# Fluid dynamics - dissipation using cylindrical coordinates

1. Apr 23, 2005

### da_willem

I have this flow field in cylindrical coordinates of wich I would like to calculate the dissipation as a function of these coordinates. Now in my fluid dynamics notes I found the following expression(s) for the dissipation:

$$2 \mu (e_{ij} -\frac{1}{3} \Delta \delta _{ij} )^2 = 2 \mu ( e_{ij}^2 - \frac{1}{3} \Delta ^2 )= \frac{d_{ij}}{2 \mu}$$

with $\mu$ the dynamic viscosity, e the rate-of-deformation tensor, [itex]\Delta[/tex] the divergence of the velocity and d the deviatoric stresses. I assume these squares express two sums required by the Einstein summation convention?

I also found some expressions for the deviatoric stresses in cylindrical coordinates, terms like $$d_{z \phi}, d_{zr}, d_{zz}$$ etc. I don't really know how to interpret these and how to proceed. Can I use the last expression and instead of summing over x,y and z sum over the three cylindrical coordinates? Or does this yield something different?

2. Apr 24, 2005

### arildno

Hi, dawillem:
If you use this formalism, you can formulate the expressions independent of the type of coordinate system you're using.

For example, I assume you're familiar with the expression:
$$\dot{e}_{ij}=\frac{1}{2}(\nabla\vec{v}+(\nabla\vec{v})^{T})$$
where T is the transpose.
For Cartesian coordinates, we have:
$$\dot{e}_{ij}=\frac{1}{2}(\frac{\partial{u}_{i}}{\partial{x}_{j}}+\frac{\partial{u}_{j}}{\partial{x}_{i}})$$

Let's calculate the matrix $$\nabla\vec{v}$$ in CYLINDRICAL coordinates:
1) We have: $$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{i}_{z}\frac{\partial}{\partial{z}}$$
Along with the relations:
$$\frac{\partial\vec{i}_{r}}{\partial{z}}=\frac{\partial\vec{i}_{\theta}}{\partial{z}}=\frac{\partial\vec{i}_{z}}{\partial{z}}=\vec{0}$$
$$\frac{\partial\vec{i}_{r}}{\partial{r}}=\frac{\partial\vec{i}_{\theta}}{\partial{r}}=\frac{\partial\vec{i}_{z}}{\partial{r}}=\vec{0}$$
$$\frac{\partial\vec{i}_{r}}{\partial{\theta}}=\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial{\theta}}=-\vec{i}_{r},\frac{\partial\vec{i}_{z}}{\partial{\theta}}=\vec{0}$$
2) Let $$\vec{v}=v_{r}\vec{i}_{r}+v_{\theta}\vec{i}_{\theta}+v_{z}\vec{i}_{z}$$
3)We therefore should have:
$$\nabla\vec{v}=\vec{i}_{r}\frac{\partial\vec{v}}{\partial{r}}+\vec{i}_{\theta}\frac{\partial\vec{v}}{r\partial{\theta}}+\vec{i}_{z}\frac{\partial\vec{v}}{\partial{z}}$$
And, for example:
$$\frac{\partial\vec{v}}{\partial{\theta}}=\frac{\partial{v}_{r}\vec{i}_{r}}{\partial\theta}+\frac{\partial{v}_{\theta}\vec{i}_{\theta}}{\partial\theta}+\frac{\partial{v}_{z}\vec{i}_{z}}{\partial\theta}$$
That is:
$$\frac{\partial\vec{v}}{\partial{\theta}}=\frac{\partial{v}_{r}}{\partial\theta}\vec{i}_{r}+v_{r}\vec{i}_{\theta}+\frac{\partial{v}_{\theta}}{\partial\theta}\vec{i}_{\theta}-v_{\theta}\vec{i}_{r}+\frac{\partial{v}_{z}}{\partial\theta}\vec{i}_{z}}$$
Thus, we can write:
$$\vec{i}_{\theta}\frac{\partial\vec{v}}{r\partial{\theta}}=\frac{\partial\vec{v}}{\partial{\theta}}=\frac{\partial{v}_{r}}{r\partial\theta}\vec{i}_{\theta}\vec{i}_{r}+\frac{v_{r}}{r}\vec{i}_{\theta}\vec{i}_{\theta}+\frac{\partial{v}_{\theta}}{r\partial\theta}\vec{i}_{\theta}\vec{i}_{\theta}-\frac{v_{\theta}}{r}\vec{i}_{\theta}\vec{i}_{r}+\frac{\partial{v}_{z}}{r\partial\theta}\vec{i}_{\theta}\vec{i}_{z}$$
A quantity of type $$\vec{i}_{\theta}\vec{i}_{\theta}$$ is called a DYAD, and we can regard it as a matrix with value 1 at row&column position $$\vec{i}_{\theta}$$ and 0 elsewhere.
(The two other "diagonal" dyads are, of course $$\vec{i}_{r}\vec{i}_{r},\vec{i}_{z}\vec{i}_{z}$$)
I'll leave the rest to you..

Last edited: Apr 24, 2005
3. Apr 24, 2005

### da_willem

I've never heard of dyads before, but most of the things you wrote I'm more or less familiar with. We haven't treated the mathematics of tensors and dyads thoroughly though. In my notes the rate of deformation tensor is written:

$$\frac{1}{2} ( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})$$

But I can see that this comes down to the expression you used. The above expression is only correct in Cartesian coordinates right, and the expression you wrote is right independent of coordinates? [If this what is called the tensor coordinate free form then how come the above expression with the indices and all reminds me more of tensors?!]

I think I know how to proceed with the rest of the terms now, but I think i'ts gonna take a while...so I'd like to find a shortcut, as I already know the expressions for the deviatoric stresses in cylindrical coordinates. The expression for the dissipation:

$$\frac{d^2_{ij}}{2 \mu}$$

if I'm not mistaken comes down to summing all the squared elements of d. Is this true as well for d in cylindrical coordinates, and why (not)?

4. Apr 24, 2005

### arildno

I'll check up in my books for the exact expressions, and get back to that.

However, you can probably design a tensor notation with a nice set of indices to get it working.
Note that, for general, 3-D curvilinear coordinates, the del-operator has the form:
$$\nabla=\vec{i}_{1}\frac{\partial}{h_{1}\partial{x}_{1}}+\vec{i}_{2}\frac{\partial}{h_{2}\partial{x}_{2}}+\vec{i}_{3}\frac{\partial}{h_{3}\partial{x}_{3}}$$
where the h's are appropriate scale factors (possibly functions of the variables), and the vectors form an orthonormal set.

5. Apr 24, 2005

### da_willem

You mean an expression for the dissipation not dependent on the particular coordinates? The expressions in my first post were the only ones I could find and they all assume Cartesian coordinates right?

I guess you could translate the part in between the square in coordinate free form as:

$$2 \mu (e_{ij} -\frac{1}{3} \Delta \delta _{ij} )^2 \rightarrow 2 \mu (\frac{1}{2}(\nabla\vec{v}+(\nabla\vec {v})^{T}) - \frac{1}{3} \nabla \cdot \vec{v} I)$$

But then I dont know what to do with the square in the expression... So I guess it boils down to find out how to translate the square (it must yield a scalar as the expression denotes the dissipation) to coordinate free form. Or find a coordinate free expression in a book somewhere.

As I interpreted it, it means a double summation:

$$d_{ij}=\Sigma^3 _{i=1} \Sigma^3 _{j} d_{ij} d_{ij} = \Sigma^3 _{i=1} \Sigma^3 _{j} d_{ij} ^2$$

This probably has got something to do with tensor invariants, and my feeling (or my laziness) says I can just do the same with the components $$d_{z \phi}, d_{zr}, d_{zz}$$ etc. But I'm not too sure.

6. Apr 25, 2005

### da_willem

I found an expression in cylindrical coordinates.... @ http://www.clarkson.edu/subramanian/ch527/supplem/dissipation.pdf [Broken]

It looks a lot like the sum of the squared elements $$d_{z \phi}, d_{zr}, d_{zz}$$ etc, but I'll have to look at it more thoroughly.

Last edited by a moderator: May 2, 2017
7. Apr 25, 2005

### dextercioby

We physicists don't worry too much about bases on tangent & cotangent spaces of a manifold,but more on how the components of (pseudo)tensor quantities behave.I'll use the column-semicolumn notation.

U wan to put this tensor (actually the components,but physicists always name "components of a tensor"="tensor")

$$e_{ij}=\frac{1}{2}v_{(i,j)}=\frac{1}{2}\left(v_{i,j}+v_{j,i}\right)$$

in covariant form.U'll need the covariant derivative

$$\tilde{e}_{ij}=\frac{1}{2}v_{(i;j)}=\frac{1}{2}\left(v_{i;j}+v_{j;i}\right)$$

where

$$v_{i;j}=v_{i,j}+\Gamma^{k} \ _{ij}v_{k}$$

All u have to do is find the Christoffel symbols.It's not difficult to find the metric tensor & its inverse,since the cylindrical system of coordinates is orthonormal...

Daniel.

8. Apr 25, 2005

### da_willem

Actually I think I know how to find, in principle, the rate-of-deformation tensor e using cylindrical coordiantes using arildno's post. But I want to know the dissipation: how is this related to e in coordinate free form?

And what exactly does 'covariant form' mean?

9. Apr 25, 2005

### dextercioby

What is the relation in euclidean/cartesian orthonormal coordinates...?

Invariant under general coordinates tranformation...

Daniel.

10. Apr 25, 2005

### da_willem

$$2 \mu (e_{ij} -\frac{1}{3} \Delta \delta _{ij} )^2 = 2 \mu ( e_{ij}^2 - \frac{1}{3} \Delta ^2 )= \frac{d^2_{ij}}{2 \mu}$$

What does invariance under general coordinates tranformation mean for the expression you wrote down. Surely the components of e change under a coordinate transormation as the Christoffel symbols change. So this invariance means for e...?!

11. Apr 25, 2005

### dextercioby

It's not tensor correct.The LHS of the last equality is a scalar,while the RHS is a II-nd rank (pseudo)tensor.

$$\hat{e}^{2}=e_{ij}e^{ij}$$

$$\Delta^{2}=\left(v^{i}\ _{,i}\right)^{2}$$

Daniel.

Last edited: Apr 25, 2005