p1/G1 = (p0/G0) ([tau]1/[tau]0)
where for air p0 = 2116.8, G0 = .08075 and [tau]0 = 492.6.
v2²/2g = {p0[tau]1[gamma]/G0[tau]0([gamma] - 1)}
{1 - (p2/p1)^([gamma] - 1)/[gamma]}; (2)
or, inserting numerical values,
v2²/2g = 183.6[tau]1 {1 - (p2/p1)^(0.29)}; (2a)
which gives the velocity of discharge v2 in terms of the pressure and
absolute temperature, p1, [tau]1, in the vessel from which the air
flows, and the pressure p2 in the vessel into which it flows.
Proceeding now as for liquids, and putting [omega] for the area of the
orifice and c for the coefficient of discharge, the volume of air
discharged per second at the pressure p2 and temperature [tau]2 is
Q2 = c[omega]v2 = c[omega] [root][(2g[gamma]p1/([gamma] - 1)G1)
(1 - (p2/p1)^([gamma] - 1)/[gamma])]
= 108.7c[omega] [root][[tau]1 {1 - (p2/p1)^(0.29)}]. (3)
If the volume discharged is measured at the pressure p1 and absolute
temperature [tau]1 in the vessel from which the air flows, let Q1 be
that volume; then
p1Q1^[gamma] = p2Q2^[gamma];
Q1 = (p2/p1)^(1/[gamma]) Q2;
Q1 = c[omega] [root][{2g[gamma]p1/([gamma] - 1)G1}
{(p2/p1)^(2/[gamma]) - (p2/p1)^([gamma] + 1)/[gamma]}].