Differentiating, we get

dH = dz + dp/G + vdv/g, (1)

for the increment of energy between Q and P. But

dz = PQ cos [phi] = ds cos [phi];

.: dH = dp/G + v dv/g + ds cos [phi], (1a)

where the last term disappears if the motion is in a horizontal plane.

Now imagine a small cylinder of section [omega] described round PQ as

an axis. This will be in equilibrium under the action of its

centrifugal force, its weight and the pressure on its ends. But its

volume is [omega] ds and its weight G[omega]ds. Hence, taking the

components of the forces parallel to PQ--

[omega]dp = Gv²[omega] ds/g[rho] - G[omega] cos [phi] ds,

where [rho] is the radius of curvature of the stream line at Q.

Consequently, introducing these values in (1),

dH = v² ds/g[rho] + v dv/g = (v/g)(v/[rho] + dv/ds) ds. (2)

Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 46 / 207 题