Converting venturimeter mercury manometer readings to water equivalent

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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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  • Printed textbook solution for a venturimeter problem: pipe diameters 30 cm and 15 cm, mercury manometer reading 20 cm, with the head formula h = x[Sh/So - 1] circled and handwritten notes beside 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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  • Handwritten working on lined paper: a U-tube manometer sketch with Hg marked in the bend, the equations p1 - p2 = rho_Hg g h and p1 - p2 = 13.6x10^3 x 9.81 x 0.2 = 26656 N/m2, then rho_w g h = 26656, h = 2.717 m, or 271.7 cm of water and 20 cm of Hg.
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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?