19.5 35.4
20 35.9
20.5 36.3
21 36.8
21.5 37.2
22 37.6
22.5 38.1
23 38.5
23.5 38.9
24 39.3
24.5 39.7
25 40.1
26 40.9
27 41.7
28 42.5
29 43.2
30 43.9
31 44.7
32 45.4
33 46.1
34 46.7
35 47.4
36 48.1
37 48.8
38 49.5
39 50.1
40 50.7
41 51.3
42 52.0
43 52.6
44 53.2
45 53.8
46 54.4
47 55.0
48 55.6
49 56.2
50 56.7
55 59.5
60 62.1
65 64.7
70 67.1
75 69.5
80 71.8
85 74.0
90 76.1
95 78.2
100 80.3
200 114.0
300 139.0
400 160.0
500 179.0
1000 254.0

_In the above example, we found that 376 cubic feet of water a minute,
under 13.5 feet head, would deliver 7.2 actual horsepower. Question:
What size wheel would it be necessary to install under such
conditions?_

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 x 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 x 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 man