Fluids Problem: Is there a Typo?

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1/2/3. Homework Statement and equations

Hi everyone,

So, I have a fluids problem to do and I think there may be a slight typo or maybe I am missing something. The image of the problem is attached.

So, the only problem I have is how L is defined. Since L is defined from below the zero line, shouldn't Poil*G*L, instead be written Poil*G*l/2? To me, it seems the system drops l/2 then rises L.

Perhaps, I am missing something...

Thanks!
 
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  • Worked solution for a manometer problem: governing equations dp/dh = rho g and rho = SG rho_water, assumptions of static, incompressible liquid, and a tank with oil in a U-tube manometer leg, with the equation for l.
    problem statement.png
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  • Textbook diagram of a rectangular open tank filled with water to 2.5 m, with a U-tube manometer connected 0.7 m above the bottom, a 0.2 m zero level marked and a deflection l shown in the manometer leg.
    ;ol.png
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BryMan92 said:
1/2/3. Homework Statement and equations

Hi everyone,

So, I have a fluids problem to do and I think there may be a slight typo or maybe I am missing something. The image of the problem is attached.

So, the only problem I have is how L is defined. Since L is defined from below the zero line, shouldn't Poil*G*L, instead be written Poil*G*l/2? To me, it seems the system drops l/2 then rises L.

Perhaps, I am missing something...

Thanks!
As you say, the system drops l/2, but that drop is in water. Hence the +l/2 term in the expression containing ρwater.

Then the rise of l is entirely in oil, so there should be a full l, not l/2, accompanying the ρoil term.

Hope that helps.
 
But, if the system has a net drop of l/2 how can the oil raise a full l? Do I not measure from the zero point, or do I have to measure from the change in height of the whole system?

Additionally, at the zero line point in t, was l/2 already above the 0 line? If so, then this would make sense. Thanks for the help!
 
It's best to measure from some point where you already know what the pressure is. Since we know the pressure is Patm = 1 atm at the top of the tank, that is where they have chosen to measure from.

So measure the change in height from the top of the tank -- but you have to account for the fact that there are two fluids, so you take the change in depth for the water and then add the change in depth for the oil.
 
I got it thanks! I just had to run through it a but more thoroughly. My professor believes that doing a problem justifies as teaching.

Cheers!