SPSolution3-5 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 3 Given: - A double U-tube connecting two horizontal pipelines - Density of sea water = 1035 kg/m 3 Find: - Pressure difference between the pipelines - Effect of air column - Effect of an oil column instead of the air column System: Assumptions: - Liquids are incompressible Basic Equation(s): atm PP g L ρ =+ Solution: (a) Let the pressure in the fresh water pipeline be P 1 and the pressure in the sea water pipeline be P 5 ; the different points are as shown in the figure above. As we move along the manometer connecting the two pipelines, pressure increases when moving in the downward direction while pressure decreases when moving in the upward direction. We can write: 1 5 freshwater freshwater mercury mercury air air seawater seawater Pg h g hg h g h P ρρ +− + = Using Table A-19, we have: 3 kg 996 m freshwater = ; 3 kg 13,529 m mercury = For STP conditions: 3 kg 1.2 m air = Sea Water Mercury Air 60 cm 10 cm 70 cm 40 cm 1 2 3 4 5 Fresh Water
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Naik, Sameer V. Fall 2009 1000 We can obtain the required pressure difference between the two pipelines to be: 15 3.36 kPa PP −= (b) We can note that the pressure equivalent of 70 cm column of air between mercury and seawater is: 8.24Pa air air gh ρ = . This is negligible compared to the total pressure difference computed above. Hence, the air column could have been neglected without causing any error in engineering analysis.
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This note was uploaded on 05/15/2010 for the course ME 200 taught by Professor Gal during the Spring '08 term at Purdue University-West Lafayette.

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SPSolution3-5 - Naik, Sameer V. Solution for Special...

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