844 



HYDRAULIC MACHINERY, FOR MINES 



of tho weight of the shrouding arms added to the weight of the water, and the square 

 root of the quotient is the radius of tho circle of gyration from tho centre of suspen- 

 sion, nearly. 



Example. Required the distance of tho centre of gyration from tho centre of 

 suspension in a water-wheel 22 feet diameter, shrouding, buckets, &c. = 18 tons, arms = 

 12 tons, and water 10 tons, 



Then V{(2 x 18 + | x 12+ 10) x H 2 -r [(18 4-12) x 2 + 10]} 



= <v/{(36 + 8 + 10) x 121 4-(30 x 2 + 10)} = <v/{(54 x 121)~70} 

 A '(6,534 ---70) = V93-34 = 9 < 7 feet from the centre of suspension. 



To find the horse-power of Water-wheels according to a method employed in 'various 

 parts of America : 



TABLE VIII. Coefficients. 



RULE. Multiply the product of the coefficients (see Table VIII.) opposite the given 

 head, the area of the opening in the sluice in square inches, the entire head in feet 

 (in case of the overshot or breast, the head by 40 and the fall by 78) by the efficiency 

 of the class of wheel, pointing off 6 figures as decimals. 



Example 1. The dimensions of a stream are 2 inches by 200 inches, the head 

 2 feet 3 inches and the fall 10 feet ; what is its horse-power applied to a breast wheel ? 



2 inches x 200 = 400 square inches opening. 



Coefficient by Table VIII. opposite 2 feet 3 inches = 64. 



Efficiency by Table VII. arising from weight = 78. 



Efficiency by Table VII. arising from impulse = 40. 



Head 2-3 = 2-25 feet. 



Product of efficiency and head 2-25 x 40 = 90. 



Product of efficiency and fall 10 x 78 = 780. 



Sum of products 780 + 90 = 870. 



Then 870 x 400 x 64 = 22-27 horses'-power. Ans. 



Example 2. The dimensions of a stream are 10 inches square, tho head 25 feet; 

 what is its horse-power applied to a good turbine ? 



Square inches in opening 10 x 10 = 100. 



Coefficient by Table VIII . opposite head (25 feet) = 213. 



Efficiency by Table VII. of turbine = -68 



Then 100 x 213 x -68 x 25 = 36-21 horses'-power. Ans. 



To find the horse-power of a single or double direct-acting pressure-engine. RULE. 

 Multiply the net area of the piston in inches by the vertical pressure in pounds per 

 square inch ; then, for a single-acting engine, by one-half tho journey made by the 

 piston per minute ; deduct for friction and absorption of power, divide by 33,000, 

 and the quotient will bo the effective horse-power. Should the engine bo double-acting, 

 double the figures of the quotient. 



