W = p0u0[Omega]/c[tau];
.: t = gdc[tau](p0³ - p1³)/6[zeta]p0³u0³,
= [zeta]^(½)l^(3/2)(p0³ - p1³)/6(gc[tau]d)^(½)(p0² - p1²)^(3/2); (15)
If [tau] = 521°, corresponding to 60° F.,
t = .001412 [zeta]^(½)l^(3/2)(p0³ - p1³)/d^(½)(p0² - p1²)^(3/2); (15a)
which gives the time of transmission in terms of the initial and final
pressures and the dimensions of the tube.
_Mean Velocity of Transmission._--The mean velocity is l/t; or, for
[tau] = 521°,
u_mean = 0.708 [root]{d(p0² - p1²)^(3/2)/[zeta]l(p0³ - p1³)}. (16)