The ``Simple'' General Case

Balloon Problem The relationship between the pressure and the volume from the physical point of view must be monotonous. Further, the relation must be also positive, increase of the pressure results in increase of the volume (as results of Hook's law. After all, in the known situations to this author pressure increase results in volume decrease (at least for ideal gas.).

In this analysis and previous analysis the initial effect of the chamber container inertia is neglected. The analysis is based only on the mass conservation and if unsteady effects are required more terms (physical quantities) have taken into account. Further, it is assumed the ideal gas applied to the gas and this assumption isn't relaxed here.

Any continuous positive monotonic function can be expressed into a polynomial function. However, as first approximation and simplified approach can be done by a single term with a different power as

$$V(t) = a P^{n}$$ (11.52)

When $n$ can be any positive value including zero, 0 . The physical meaning of $n =0$ is that the tank is rigid. In reality the value of $n$ lays between zero to one. When $n$ is approaching to zero the chamber is approaches to a rigid tank and vis versa when the $n \rightarrow 1$ the chamber is flexible like a balloon.

There isn't a real critical value to $n$ . Yet, it is convenient for engineers to further study the point where the relationship between the reduced time and the reduced pressure are linear^11.6 Value of $n$ above it will Convex and below it concave.

$${d \over d \bar{t}} \left( \bar{P}^{1 + nk -k \over k} \right) - \bar{P}^{k +1 \over 2k} \bar{M} f[M] = 0$$ (11.53)

Notice that when $n=1$ equation (11.49) reduced to equation (11.43).

After carrying-out differentiation results

$${1 + nk -k \over k} \bar{P}^{1 +nk -2k \over k} {d \bar{P} \over d \bar{t}} - \bar{P}^{k +1 \over 2k} \bar{M} f[M] = 0$$ (11.54)

Again, similarly as before, variables are separated and integrated as follows

$$\int_0^{\bar{t}} dt = {1+ nk -k\over k}\int _{1}^{\bar{P}} {\bar{P}^ {1 +2nk -5k \over 2k} d\bar{P} \over \bar{M} f[M] }$$ (11.55)

Carrying-out the integration for the initial part if exit results in

$$\bar{t} = {2k^2 \over \bar{M} f[M] (3k -2nk - 1) (1+k)} \left[ 1 - \bar{P}^{3k -2nk -1 \over 2k} \right]$$ (11.56)

The linear condition are obtain when

$$3k -2nk -1 = 1 \longrightarrow n = { 3k -2 \over 2k}$$ (11.57)

That is just bellow 1 ( $n = 0.785714286$ ) for $k=1.4$ .