WDH dt = -(c[tau]W/p) dp dt - [zeta](H/m)W dl dt

dH + (c[tau]/p) dp + [zeta](H/m) dl = 0,

dH/H + (c[tau]/Hp) dp + [zeta]dl/m) = 0 (4)

Encyclopaedia Britannica, 11th Edition, "Husband" to "Hydrolysis" Volume 14, Slice 1 · 第 109 / 207 题