Changing height in tank depending on time

[havtorn]

Hi, I have a mixing tank with 2 inputs and 1 output
The volume in the tank is not constant
The cross section area is constant
How can I build a mathematical formula for changing high h in the tank depending on the time?

I have done this:
(V/dt)in − (V/dt)out =A*d(h)/dt

 

[I like Serena]

Hi, I have a mixing tank with 2 inputs and 1 output
The volume in the tank is not constant
The cross section area is constant
How can I build a mathematical formula for changing high h in the tank depending on the time?

I have done this:
(V/dt)in − (V/dt)out =A*d(h)/dt

Hi havtorn, welcome to MHB! (Wave)

Your formula seems fine to me.
Since you have 2 inputs, shouldn't it be:
$$\d {V_{in,1}}t + \d {V_{in,2}}t - \d {V_{out}}t = A\d h t$$
though?

To find $h(t)$, we can integrate and find:
$$h(t) = \frac 1A \int \left(\d {V_{in,1}}t + \d {V_{in,2}}t - \d {V_{out}}t\right)\, dt$$
And if we write $Q$ for the volumetric flow $\d V t$, we can write it as:
$$h(t) = \frac 1A \int \left(Q_{in,1} +Q_{in,2} - Q_{out}\right)\, dt$$