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Old Dec 9th 2017, 07:01 AM   #1
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Fluid Mechanics help.

Need help in this problem.

I tried doing it using my only limited knowledge (1st time student of fluid mechanics) and I cannot get the answer written.

No prof feedback either. this is all i got as a given.

At least a hint would be appreciated.
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Last edited by lazz; Dec 9th 2017 at 07:24 AM.
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Old Dec 10th 2017, 04:58 AM   #2
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Before getting into the gory details of this, I have one question for you: did you remember to include the effects of atmospheric pressure?
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Old Dec 10th 2017, 08:35 PM   #3
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Originally Posted by ChipB View Post
Before getting into the gory details of this, I have one question for you: did you remember to include the effects of atmospheric pressure?
I can't see that comes into it ...manometer sealed at one end , pipe at the other ....

I nearly finished this yesterday , then a computer glitch lost it all , can't be arsed to do it all again ...question gives Hg density as 3.6 !! it should be 13.6, also no indication of what h (40) is measured in !!
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Old Dec 11th 2017, 07:46 AM   #4
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I don't know how to do this one for sure, but according to wikipedia, the difference in pressure gives rise to a change in height in a manometer related by

$\displaystyle \Delta h = \frac{\Delta P}{\rho g}$

$\displaystyle \Delta h = \frac{P_{atm} - P_{pipe}}{\rho g}$

$\displaystyle P_{pipe}$ is what you're looking for, but you might need to take into account the difference in area in the left-hand tube by equating force:

$\displaystyle F = PA$

$\displaystyle P_1 A_1 = P_2 A_2$

So you'd get something like

$\displaystyle \Delta h = \frac{P_{atm} - P_2}{\rho g}$

where

$\displaystyle P_2 = \frac{P_{pipe} A_{pipe}}{A_{2}} = 100 P_{pipe}$

So maybe try something like that? Have a go and see what you get

EDIT: Okay, so the above is probably wrong... I'll keep investigating.

Last edited by benit13; Dec 11th 2017 at 09:05 AM.
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