$ \frac{dz}{dt} $ ascent velocity of ballcon (ft./sec.)
$ \frac{V}{27800} $ volume of gas lost per foot of ascent (ft.3/ft.)
$ A_{a} $ area of appendix opening (ft.2)
The back pressure caused by this velocity:
1) $ \Delta p=\left(\frac{V}{1333}\right)^{2}\cdot14.7\frac{d g}{d a_{0}} $
$ \Delta p $ = back pressure induced (psi)
$ V $ = velocity of escape of gas (ft./sec.)
$ d_{g} $ = density of lifting gas at altitude of balloon (lb./ft. $ ^{3} $ )
$ d_{a_{0}} $ = density of air at sea level (lb./ft. $ ^{3} $ )
14.7 = pressure of air at sea level (psi)
1333 = escape velocity of air to produce back pressure of 1 atmosphere at sea level ( ft/sec )
or, combining equation (8) and (9):
$$
\Delta p = \frac {\left(\frac {d z}{d t} \cdot \frac {V}{2 7 8 0 0} \cdot \frac {1}{A _ {0}}\right) ^ {2}}{(1 3 3 3) ^ {2}} \cdot 1 4. 7 \frac {d a}{d a _ {0}} \quad p s i
$$
As an example, let us find the back pressure induced in a 20' diameter balloon with a 1' diameter opening ascending at 800 ft./minute, as it becomes full at 30,000 ft. (density of helium @ 30,000 ft. $ = \frac{300}{1013}\cdot \frac{290}{232}\cdot 0.138\mathrm{d}_{a_{0}} $ )
$$
\Delta p _ {2 0} = \frac {\left(\frac {8 0 0}{6 0} \cdot \frac {\pi \cdot 2 0 ^ {3}}{6 \cdot 2 7 8 0 0} \cdot \frac {4}{\pi}\right) ^ {2}}{1 3 3 3 ^ {2}} \cdot 1 4. 7 \cdot 0. 0 5 1 = . 2 7 5 \times 1 0 ^ {- 4} p s i
$$
It is to be noted that equation (5) can be arrived at by more simple reconstruction of the standard equation for fluid flow:
$$
\frac {d V}{d t} = C _ {d} A _ {a} \sqrt {2 g h}
$$
$$
\frac {d V}{d t} = \text {v o l u m e r a t e o f f l o w (f t .} ^ {3} / \sec .)
$$
$ C_{d} $ a constant of flow