AND PUMPS.
The following is an old rule for obtaining the power of condensing engines:— Square the diameter of the cylinder and multiply the sum by .7854, the product by 10, and the whole product by 144, which will show the number of pounds the engine will lift one foot high in a minute. Divide the number of pounds by 33,000, and the result will be the horse-power. In the above calculation, the pressure per square inch is estimated at 10, independent of friction, with 9 revolutions per minute and 8 feet stroke.
The example is thus:— What is the power of a steam-engine the cylinder of which is 50 inches in diameter, and the stroke 8 feet? 50 X 50 = 2500 .7854 1963.5000 10 19635.0000 144 33000)2727440.0000(82,344 Ans. 264000 87440 66000 1440 2,727,440 pounds lifted one foot high per minute, equal 82 horse-power, with a frac tion over.
A better rule for estimating the nominal horse-po'er of a condensing engine of a constant uniform pressure of seven pounds per square inch is thus:— Example.
Square the diameter of the cylinder in inches, and multiply by the cube root of the stroke in feet, and the result by .0213; the product is the nominal horse-power. What is the power of an engine with 30 inch cylinder and 6 feet stroke? = 900 and 1/ = 1.817: hence 900 + 1817 X .0213 = 34.8 horse-power.
The effective power of any engine may be obtained by this rule:— Find the area of the piston in inches, and multiply the same by the uniform pressure of the steam in pounds per square inch, and the result by the velocity of the piston in feet per minute. Deduct 21 pounds per square inch for friction, and divide the re mainder by 33,000: the quotient is the effective horse-power.
Example.
What is the effective power of an engine whose cylinder is 40 inches in diameter, the piston's velocity 200 feet per minute, and the pressure 40 pounds per square inch? X .7854 = 1256.64 inches area and 40 pounds pressure. 1256.64 X 40 X 200 = 10049982, with friction deducted = 33000)10049982(3042 horse-power.
Having a certain amount of work to accomplish, or a certain load to lift, required the power to lift it at a certain rate per minute.
RULE.—Reduce the weight to be lifted to pounds, multiply by the feet through which the load is to be lifted per minute, and divide by 33,000: the quotient gives the horse-power required.
Example.
Required the power to lift 5 tons 500 feet per minute.
2240 lbs. X 5 = 11200 lbs. X 500 = 33000)5600000(169.03 horse-power.