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?

  • $\begingroup$ 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

1 Answer 1


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.