$$
\frac{dz}{dt} = \text{ascent velocity of balloon (ft./sec.)}
$$
$$
\frac{V}{27800} = \text{volume of gas lost per foot of ascent (ft.}^3\text{/ft.)}
$$
$$
A_a = \text{area of appendix opening (ft.}^2\text{)}
$$
The back pressure caused by this velocity:
(9)
$$
\Delta p = \left( \frac{V}{1333} \right)^2 \cdot 14.7 \cdot \frac{d_g}{d_{a0}}
$$
$\Delta p =$ back pressure induced (psi)
$V =$ velocity of escape of gas (ft./sec.)
$d_g =$ density of lifting gas at altitude of balloon (lb./ft. $^3$ )
$d_{a0} =$ 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):
(10)
$$
\Delta p = \frac{\left( \frac{dz}{dt} \cdot \frac{V}{27800} \cdot \frac{1}{A_a} \right)^2}{(1333)^2} \cdot 14.7 \cdot \frac{d_g}{d_{a0}} \quad \text{psi}
$$
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. = $\frac{300}{1015} \cdot \frac{290}{232} \cdot 0.138 \, d_{a0}$ )
$$
\Delta p_{20} = \frac{\left( \frac{800}{60} \cdot \frac{\pi \cdot 20^3}{6 \cdot 27800} \cdot \frac{4}{\pi} \right)^2}{1333^2} \cdot 14.7 \cdot 0.051 = .275 \times 10^{-4} \text{ psi}
$$
It is to be noted that equation (5) can be arrived at by more simple reconstruction of the standard equation for fluid flow:
(11)
$$
\frac{dV}{dt} = C_d \, A_a \sqrt{2gh}
$$
$\frac{dV}{dt} =$ volume rate of flow (ft. $^3$ /sec.)
$C_d =$ a constant of flow
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