xamples of
wheels of the noria class from 30 to 90 feet in diameter; the term
_noria_ having been applied to water wheels carrying buckets for raising
water; the Spanish _noria_ having buckets on an endless chain.
Records of a Chinese noria, of 30 feet diameter, made of bamboo, show a
lifting capacity of 300 tons of water per day to a height of 3/4 of the
diameter of the wheel--velocity of current not stated.
For less quantity and greater elevation, these forms of wheel may have
pumps attached to the shaft, by crank, that will give a fair duty for a
high water supply.
For power purposes, as in the plain current wheel, Fig. 23, there are
two principal factors in the problem of power--the velocity of the
current and the area of the buckets or blades.
[Illustration: Fig. 23]
Their efficiency is very low, from 25 to 36 per cent., according to
their lightness of make and form of buckets. A slightly curved plate
iron bucket gives the highest efficiency, thus ( to the current, and an
additional value may also be given by slightly shrouding the ends of the
buckets.
The relative velocity of the periphery of the wheel to the velocity of
the current should be 50 per cent. with curved blades for best effect.
The most useful and convenient sizes for power purposes are from 10 to
20 feet, and from 2 to 20 feet wide, although, as before stated, there
is scarcely a limit under 100 feet diameter for special purposes.
In designing this class of wheels special attention should be given to
the concentration and increase of the velocity of the current by wing
dams or by the narrowing of shallow streams; always bearing in mind that
any increase in the velocity of the current is economy in increased
power, as well as in the size and cost of a wheel for a given power.
The blades in the smaller size wheels should be 1/4 of the radius in
width, and for the larger sizes up to 20 feet, 1/5 to 1/6 of the radius
in width and spaced equal to from 1/4 to 1/3 of the radius.
They should be completely submerged at the lowest point.
For obtaining the horse power of a current wheel, the formula is
Area of 1 blade x velocity of the current in ft. per sec.
----------------------------------------------------------
400
x by the square of difference of velocities of current and wheel
periphery = the horse power; or
A x V 2
------ x (V - v) = h. p.
400
[TEX: \frac{A \times V}{400}
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