Modeling the flow from a gate at the bottom of the reservior

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SUMMARY

This discussion focuses on modeling water flow from a reservoir through a gate with a time-dependent opening function, specifically defined as f(t) = 0.01t². The key equations derived include Bernoulli's principle, which relates the height of the water in the reservoir to the speed of the water exiting the gate. The resulting differential equations are v = √(2gh) and dh/dt = (w/A)√(2gh)f(t), where h is the height of the water, w is the width of the gate, and A is the cross-sectional area of the reservoir. These equations facilitate the analysis of the varying area of the gate over time.

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sumanraj
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I need to model a system of water flow from a gate whose opening is a function of time
the height of the reservoir = H
width of the gate = w
opening of the gate = f(t) = 0.01t^2

how can i create a differential equation that takes an account of the varying area of the gate?
 
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sumanraj said:
I need to model a system of water flow from a gate whose opening is a function of time
the height of the reservoir = H
width of the gate = w
opening of the gate = f(t) = 0.01t^2

how can i create a differential equation that takes an account of the varying area of the gate?

Hi sumanraj! Welcome to MHB! (Wave)

What kind of gate and reservoir are we talking about?

Some gate that opens from the bottom up?
What is f(t)? The height that the gate has been lifted perhaps?

Does the reservoir empty itself?
Or should we assume that it's always fully filled?

Let me make a couple of assumptions.
Suppose we have a reservoir with cross section area A.
And it's filled up to height h(t) at time t.
Let v(t) be the speed of the water going through the gate.
Then from Bernoulli (assuming perfect conditions) we get that:
$$g h(t) = \frac 12 v(t)^2$$

After an infinitesimal time $dt$, the height $h$ lowers by $dh$, and a volume of $A\cdot dh$ is transported through the gate.
At the gate itself, we have an area of $w\cdot f(t)$ where the water leaves at speed $v$.
That means that in the same time $dt$, a volume of $w\cdot f(t) \cdot vdt$ leaves the gate.
So with the current assumptions we get:
$$\begin{cases}
g h = \frac 12 v^2 \\
A\cdot dh = w\cdot f(t) \cdot vdt
\end{cases}\Rightarrow
\begin{cases}
v=\sqrt{2gh} \\
\d ht = \frac wA \sqrt{2gh} f(t)
\end{cases}
$$
 

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