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The volume for an open tank is given: <$V = L \cdot L \cdot h(t)+ h(t)^2 \cdot tan(\theta)$>

Where h is the volume of the liquid within the tank.

I want to calculate the change in height as a function of flow thus the continuity equation is considered

<$Q_{in}-Q_{out}= \dot{V}$>

Is it correct understood that I should insert the volume differentiated with respect to time

<$\dot{V}= L^2 \dot{h}(t)+ h(t) \cdot tan(\theta) \cdot \dot{h}(t)$>

Or should I just replace the h with <$\dot{h} $> in the volume equation ?

And if I want to linearize and put it on laplace form how should I proceed?

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  • $\begingroup$ The last term in the first-reproduced equation does not have the same dimensions as the other terms. Something is wrong. $\endgroup$ Commented Mar 11, 2023 at 18:00

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Volume differentiation should work, you just missed 2 when differentiating the $h(t)^2$ term:

$$\dot{V}= L^2 \dot{h}(t)+ 2h(t) \cdot tan(\theta) \cdot \dot{h}(t)$$

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