Explanation Given information: The exit pressure of the nozzle is 101 kPa and the constant flow rate of 160 gal / min . The friction is neglected in the pipe flow. Write the expression for velocity at section (1). V 1 = Q A 1 .... (I) Here, the velocity is V 1 , the area is A 1 and the flow rate is Q . Write the expression for the area of section (1). A 1 = π 4 D 1 2 Here, the diameter is D 1 . Substitute π 4 D 1 2 for A 1 in Equation (I). V 1 = Q π 4 D 1 2 = 4 Q π D 1 2 .... (II) Write the expression for velocity at section (2). V 2 = Q A 2 .... (III) Here, the velocity is V 2 and the area is A 2 . Write the expression for the area of section (2). A 2 = π 4 D 2 2 Here, the diameter is D 2 . Substitute π 4 D 2 2 for A 2 in Equation (III). V 2 = Q π 4 D 2 2 = 4 Q π D 2 2 .... (IV) Write the expression for the Bernoulli’s equation between two sections. p 1 ρ g + V 1 2 2 g + z 1 = p 2 ρ g + V 2 2 2 g + z 2 ( p 1 ρ g − p 2 ρ g ) + ( V 1 2 2 g − V 2 2 2 g ) + ( z 1 − z 2 ) = 0 ( p 1 − p 2 ρ g ) + ( V 1 2 2 g − V 2 2 2 g ) + ( z 1 − z 2 ) = 0 .... (V) 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 oil is ρ and the acceleration due to gravity is g . Since both the section is at the same level from the datum plane, so the difference between the datum height at the two sections is zero. z 1 − z 2 = 0 Since the pressure at section (2) is atmospheric, so the gage pressure at section (2) is zero. p 2 = 0 Substitute 0 for z 1 − z 2 and 0 for p 2 in Equation (V). ( p 1 − 0 ρ g ) + ( V 1 2 2 g − V 2 2 2 g ) + 0 = 0 ( p 1 ρ g ) + ( V 1 2 2 g − V 2 2 2 g ) = 0 ( p 1 ρ ) + ( V 1 2 − V 2 2 2 ) = 0

8th Edition

ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 3, Problem 3.2FEEP

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The gage pressure at section (1).

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