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}$
