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How did he obtain the differential area of flow as dA= Wdy ?

How did he obtain the differential area of flow as dA= Wdy ?

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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How did he obtain the differential area of flow as dA= Wdy ?
$ 4 Example-1.1: Calculation of flow rate using velocity distribution in an open channel flow R A liquid flows through an open channel as shown in the figure below. The variation of velocity in the vertical direction (v) can be expressed by: x = 2y12, Calculate the volumetric flow rate and the average velocity in the channel? Solution: The total flow rate is given by Eq. (1.3): % Atotal W=1.5 m and Y= 1.2 m 5 The average velocity is given by Eq. (1.4), Q Atotal 2.63 1.5x1.2 V = F6 Q = fundA dA = Wdy The differential area of flow for the above geometry is: Replacing dA by Wdy and ux by 2y¹2 in Eq. (1.3), and putting the limits of integration, Q=²2y¹/2 Wdy 6 → V = →Q= F7 & 7 Atotal = 1.46 W F8 8 m s F9 →>>> Q2.63 m³s-1 61 Ux=2 9 2¹/2 Ux(y) E F10 O. F11 (1.4) 0 F12
Transcribed Image Text:$ 4 Example-1.1: Calculation of flow rate using velocity distribution in an open channel flow R A liquid flows through an open channel as shown in the figure below. The variation of velocity in the vertical direction (v) can be expressed by: x = 2y12, Calculate the volumetric flow rate and the average velocity in the channel? Solution: The total flow rate is given by Eq. (1.3): % Atotal W=1.5 m and Y= 1.2 m 5 The average velocity is given by Eq. (1.4), Q Atotal 2.63 1.5x1.2 V = F6 Q = fundA dA = Wdy The differential area of flow for the above geometry is: Replacing dA by Wdy and ux by 2y¹2 in Eq. (1.3), and putting the limits of integration, Q=²2y¹/2 Wdy 6 → V = →Q= F7 & 7 Atotal = 1.46 W F8 8 m s F9 →>>> Q2.63 m³s-1 61 Ux=2 9 2¹/2 Ux(y) E F10 O. F11 (1.4) 0 F12
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