## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 15 - Exercises and Problems - Page 281: 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: $v = Av = \pi (.01)^2 (1.5 \ m/s)=4.73 \times 10^{-4}\frac{m^3}{s}$

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