Answer:
Volume flow rate [tex]= 1.81 * 10^{-2}[/tex] 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
[tex]P_1 = P_2[/tex]
and
[tex]A_1 V_1 = A_2 V_2[/tex]
Re arranging the file, we get -
[tex]V_1 = \frac{A_2 V_2}{A_1}[/tex]
The flow equation is
[tex]\frac{1}{2}\rho * V_1^2 = \frac{1}{2}\rho * V_2^2 + \rho * g * h\\[/tex]
Substituting the value of [tex]V_1[/tex] in above equation, we get -
[tex]V_2 = \sqrt{\frac{2gh}{(\frac{A_2}{A_1})^2-1} }[/tex]
Substituting the given values in above equation we get
[tex]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[/tex]
Volume flow rate
[tex]Q_2 = A_2 V_2\\= \pi r_2^2V_2^2\\= 3.14 * 0.04^2 * 3.61 \\= 1.81 * 10^{-2}[/tex]
