Explanation Given information: The reading of pressure gage is 330 kPa specific gravity of oil is 0.79 , specific gravity of mercury is 13.6 , and specific gravity of gasoline is 0.7 . Write the expression for the gage pressure. P g a g e = ρ w g h w − ρ o g h o + ρ m g h m + ρ g g h g + P g P g = − ( ρ w g h w − ρ o g h o + ρ m g h m + ρ g g h g ) + P g a g e P g = − [ g ρ w ( h w − ρ o ρ w h o + ρ m ρ w h m + ρ g ρ w h g ) ] + P g a g e ....... (I) Here, the gage pressure is P g a g e ,the density of water is ρ w , the manometric head of water column is h w , acceleration due to gravity is g , the density of oil is ρ o , the manometric head of oil h o , is the density of mercury ρ m , the manometric head of mercury is h m , the density of gasoline is ρ g , the manometric head of gasoline is h g , and the pressure exerted by gasoline P g . Substitute S G o for ρ o ρ w , S G m for ρ m ρ w , and S G g for ρ g ρ w in equation (I). P g = − [ g ρ w ( h w − S G o h o + S G m h m + S G g h g ) ] + P g a g e ....... (II) Here the specific gravity of oil is S G o , the specific gravity of mercury is S G m , and the specific gravity of gasoline is S G g . Calculation: Substitute 0.79 for S G o , 13.6 for S G m and 0.7 for S G g , 1000 kg / m 3 for ρ w , 9.81 m / s 2 for g , 45 cm for h w , 50 cm for h o , 330 kPa for P g a g e , 10 cm for h m in equation (II)

Fluid Mechanics Fundamentals And Applications

3rd Edition

ISBN: 9780073380322

Author: Yunus Cengel, John Cimbala

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 3, Problem 154P

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