Water wheels of various forms for this purpose have been used from time immemorial in Europe, Asia and Egypt, where the record gives examples 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} \times (V — v)^2 = h. p.]

in which A equals the area of blade in square feet, V and v velocities of current and wheel periphery respectively, in feet per second.  Thus, for example, a wheel 10 feet in diameter with blades 6 feet long and 1 foot in width, running in a stream of 5 feet per second—­assuming the wheel to be giving as much power as will reduce its velocity to one half that of the stream—­the figures will be