Question 8.12: A circular pipe of radius a and length L is attached to a sm...
A circular pipe of radius a and length L is attached to a smoothly rounded outlet of a liquid reservoir by means of flanges and bolts as shown in Fig. 8.18. At the flange section the velocity is uniform over the cross-section with magnitude V_{0}. At the outlet, which discharges into the atmosphere, the velocity profile is parabolic because of the friction in the pipe. What force must be supplied by the bolts to hold the pipe in place?
For the control volume as shown, continuity equation gives (for steady flow)
\int_{C S} \rho \vec{V} \cdot d \vec{A}=0 \quad \text { or, } \quad-V_{0} \pi a^{2}+\int_{0}^{a} 2 V_{e} \pi r d r=0, \text { So, } \int_{0}^{a} V_{e} r d r=\frac{V_{0} a^{2}}{2}
since V_{e} is parabolic, let V_{e}=a_{0}+a_{1} r+a_{2} r^{2}
\text { at } r=a, V_{e}=0 \rightarrow a_{1} a+a_{2} a^{2}+a_{0}=0
\text { at } r=0, \quad \frac{ d V_{e}}{ d r}=0 \rightarrow a_{1}=0
This will give
\int_{0}^{a}\left(a_{0}+a_{2} r^{2}\right) r d r=\frac{V_{0} a^{2}}{2}
or a_{0} \cdot \frac{a^{2}}{2}+a_{2} \frac{a^{4}}{4}=V_{0} \frac{a^{2}}{2} \quad \therefore \quad a_{0}+\frac{a^{2}}{2} a_{2}=V_{0}
Again, we know a_{0}+a^{2} a_{2}=0
Combining the above two expressions, we get,
\frac{a^{2}}{2} a_{2}=-V_{0} \quad \therefore \quad a_{2}=-\frac{2 V_{0}}{a^{2}} \quad \text { and } \quad a_{0}=+2 V_{0}
\therefore \quad V_{e}=2 V_{0}\left[1-\frac{r^{2}}{a^{2}}\right]
Now, applying momentum equation to the control volume in the flow direction (let F be the force as shown on the control volume; same force must be supplied by bolts):
F+\left(p_{1}-p_{2}\right) \pi a^{2}=-\rho V_{0} \pi a^{2} \cdot V_{0}+\rho \int_{0}^{a} 2 \pi r d r V_{e}^{2}
=-\pi \rho V_{0}^{2} a^{2}+\pi \rho V_{0}^{2} \int_{0}^{a} 8 r\left(1-\frac{r^{2}}{a^{2}}\right)^{2} d r
=\pi \rho V_{0}^{2} a^{2}\left[-1+\int_{0}^{1} 8 \frac{r}{a}\left(1+\frac{r^{4}}{a^{4}}-2 \frac{r^{2}}{a^{2}}\right) d\left(\frac{r}{a}\right)\right]
=\pi \rho V_{0}^{2} a^{2}\left[-1+\left(4+\frac{4}{3}-4\right)\right]
=\frac{1}{3} \pi \rho a^{2} V_{0}^{2}
\therefore F=\frac{1}{3} \pi \rho a^{2} V_{0}^{2}-\left(p_{1}-p_{2}\right) \pi a^{2} in horizontal direction only (gravity is in vertical direction).