HYDRAULIC MACHINERY, FOR MINES 843 



water in feet per second to the head of water, or distance between the centre of the 

 (sluice aperture and the point where the water strikes upon the wheel. 



Example. Required the head of water necessary for a wheel 25 feet diameter, 

 moving with a velocity of 5 feet per second. 



5-f-f =5 x ij = 7'5, velocity of the water in feet. 



259 : 4::7-5 2 : '87, or head of water 10^ inches. Ans. 



For wheels of from 15 to 20 feet diameter, add ^th of the diameter 1 foot. For 

 wheels from 20 to 30 feet diameter add ith of the diameter. This additional head is 

 intended to compensate for the friction of water in the aperture of the sluice to keep 

 the velocity as 3 of water to 2 of the wheel ; thus, in place of '87 feet of head for a 

 25-feet wheel, it will be -87 = 10^ inches + 25||= 10^ inches + 1'3 = 2 feet l^j inch, 

 head of water. 



RULE. (1.) Find the velocity of the water per minute by extracting the square 

 root of the height of the head of the water, viz. from the surface to the middle of the 

 sluice-aperture, and multiply by 8, unless when the opening is small, and the head of 

 water great, or proportionally so, when use from 5'5 to 8 for a multiplier. 



(2.) Find the number of cubic feet and pounds delivered per minute; ascertaining 

 the area of the sluice-aperture in feet, then multiply by the velocity of the water in 

 feet per minute ; divide the product by 1,728, and multiply the quotient by 62'5 Ibs. 



(3.) Multiply the number of pounds delivered per minute by '35, and divide by 

 33,000 ; the quotient is the number of horse-power. 



Example. What is the power of an undershot wheel applied to a stream 2 feet x 

 80 inches from a head 25 feet ? 



(1.) V25 x 6-5 x 60 = 1,950 feet, velocity of water per minute. 

 (2.) 2 x 80 x 1,950 x 12-1,728 = 2166-6 cubic feet delivered per minute. 

 Then 2166-6 x 62-5 = 135,412 Ibs. of water discharged in 1 minute. 

 (3.) 135,412 x -35 -j-33,000 = 1-4 effective horse-power. Ans. 



To find the power of a breast wheel. RULE. Find the effect of an undershot wheel, 

 the head of water of which is the difference of level between the surface and where it 

 strikes the wheel (breast), and add to it the effect of that of an overshot wheel, the 

 height of the head of which is equal to the difference between where the water strikes 

 the wheel and the tail- water ; the sum is the effective horse-power. 



Example. What would be the power of a breast wheel applied to a stream 2 x 80 

 inches 14 feet from the surface, the rest of the fall being 13 feet ? 



V14 x 6-5 x 60 = 1,458-6 feet, velocity of water per minute, 



and 2x80x1,458x12 * 1,728 = 1,620 cubic feet x 62'5 = 101,250 Ibs. water dis- 

 charged in 1 minute. 

 Then 101,250 x -35= 1-07 horse-power as an undershot 



V 11 x 6-5 x 60= 1,290 feet, velocity of water per minute, 



and 2 x 80 x 1,290 x 12 -=-1,728= 1,433 cubic feet x 62*5 = 89,562 Ibs. of water dis- 

 charged in 1 minute. 

 xll height of fall x '35 -r- 33,000 = 20 -3 horses, which added to the above = 



21-37. Ans. 



To find the power of an overshot wheel. RULE. Multiply the weight of water in 

 pounds discharged upon the wheel in one minute by the height or distance in feet 

 from the centre of the sluice-opening to the lower edge of the wheel, then multiply 

 by -68 and divide the product by 33,000, the quotient will be the number of horses'- 

 power. 



Example. The weight of water discharged on a wheel per minute is 50,000 pounds, 

 the height of fall is 26 feet, and diameter of wheel 25 feet, what is the power of the 

 wheel ? 



Total height of fall from centre of sluice opening to bottom of tail-race, 26 ft. ; 

 deduct clearance below, 8 in. Total, 25 ft. 4 in. 



50,000 x 25-4 x -68-5-33,000 = 26^ effective horses'-power. Ans. 



Circle of gyration. The centre or circle of gyration is that point or circle in a wheel 

 into which, if the whole quantity of matter were collected, the same moving force would 

 generate the same angular velocity. Motion ought always to be communicated from 

 that centre whenever it is practible. 



To find the circle of gyration. RULE. Add together twice the weight of the 

 shrouding, buckets, &c., rds of the weight of the arms and the weight of the water ; 

 multiply the sum by the square of the radius ; divide the product by twice the sum 



