Finding change in pressure when mercury level drops

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Homework Help Overview

The discussion revolves around a problem involving the change in pressure when the mercury level in a manometer drops, specifically addressing the implications of this change on gauge pressure calculations. The subject area includes fluid mechanics and pressure measurement principles.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to account for the density of water in addition to mercury when calculating pressure changes. There are questions about the "real balance equation" and how to incorporate the effects of both mercury and water levels in the pressure calculations.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications about the relationship between the heights of mercury and water in the context of pressure changes. Some participants express uncertainty about the concepts being discussed, while others offer corrections and suggestions for approaching the problem.

Contextual Notes

There is mention of the problem being part of a 20-mark question, indicating a significant academic weight. Participants also reflect on their learning experiences and the challenges of understanding this type of problem.

confusedperson
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I don't know what I did wrong for this question, but I assume something is wrong as my solution is extremely simple and this is a 20 marks question...
1. Homework Statement

Refer to uploaded image.

Homework Equations


Pressure = h ( density in kg/m^3 ) ( gravitational acceleration)

The Attempt at a Solution


Pressure change at A = -0.04 ( 13600 ) ( 9.81 )[/B]
Final gauge pressure = 150.3kPa - 0.04 (13600) (9.81) - 1.025 * 10^5 Pa
 

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Hi conf,

You want to make a real balance equation and take the density of water into account as well...
 
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Sorry I do not understand what is the real balance equation...
 
4 cm of Hg took the place of 4 cm of water. That 4 cm of water exercised a pressure as well
 
Thank you. I'm not entirely sure how that works but I assume I have to add 4cm of water back into the final gauge pressure?
 
And don't forget the 4 cm of mercury that were there. The mercury level changes on both sides, the water level changes on one side.
 
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Huh. That was not expected...I simply have not come across a problem of this type before. There goes at least 10 marks for my final exam. However, lesson learnt. Thank you guys for enlightening me on this new knowledge.
 
That's the idea.
mfb said:
And don't forget the 4 cm of mercury that were there. The mercury level changes on both sides, the water level changes on one side.
I thought he/she did take the Hg, but only forgot the water ? So with (13600 - 1000) instead of just 13600 kg/m3 it would have been alright ?
 
BvU said:
I thought he/she did take the Hg, but only forgot the water ? So with (13600 - 1000) instead of just 13600 kg/m3 it would have been alright ?
@mfb is pointing out that if the level of Hg on the right drops by x then the level on the left rises by x, so the difference in heights drops by 2x. The pressure exerted comes from the height difference.
 
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  • #10
Oh boy, have I been sleeping o:) o:) ! Fixed on the water :smile: Thanks for the wake-up call ! But nevertheless I'm off to bed.:sleep:
 

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