Section 4.4.2.1: Analytic solution of gravity outflow problem with constant feed

The dynamic material balance is

$\begin{displaymath}dM/dt = a_{1} - k \sqrt M \end{displaymath}$

Separation of variables gives

$\begin{displaymath}dM / ( a_{1} - k \sqrt M ) = dt \end{displaymath}$

Then integrate from M = M₀ and t = 0 to get

$\begin{displaymath}\int_{M_{0}} dM / ( a_{1} - k \sqrt M ) = \int_{0} dt = t \end{displaymath}$

Try a change of variable of integration to u, where

$\begin{displaymath}u = a_{1} - k \sqrt M \end{displaymath}$

and

$\begin{displaymath}du = - (k/2 \sqrt M) dM \end{displaymath}$

$\begin{displaymath}dM = - (2 \sqrt M / k) du = - (2 (a_{1} - u) / k^{2}) du = 2 ((u - a_{1}) / k^{2}) du \end{displaymath}$

With this substitution the integral over M is transformed to

$\begin{displaymath}\int_{u_{0}} du (2 (u - a_{1})) / (k^{2} u) = (2 / k^{2}) \int_{u_{0}} du ( 1 - a_{1} / u) \end{displaymath}$

where $u_{0} = a_{1} - k \sqrt M_{0}$ .

The integral is

(2/k²) ( u - u₀ - a₁ ( ln u - ln u₀ )

(2/k²) ( u - u₀ - a₁ ln (u / u₀ )

This integral is equal to t - 0, or just t. Thus with a little rearrangement we get

k²t/2 = u - u₀ - a₁ ln (u/u₀)

or expressed in terms of M again,

$\begin{displaymath}k^{2}t/2 = k (\sqrt M_{0} - \sqrt M) - a_{1} ln (\frac{a_{1} - k \sqrt M}{a_{1} - k \sqrt M_{0}} ) \end{displaymath}$

Last modified: Tue Aug 4 12:53:22 BST