TP1:Example Problem 25

• Define a problem
• Explain the importance of the problem.
• Include a figure with flow (velocity) vectors demonstrating fluid flow in the radial direction
• Include both SI and Imperial Units|RyanToomey|Natthakan Welaha and Jordan Kubus}}

Radial Flow in Spherical Coordinates

Many engineering models make use of spherical coordinates. A sphere can represent a multitude of engineering systems, from spherical tanks/reservoirs, to spherical pills being dissolved. Here, we show how to apply this coordinate system to the general microscopic balance to solve for a radial velocity and a volumetric flow rate.

Real Life Example: New Designs for Hydrocracking in the Petroleum Industry

Radial flow in a spherical packed bed used in the hydrocracking process of petroleum refining. Hydrocracking is a process that takes high molecular weight feedstock such as gas oil (pre-processed oil to make gasoline) converting and hydrogenating into products lower in molecular weight. Hence the name "cracking". This catalytic reaction upgrades the low-quality gas oil into higher quality fuels.

The new design tests the effectiveness of a spherical shaped packed bed reactor versus the traditional cylindrical shape. It was shown that although the pressure drop difference between the two shapes are negligible, the product yield can be easily manipulated in the spherical shape by adding more catalyst. This is possible due to the lower temperature achieved at the center of the spherical shaped reactor and prevents hot zone formations. The spherical nature allows for even heating, thus better reaction yield. This is impossible in the cylindrical reactor design.

Spherical Coordinates

Learning Outcome

-      Express the continuity equation in terms of the incompressible approximation

-      Highlight the difference between Lagrangian and Eulerian perspectives of derivative

-      Determine volumetric flow rate given a velocity profile

Example Question 1

A viscous fluid such as water is flowing in the radial direction in spherical coordinates. Use the continuity equation to show how the radial component of the velocity varies with radius at steady state. Given that radial velocity is 1 m/s at $r$ = 5 cm, what is the velocity at $r$ = 20 cm?

Question 1 Solution

Velocity Vector for Spherical Coordinates:

$$\overrightarrow{V}=V_r(r,\theta ,\varphi )+V_{\theta }(r,\theta ,\varphi )+V_{\varphi }(r,\theta ,\varphi )$$
1. Reduce to Radial Flow Only: We only have flow in the radial direction, so the theta and phi components are zero. Also, the radial velocity is only a function of radial position due to symmetry

$$\overrightarrow{V}=V_r(r)$$

General Microscopic Balance: We will set this up and solve it for our given conditions.

$$\frac{\delta \psi }{\delta t}=-(▽\cdot \mathrm{\Psi })+{\dot{r}}_{\psi }$$ Here, $\mathrm{\Psi }$ denotes Flux of transport property in question and $\psi$ will represent the intensive property.

In our example, it is the flux of the velocity field we are interested in. The flux of the velocity field is composed of bulk flow only:

$$\mathrm{\Psi }=\rho \overrightarrow{v}+{\overrightarrow{j}}_{\mathrm{\Psi }}$$$$\mathrm{\Psi }=\rho \overrightarrow{v}$$

There is also no non-conservative term, general microscopic balance becomes:$$\frac{\delta \rho }{\delta t}=-▽\cdot \rho \overrightarrow{v}$$
From here, we can apply the chain rule and rearrange:$$\frac{\delta \rho }{\delta t}=-(\overrightarrow{v}\cdot ▽\rho )-\rho (▽\cdot \overrightarrow{v})$$$$\frac{\delta \rho }{\delta t}+(\overrightarrow{v}\cdot ▽\rho )=-\rho (▽\cdot \overrightarrow{v})$$
The left-hand side of this is equation represents the Total Derivative, can be re-written as:

$$\frac{D\rho }{Dt}=-\rho (▽\cdot \overrightarrow{v})$$
At this point, apply the steady state assumption:$$0=-\rho (▽\cdot \overrightarrow{v})$$
Since the density of water is non-zero, this suggests that the term in parentheses $(▽\cdot \overrightarrow{v})$ must be zero. This term represents the Divergence of the velocity field. It is zero for an incompressible fluid flowing at steady state, in the absence of any non-conservative mechanisms. It is this relationship, coupled with the divergence of a vector field in spherical coordinates, that allows us to solve for a velocity profile.

Divergence of a velocity field in spherical coordinates:

$$\frac{1}{r^2}\frac{\delta (r^2{V}_r)}{\delta r}+\frac{1}{rsin(\theta )}\frac{\delta }{\delta \theta }(V_{\theta }sin\theta )+\frac{1}{rsin(\theta )}\frac{\delta V_{\varphi }}{\delta \varphi }$$
Cross out the velocity components in which there is no flow: $V_{\theta }=V_{\varphi }=0$ The Divergence, which is equal to zero for a steady-state incompressible fluid, now looks like:

$$\frac{1}{r^2}\frac{\delta (r^2{V}_r)}{\delta r}=0$$
Multiply each side by $r^2$:

$$\frac{\delta (r^2{V}_r)}{\delta r}=0$$
Integrate with respect to radius:

$$\int \delta (r^2{V}_r)=\int \delta r$$$$r^2{V}_r=C_1$$$$V_r=\frac{C_1}{r^2}$$
We now have radial velocity as a function of radius. Apply initial condition to get value for C1:

$$V_r(0.05m)=1\frac{m}{s}$$$$1=\frac{C_1}{0.05^2}\to C_1=0.0025\frac{m^3}{s}$$
Solve for the radial velocity at 20cm:$$V_r(0.20m)=\frac{0.0025}{0.20^2}\to V_r(0.20m)=0.0625\frac{m}{s}$$

Example Question 2

Why does radial velocity decrease with $r^2$? Show that the reason why is so that the volumetric flow rate is constant, independent of position.

Question 2 Solution

To get the volumetric flow rate $\dot{V}$ from the velocity profile, solve the surface integral:

$$\dot{V}=\int \int (\overrightarrow{v}\cdot \hat{n})dA$$
Start by determining the correct $\hat{n}$ and $dA$ terms. We are looking for radial flow rate, so $\hat{n}$ = $\hat{r}$. Differential Area for spherical coordinates:

$$dA=(r^{2}sin(\varphi )d\theta d\varphi )\hat{r}+(rsin\varphi drd\varphi )\hat{\theta }+(rdrd\theta )\hat{\varphi }$$

The only term in the Differential Area to survive the dot product with $\hat{r}$ is:

$$r^{2}sin(\varphi )d\theta d\varphi$$

We can now combine all of this into the surface integral above:$$\dot{V}=\underset{0}{\overset{\pi }{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{0.0025}{r^2}{r}^{2}sin\varphi d\theta d\varphi$$

The inner integral sweeps fully around $\theta$ from 0 to $2\pi$, while the outer integral only sweeps halfway around $\varphi$. If phi went from 0 to $2\pi$, then the surface integral would solve to zero. Notice that the $r^2$ terms will cancel in the integral, leaving the volumetric flow rate to be independent of radial position.

Integral of $d\theta$ from 0 to $2\pi$:$$\underset{0}{\overset{2\pi }{\int }}d\theta =2\pi$$

Surface Integral becomes:$$\dot{V}=0.0025(2\pi )\underset{0}{\overset{\pi }{\int }}sin(\varphi )d\varphi$$

Solve the remaining integral:$$\underset{0}{\overset{\pi }{\int }}sin(\varphi )d\varphi =-cos(\pi )-(-cos(0))$$$$\underset{0}{\overset{\pi }{\int }}sin(\varphi )d\varphi =0-(-1)$$

Solve for Volumetric Flow Rate:$$\dot{V}=0.0025(2\pi )(1)$$$$\dot{V}=0.0157\frac{m^3}{s}$$

This shows that the Volumetric Flow rate is a constant at steady state. It is independent of radial position.

Example Problem

A viscous fluid such as water is flowing in the radial direction in spherical coordinates. Use the continuity equation to show how the radial component of the velocity varies with $r$ at steady state. What is the reason for this variation? Show that the reason why is so that the volumetric flow rate is constant, independent of position $r$. You can take a velocity at one radius (say 1 m/s at $r$ = 5 cm. What is the velocity at $r$ = 20 cm? Show then that the volumetric flow rate is the same at those two different radii. Here take the fluid height to be constant (i.e 10 cm) and do not consider any edge effects.