- #1

- 2

- 0

Thanks!

- Thread starter Jade Sola
- Start date

- #1

- 2

- 0

Thanks!

- #2

- 27

- 4

Um.. I think you need to be a bit more specific. What form are you working with?

- #3

- 2

- 0

cylindrical coordintates....so this is what the velocity field isUm.. I think you need to be a bit more specific. What form are you working with?

what I am confused about is what exactly is the equation for viscous dissipation ... My professor said in class that it is 2muS:S (from Navierstokes eq.3) but I am seeing differen things online and I also heard from another classmate of mine that it has something to do with Sxtau components. Im just confused ....sorry if Im confusing the matter here.

- #4

Chestermiller

Mentor

- 20,773

- 4,502

If S is the rate of deformation tensor (which it certainly appears to be), then:cylindrical coordintates....so this is what the velocity field is

View attachment 75868

what I am confused about is what exactly is the equation for viscous dissipation ... My professor said in class that it is 2muS:S (from Navierstokes eq.3) but I am seeing differen things online and I also heard from another classmate of mine that it has something to do with Sxtau components. Im just confused ....sorry if Im confusing the matter here.

S = (∇V+∇V

and

τ = 2μS

So, the rate of viscous dissipation is τ:∇V = τ:S = 2μS:S. All the relations you wrote are the same, and should give the same results.

Chet

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 9K

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 10

- Views
- 7K

Are the viscous hydrodynamic forces and the nonlinear viscous hydrodynamic forces applied on offshor

- Last Post

- Replies
- 4

- Views
- 804

- Last Post

- Replies
- 1

- Views
- 805

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 5

- Views
- 7K

- Last Post

- Replies
- 4

- Views
- 2K