Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 15 - Section 15.5 - Applications of Fluid Dynamics - Example: 15.7

Answer

$v_1 = \sqrt{\frac{2\Delta p}{\rho(b^2-1)}}$

Work Step by Step

We simplify the Bernoulli equation using the known conditions to find: $p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ We call $\Delta p=p_1 -p_2$ to find: $ \Delta p + \frac{1}{2} \rho v_1^2 = \frac{1}{2}\rho v_2^2$ We know that $\Delta p = \frac{1}{2}\rho b^2 v_1^2 - \frac{1}{2}\rho v_1^2$. Simplifying using the known conditions as well as math, we find: $v_1 = \sqrt{\frac{2\Delta p}{\rho(b^2-1)}}$
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