Explanation Given information: The atmospheric pressure is 14.5 psia specific gravity of oil is 0.80 and Specific gravity of mercury is 13.6 . Write the expression for the absolute pressure. P = ρ w g h w + ρ o 1 g h o 2 + ρ m g h m + ρ o 2 g h o 2 + P a t m = [ g ( ρ w h w + ρ o h o + ρ m h m + ρ o 2 h o 2 ) ] + P a t m ....... (I) Here, the atmospheric pressure is P a t m , the density of water is ρ w , the manometric head of water column h w , the acceleration due to gravity is g , the density of oil ρ o , the manometric head of oil is h o , the density of mercury is ρ m , the manometric head of mercury is h m , the density of oil is ρ 02 , and the manometric head of oil at right side is h o 2 . Write the expression for the density of oil. ρ o = ρ w S G o ....... (II) Here, the specific gravity of oil is S G o . Write the expression for the density of mercury. ρ m = ρ w S G m ....... (III) Here, the specific gravity of mercury is S G m . Calculation: Substitute 62.3 lbm / ft 3 for ρ w and 0.8 for S G o in Equation (II). ρ o = 62.3 lbm / ft 3 × 0.8 = 49.84 lbm / ft 3 Substitute 62.3 lbm / ft 3 for ρ w and 13.6 for S G m in Equation (III). ρ m = 62.3 lbm / ft 3 ×13 .6 = 847.28 lbm / ft 3 Substitute 49.84 lbm / ft 3 for ρ o , 847.28 lbm / ft 3 for ρ m and 62.3 lbm / ft 3 lbs / in 3 for ρ w , 32

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 155EP

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The absolute pressure at centre of pipe.

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Chapter 3, Problem 154P

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Fluid Mechanics Fundamentals And Applications

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The absolute pressure at centre of pipe.

Solution Summary: The author explains the absolute pressure at the centre of pipe, the density of water, and the acceleration due to gravity.

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The absolute pressure at centre of pipe.

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