Filed under: Water Power | Tags: measure water power, water head, water wheels
Let us attack the problem of water-power in another way. _A farmer
wishes to install a water wheel that will deliver 10 horsepower on the
shaft, and he finds his stream delivers 400 cubic feet of water a
minute. How many feet fall is required?_ Formula:
33,000 x horsepower required
(C) Head in feet = ——————————
Cu. Ft. per minute x 62.5
Since a theoretical horsepower is only 75 per cent efficient, he would
require 10 x 4/3 = 13.33 theoretical horsepower of water, in this
instance. Substituting the values of the problem in the formula, we
have:
33,000 x 13.33
Answer: Head = —————- = 17.6 feet fall required.
400 x 62.5
_What capacity of wheel would this prospect (400 cubic feet of water a
minute falling 17.6 feet, and developing 13.33 horsepower) require?_
By referring to the table of velocities, we find that the velocity for
17.5 feet head (nearly) is 33.6 feet a second. Four hundred feet of
water a minute is 400/60 = 6.67 cu. ft. a second. Substituting these
values, in formula (B) then, we have:
Answer: Capacity of wheel =
144 x 6.67
———- = 28.6 square inches of water.
33.6
