By referring to the table of velocity above, (or by using the formula), we find that water under a head of 13.5 feet, has a spouting velocity of 29.5 feet a second. This means that a solid stream of water 29.5 feet long would pass through the wheel in one second. _What should be the diameter of such a stream, to make its cubical contents 376 cubic feet a minute or 376/60 = 6.27 cubic feet a second?_ The following formula should be used to determine this: 144 × cu. ft. per second (B) Sq. Inches of wheel = -------------------------- Velocity in ft. per sec. Substituting values, in the above instance, we have: Answer: Sq. Inches of wheel = 144 × 6.27 (Cu. Ft. Sec.) --------------------------- = 30.6 sq. in. 29.5 (Vel. in feet.) That is, a wheel capable of using 30.6 square inches of water would meet these conditions. _What Head is Required_ 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 × horsepower required (C) Head in feet = ------------------------------ Cu. Ft. per minute × 62.5 Since a theoretical horsepower is only 75 per cent efficient, he would require 10 × 4/3 = 13.33 theoretical horsepower of water, in this instance. Substituting the values of the problem in the formula, we have: 33,000 × 13.33 Answer: Head = ---------------- = 17.6 feet fall required. 400 × 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 × 6.67 ---------- = 28.6 square inches of water. 33.6 _Quantity of Water_ Let us take still another problem which the prospector may be called on to solve: _A man finds that he can conveniently get a fall of 27 feet. He desires 20 actual horsepower. What quantity of water will be necessary, and what capacity wheel?_ Twenty actual horsepower will be 20 × 4/3 = 26.67 theoretical horsepower. Formula: 33,000 × Hp. required (D) Cubic feet per minute = --------------------- (Head in feet × 62.5) Substituting values, then, we have: Cu. ft. per minute = 33,000 × 26.67 -------------- = 521.5 cubic feet a minute. 27 × 62.5