Question

A liquid is flowing through a horizontal pipe whose radius is m. The pipe bends straight upward through a height of 10.6 m and joins another horizontal pipe whose radius is m. What volume flow rate will keep the pressures in the two horizontal pipes the same if the pressure in both horizontal pipes is the same

Answer 1

Answer:

Volume flow rate $= 1.81 * 10^{-2}$ meter cube per second

Explanation:

As we know that the

Pressure at the two ends would be the same along with volume of flow.

i.e

$P_1 = P_2$

and

$A_1 V_1 = A_2 V_2$

Re arranging the file, we get -

$V_1 = \frac{A_2 V_2}{A_1}$

The flow equation is

$\frac{1}{2}\rho * V_1^2 = \frac{1}{2}\rho * V_2^2 + \rho * g * h$

Substituting the value of $V_1$ in above equation, we get -

$V_2 = \sqrt{\frac{2gh}{(\frac{A_2}{A_1})^2-1} }$

Substituting the given values in above equation we get

$V_2 = \sqrt{\frac{2*9.8*10.6}{(\frac{\pi 0.04^2}{\pi 0.02^2} )^2 -1} }\\ V_2 = 3.61 m$

Volume flow rate

$Q_2 = A_2 V_2\\= \pi r_2^2V_2^2\\= 3.14 * 0.04^2 * 3.61 \\= 1.81 * 10^{-2}$