in cfx 4, it is found the interphase raidation calculation is updated on 2005.07.04, by dividted by a factor of 4. which is believed to make it consistent with cfx5 help? but why? can anyone know the solid reference of this change, because this is a big change. Anyone know about this? Pls give the title of the book/journal/conf and page. Thanks.

Dear Ben,

This change was not only to make it consistent with ANSYS CFX-5.7 (at the time), but also consistent with CFX-TASCflow . It was a bug in the CFX-4.x source code.

The main concept is that the radiative flux to the particle is proportional to the area of its cross section, and not the surface area. Here is a simplified version of the issue (after a few solid angle integrals)

Q_rad_p = Emissivity_p * A_p * (Incident Radiation - Emission_p)

where

A_p = (pi * D_p**2 / 4)

Incident Radiation = 4 * pi * Radiation Intensity

Emission_p = 4 * n**2 * sigma * T_p ** 4

Therefore,

Q_rad_p = Emissivity_p * (pi * D_p**2) * ( pi * Radiation Intensity - n**2 * sigma * T_p**4)

Notice that the factor of 4 is gone (written using D_p), or it will show up if using r_p = D_p / 2.

Also, you can check the basics by setting a single particle at temperature T_p within a box with wall temperature of 0 [K], or closer. Check that the irradiation to the walls matches the emission of the particle; otherwise, where did the energy go?.

Hope this helps,

Opaque

Dear Opaque,

Thanks a lot for your help.

Actually, I was told it is projective surface, rather than full surface before. But I just don't know why, because there is no apparent wrong if using full surface area. I have to argue with my collegue on this issue using a scientific evidence. I need a solid reference on this area determination issue, such as a book chapter or journal paper and its papge number. I cannot find one like that with a quarter. Could you find one for me please.

Thanks a lot in advance.

BEN

Hi Ben,

It is the simple definition of radiation intensity:

I = radiative energy flow per :

- time

- area normal to rays (projective as you called above)

- solid angle

- wavelength (wavenumber or frequency)

Therefore, the total amount of energy arriving is the integral over the "projected" area, over the sphere (solid angle), over the spectrum.

Any radiation transfer book will explain that in the introduction of radiation terminology.

Radiative Heat Transfer, M. Modest

Thermal Radiation Heat Transfer, R. Siegel and J.R. Howell,

Thermal Radiative Transfer and Properties, M. Brewster

Radiative Transfer, S. Chandrasekhar

Hope this helps,

Opaque.

Dear Opaque,

Really appreciate your info.

Could you do me one more favor? I just cannot find these books in my uni library. Could you please find one reference showing prjective surface issue for interphase radiation, with details of book title and page number. Thanks a lot for your help.

Regards,

BEN

My collegue just found a book chapter, saying that A is surface area, not projective area. It is exactly opposite to your point. <Heat transfer, J.P. Holman, 5th edition, McGRAW-HILL>

what do u think about this?

Dear opaque,

back to your first reply: "Notice that the factor of 4 is gone (written using D_p), or it will show up if using r_p = D_p / 2. "

HOWEVER, in CFX5 help, 1/4 shows when using Dp, as shown below,

Qr=0.25*e*pi*d**2(I-sigma*Tp**4) CFX5

looks inconsistent...?

Dear Ben,

That is a typo.. Please contact ANSYS CFX help desk, and they should be able to confirm that.

Regards, Opaque.

Dear Ben,

A is the surface area in which context? for the definition of radiation intensity, or radiative heat flux on the particle?

If it is for radiation intensity, would you mind naming the book/author? I would like to read the context that A is used in before commenting any further on that.

Just try to define the concept of radiation intensity when traveling in straight lines. Start with the energy flowing and define a concept that is indifferent to the view angle towards the surface, and how far you are from the surface. You will end up using some form of a projected surface area and a solid angle.

Opaque.

Dear opaque,

my collegue and me are arguing about which surface area should be used, full surface or projected surface. He just named a journal paper, saying that it is full surface for intensity.

R.Kurose et al, FUEL, 83, p696

cheers,

Ben

Dear Ben,

Here you can find some amunition to support your position:

http://ctr.stanford.edu/Summer06/watanabe.pdf

Authored by Kurose himself. Please check equation 2.15 where the radiation on the particle is computed as

A_d * epsilon_d * (pi * I - sigma * T_d **4 )

and clearly states: "A_d is the projected area of the droplet"

As a last resort, you and your colleague should get a basic radiative heat transfer book and work out the radiation interaction with dispersed particle. Formulate the integrals based on first principles (conservation of energy + definition of radiation intensity) and when you come across the term "cos theta sin theta dtheta dphi dA" you will realize that the projected area is involved.

Leave it with you, enjoy

Opaque