Question 8.9: Consider steady flow of an incompressible Newtonian fluid (d...

Infinitely long cylinder implies \frac{\partial}{\partial z} (any variable)= 0.

Since, the flow is rotational symmetric, we have( \frac{\partial}{\partial z} (any variable)=0)

The continuity equation in cylindrical coordinates under the above conditions is given by

which implies that r \mathcal{v}_{r} is not a function of r. Now, by virtue of no penetration boundary condition v_{r}=0  at r=a, b , which implies that v_{r}=0  everywhere in the flow field (except at r = 0, which is a singularity in this case).
From the z momentum equation in cylindrical coordinates (Eq. 8.28f), we get, using the above simplifying considerations

\rho\left(\frac{\partial v_{z}}{\partial t}+v_{r} \frac{\partial v_{z}}{\partial r}+v_{\theta} \frac{\partial v_{z}}{\partial \theta}+v_{z} \frac{\partial v_{z}}{\partial z}\right)=-\frac{\partial p}{\partial z}  +\mu\left[\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{z}}{\partial \theta^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}}\right]+\rho b_{z}                     (8.28f)

Integrating the above equation (noting that v_{z} is a function of r only), we have

v_{z}=C_{1} \operatorname{In} r+C_{2}          (8.73a)

where C_{1}  and C_{2}  are constants of integration.

Equation (8.73a) is subjected to the following boundary conditions:

v_{z}=U_{i}      at      r=a   and   v_{z}=0         at      r=b

The constants of integrations are accordingly found to be

C_{1}=\frac{U_{i}}{\ln \frac{a}{b}} and C_{1}=-\frac{U_{i}}{\ln \frac{a}{b}} \ln b

Thus, Eq. (8.73a) becomes

From the \theta momentum equation in cylindrical coordinates (Eq. 8.28e) and using the simplifying considerations specified as before, we get

\rho\left(\frac{\partial v_{\theta}}{\partial t}+v_{r} \frac{\partial v_{\theta}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{v_{r} v_{\theta}}{r}+v_{z} \frac{\partial v_{\theta}}{\partial z}\right)=-\frac{1}{r} \frac{\partial p}{\partial \theta}  +\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{\theta}\right)\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{\theta}}{\partial \theta^{2}}+\frac{\partial^{2} v_{\theta}}{\partial z^{2}}+\frac{2}{r^{2}} \frac{\partial v_{r}}{\partial \theta}\right]+\rho b_{\theta}                                  (8.28e)

Integrating the above equation (noting that v_{\theta} is a function of r only), we have

v_{\theta}=C_{3} \frac{r}{2}+\frac{C_{4}}{r}      (8.73b)

Equation (8.73b) is subjected to the following boundary conditions:

At r=a, v_{\theta}=a \omega   and at r=b, v_{\theta}=b \omega .

The constant of integrations are found to be
C_{3}=2 \omega        and   C_{4}=0

Finally, Eq. (8.73b) becomes
v_{\theta}=\omega r

(ii) The r momentum equation in cylindrical co-ordinates (Eq. 8.28d),  using the simplifying considerations specified as before, reads

\rho\left(\frac{\partial v_{r}}{\partial t}+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{r}}{\partial \theta}-\frac{v_{\theta}^{2}}{r}+v_{z} \frac{\partial v_{r}}{\partial z}\right)=-\frac{\partial p}{\partial r} +\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{r}\right)\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{r}}{\partial \theta^{2}}+\frac{\partial^{2} v_{r}}{\partial z^{2}}-\frac{2}{r^{2}} \frac{\partial v_{\theta}}{\partial \theta}\right]+\rho b_{r}                           (Eq. 8.28d)

Or     \frac{d p}{d r}=\rho \frac{v_{\theta}^{2}}{r} ( Since p is a function of r only)

Or     \frac{d p}{d r}=\rho \frac{\omega^{2} r^{2}}{r}

Integrating the above equation, we have
\int_{p_{i}}^{p} d p=\int_{a}^{r} \rho \omega^{2} r d r