[Thermodynamics (More like fluid actually)], Calculating absolute pressure

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SUMMARY

The discussion focuses on calculating absolute pressure in fluid systems, specifically involving mercury and brine. Key equations include the force-area relationship (F1/A1 = F2/A2) and the pressure formula (Pressure = Density * Height * Gravity). Participants emphasize the importance of understanding how changes in air pressure affect liquid columns and the static head due to mercury. The relationship between the heights of mercury (h1 and h2) and their corresponding areas (A1 and A2) is crucial for solving the problem.

PREREQUISITES
  • Understanding of fluid mechanics principles
  • Familiarity with pressure equations in fluid dynamics
  • Knowledge of specific gravity (s.g.) concepts
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of hydrostatic pressure in fluids
  • Learn about incompressible fluid assumptions in fluid mechanics
  • Explore the effects of specific gravity on fluid displacement
  • Investigate the relationship between pressure and liquid column height
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Students studying fluid mechanics, engineers working with hydraulic systems, and anyone involved in pressure calculations in fluid dynamics.

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Homework Statement


http://img409.imageshack.us/img409/771/fluidcm5.jpg

Homework Equations


F1/A1 = F2/A2

Pressure = Density * Height * Gravity


The Attempt at a Solution


I do not have a clue of how to solve this problem.

Can anyone start me off and provide me couple of hints?
 
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To solve this problem, think about what changes.

The air pressure decreases, so the force/pressure on the liquid columns changes. The pressure in the brine pipe does not change, and the pressure due to the water between the air and the mercury does not change. The levels of mercury do change, but the volume of mercury does not (one can assume it is incompressible).

The static head due to the mercury changes, because the elevations of the interfaces change.

Since the volume of mercury does not change, the change in mercury volume in pipe A2 must equal the change in volume in the tank A1, so one has

V2 = V1 or A2h2 = A1h1, so one has a relationship between h2 and h1 in terms of the areas.

Now, one needs a second equation, which comes from the head, H.

H = h2 + h1 for the mercury.

The static head is simply the pressure due to a height, z, of liquid and is given by \rhogz, where \rho is the fluid density (which is related to s.g.) and g is the acceleration of gravity.

Also think about the water (s.g. =1)/brine (s.g.=1.1) displacement as well. What is the effect of that compared to the mercury.
 

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