Explanation Write the expression for the energy balance equation for the given system including the minor losses. p 1 γ + V 1 2 2 g + z 1 = p 2 γ + V 2 2 2 g + z 2 + K L V 2 2 2 g (I) Here, the pressure at point 1 is p 1 , the velocity at point 1 is V 1 , the elevation at point 1 is z 1 , the pressure at point 2 is p 2 , the velocity at point 2 is V 2 , the elevation at point 2 is z 2 and loss coefficient is K L . Rearrange the equation in terms of pressure drop by neglecting the datum energy. ( p 1 − p 2 γ ) = ( V 2 2 2 g − V 1 2 2 g ) + ( K L V 2 2 2 g ) p 1 − p 2 = ρ 2 [ ( V 2 2 − V 1 2 ) + ( K L V 2 2 ) ] Δ p = ρ 2 [ ( K L V 2 2 ) + ( V 2 2 − V 1 2 ) ] (II) Write the expression to calculate the area of pipe. A = π 4 D 2 (III) Here, the diameter of pipe is D . Write the expression for the area ratio of the pipe. A 2 A 1 = ( D 2 D 1 ) 2 (IV) Here, section (1) diameter is D 1 and section (2) diameter is D 2 . Write the expression for the flow rate using the continuity equation. Q = A V V = Q A (V) Here, the flow rate is Q , area of cross section of pipe is A , and velocity of pipe at point is V . Conclusion: Consider the larger pipe as section (1) and smaller pipe as section (2). Calculate the area of the section (1) using the equation (III). Substitute 0.12 m for D in equation (III). A 1 = π 4 ( 0.12 m ) 2 = 0 .01131 m 2 Calculate the area of the section (2) using the equation (III). Substitute 0.06 m for D in equation (III). A 2 = π 4 ( 0

Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version

8th Edition

ISBN: 9781119080701

Author: Philip M. Gerhart, Andrew L. Gerhart, John I. Hochstein

Publisher: WILEY

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Chapter 8.4, Problem 53P

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The pressure drops across the contraction section and identifies the pressure drop due to losses and change in kinetic energy individually.

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Chapter 8.4, Problem 52P

Chapter 8.4, Problem 54P

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