Compressible Flow Free Textbook

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With Fanno Flow

The relative Volume, $\bar{V}(t) = 1$ , is constant and equal one for a completely rigid tank. In such case, the general equation (11.17) ``shrinks'' and doesn't contain the relative volume term.

A reasonable model for the tank is isentropic (can be replaced polytropic relationship) and Fanno flow are assumed for the flow in the tube. Thus, the specific governing equation is

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

(11.37)

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 (11.32) is obtained by noticing that $\bar{M}$ is not a function of time and by variables separation results in

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

(11.38)

direct integration of equation (11.33) results in

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

(11.39)

It has to be realized that this is ``reversed'' function i.e. $\bar{t}$ is a function of P and can be reversed for case. But for the chocked case it appears as

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

(11.40)

The function is drawn as shown here in Figure (11.5).

**Figure 11.5:** The reduced time as a function of the modified reduced pressure

The Figure (11.5) 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 certain point the flow becomes chokeless flow (unless the back pressure is complete vacuum). The transition point is denoted here as The big struggle look for suggestion for better notation.. Thus, equation (11.34) has to include the entrance Mach under the integration sign as

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

(11.41)

For practical purposes if the flow is choked for more than 30% of the charecteristic time the choking equation can be used for the whole range, unless extra long time or extra low pressure is calculated/needed. Further, when the flow became chokeless the entrance Mach number does not change much from the choking condition.

Again, for the special cases where the choked equation is not applicable the integration has to be separated into zones: choked and chokeless flow regions. And in the choke region the calculations can use the choking formula and numerical calculations for the rest.
$\begin{examl} A chamber with volume of 0.1[$m^3$] is filled with air at pressure... ...ubber tube with $f = 0.025$, $d=0.01[m]$ and length of $L = 5.0[m]$ \end{examl}$
Solution

Next: Filling Process Up: Rapid evacuating of a Previous: Rapid evacuating of a Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21