2 
Transactions Texas Academy of Science. 
[48] 
If we know the flow in cubic feet (Q) and the fall h in feet, the work 
(W) that can he done per second hy the total flow can be found from 
the following : 
W=62.5 Qh foot-pounds. 
To reduce this to horse power divide by 550. 
No. of H. P . = 62 , 5 Qh ^ 5 _Qh - 
550 44 
No machine can develop all of the' power in the water on account of 
frictional resistances. They develop only a certain fraction of the the- 
oretic power. This fraction reduced to per cent, is called the efficiency 
of the machine. A first-class turbine can develop f or 80 per cent, of the 
theoretic power. Then, we have 
No. of H. P. = -*-Qh = ^ (A). 
44 o J l 
Rule: To find the H. P. that can be developed under favorable con- 
ditions multiply the flow in Q cubic feet per second by the effective head 
h, and divide the result by 11. 
If the flow is given in gallons per minute, which is often the case, let 
G=gallons per minute. As each gallon weighs 8^ pounds, the work in 
foot-pounds per minute will he given by 
W=Grx8Jxh foot-pounds. 
25 Gh 4 Gh 
No. of H. P. 
No. of H. P. 
3 X 33000 X 5 
Gh 
4950 
5000 
nearly (B). 
For an efficiency of f, the last formula becomes 
G h 
No. of H. P. 
5280 
(C). 
This formula can be easily remembered: Multiply the number of 
gallons per minute by the head in feet and divide the product by the 
number of feet in a mile. 
It often happens that for easy transmission the water power is con- 
verted first into electric energy and then transmitted to the locality where 
it is needed. One horse power is equal to 746 watts, the unit of work in 
electrical measurements. 
One kilowatt equals 737 foot-pounds per second. 
One kilowatt = J T 0 4 0 /-X one II. P* = f one H. P., nearly. 
Substituting this value in (A) and (B) we have 
No. of K. W.S~ (D). 
For an efficiency of 78 per cent. : 
No. kilowatts = ^ (E). 
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