Water flowing out of a rotating vessel

[Bling Fizikst]

Homework Statement

refer to the image

Relevant Equations

refer to the image

Let's say at the steady state the vertex of the parabola (paraboloid) is at the origin . Then the eqn of the formed parabola would be $$y=\frac{\omega^2x^2}{2g}$$ Now , initial volume of liquid is $\pi R^2h$ . As the liquid flows out of the orifice , the surface would maintain it's structure (paraboloid) till the vertex reaches the orfice . After that , liquid should stop flowing out .

Unable to think beyond this . I tried bernoulli but not sure how i would apply it as pressure is constantly varying with $x$

 

[jbriggs444]

Unable to think beyond this . I tried bernoulli but not sure how i would apply it as pressure is constantly varying with $x$

Pressure? You are not concerned with the rate at which the contents drain out. You are concerned with the volume at the final state. More specifically, you are concerned with the volume of those little blue-shaded paraboloid sections that you have drawn.

You already have a function for the height of the surface ($y$) as a function of distance from the center ($x$). Do you know how to integrate?

 

[Bling Fizikst]

Pressure? You are not concerned with the rate at which the contents drain out. You are concerned with the volume at the final state. More specifically, you are concerned with the volume of those little blue-shaded paraboloid sections that you have drawn.

You already have a function for the height of the surface ($y$) as a function of distance from the center ($x$). Do you know how to integrate?

I am not sure how i will integrate to find the volume . Tell me how to proceed .

 

[jbriggs444]

I am not sure how i will integrate to find the volume . Tell me how to proceed .

Do you know integral calculus at all?

 

[Bling Fizikst]

Do you know integral calculus at all?

high school level? yes

 

[erobz]

Homework Statement: refer to the image
Relevant Equations: refer to the image

View attachment 344304

Let's say at the steady state the vertex of the parabola (paraboloid) is at the origin . Then the eqn of the formed parabola would be $$y=\frac{\omega^2x^2}{2g}$$ Now , initial volume of liquid is $\pi R^2h$ . As the liquid flows out of the orifice , the surface would maintain it's structure (paraboloid) till the vertex reaches the orfice . After that , liquid should stop flowing out .


View attachment 344305

Unable to think beyond this . I tried bernoulli but not sure how i would apply it as pressure is constantly varying with $x$

I don't think your function fits the picture? The volume doesn't start out as a paraboloid, its initially a cylinder, the thing starts spinning and some of the water flies out leaving a parabolic cut out.

if you substitute $x=0$, you get $y=0$. Did they give you this function in the problem statement?

 

[jbriggs444]

high school level? yes

OK

You've already written down an equation for the height of the surface of the paraboloid ($y$) as a function of distance from the center $x$, right?$$y=\frac{\omega^2x^2}{2g}$$What would be the volume of a thin cylindrical ring of thickness $dx$, height $y$ and circumference $2\pi x$?

 

[erobz]

OK

You've already written down an equation for the height of the surface of the paraboloid ($y$) as a function of distance from the center $x$, right?$$y=\frac{\omega^2x^2}{2g}$$What would be the volume of a thin cylindrical ring of thickness $dx$, height $y$ and circumference $2\pi x$?

The equation doesn't fit the problem. At steady state no portion of the bottom is dry...

Its going to need to be something like $y = kx^2 +C(\omega)$

 

[jbriggs444]

The equation doesn't fit the problem. At steady state no portion of the bottom is dry...

It fits the problem. You need to follow the strategy that OP has already laid out.

We are finding the volume of the portion of fluid above the bottom of the paraboloid at steady state.

 

[erobz]

It fits the problem. You need to follow the strategy that OP has already laid out.

We are finding the volume of the portion of fluid above the bottom of the paraboloid at steady state.

$x = 0, y = 0$? that is clearly not what is presented in the diagram. It ask how much fluid can drain out after a hole is drilled.

 

[jbriggs444]

$x = 0, y = 0$? that is clearly not what is presented in the diagram.

Note the similar shapes in the drawing. If we do the integration then we will get a useful volume as a result.

 

[erobz]

Note the similar shapes in the drawing.

Ok, they put the origin at the vertex. They need to figure out the offset, I think I see now. Sorry.

 

[Bling Fizikst]

You've already written down an equation for the height of the surface of the paraboloid ($y$) as a function of distance from the center $x$, right?$$y=\frac{\omega^2x^2}{2g}$$What would be the volume of a thin cylindrical ring of thickness $dx$, height $y$ and circumference $2\pi x$?

$dV=2\pi x y dx=\frac{\pi\omega^2}{g}x^3\cdot dx$ . So , my volume would be $$V=\int_{r}^{R}\frac{\pi\omega^2}{g}x^3\cdot dx=\frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$ So , final answer should be : $$V_{\text{out}}=\pi R^2 h - \frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$

 

[erobz]

$dV=2\pi x y dx=\frac{\pi\omega^2}{g}x^3\cdot dx$ . So , my volume would be $$V=\int_{r}^{R}\frac{\pi\omega^2}{g}x^3\cdot dx=\frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$ So , final answer should be : $$V_{\text{out}}=\pi R^2 h - \frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$

$r$ being?

 

[Bling Fizikst]

$r$ being?

$r$ is the radius of orifice . Read the question carefully .

 

[erobz]

$r$ is the radius of orifice . Read the question carefully .

Why does the orifice suddenly opened effect the volume of the liquid in the container after the bucket rotating at constant angular velocity?

Maybe you should read the question carefully...

 

[Orodruin]

Why does the orifice suddenly opened effect the volume of the liquid in the container after the bucket rotating at constant angular velocity?

Because the liquid will drain out of the container until the orifice is no longer covered by the liquid.

Maybe you should read the question carefully...

…

 

[erobz]

Because the liquid will drain out of the container until the orifice is no longer covered by the liquid.


…

But it says the liquid is filled up in the bucket to a height $h$ above the flat bottom. Then it spins at some $\omega$ and some amount spills out. Then the drain is opened. $h$ is the initial height of the unspinning volume. Don't they need to fine the volume left after it is spinning, before the plug is opened? "How much liquid can flow out"

 

[jbriggs444]

Then it spins at some $\omega$ and some amount spills out.

The problem says "the liquid does not overflow".

 

[erobz]

The problem says "the liquid does not overflow".

Brutal... I'm getting all kinds of messed up. Good thing this kid is smarter than me!

 

[Orodruin]

But it says the liquid is filled up in the bucket to a height $h$ above the flat bottom. Then it spins at some $\omega$ and some amount spills out. Then the drain is opened. $h$ is the initial height of the unspinning volume. Don't they need to fine the volume left after it is spinning, before the plug is opened?

No. It is the same as the initial volume, which is given by the volume of a cylinder with radius $R$ and height $h$.

Volume flowing out = initial volume - final volume

"How much liquid can flow out"

Exactly.

 

[Orodruin]

Brutal... I'm getting all kinds of messed up. Good thing this kid is smarter than me!

Reading the problem statement helps …

 

[jbriggs444]

The niggling concern that I had with the problem statement was the possibility that the assembly would not remain at a constant rotation rate while the fluid was being drained. We are told that the rotation rate is steady until equilibrium is reached. But we are not explicitly told that it is steady thereafter.

But since we were given no information on the moment of inertia of the container nor a good way to compute the angular momentum dumped through the finite sized drain, it seemed reasonable to avoid that complication by assuming a continued constant rotation rate.

 

[erobz]

No. It is the same as the initial volume, which is given by the volume of a cylinder with radius $R$ and height $h$.

Volume flowing out = initial volume - final volume


Exactly.

Someone please take my HH badge away.

 

[Orodruin]

But we are not explicitly told that it is steady thereafter.

This is likely to be assumed from the statement that r is small. This implies a slow flow and therefore close to steady state.

 

[jbriggs444]

This is likely to be assumed from the statement that r is small. This implies a slow flow and therefore close to steady state.

That would lead me to the alternate assumption that the drain of angular momentum would be negligible and that the rotation rate would, therefore, accelerate.

 

[Orodruin]

$dV=2\pi x y dx=\frac{\pi\omega^2}{g}x^3\cdot dx$ . So , my volume would be $$V=\int_{r}^{R}\frac{\pi\omega^2}{g}x^3\cdot dx=\frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$ So , final answer should be : $$V_{\text{out}}=\pi R^2 h - \frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$

This is not completely correct. It is a small effect, but you have assumed that the height of the surface above the bottom is $\frac{\pi\omega^2r^2}{2g}$ at $x=r$. At that distance, the height above the bottom at $x=r$ when nothing flows out should be zero.

 

[Orodruin]

That would lead me to the alternate assumption that the drain of angular momentum would be negligible and that the rotation rate would, therefore, accelerate.

The way I read the problem, the container maintains fixed angular speed. Assuming a slow flow out of the cylinder, the viscosity of the fluid and no slip conditions at the walls ensure the fluid continues spinning essentially at $\omega$.

 

[Bling Fizikst]

This is not completely correct. It is a small effect, but you have assumed that the height of the surface above the bottom is $\frac{\pi\omega^2r^2}{2g}$ at $x=r$. At that distance, the height above the bottom at $x=r$ when nothing flows out should be zero.

It should be 'approximately' true i guess as $r$ is given to be small . Hence , the second order term $r^2\approx 0$ . So , i guess we can only get an approximate answer .

 

[jbriggs444]

It should be 'approximately' true i guess as $r$ is given to be small . Hence , the second order term $r^2\approx 0$ . So , i guess we can only get an approximate answer .

The correct answer can be computed. There are a few ways to proceed. Let me try to suggest one.

You had computed the volume of the exterior of a paraboloid that sits a little bit too high. How much too high? If you slide the paraboloid down so that it rests on the rim of the hole rather than sitting on the imaginary point at the hole's center, how far down did you have to slide it? How much fluid is that?

Another approach is to concentrate on the volume of the shallow divot where the paraboloid sits below the rim of the hole. If you know how far the original paraboloid had to slide down to get there, that gives you a preliminary estimate for the volume of water that drains out. If you know the volume of the divot, that will tell you the error in that estimate.

Both of the above approaches should yield equivalent calculations and identical results. You're just changing up the inclusion/exclusion rule and changing the bounds of the integral. Either way you still have to figure out the depth of the divot.

 

[Orodruin]

It should be 'approximately' true i guess as $r$ is given to be small . Hence , the second order term $r^2\approx 0$ . So , i guess we can only get an approximate answer .

If you just want an approximate answer, you should just send r to zero. The answer you gave contains a term proportional to $r^4$. This term is negligible compared to the term you are missing due to not including the offset.