Filling Process

The governing equation is

$${d \bar{P} \over d \bar{t}} - k \bar{M} f[M] \bar{P} ^{3k -1 \over 2k} = 0$$ (11.42)

For a choked flow the entrance Mach number to the tube is at its maximum, $M_{max}$ and therefore $\bar{M} = 1$ . The solution of equation (11.37) is obtained by noticing that $\bar{M}$ is not a function of time and by variable separation results in

$$\int _{0}^{\bar{t}} {d\bar{t}} = \int _{1}^{\bar{P}} {d\bar{P} \o... ...ver {k \bar{M} f[M]}} \int _{1}^{\bar{P}} {{\bar{P}^{1-3k \over 2k}} d\bar{P} }$$ (11.43)

direct integration of equation (11.38) results in

$$\bar{t} = {2 \over (k-1) \bar{M} f[M]} \left[ \bar{P}^{1 -k \over 2k} -1 \right]$$ (11.44)

It has to be realized that this is a reversed function. Nevertheless, with today computer this should not be a problem and easily can be drawn as shown here in Figure (11.5).

Figure 11.6: The reduced time as a function of the modified reduced pressure

The Figure shows that when the modified reduced pressure equal to one the reduced time is zero. The reduced time increases with decrease of the pressure in the tank.

At some point the flow becomes chokeless flow (unless the back pressure is a complete vacuum). The transition point is denoted here as $chT$ . Thus, equation (11.39) has to include the entrance Mach under the integration sign as

$$\bar{t} -\bar{t}_{chT} = \int _{P_{chT}}^{\bar{P}} {1 \over {k \bar{M} f[M]}} {{\bar{P}^{1-3k \over 2k}} d\bar{P} }$$ (11.45)