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

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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