Answer
$A_3=3.33 m^{2}$
Work Step by Step
$Q_{in}=Q_{out}$
$Q_1=Q_2+Q_3$
$40 \frac{m^{3}}{s}=20 \frac{m^{3}}{s}+Q_3$
$Q_3=20 \frac{m^{3}}{s}$
$Q=vA$
Since $Q_3=20 \frac{m^{3}}{s}$ and $v_3=6 \frac{m}{s}$
$20=6A_3$
$A_3\approx3.33 m^{2}$
Numerical Methods for Engineers
by Chapra, Steven; Raymond, Canale;
Published by
McGraw-Hill Science/Engineering/Math
ISBN 10:
0073401064
ISBN 13:
978-0-07340-106-5
Chapter 1 - Mathematical Modeling and Engineering Problem Solving - Problems - Page 22: 1.9
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