Understanding the Conversion Formula for Venturimeter Readings

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Discussion Overview

The discussion revolves around understanding the conversion formula for venturimeter readings, specifically the conversion of differential manometer readings from mercury to water. Participants are addressing a numerical problem related to this conversion and exploring the underlying principles and calculations involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion regarding the conversion formula from mercury to water in a venturimeter context.
  • Another participant suggests drawing a U-tube manometer to visualize the pressure differences and apply the ρgh formula to derive the pressure differential.
  • A participant notes a discrepancy between their calculated conversion (20 cm of Hg = 272 cm of H2O) and the provided solution (252 cm of H2O).
  • Another reply emphasizes the importance of considering the water column in the U-tube and encourages maintaining an algebraic approach without substituting numerical values.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are differing calculations and interpretations of the conversion formula. The discussion remains unresolved regarding the correct conversion values.

Contextual Notes

Limitations include potential oversight of the water column's impact on pressure differentials and the need for clarity in the algebraic derivation of the conversion formula.

ujjwal kapil
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Hi all

I was trying to solve some numerical problems on venturimeter. I got stuck due to a formula, that I was not able to make sense of. It is a conversion of reading of differential manometer in mercury to corresponding reading in water. I've attached the question, please help me understand it.
 

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Hi ujjwal kapil. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Draw a U-tube manometer showing a difference in mercury levels. Now, add water above both surfaces of mercury, and show pressures of P1 and P2 acting on that water.

Apply your ρgh formula, etc., and derive the pressure differential P2 - P1 in terms of the other quantities present. Write your working here.
 
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I have found an answer, but not in accordance with the solution.
20cm of Hg= 272cm of H2O, but it is given as 252 in the solution.
 

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Keep it simple, draw a symmetrical U-tube.

You overlooked the water! On one side you have P1 and a column of h metres of H2O balanced on the other side by P2 and a column of h metres of Hg.

So try this again. Keep it as algebra, don't substitute numbers. We're aiming towards that expression you circled, remember?
 

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