The diameter of the pipe is d = 0.768 m
The diameter of smooth constriction is d' = 0.4608 m
The density of oil is
[tex]\rho\text{ = 821 }\frac{kg}{m^3}[/tex]The pressure in the pipe is, P = 7120 N/m^2
The pressure in the constricted section is P' = 5340 N/m^2
We have to find the rate of the flowing.
The area of the pipe will be
[tex]\begin{gathered} a=\pi\text{ (}\frac{d}{2})^2 \\ =\text{ 3.14}\times(\frac{0.768}{2})^2 \\ =0.463m^2 \end{gathered}[/tex]
The area of the constricted section will be
[tex]\begin{gathered} a^{\prime}=\text{ }\pi\times(\frac{d^{\prime}}{2})^2 \\ =3.14\times(\frac{0.4608}{2})^2 \\ =0.1667m^2 \end{gathered}[/tex]The formula to find the rate of flow is
[tex]V=\text{ aa'}\sqrt[]{\frac{2(P-P^{\prime})}{\rho(a^2-a^{\prime2})}}[/tex]Substituting the values, the rate of flow will be
[tex]\begin{gathered} V=0.463\times0.1667\times\sqrt[]{\frac{2\times(7120-5340)}{821\lbrack(0.463)^2-(0.1667)^2\rbrack}} \\ =\frac{0.3721m^3}{s} \end{gathered}[/tex]
