Consider a short length dl of the pipe limited by sections A0, A1 at a distance dl (fig. 99). Let p, u be the pressure and velocity at A0, p + dp and u + du those at A1. Further, suppose that in a very short time dt the mass of air between A0A1 comes to A′0A′1 so that A0A′0 = udt and A1A′1 = (u + du) dt1. Let Ω be the section, and m the hydraulic mean radius of the pipe, and W the weight of air flowing through the pipe per second.
From the steadiness of the motion the weight of air between the sections A0A′0, and A1A′1 is the same. That is,
W dt = GΩu dt = GΩ (u + du) dt.
By analogy with liquids the head lost in friction is, for the length dl (see § 72, eq. 3), ζ (u2/2g) (dl/m). Let H = u2/2g. Then the head lost is ζ(H/m)dl; and, since Wdt ℔ of air flow through the pipe in the time considered, the work expended in friction is −ζ (H/m)W dl dt. The change of kinetic energy in dt seconds is the difference of the kinetic energy of A0A′0 and A1A′1, that is,
(W/g) dt {(u + du)2 − u2} / 2 = (W/g) u du dt = W dH dt.
The work of expansion when Ωudt cub. ft. of air at a pressure p expand to Ω(u + du) dt cub. ft. is Ωp du dt. But from (3a) u = cτW/Ωp, and therefore
du / dp = −cτW / Ωp2.
And the work done by expansion is −(cτW/p) dp dt.
The work done by gravity on the mass between A0 and A1 is zero if the pipe is horizontal, and may in other cases be neglected without great error. The work of the pressures at the sections A0A1 is
pΩu dt − (p + dp) Ω (u + du) dt
= −(p du + u dp) Ω dt