QUESTIONS RELATING TO PUMPS, &c. 260. Given the diameter and length of the stroke of a brine-pump, also the number of strokes per minute, and the ratio of the fulness of the pump, to find how many cubic feet of water it will extract in an hour. RULE CVI. 1o. Find the area of a horizontal section of the pump by multiplying the square of the diameter by 7854. 2°. Multiply this by the inches in the stroke; the product is the capacity of the ритр. 3°. Next, multiply this product (No 2°) by the fraction it is full; the result is the quantity delivered at one stroke. 4°. Next, find the quantity delivered in an hour by multiplying the last result (No. 3°) by the strokes per minute, and by (60) the number of minutes in an hour. 5°. Divide the cubic inches delivered in an hour as found above (No. 4°) by the number of cubic inches in a cubic foot (1728); the result is the cubic feet of water discharged by the pump in an hour. EXAMPLES. Ex. 1. How many cubic feet of water will be extracted in an hour by a brine-pump 3 inches in diameter and 10 inches stroke, making 18 strokes per minute; the pump being full each stroke? 1. Capacity of pump = 7854 X 32 X 1070-686 cubic inches. 2. .. quantity delivered at each stroke = X 70-686 = 47°124 cubic inches. 3. And quantity delivered in an hour 18 X 60 X 47°124 = 50893'92. 4. But a cubic foot is equal to 1728 cubic inches .. cubic feet discharged 49803:22. 1728 An examination of the foregoing solution will show that the quantities 7854 and 60 occur as constant multipliers, and that 1728 is a constant divisor. Now, if we divide the product of 7854 X 60 by 1728, the constant multiplier 02727 is obtained, and the work will be somewhat abridged. The work will stand thus: 32 X 10 X X 18 X *02727 = 9 × 10 = (90) × } = (60) × 18 = (1080) X *02727= 29'4516 cubic feet. Ex. 2. How many cubic feet of water will be extracted in an hour by a brine-pump 3 inches in diameter and 12 inches stroke, making 15 strokes a minute; the pump being full at each stroke? 34 inches 325 and 3.252 = 10.5625. 1. Capacity of pump = 7854 X 10'5625 X 12 = 99*54945. 2. .. quantity delivered at each stroke X 99'54945 = 74'66209. 3. And quantity delivered in an hour 15 X 60 X 74·6621 = 67195*8900. 4. But a cubic foot equals 1728 cubic inches... cubic feet discharged 67195-89 261. Given the diameter and the length of the stroke of a pump, also the number of revolutions per minute, and the ratio of the fulness of the pump, to find how long it will take to pump a ballast tank out containing a given number of tons of sea water. RULE CVII. 1°. Proceed according to Rule CVI, Nos. 1° to 5o, inclusive. 2°. Reduce the tons of water contained in the tank to cubic feet by multiplying them by the number of cubic feet (35) in a ton. 3°. Divide this last product by the cubic feet of water discharged by the pump in an hour (No. 2°); the quotient is the time required to discharge the tank. EXAMPLES. Ex. 1. How long will it take to pump a ballast tank out containing 150 tons of sea water, the diameter of the ballast donkey pump is 5 inches, and length of stroke 10 inches (double-acting), making 100 strokes per minute, of the pump being empty space? NOTE.-The question as worded at the examinations is 100 strokes, but it is properly 100 revolutions, and as it is double-acting it must be doubled. 1. Capacity of pump=7854 X 52 X 10 = 196.35 cubic inches. 2. .. quantity delivered at each stroke = × 196*35 = 147′2625. 3. And quantity delivered in an hour = (100 X 2) X 60 X 147′2625 = 1767150. 4. But a cubic foot is equal to 1728 cubic inches, .. cubic feet discharged in an hour = 1767160 1728 =1022.656. 5. Cubic feet contained in tank = 150 X 35 = 6. Time required to discharge water in tank = 6250.000 5250 cubic feet. 2. Ex. 2. How long will it take to pump a ballast tank out containing 165 tons of water the diameter of the ballast donkey-pump is 6 inches, and length of stroke 12 inches (double acting), making 120 strokes per minute, of the pump being empty space. 62 × 7854 X 12 X X 60 X 120 X 2 165 70.888 1728 × 35 70 686 tons pumped per hour. 888 2h 20m 3o, time required to pump the tank out. Ans.: 6 × 6 × •7854 (= 28°2744) X 12 × 7 (= 296.8812) × 60 × 120 (= 213754464) X 2 = 4275089°28÷ 1728 ÷ 35 = 70-686 tons pumped per hour. 165 tons 70-686 = 2h 20m 3o, the time required to pump the tank out. 2. 9" 14 120 231'43", EXAMPLES. Ans.: Time required to pump the tank out. 3b 37m 188 I 27 18 Ex. I. The water ballast donkey working alone can discharge the water ballast in 2 hours; the boiler donkey working alone can discharge it in 8 hours: in what time will it be discharged if both pumps are working? In 1 hour ist empties of tank. 2nd "" Hence in 1 hour both together discharge }+}=}+} = § of tank, and the whole work I divided by § gives the hours, &c., they will require to do it jointly; thus: NOTE.—To divide by a fraction, învert the divisor and multiply. Or, stating the question directly we should say # : I :: time required : But the greater the fraction representing the hourly work done, the smaller must be the time required for any given quantity of work. Hence : 1h time required = § of 1h, and § of 1h 1h 36m. : Or again, here 1 in which the fraction is done is obviously to be increased in that ratio which will turn § into 1, or the whole, and this ratio is, for X = 1. The following rule will perhaps be preferred by some as being simpler than the preceding methods. RULE CVIII. Take the sum of the times in which they separately discharge the tank. 2o. Find the product of these times. 3°. Divide the product found by No. 2° by the sum found in No. 1°; the quotient is the time they will require to do it jointly. Then (ex. above) since they separately discharge the tank in 2 hours and 8 hours respectively, we have Ex. 2. The water ballast donkey working alone can discharge the water ballast in 2 hours, the boiler donkey working alone can discharge it in 9 hours, in what time will it be discharged if both pumps are working? Then since they both (together) discharge of the tank in 1 hour, the whole work 1 divided by gives the hours, &c., they will require to do it jointly. Here, again Or, since they separately discharge the tank in 2 hours and 9 hours respectively, we have EXAMPLES FOR PRACTICE. In each of the following examples given the times in which the ballast donkey and boiler donkey can separately discharge the water ballast tank, to find the time in which both can do the work. 262. Given the diameter of cylinder, number of revolutions, length of stroke, and the cut-off of a pair of engines, to find the cubic feet of steam used. RULE CIX. 1o. Square the diameter of cylinder, multiply this by 7854, by the number of revolutions per minute, by 60, by double the stroke in inches, by the cut-off, and by 2 (being a pair of engines), the result is cubic inches of steam used. 2o. Divide this last by 1728, the quotient is the cubic feet of steam used. EXAMPLES. Ex. 1. How many cubic feet of steam will be used per hour by a pair of engines making 47 revolutions per minute, the diameter of the cylinders being 42 inches, the stroke 42 inches, the cut-off being at of the stroke from the beginning? |