Solving a Thermodynamics Problem

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The discussion revolves around solving a thermodynamics problem involving water and mercury in a U-tube. The user is trying to determine the height 'h' given the pressure at point X and the height 'a' of water. Key equations involve converting gauge pressure to absolute pressure and using fluid density to find the height difference. A critical point raised is the need to ensure correct unit conversions, particularly between kPa and Pa. The expected solution for 'h' is 1.145m, indicating a potential miscalculation in the user's approach.
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Kinda lost...please help !

Hello,

There's this basic problem in thermodynamics that I'm stuck on. The question is

Fluid 'A' is water and 'B' is mercury. Determind the value of 'h' if the pressure at 'X' is 138 kN/m2 and 'a' is 1.5m
rhow = 1000 kg/m3
rhom = 13600 kg/m3

The main equation used is:
Absolute pressure = Gauge pressure + atmospheric pressure
Gauge pressure = rho * g * h

The value that you're supposed to get is 1.145m. Please help, I'm stuck on it for hours now and I'm sure its somethin basic that I've forgotten.

Here's a figure that might help with the problem:

http://s155.photobucket.com/albums/s300/br1ttt/?action=view&current=thermo.jpg

Many thanks,
-Brittt
 
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You have atmospheric pressure outside, you have pressure X that's a bit higher in the vessel. Now just find the height difference (h-a) that corresponds to the pressure difference according to the formula you already wrote above.
 
The given info is where you need to start. At X you have a given static pressure. Then you travel down the U tube to the bottom of dimension A. At that point, for the water you have the combination of the pressure at X and the rho*g*a contribution. That will give you a new static pressure at that point. At the point that coincides with the bottom of dimensions A and H, the pressure has to be equal. So you take your new pressure that you just calculated and set that equal to the rho*g*h for the mercury to calculate h.
 
FredGarvin said:
The given info is where you need to start. At X you have a given static pressure. Then you travel down the U tube to the bottom of dimension A. At that point, for the water you have the combination of the pressure at X and the rho*g*a contribution. That will give you a new static pressure at that point. At the point that coincides with the bottom of dimensions A and H, the pressure has to be equal. So you take your new pressure that you just calculated and set that equal to the rho*g*h for the mercury to calculate h.

If i calculate as per your instructions, it would be:
X + rho * g * a
138 + 1000 * 9.81 * 1.5
14853 N/m2

Now,
14853 = rho * g * h
h = 14853 / rho * g
h = 14853 / (13600 *9.81)
Therefore, h = 0.111328

But the solution is 1.145m :(

Thanks for trying to help...any other ideas?

Cheers,
-Brittt
 
Aero Stud said:
You have atmospheric pressure outside, you have pressure X that's a bit higher in the vessel. Now just find the height difference (h-a) that corresponds to the pressure difference according to the formula you already wrote above.

In order to find the height difference, we need to know both the a and h, and 'h' is what we're expected to calculate.
 
You have 138 KN/m^2=138,000 N/m^2 or 138 kPa. Put it in what you wrote already and you get the answer.

p.s. Your drawing doesn't make clear which material is where, so at first I thought you have water only in the ball so it didn't matter.
 
The pressure at X is in kPa, NOT Pa. Your units are incorrect by a factor of 1000.

h=\frac{138x10^3 + (9.81*1000*1.5)}{9.81*13.6x10^3}
 
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