398--for the pressure of a current upon a solid body immersed in it.
This equation, F = 1.8 m A v squared / 2g, where m is the weight of a unit of
volume of the fluid--say 62 lb.--A is the area exposed, and v the relative
velocity of the current. Mr. Jagn finds that the maximum of efficiency is
obtained when the rope moves at one-third the velocity of the stream. If
this velocity be 3 feet per second, we shall have v = 2. and we then get F
= 7 lb. per sq. ft. very nearly. Now 1 sq. meter = 10.76 sq. ft., and a
speed of 1 ft. per second (which is that of the rope) is 60 ft. per
minute. Hence the H.P. realized in the same case as that taken above will
be 7 x 10.76 x 60 / 33,000 = 0.137 H.P. The difference between the two
values is very large, but Rankine, of course, depends entirely on the
value of the constant 1.8, which is quite empirical, and is for a flat
band instead of a hollow parachute. Taking, however, his smaller figure,
and an area of 544 square inches, which Mr. Jagn has actually employed, we
get a gross power of = 0.137 x 544 = 7.43 H.P. Hence it will be seen that
the amount of power which can be realized by the system is far from being
inconsiderable.
Lastly, we may point out that the durability of the apparatus will be
considerable.
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