Calculating Gauge Pressure at the Bottom of a Test Tube

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Homework Help Overview

The problem involves calculating the gauge pressure at the bottom of a test tube containing two fluids: oil and water. The context is within fluid mechanics, specifically focusing on pressure calculations involving different fluid densities and heights.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the hydrostatic pressure equation to find the gauge pressure. Some participants question the notation used in the pressure equation and whether atmospheric pressure should be considered in the calculations.

Discussion Status

There is a mix of agreement on the calculations presented, with some participants confirming the original poster's result. However, questions about the inclusion of atmospheric pressure and the notation used indicate that multiple interpretations are being explored.

Contextual Notes

Participants discuss the implications of using gauge pressure versus absolute pressure and the notation differences in the equations presented. There is also mention of the professor's method of teaching, which may influence the understanding of the problem.

DrMcDreamy
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Homework Statement



A test tube standing vertically in a test tube rack contains 3.8 cm of oil, whose density is 0.81 g/cm3 and 6.4 cm of water. What is the gauge pressure on the bottom of the tube? The acceleration of gravity is 9.8 m/s2. Answer in units of Pa

Homework Equations



P1-Po=[tex]\rho[/tex]oilgh1+[tex]\rho[/tex]H2Ogh2

The Attempt at a Solution



P1-Po= (810 kg/m3)(9.80 kg m /s2)(0.038 m) + (1000 kg/m3)(9.80 kg m/s2)(0.064 m)

P1-Po= 929 Pa

Is the work and answer right? TIA
 
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They are indeed correct.
 
Thank you! :smile:
 
Actually, I wonder whether you should also consider atmospheric pressure.

Also, why is it P1-P0 and not +?
 
^It came out right with 929 Pa.
The way the prof had shown it in class is P1-P0:

P2=Po+[tex]\rho[/tex]oilgh1

P1=P2+[tex]\rho[/tex]H2Ogh2

P1=Po+[tex]\rho[/tex]gh1+[tex]\rho[/tex]gh2

P1-Po=[tex]\rho[/tex]gh1+[tex]\rho[/tex]gh2
 
Oh ok, it was down to notation, then.
When I thought you might need the atmospheric pressure is because I was calculating P1, but you needed P1-P0. :)
 

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