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

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The discussion focuses on modeling water flow from a gate at the bottom of a reservoir, where the gate's opening is defined by the function f(t) = 0.01t^2. To create a differential equation that accounts for the varying area of the gate, assumptions about the reservoir's cross-sectional area and the height of water over time are made. Using Bernoulli's principle, the relationship between the height of the water, the speed of flow, and the volume of water exiting through the gate is established. The resulting equations describe how the height of the reservoir changes over time based on the gate's opening and water flow dynamics. This approach provides a framework for analyzing the system's behavior under the given conditions.
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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