Testing the Horse Power of an Engine

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 screwing up the bolt the wood blocks are brought to press against the circumference 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 circumference 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

1 080

27 00

radius of pulley in feet 1.5 ) 28 35.00 ( 1890 lbs. at pulley perimeter.

15

13 3

12 0

1 35

1 35

. ..0

Then

3.141 6

3 diameter of pulley in feet.

9.424 8 circumference of pulley in feet.

1 50 revolutions per minute.

4712 400

9424 8

1413.7 200 velocity of pulley perimeter.

18 90 pounds at pulley perimeter.

12723 480

113097 6

141372

267193 0.80 foot lbs. per minute.

Then

33000 ) 267193 0.80 ( 80.9

264000

3193 08

2970 00

223 080

Answer, 80⁄10 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 circumference, or, what is the same thing, the perimeter of the brake pulley.

By multiplying the circumference 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.