The old reaction wheel consisted of a vertical pipe balanced on a vertical axis, and supplied with water (fig. 183). From the bottom of the vertical pipe two or more hollow horizontal arms extended, at the ends of which were orifices from which the water was discharged. The reaction of the jets caused the rotation of the machine.
Let H be the available fall measured from the level of the water in the vertical pipe to the centres cf the orifices, r the radius from the axis of rotation to the centres of the orifices, v the velocity of discharge through the jets, α the angular velocity of the machine. When the machine is at rest the water issues from the orifices with the velocity √ (2gH) (friction being neglected). But when the machine rotates the water in the arms rotates also, and is in the condition of a forced vortex, all the particles having the same angular velocity. Consequently the pressure in the arms at the orifices is H + α2r2/2g ft. of water, and the velocity of discharge through the orifices is v = √ (2gH + α2r2). If the total area of the orifices is ω, the quantity discharged from the wheel per second is
Q = ωv = ω √ (2gH + α2r2).
While the water passes through the orifices with the velocity v, the orifices are moving in the opposite direction with the velocity αr. The absolute velocity of the water is therefore
v − αr = √ (2gH + α2r2) − αr.
The momentum generated per second is (GQ/g)(v − αr), which is numerically equal to the force driving the motor at the radius r. The work done by the water in rotating the wheel is therefore
(GQ/g) (v − αr) αr foot-pounds per sec.
The work expended by the water fall is GQH foot-pounds per second. Consequently the efficiency of the motor is
| η = | (v − αr) αr | = | {√ (2gH + α2r2) − αr} αr | . |
| gH | gH |
Let