Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems - Page 288: 62

Answer

1200 Pa.

Work Step by Step

Use Poiseuille’s equation, 10-9, to find the pressure difference. $$Q=\frac{\pi R^4 (P_1-P_2)}{8 \eta \mathcal{l}}$$ $$ P_1-P_2=\frac{8Q \eta \mathcal{l}}{\pi R^4}$$ Convert the flow rate Q to $6.5 \times10^{-4}\;m^3/s$. $$ P_1-P_2=\frac{8(6.5 \times10^{-4}\;m^3/s) (0.2\;Pa \cdot s) (1600\;m)}{\pi (0.145\;m)^4}$$ $$ P_1-P_2=1198\;Pa\approx1200\;Pa$$
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