Equation of continuity of water depth

[etothey]

Homework Statement

A can of height h and cross-sectional Area Ao is initially full of water. A small hole of area A1<<Ao is cut in the bottom of the can. Find an expression for the time it takes all the water to drain from the Can. Hint: Call the water depth y use the continuity equation to relate dy/dt to the outflow sped at the hole and then integrate.

Homework Equations

A1v1=A2v2

The Attempt at a Solution

I tried to solve this by my own, managed to get stuck, and then tried to understand the solution from cramster. There they start mentioning the variables and one of the things they say is, Rate of change in volume through the orifice = A1V=s*(2gh)^1/2. This is the part i don't understand. Afterwards they set Ao*dh/dt equal to A1*(2gh)^1/2. I don't understand where (2gh)^1/2 comes from so that i can solve this.
Thankful for help!

 

[tiny-tim]

hi etothey!

(have a square-root: √ and try using the X² icon just above the Reply box )

it comes from Bernoulli's equation (conservation of energy) …

1/2 ρv² = ρgh ​

 

[Metaleer]

You need to use Bernoulli's equation and the continuity equation to solve this problem. Applying the Bernoulli equation between the top of the fluid and the bottom of the can, we have

$$\frac{1}{2}\rho v_1^2 + \rho g y = \frac{1}{2}\rho v_2^2,$$​

that is,

$$v_2^2 = v_1^2 + 2 g y.$$​

But the continuity equation tells us that $$A_0 v_1 = A_1 v_2$$, so plugging into the Bernoulli equation, we have

$$v_2^2 = v_1^2 + 2 g y = \left(\frac{A_1}{A_0}\right)^2 v_2^2 + 2 g y$$​

but they tell us that $$A_0$$ is much greater than $$A_1$$, so $$\left(\frac{A_1}{A_0}\right)^2 \approx 0$$, so we are left with

$$v_2^2 = 2 g y \Rightarrow v_2 = \sqrt{2 g y}.$$​

Going back to the continuity equation, we have that

$$A_0 v_1 = A_1 v_2,$$​

but we have seen that $$v_2 = \sqrt{2 g y}$$, and $$v_1$$ is the rate of change of the height as the fluid is drained, that is $$v_1 = - \frac{dy}{dt}$$ (notice the minus sign; $$v_1$$ is a positive function, but since the height $$y$$ is decreasing with time, its time derivative is negative, so we need to fix this by adding a minus sign). So, we have the following differential equation:

$$A_0 v_1 = A_1 v_2 \Rightarrow - A_0 \frac{dy}{dt} = A_1\sqrt{2 g y},$$​

and you separate variables to find the time required for the can to fully drain.

Hope this helps.

 

[etothey]

hi etothey!

(have a square-root: √ and try using the X² icon just above the Reply box )

it comes from Bernoulli's equation (conservation of energy) …

1/2 ρv² = ρgh ​

sexy!

 

[etothey]

You need to use Bernoulli's equation and the continuity equation to solve this problem. Applying the Bernoulli equation between the top of the fluid and the bottom of the can, we have

$$\frac{1}{2}\rho v_1^2 + \rho g y = \frac{1}{2}\rho v_2^2,$$​

that is,

$$v_2^2 = v_1^2 + 2 g y.$$​

But the continuity equation tells us that $$A_0 v_1 = A_1 v_2$$, so plugging into the Bernoulli equation, we have

$$v_2^2 = v_1^2 + 2 g y = \left(\frac{A_1}{A_0}\right)^2 v_2^2 + 2 g y$$​

but they tell us that $$A_0$$ is much greater than $$A_1$$, so $$\left(\frac{A_1}{A_0}\right)^2 \approx 0$$, so we are left with

$$v_2^2 = 2 g y \Rightarrow v_2 = \sqrt{2 g y}.$$​

Going back to the continuity equation, we have that

$$A_0 v_1 = A_1 v_2,$$​

but we have seen that $$v_2 = \sqrt{2 g y}$$, and $$v_1$$ is the rate of change of the height as the fluid is drained, that is $$v_1 = - \frac{dy}{dt}$$ (notice the minus sign; $$v_1$$ is a positive function, but since the height $$y$$ is decreasing with time, its time derivative is negative, so we need to fix this by adding a minus sign). So, we have the following differential equation:

$$A_0 v_1 = A_1 v_2 \Rightarrow - A_0 \frac{dy}{dt} = A_1\sqrt{2 g y},$$​

and you separate variables to find the time required for the can to fully drain.

Hope this helps.

Thank you very much!