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 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Dec 9th 2017, 07:01 AM #1 Junior Member   Join Date: Dec 2017 Posts: 4 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. Attached Thumbnails   Last edited by lazz; Dec 9th 2017 at 07:24 AM.
 Dec 10th 2017, 04:58 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,286 Before getting into the gory details of this, I have one question for you: did you remember to include the effects of atmospheric pressure?
Dec 10th 2017, 08:35 PM   #3
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Join Date: Apr 2017
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 Originally Posted by ChipB 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 !!

 Dec 11th 2017, 07:46 AM #4 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 166 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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