To determine The volumetric flow of water through the meter and Plot the HGL and EGL. Explanation Given: Specific weight of the mercury ( γ H g ) is 846 lb/ft 3 . Calculation: Draw the schematic diagram of the manometer as shown in figure (1). Refer the figure (1), Apply the manometer rule from A to B. p A + γ w h A D − γ Hg h D E − γ w h B E = p B p A + ( 62.4 lb / ft 3 ) ( 1 ft ) − ( 846 lb / ft 3 ) ( 0.5 ft ) ( 62.4 lb / ft 3 ) ( 0.5 ft ) = p B p A − p B = 391.8 lb/ft 2 (I) Draw the schematic diagram of the piezometer as shown in figure (2). Refer figure (2), Apply the manometer rule from C to G. p C − γ w h C F − γ Hg h F G = p atm = 0 p C − ( 62.4 lb / ft 3 ) ( 1.5 ft ) − ( 846 lb / ft 3 ) ( 0.75 ft ) = 0 p C = 728.1 lb/ft 2 Consider the water as incompressible and inviscid and the flow is steady. The datum line is [ass through the point A in a horizontal pipe, so the datum levels is z A = z B = 0 . Apply Bernoulli equation between A and B . p A ρ g a s + V A 2 2 + g z A = p B ρ g a s + V B 2 2 + g z B p A γ w + V A 2 2 g = p B γ w + V B 2 2 g p A − p B γ w = 1 2 g ( V B 2 − V A 2 ) p A − p B = 1 2 ( 62.4 lb / ft 3 32.2 ft / s 2 ) ( V B 2 − V A 2 ) p A − p B = 0.9689 ( V B 2 − V A 2 ) (II) Consider the density of the water is constant and apply the continuity equation. ∂ ∂ t ∫ cv ρ d V + ∫ ∞ ρ V ⋅ d A 0 − V A A A + V B A B = 0 − V A [ π ( 2 12 ft ) 2 ] + V B [ π ( 0.5 12 ft ) 2 ] = 0 V B = 16 V A (III) Substitute Equation (I) in Equation (II)

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 5, Problem 78P

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Chapter 5, Problem 77P

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