STATIONARY ENGINES.

301

RULES FOR PUMPING ENGINES.

Horse-power .—The Standard fixed upon to represent the workof one horse is 33,000 lbs. raised 1 foot high in one minute. Tofind the horse-power, the quantity of water to be raised is reducedto lbs. and multiplied by the height in feet, and the product dividedby 33,000 expresses the horse-power. A gallon of water weighsexactly 10 lbs., thus any number of gallons can be expressed inlbs. by adding a cipher. Hence the following formula:

Gallons to be raised per minute x iox height ,

- r - -— horse-power.

33000

But in practice about one-fifth must be added for the friction of theengine. Examples:—

Supposing 1000 gallons of water per minute is required to bepumped through a line of piping to a height of 120 feet, and theallowance made for the friction in the pipes is equivalent to a headof water of 150 feet: required the horse-power. Thus we have—

Gals. per minute. Lbs. Height in feet.

IOOO X 10 x 150

-— = 45 '45

33000

to which add Jth for the friction of the engine = 9*09

54*54 horse-power.

Again, supposing 1,440,000 gallons of water is required to bepumped up in the 24 hours the same height, we have—

Lbs. raised1 foot high per

Gals. in 24 hours. Lbs. Height. minute = h.p. mins. hours.

1,440,000 x 10 x 150 -T- 33000 x (60 x 24) = 45*45adding as before £th for friction of the engine. = 9*09

54*54 horse-power.

Another method gives the horse-power as follows:—■

Gals. in 24 hours. Height. Constant.

1,440,000 x 150 4- 4,752,000 = 45*45•gth added = 9.09

54*54 horse-power.

Supposing the quantity is given in cubic feet to be delivered inthe 24 hours, at the same height as before, we have—