where [Omega] is the section of the pipe and u the velocity of the
air. Combining (1) and (3),
[Omega]up/W = c[tau] = constant. (3a)
[Illustration: FIG. 99.]
Since the work done by gravity on the air during its flow through a
pipe due to variations of its level is generally small compared with
the work done by changes of pressure, the former may in many cases be
neglected.
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 [Omega] 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 = G[Omega]u dt = G[Omega](u + du) dt.
By analogy with liquids the head lost in friction is, for the length
dl (see § 72, eq. 3), [zeta](u²/2g)(dl/m). Let H = u²/2g. Then the
head lost is [zeta](H/m)dl; and, since Wdt lb. of air flow through the
pipe in the time considered, the work expended in friction is
-[zeta](H/m)Wdl 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 [Omega]udt cub. ft. of air at a pressure p
expand to [Omega](u + du)dt cub. ft. is [Omega]p du dt. But from (3a)
u = c[tau]W/[Omega]p, and therefore
du/dp = -c[tau]W/[Omega]p².
And the work done by expansion is -(c[tau]W/p)dpdt.
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
p[Omega]u dt - (p + dp)[Omega](u + du) dt
= -(pdu + udp)[Omega] dt