third, because the live steam in the port and pa.s.sage at the time the cut off occurs gives out some power during the period of expansion.

Fourth, because there is some condensation of the steam in the cylinder after the point of cut off, and there is no means of finding by calculation how much loss there may be from this cause.

During the live steam period there is also loss from condensation in the cylinder, but this is made up for by steam from the steam chest.

Fifth, the loss from condensation after the cut off has occurred will vary with the speed of the engine, and is greater in proportion as the piston speed is less, because there is more time for the condensation to occur in.

Sixth, there is some pressure on the piston between the time that the exhaust begins and the piston ends its stroke.

Seventh, because the compression absorbs some of the piston power.

a.s.suming the average pressure on the piston to be known, however, we may calculate the horse power as follows:

_Example._--What is the horse power of an engine whose piston is 20 inches in diameter, and stroke 30, the revolutions per minute being 120, and the average pressure on the piston 60 lbs. per square inch?

Diameter of piston 20 Diameter of piston 20 --- Diameter of piston squared 400 --- .7854 400 ------------ Area of piston = 314.1600 ([<--] these="" two="" ciphers="" neglected.)="" 60="" average="" steam="">

-------- lbs. pressure on piston 18849.60 ([<--] this="" cipher="" neglected.)="" 30="" length="" of="" stroke="" in="">

-------- 565488.0 inch lbs. per stroke.

2 two piston strokes per revolution.

------- 12 ) 1130976 inch lbs. per revolution.

------- 94248 foot lbs. per revolution.

120 revolutions per minute.

------- 1884960 94248 -------- 11309760 foot lbs. per minute.

33000 ) 11309760 ( 342.72 = horse power of engine.

99000 140976 132000 ------ 89760 66000 ----- 237600 231000 ------ 66000 66000 -----

In working out the calculation, the ciphers that are decimals and are on the right hand are neglected or taken no account of, because they represent no value and may therefore be discarded.

[Ill.u.s.tration: Fig. 3355.]

Thus the area of the piston is 314.1600 inches, the two right hand ciphers having no value. Again the lbs. pressure on the piston is 18849.60 lbs., and the right hand cipher, having no value, is discarded.

The inch lbs. per stroke is 565488.0, and the decimal cipher, representing nothing, is discarded when multiplying by the 2.

We have in this case taken no account of the fact that the piston rod prevents the steam from acting against a part of the piston area during one stroke; hence for correct results we must subtract from the area of the piston one half the area of the piston rod.

The horse power thus obtained is that which the engine receives from the steam, and is more than the engine is capable of exerting to drive machinery, because a part of this power is consumed in overcoming the friction of the working parts of the engine.

TESTING THE HORSE POWER OF AN ENGINE.

[Ill.u.s.tration: Fig. 3356.]

The useful horse power of a stationary engine may be readily and accurately obtained by means of a pair of scales, and a brake, as shown in Fig. 3356, which is constructed and used as follows:

On the crank shaft of the engine is a pulley enveloped by a friction brake, which consists of an iron band, to which wooden blocks are fastened.

The ends of the iron band do not meet, but are secured together by a bolt as shown.

By s.c.r.e.w.i.n.g up the bolt the wood blocks are brought to press against the circ.u.mference of the wheel.

This forms a friction brake that would revolve with the wheel, were it not for two arms that are secured to the brake, and rest at the other end upon a block placed upon a pair of scales.

The principle of action of this device is that the amount of friction between the brake and the wheel is weighed upon the scales, and this amount, multiplied by the velocity of the wheel at its circ.u.mference and divided by 33,000, is the horse-power of the engine.

It is necessary, in arranging this brake, to have its end rest upon the scale at the same height from the floor as the centre of the crank shaft, so that the line marked 5' 3" (5 feet 3 inches), which represents the length of the lever, shall stand parallel with the surface of the platform of the scale.

To test the horse-power, we proceed as follows:

Suppose the pressure of the end of the lever on the scale is found by the weight on the scale beam to be 540 lbs., the diameter on which the brake blocks act being 3 feet, the length of the leverage being 5 feet 3 inches, as marked, and the engine making 150 revolutions per minute, and the calculation is as follows:

540 lbs. on scale.

5.25 leverage in feet.

----- 2700 1080 2700 ------ radius of pulley in feet 1.5 ) 2835.00 ( 1890 lbs. at pulley perimeter.

15 --- 133 120 --- 135 135 ---- ...0 ====

Then

3.1416 3 diameter of pulley in feet.

------ 9.4248 circ.u.mference of pulley in feet.

150 revolutions per minute.

------- 4712400 94248 --------- 1413.7200 velocity of pulley perimeter.

1890 pounds at pulley perimeter.

--------- 12723480 1130976 141372 ---------- 2671930.80 foot lbs. per minute.

Then

33000 ) 2671930.80 ( 80.9 264000 ------ 319308 297000 ------ 223080 ======

Answer, 80-9/10 horse power.

In this calculation we have nothing to do with the size of the cylinder or the steam pressure, because the scale beam tells us how many lbs. the brake exerts on the scale, and we treat the brake and brake pulley as levers. Thus by multiplying the lbs. on the scale by the leverage of the brake arm we get the number of lbs. exerted at the centre of the crank shaft, and by dividing this by the radius of the brake pulley we get the number of lbs. on the circ.u.mference, or, what is the same thing, the perimeter of the brake pulley.

By multiplying the circ.u.mference of the pulley in feet by the revolutions per minute, we get the speed at which the pounds travel, and by multiplying this speed by the number of lbs. we get the foot lbs.

per minute, which, divided by 33,000, gives us the effective horse power of the engine.

This effective horse power is correct, because in loading the engine by the brake the crank pin, the cross head guides, etc., are all placed under the same friction as they would be if it was a circular saw, or some other piece of machinery or machine that the engine was driving.

SAFETY VALVE CALCULATIONS.