Is Turbulence at Point A Isotropic? Calculation Help

[nightingale]

Both the viscous stress and the turbulent stress you calculated are on a surface of constant r in the z direction, but also on a surface of constant z in the r direction.

Your questions are indicating that your fundamental understanding of turbulence and turbulent stresses needs beefing up. You need to understand the fundamentals before you start trying to apply it to problems. I'm going to recommend another book that I hold in high regard, and hope you will consider using it: Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 5. Physics Forums is just not structured to present a complete primer on Turbulent Flow.

Chet

Thank you Sir.

I have borrowed the book you recommend and read chapter 5 and chapter 1 which talk about kronecker delta. I really helps me to understand turbulence better.
I'm still intrigued on why Pope (2000) simply said V_T as 'eddy viscosity' when it is, I think, 'kinematic eddy viscosity' but I decided I'll ask my teacher next time I met her.

I now know that the formula I should be using is:

-ρu'v' = μ_T * dU/dy

Where the μ_T here is the eddy viscosity.

Earlier the turbulent shear stress is:
τ = -ρu'v'
τ = - 1.2 * ((0.0625)(0.3125)+(0.4625)(0.1125)+(0.5624)(-0.2875)+...)/8
τ = 0.11343

and the velocity gradient is 0.4219

Thus the eddy viscosity is:
0.11343 = μ * 0.4219
μ = 0.27
Did I do right? Thank you very much.

 

[Chestermiller]

Thank you Sir.

I have borrowed the book you recommend and read chapter 5 and chapter 1 which talk about kronecker delta. I really helps me to understand turbulence better.
I'm still intrigued on why Pope (2000) simply said V_T as 'eddy viscosity' when it is, I think, 'kinematic eddy viscosity' but I decided I'll ask my teacher next time I met her.

I now know that the formula I should be using is:

-ρu'v' = μ_T * dU/dy

Where the μ_T here is the eddy viscosity.

Earlier the turbulent shear stress is:
τ = -ρu'v'
τ = - 1.2 * ((0.0625)(0.3125)+(0.4625)(0.1125)+(0.5624)(-0.2875)+...)/8
τ = 0.11343

and the velocity gradient is 0.4219

Thus the eddy viscosity is:
0.11343 = μ * 0.4219
μ = 0.27
Did I do right? Thank you very much.

Yes. But please make sure the you show the units.

 

[nightingale]

Yes. But please make sure the you show the units.

Thank you, Sir. I take the unit of the eddy viscosity as Pa.s.

I attempt to do the last question, which asked me to determine the production of the turbulence energy at B.
I found that the corresponding formula is:

Production = -2 u'v'*(dU/dy)
and thus the production is:
Production = -2 *((0.0625)(0.3125)+(0.4625)(0.1125)+(0.5624)(-0.2875)+...)/8 * (0.4219)
Production = -2 * -0.094525 * 0.4219
Production = 0.0798

Did I do it correctly?
Thank you very much for the guidance.

 

[Chestermiller]

I can't help you here. I'm not familiar with the equation you wrote for the rate of production of turbulent energy. But, it doesn't seem to have the right units.

Chet

 

[nightingale]

I can't help you here. I'm not familiar with the equation you wrote for the rate of production of turbulent energy. But, it doesn't seem to have the right units.

Chet

I understand, I will continue to read about the topic. Thank you very much for the patience and the guidance all this time Chet! You are a wonderful teacher!