Finding the velocity of fluid through orifice

  • Thread starter Saitama
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  • #1
Saitama
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Homework Statement


A horizontally oriented tube AB of length ##l## rotate with a constant angular velocity ##\omega## about a stationary axis OO' passing through the end A. The tube is filled with an ideal fluid. The end A of the tube is open, the closed end B has a very small orifice. Find the velocity of the fluid relative to the tube as a function of the column "height" h.


Homework Equations


$$P+\rho gh+\frac{1}{2}\rho v^2=\text{constant}$$


The Attempt at a Solution


The book I refer to states that the Bernoulli equation (stated in relevant equations) is applicable in irrotational flow. But in the given problem, the fluid rotates too. I am thinking of switching to rotating frame where the liquid moves radially so there is no problem of "irrotational flow". Is this is a valid step?

Applying Bernoulli at the open surface and near the orifice,
$$P_o=P_o+\int_{l-h}^l \rho \omega^2x\,dx+\frac{1}{2}\rho v^2$$
Solving for ##v## gives
$$v=\omega\sqrt{h^2-2hl}$$
According to the answer key, the above is incorrect. The correct answer is ##v=\omega\sqrt{2hl-h^2}##. What is wrong with my method? :confused:

Also, is it possible to find the position of CoM of remaining liquid as a function of time?

Any help is appreciated. Thanks!
 

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Answers and Replies

  • #2
voko
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Your starting equation is obviously incorrect. Cancel the ##P_0## terms, and you have that either the integral of pressure due to rotation is negative, or the square of velocity is negative. Neither can be true.

What you need to do is compare the conditions just inside and outside the orifice.
 
  • #3
Saitama
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Your starting equation is obviously incorrect. Cancel the ##P_0## terms, and you have that either the integral of pressure due to rotation is negative, or the square of velocity is negative. Neither can be true.

So where is the error? I mean, what's wrong with applying Bernoulli's equation at those two points?

What you need to do is compare the conditions just inside and outside the orifice.

Something like this:
$$P_o+\int_{l-h}^l\rho \omega^2x\,\,dx=P_o+\frac{1}{2}\rho v^2$$
?
 
  • #4
voko
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That makes more sense.

As for the error, what two points did you have in mind? Are they possible states of the same fluid element?
 
  • #5
Saitama
4,244
93
As for the error, what two points did you have in mind? Are they possible states of the same fluid element?

I was considering two points. One on the open surface (left) and the other just on the left of orifice which is wrong. I found the error, thank you very much voko! :)

As I asked in post #1, is it possible to find the CoM of remaining liquid as a function of time?
 
  • #6
voko
6,054
391
You would need to know the area of the orifice. Given that, you can find out the volumetric rate of discharge, and with that, the amount of the remaining fluid.
 
  • #7
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So where is the error? I mean, what's wrong with applying Bernoulli's equation at those two points?



Something like this:
$$P_o+\int_{l-h}^l\rho \omega^2x\,\,dx=P_o+\frac{1}{2}\rho v^2$$
?

Yes. This final equation looks correct. The left hand side represents the pressure inside the tube at point B (where the radial velocity in the tube is essentially zero). The right hand side represents the Bernoulli expression immediately outside the orifice. Essentially, as a result of the rotation, you have created artificial gravity within the tube, and the pressure at B is essentially hydrostatic (of course, allowing for the radially varying "gravitational acceleration."
 

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