# stirred tank model; DAE versus ODE model

I do have a stirred tank reactor with two inlets and one outlet. Several components enter the reactor at inlet 0 and particles at inlet 1. All component from inlet 0 adsorb on the particles from inlet 1.

I can write a mass balance like:

$$C \cdot V_{liquid} + Q \cdot V_{particles} = 0$$

C is the concentration in the liquid, Q is the concentration on the particles, V are volumes (or flow rates).

The ODE may look like (k,l,m are parameters):

$$\frac{dQ}{dt} = k \, (1 - Q[0])^l * m$$

Accordingly I can write

$$\frac{dC}{dt} = - V_{particles}/V_{liquid} *\frac{dQ}{dt}$$

So I have two unknowns for every components: C[2] and Q[2] which are the concentrations in the reactor (or at the outlet of the reactor). The ODE model adds two equations for every component, therefore the system is not under determined.

If I write all equations together, it is an implicit DAE model. However, if I say, that the components are mixed instantaneously at time = 0 and adsorption happens afterwards, I can calculate all initial conditions necessary for an ODE system. It seems to work but I am not sure whether this trick is a pitfall or not.

Is this assumption correct ? Or do I miss something ?

As you say, if the mass balance equation is satisfied at time $t = 0$, then it will be satisfied for all $t > 0$ as well, which implies that you do not need to regard your system as a differential-algebraic system because the algebraic equation is redundant. To see this, let $$M(t) = C(t) \, V_{\rm liquid} + Q(t) \, V_{\rm particles}.$$
Taking derivatives with respect to time, we get $$\frac{d M}{d t} = \frac{d C}{d t} \, V_{\rm liquid} + \frac{d Q}{d t} \, V_{\rm particles}.$$
Now, using $$\frac{dC}{dt} = - \frac{V_{\rm particles}}{V_{\rm liquid}} \frac{d Q}{d t},$$ we see that $\frac{d M}{d t} = 0$, which in turn implies that $M(t)$ is constant for all $t$. If $M(0) = 0$, you can conclude that the mass balance equation is satisfied for all times.