But

u = c[tau]W/[Omega]p,

and

H = u²/2g = c²[tau]²W²/2g[Omega]²p²,

.: dH/H + (2g[Omega]²p/c[tau]W²) dp + [zeta] dl/m = 0. (4a)

For tubes of uniform section m is constant; for steady motion W is

constant; and for isothermal expansion [tau] is constant. Integrating,

log H + g[Omega]²p²/W²c[tau] + [zeta]l/m = constant; (5)

for

l = 0, let H = H0, and p = p0;

and for

l = l, let H = H1, and p = p1.

log (H1/H0) + (g[Omega]²}/W²c[tau]) (p1² - p0²) + [zeta]l/m = 0. (5a)

where p0 is the greater pressure and p1 the less, and the flow is from

A0 towards A1.

By replacing W and H,

log (p0/p1) + (gc[tau]/u0²p0²)(p1² - p0² + [zeta]l/m = 0 (6)

Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 110 / 207 题