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I have a system like this:
The upper tank has an input ($q_i$) and an output, that is the input for the lower tank ($q_{12}$). The lower tank has an output ($q_o$).
Does anyone know how to get the equation that describes that system? I know that I will have two poles.
Gravity-fed outflow is proportional to the square root of the liquid level (see Bernoulli's equation ). Let's call $h_i$ the height of the upper (inflow) tank, and $h_o$ the height of the lower (outflow) tank. $q_{12} = k_{q_{12}} \sqrt(h_i) $ and $(dh_i/dt) = k_{h_i} (q_i - q_{12})$.
For the system outflow, $q_o = k_{q_o} \sqrt(h_o)$, and $(dh_o/dt) = k_{h_o} (q_{12} - q_o)$.
Valve assumptions: linear and fast. So the steady-state flowrate of $q_i$ is proportional to the valve opening via $k_{v_i}$, and the dynamics of the valve are much much much faster than the tank level dynamics, so no differential equation is needed - algebra only.
Under P-only control, $q_i = k_{v_i} k_c (h_{o_sp} - h_o)$
You might want to linearize those square root terms via Taylor expansion to make the Laplace transforms a bit less hairy, if you're working in the $s$ or $z^{-1}$ domain.
Of course, in industry (and assuming that the level of this tank is not critical - it rarely is) you'd just set it up to P-only control with a dimensionless gain $k_c = 1.67$. This way, that if the setpoint is 50%, the valve is wide open when the lower tank level is at 20%, and fully shut when the lower tank is at 80%. And then you'd move on to the next crisis. :)
