Explanation Given information: The manometer fluid is mercury. The temperature of water and the mercury is 20 ° C . The height of section (2) with respect to section (1) is 10 ft . The height of water level in the manometer from the centerline of the pipe is 2 ft . The diameter of the pipe at section (1) is 3 in and the diameter of the pipe at section (2) is 1 in . The velocity at section (1) is 1.7 ft / s . Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system. p 1 ρ g + V 1 2 2 g + z 1 = p 2 ρ g + V 2 2 2 g + z 2 ( p 1 − p 2 ρ g ) + ( V 1 2 2 g − V 2 2 2 g ) + ( z 1 − z 2 ) = 0 …… (I) Here, the pressure at section (1) is p 1 , the velocity at section (1) is V 1 , the height of the section (1) from the datum is z 1 , the pressure at section (2) is p 2 , the velocity at section (2) is V 2 , the height of the section (2) from the datum is z 2 , the density of water is ρ , and the acceleration due to gravity is g . Write the expression for the specific weight for water. γ = ρ g Here, the specific weight is γ . Substitute ρ g for γ in Equation (I). ( p 1 − p 2 γ ) + ( V 1 2 2 g − V 2 2 2 g ) + ( z 1 − z 2 ) = 0 …… (II) Write the expression for the flow rate at section (1). Q = A 1 V 1 …… (III) Here, the area at section (1) is A 1 and the velocity at section (1) is V 1 . Write the expression for the area at section (1). A 1 = π 4 D 1 2 Here, the diameter at section (1) is D 1 . Substitute π 4 D 1 2 for A 1 in Equation (III). Q = π 4 D 1 2 V 1 …… (IV) Write the expression for the flow rate at section (2). Q = A 2 V 2 …… (V) Here, the flow rate is Q , the jet area at section (2) is A 2 and the velocity at section (2) is V 2 . Write the expression for the area at section (2). A 2 = π 4 D 2 2 Here, the diameter at section (2) is D 2 . Substitute π 4 D 2 2 for A 2 in Equation (V). Q = π 4 D 2 2 V 2 …… (VI) Write the expression for the pressure difference from manometer differential pressure measurement. p 1 − p 2 = ( ρ M − ρ ) g h = ρ M g h − ρ g h …… (VII) Here, the density of manometer is ρ M and the manometer reading is h . Write the expression for the specific weight for mercury. γ M = ρ M g Here, the specific weight is γ . Substitute ρ M g for γ M and ρ g for γ in Equation (VII). p 1 − p 2 = γ M h − γ h = ( γ M − γ ) h …… (VIII) Calculation: Refer to the property table to obtain the specific weight of water at temperature 20 ° C as, γ = 64.2 slug / ft 3 Refer to the property table to obtain the specific weight of mercury at temperature 20 ° C as, γ M = 846 slug / ft 3 Substitute 64

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ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 3, Problem 3.131P

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