To determine The gage pressure at section (1). Explanation Given information: The fluid is gasoline at temperature 20 ° C . The weight flow is 120 N / s . The diameter of pipe at section (1) is 8 cm and the diameter of the open jet is 5 cm . The height of section (2) with respect to section (1) is 12 m . The pressure at section (2) is atmospheric. 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 weight flow. w ˙ = ρ Q g Q = w ˙ ρ g …… (II) Here, the flow rate is Q , the weight flow is w ˙ , the density of gasoline is ρ 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) Calculation: Refer to the property table to obtain the density of gasoline at 20 ° C temperature as, ρ = 680 kg / m 3 Substitute 120 N / s for w ˙ , 680 kg / m 3 for ρ and 9.81 m / s 2 for g in Equation (II). Q = ( 120 N / s ) ( 680 kg / m 3 ) ( 9.81 m / s 2 ) = ( 120 N / s ) 6670.8 kg / m 2 s 2 = 0.01798 m 3 / s Substitute 0.01798 m 3 / s for Q and 8 cm for D 1 in Equation (IV). ( 0.01798 m 3 / s ) = π 4 ( 8 cm ) 2 V 1 ( 0

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ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 3, Problem 3.130P

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The gage pressure at section (1).

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Chapter 3, Problem 3.129P

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Fluid Mechanics, 8 Ed

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