Mixture problem. How to solve for C?

[Jeff12341234]

I need to solve for C. I know it's probably simple but i don't remember how to. This is what I have so far:

 

[Chestermiller]

Because the volume flow rate entering is different from the volume flow rate leaving, you need to write down two differential equations, rather than 1:

Volume Input - Volume Output = accumulation for the volume of fluid in the tank

Chemical X Input - chemical X Output = accumulation for chemical X in the tank

If V(t) is the volume of fluid in the tank at time t, f_in is the volumetric flow rate of fluid in, and f_out is the volumetric flow rate of fluid out, what is the differential equation for V?

If C(t) is the concentration of chemical X within the tank at time t, and C_in is the concentration of chemical X in the feed to the tank, what is the differential equation for the rate of change of total chemical X in the tank?

The next step is to multiply the differential equation for V by C, and subtract the resulting relationship from the mass balance on chemical X.

 

[Jeff12341234]

I don't follow.. The way I did it is the way the professor instructed us and the steps match the steps in his example. To solve for C, I now realize from an example in the book that A(0)=35. With that information, I can solve for C.

 

[Chestermiller]

OK. I see what you did, and, of course, it is right. But, here's my alternate version to consider:

$$\frac{dV}{dt}=f_{in}-f_{out}$$
$$\frac{d(VC)}{dt}=f_{in}C_{in}-f_{out}C$$
Multiply the first equation by C and subtract it from the second equation:

$$V\frac{dC}{dt}=f_{in}(C_{in}-C)$$

where $V=V_0+(f_{in}-f_{out})t$
So,

$$\frac{dC}{(C_{in}-C)}=f_{in}\frac{dt}{V_0+(f_{in}-f_{out})t}$$