But from (3a)
pu = constant,
p du + u dp = 0,
and the work of the pressures is zero. Adding together the quantities of work, and equating them to the change of kinetic energy,
W dH dt = −(cτW/p) dp dt − ζ (H/m) W dl dt
dH + (cτ/p) dp + ζ (H/m) dl = 0,
dH/H + (cτ/Hp) dp + ζdl / m = 0
(4)
But
u = cτW / Ωp,
and
H = u2/2g = c2τ2W2 / 2gΩ2p2,
∴ dH/H + (2gΩ2p / cτW2) dp + ζdl / m = 0.