Page:On Governors.pdf/9

Let $\scriptstyle \ \phi$ be the angle of position of the tube about the vertical axis, then the moment of momentum of the liquid in the tube is

(35)	$H=A{\frac {d\phi }{dt}}+BQ.$

The moment of momentum of the liquid thrown out of the tube in unit of time is

(36)	${\frac {dH'}{dt}}=\rho r^{2}Q{\frac {d\phi }{dt}}+\rho {\frac {r}{k}}Q^{2}cos(\alpha ),$

where $\scriptstyle \ r$ is the radius at the orifice, $\scriptstyle \ k$ its section, and $\scriptstyle \ \alpha$ the angle between the direction of the tube there and the direction of motion.

The energy of motion of the fluid in the tube is

(37)	$W={\tfrac {1}{2}}A\left[{\frac {d\phi }{dt}}\right]^{2}+BQ{\frac {d\phi }{dt}}+{\tfrac {1}{2}}CQ^{2}.$

The energy of the fluid which escapes in unit of time is

(38)	${\frac {W'}{dt}}=\rho gQ(h+z)+{\tfrac {1}{2}}\rho r^{2}Q\left[{\frac {d\phi }{dt}}\right]^{2}+\rho {\frac {r}{k}}Q^{2}cos(\alpha ){\frac {d\phi }{dt}}+{\tfrac {1}{2}}\rho {\frac {r}{k^{2}}}Q^{3}.$

The work done by the prime mover in turning the shaft in unit of time is

(39)	$L{\frac {d\phi }{dt}}={\frac {d\phi }{dt}}\left({\frac {dH}{dt}}+{\frac {dH'}{dt}}\right).$

The work spent on the liquid in unit of time is

(40)	${\frac {dW}{dt}}+{\frac {dW'}{dt}}.$

Equating this to the work done, we obtain the equations of motion

(41)	${\begin{aligned}A{\frac {d^{2}\phi }{dt^{2}}}+B{\frac {dQ}{dt}}+\rho r^{2}Q{\frac {d\phi }{dt}}+\rho {\frac {r}{k}}cos\alpha Q^{2}=L\\B{\frac {d^{2}\phi }{dt^{2}}}+C{\frac {dQ}{dt}}+{\tfrac {1}{2}}\rho {\frac {\rho }{k^{2}}}Q^{2}+\rho gQ(h+z)-{\tfrac {1}{2}}\rho r^{2}\left[{\frac {d\phi }{dt}}\right]^{2}=0\end{aligned}}$
(42)

These equations apply to a tube of given section throughout. If the fluid is in open channels, the values of $\scriptstyle \ A$ and $\scriptstyle \ C$ will depend on the depth to which the channels are filled at each point, and that of $\scriptstyle \ k$ will depend on the depth at the overflow.

In the governor described by Mr. C. W. Siemens in the paper already referred to, the discharge is practically limited by the depth of the fluid at the brim of the cup.

The resultant force at the brim is $\scriptstyle \ f={\sqrt {g^{2}+\omega ^{4}r^{2}}}$.

If the brim is perfectly horizontal, the overflow will be proportional to $\scriptstyle \ x^{n}$ (where $\scriptstyle \ x$ is the depth at the brim), and the mean square of the velocity relative to the brim will be proportional to $\scriptstyle \ x$, or to $\scriptstyle \ Q^{2/3}$.

If the breadth of overflow at the surface is proportional to $\scriptstyle \ x^{n}$ where $\scriptstyle \ x$ is the height above the lowest point of overflow, then $\scriptstyle \ Q$ will vary as $\scriptstyle \ x^{n+3/2}$ and the mean square of the velocity of overflow relative to the cup as $\scriptstyle \ x$ or as $\scriptstyle \ 1/Q^{n+3/2}$.

If $\scriptstyle \ n=-1/2$, then the overflow and the mean square of the velocity are both proportional to $\scriptstyle \ x$.

From the second equation we find for the mean square of velocity

(43)	${\frac {Q^{2}}{k^{2}}}=-{\frac {2}{\rho }}\left(B{\frac {d^{2}\phi }{dt^{2}}}+C{\frac {dQ}{dt}}\right)+r^{2}\left[{\frac {d\phi }{dt}}\right]^{2}-2g(h+z)$