Solving Manometer Height: Oil & Water

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Homework Help Overview

The discussion revolves around a manometer problem involving two liquids, oil and water, and their respective densities. The original poster is tasked with determining the height of the water in the manometer given specific pressures and densities.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the hydrostatic pressure equation and question the validity of the given pressures and setup. There is an exploration of potential typos in the problem statement regarding the gas pressure.

Discussion Status

Some participants have offered guidance on reconsidering the pressure values used in the calculations. There is acknowledgment of confusion regarding the physical arrangement of the liquids in the manometer, with multiple interpretations being explored.

Contextual Notes

Participants note that the problem may not be solvable with the provided values, raising concerns about the accuracy of the problem statement and the feasibility of the liquid arrangement described.

samantha.
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Homework Statement



Consider a manometer with a glass bulb (Pgas=88kPa) at one end and which is open to the atmosphere (P =101 kPa) at the other. The manometer has two liquids oil and water in it, as shown. If the oil has a density of 823 kg/m3 and water a density of 1.00x10^3 kg/m3 find the height of the water.

http://s16.postimage.org/9ihj14zir/Screen_shot_2012_04_12_at_3_22_43_PM.png

Homework Equations



P=pgh

The Attempt at a Solution



I keep deducing the equation to be;

Patm + ρwater*g*H = Pgas + ρoil*g*0.699m

Solving for H, I get 0.750m. The answer though, is 0.269m.

Where am I going wrong?
 
Last edited:
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samantha. said:

Homework Statement



Consider a manometer with a glass bulb (Pgas=88kPa) at one end and which is open to the atmosphere (P =101 kPa) at the other. The manometer has two liquids oil and water in it, as shown. If the oil has a density of 823 kg/m3 and water a density of 1.00x10^3 kg/m3 find the height of the water.

http://s16.postimage.org/9ihj14zir/Screen_shot_2012_04_12_at_3_22_43_PM.png

Homework Equations



P=pgh

The Attempt at a Solution



I keep deducing the equation to be;

Patm + ρwater*g*H = Pgas + ρoil*g*0.699m

Solving for H, I get 0.750m. The answer though, is 0.269m.

Where am I going wrong?
With the values you have (and the figure as it is), the problem is not solvable. No amount of water would ever cause the height of the oil to be 69.9 cm, as depicted in the figure.

I'm guessing the pressure of the gas should be 98 kPa, not 88. It was probably a typo of some sort.

(And just to be nit-picky about this problem, what is the water doing "floating" on oil? Oil is lighter that water. The water would sink to the bottom, and the oil would slurp up replacing the water at the top. Well, unless there's some sort of hidden barrier separating them. Anyway, good luck with it. :wink:)
 
88kPa is the value given. If it was a typo it wasn't on my part. I noticed the water floating on top of oil.. strange yes. Haha.
I spent half of my night panicking over my exam tomorrow, trying to figure out where I was going wrong in all of these problems that my professor gave me when in fact there's an issue with almost every question.

Thanks for the reply though. :)
 
By the way, I think your approach to this problem was valid. Try replacing 88 kPa with 98 kPa, and see what happens.

Good luck on your exam tomorrow! :smile:
 
Thank you, Sir. :)
 

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