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, 809⁄10 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.
SAFETY VALVE CALCULATIONS.
Among the most frequent questions asked in an engineer’s examination are those relating to the safety valves of boilers.
These questions may be easily answered from a study of the following: