Question

Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to another horizontal pipe with a radius of 0.0420 m on a floor of the factory that is 12.6 m higher. The connection is made with a vertical section of pipe and an expansion joint. Determine the volume flow rate that will keep the pressure in the two horizontal pipes the same.

Answer 1

Answer:

$0.0168 m^3/s$

Explanation:

We are given that

$r_1=0.0183 m$

$h_1=0$

$r_2=0.0420 m$

$h_2=12.6 m$

Let $P_1=P_2=P$

By using Bernoulli theorem

$P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2$

$\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2$

$v^2_1+2gh_1=v^2_2+2gh_2$

$A_1v_1=A_2v_2$

$v_1=\frac{A_2v_2}{A_1}$

$(\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6$

$(\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96$

$v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96$

$v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}$

$v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}$

$v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}$

$v_2=3.038 m/s$

Volume flow rate =$A_2v_2$

Volume flow rate =$\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s$