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Why doesn't Bernoulli's Equation apply in this problem? |
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| Nov27-09, 04:45 PM | #1 |
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Why doesn't Bernoulli's Equation apply in this problem?
1. The problem statement, all variables and given/known data
A cylindrical bucket of liquid (density ρ) is rotated about its symmetry axis, which is vertical. If the angular velocity is ω, show that the pressure at a distance r from the rotation axis is [tex]P = P_0 + \frac{1}{2} \rho \omega^2 r^2[/tex] where P0 is the pressure at r = 0. 2. Relevant equations P = F/A 3. The attempt at a solution I was able to get the correct answer by considering the net force on a mass element dm since it is undergoing centripetal acceleration. However, I was wondering what about this problem made Bernoulli's Equation not applicable? Bernoulli's equation yields: [tex]P = P_0 - \frac{1}{2} \rho \omega^2 r^2[/tex] |
| Nov28-09, 04:08 AM | #2 |
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Recognitions:
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Bernoulli's equation works along a streamline, which means the path of fluid flow. In this case, you took the radial pressure difference but used the tangential velocity v instead of the radial velocity of 0. If you use 0 and put back the gravitational potential term, because centripetal acceleration is equivalent to gravity, you'll get the right answer.
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| Nov29-09, 09:23 PM | #3 |
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I'm sorry. I've been thinking about what you said but I do not understand what to do. I tried to rewrite ρgy as ρ(v2/r) but this isn't working either.
Also, to make sure I understand what streamline means, would this be a proper application of Bernoulli's equation? If you consider a small portion of the water dm traveling along its path of flow (in a circle), then v1 = ωr = v2 and y1 = y2. Thus P1 = P2, so it's pressure remains constant. |
| Nov30-09, 01:00 PM | #4 |
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Recognitions:
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Why doesn't Bernoulli's Equation apply in this problem? |
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