Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 15 - Exercises and Problems: 57

Answer

a) 1.5 m/s b) $4.73 \times 10^{-4}\frac{m^3}{s}$ L/s

Work Step by Step

a) In example 15.7, we found the equation for the flow speed in the pipe: $v_1 = \sqrt{\frac{2\Delta P}{\rho(b^2 - 1)}}$ This can be simplified to: $v_1 = \sqrt{\frac{2\Delta P}{\rho((\frac{A_1}{A_2})^2 - 1)}}$ $v_1 = \sqrt{\frac{2\Delta P}{\rho((\frac{\pi r_1^2}{\pi r_2^2})^2 - 1)}}$ $v_1 = \sqrt{\frac{2\Delta P}{\rho((\frac{\pi r_1^2}{\pi r_2^2})^2 - 1)}}$ $v_1 = \sqrt{\frac{2\Delta P}{\rho((\frac{r_1^2}{r_2^2})^2 - 1)}}$ $v_1 = \sqrt{\frac{2\times17,000}{1000((\frac{(.01)^2}{(.005)^2})^2 - 1)}}$ $v_1=1.5 \ m/s$ b) We can now find the volume flow rate: $ v = Av = \pi (.01)^2 (1.5 \ m/s)=4.73 \times 10^{-4}\frac{m^3}{s}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.