Question 5.2: Water flows through a right-angled bend (Fig. 5.29) formed b...

Here the streamlines are concentric circular arcs, and hence the velocity of fluid is in the tangential direction only. Moreover, the velocity field satisfies the equation of free vortex as

where, c is a constant.

We have V_{\theta}=3 m / s \text { at } r=\frac{0.15+0.45}{2}=0.3 m

Hence, c=3 \times 0.3=0.9 m ^{2} / s

Therefore, velocities at the inner and outer radii are

\left(V_{\theta}\right)_{ at r=r_{1}}=\frac{0.9}{0.15}=6 m/s

\left(V_{\theta}\right)_{\text {at } r=r_{2}}=\frac{0.9}{0.45}=2 m/s

(a) The accelerations along the streamline and normal to it can be written as a_{s}

(acceleration along the streamline) =V_{\theta} \frac{\partial V_{\theta}}{r \partial \theta}=0

a_{n} (acceleration normal to streamline) =\frac{-V_{\theta}^{2}}{r}

Therefore, \left(a_{n}\right)_{r=r_{1}}=\frac{-6 \times 6}{0.15}=-240 m / s ^{2}

\left(a_{n}\right)_{r=r_{2}}=\frac{-2 \times 2}{0.45}=-8.89 m / s ^{2}

Minus sign indicates that the accelerations are radially inwards. (b) The pressure gradient normal to the streamline is given by Eq. (5.22) as

\frac{1}{\rho} \frac{ d p}{ d r}=\frac{V_{\theta}^{2}}{r}-g \frac{ d z}{ d r} (5.22)

Therefore, \left(\frac{ d p}{ d r}\right)_{\text {at } r=r_{1}}=1000 \times \frac{6 \times 6}{0 \cdot 15}=240 \times 10^{3} N / m ^{2}=240 kN/m ^{2}

and \left(\frac{ d p}{ d r}\right)_{ at r=r_{2}}=1000 \times \frac{2 \times 2}{0.45}=8.89 \times 10^{3} N / m ^{2}=8.89 kN/m ^{2}

(c) At any radius r,

Hence, \int_{r_{1}}^{r_{2}} \frac{ d p}{ d r} d r=\int_{r_{1}}^{r_{2}} \rho \frac{c^{2}}{r^{3}} d r

or p_{2}-p_{1}=\frac{\rho}{2}\left(c^{2} / r_{1}^{2}-c^{2} / r_{2}^{2}\right)=\frac{\rho}{2}\left(V_{\theta_{1}}^{2}-V_{\theta_{2}}^{2}\right)

=\frac{1000}{2} \times(36-4) N / m ^{2}=16 kN / m ^{2}