Find the view factor of a tillted cylinder-shaped poolfire?

  • Thread starter isabella
  • Start date
Does anybody out there know how to find the view factor of a tillted cylinder-shaped poolfire?
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
What text book are you using?

Other terms are configuration factor (or view factor), which is the product of the emissivity factor and geometry or shape factor.

The theory is explained at - http://www.me.utexas.edu/~howell/intro.html.
 
Last edited:
I've checked the website and the formula is similar to the formula in the book that I'm reading(Fire protecting Engineering). But I still don't understand. Could you explain?
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
There are different configuration/view factors here -

http://www.me.utexas.edu/~howell/tablecon.html#C2

You need to check which geometry is applicable.


The radiated energy received is a function of the surface area receiving the radiation and the surface area emitting the radiation.

The configuration or view factor is a product of the emissivity factor and arrangement or geometry factor. The geometry factor is dependent on the distance between the two surfaces of interest and the orientation.

I also wonder - what is the shape and orientation of the absorbing area. Also, when considering a cylindrical fire, does one consider the planar surface of the fire (like a disk) and the sides of the cylinder?
 
Last edited:
the shape of the absorbing area is a very small rectangle. the emmiting area is a sheared tilted cylinder.but it retains its circular shape.say the absorbing area is a person, the view factor needed does not include the top and bottom part of the cylinder.only part of the side of the cylinder that can be viewed by the person is taken into account
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
Basically the problem involves a double surface integral. Depending on geometry - distance, shape and orienation, the integrals can be difficult, and few will be solved analytically for most cases. Some special cases can be solved analytically.

Start at page 61 of this document - Heat Transfer Manual - www.me.memphis.edu/Faculty/janna/Heat_Tr_Manual.pdf[/URL]

Then go back to page 54 and read through page 61 for some additional background.
 
Last edited by a moderator:
I have the max and min angles that determine the area that the person can view. I also have the max length of the cylinder and the area of the cylinder. How can I use the formula using all the variable that i have?

Do you know how to find the area of the visible part using Numerical integration?
 
I have the max and min angles that determine the area that the person can view. I also have the max length of the cylinder and the area of the cylinder. How can I use the formula using all the variable that i have?

Do you know how to find the area of the visible part using Numerical integration?
 
ok so phi1 and phi 2 give the angles of the two lines from the target to the cylinder which are tangent to the cylinder.they are the angles of the edges of the visible part of the cylinder.
phi1=3.3386
phi2=6.1454
zmin=0
zmax=19.2
diameter of cylinder,D=10m
visible area=integration integration(D/(2cos(theta))SQRT(cos^2(theta)-(cos^2(theta)cos^2(phi))+cos^2(phi))dphi dz

the target is a person standing at at a distance of 46.78m away from the surface of the cylinder.
 
view factor diagram

this is the diagram of the cylinder and the smaller surface
 

Attachments

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
The view (or configuration) factor F[usb]ij[/sub] is defined as the fraction of radiation leaving surface i, which is then intercepted by surface j. Consider two arbirarily oriented surfaces Ai and Aj. Then consider the elemental areas designated as dAi and dAj, which are connected by a line of length R, which forms polar angles [itex]\theta_i[/itex] and [itex]\theta_j[/tex] with respect to the surface normals ni and ni, respectively. Note the values of R, [itex]\theta_i[/itex] and [itex]\theta_j[/tex] vary with position and orientation (relative to both areas) of the two areas.

Let [itex]\large dq_{i\rightarrow j}[/itex] be the rate at which radiation leaves dAi and is intercepted by dAj, then

[itex]\large dq_{i\rightarrow j} = I_i cos \theta_i dA_i d\omega_{j-i}[/itex]

where [itex]I_i[/itex] is the radiation intensity leaving surface i and [itex]d\omega_{j-i}[/itex] is the solid angle subtended by dAj when viewed from dAi.

By definition of the solid angle,

[itex]\large d\omega_{j-i}[/itex] = [itex] ( cos \theta_j dA_j) / R^2 [/itex], then it follows that

[itex]\large dq_{i\rightarrow j} = I_i cos \frac{\theta_i \theta_j}{R^2} dA_i dA_j[/itex] and letting [itex]J_i = \pi I_i[/tex], this becomes

[itex]\large dq_{i\rightarrow j} = I_i cos \frac{\theta_i \theta_j}{\pi R^2} dA_i dA_j[/itex].

Then integrating over surfaces Ai and Aj, the total rate at which radiation leaves surface i and is intercepted by j is given by,

[itex]\large q_{i\rightarrow j}\,=\,J_i \int_{A_i}\int_{A_j} \frac{\theta_i \theta_j}{R^2} dA_i dA_j[/itex]

and from the definition of the view factor,

Fij = [itex]\large \frac{q_{i\rightarrow j}}{A_i\,J_i}[/itex]

or

Fij = [itex]\large \frac{1}{A_i} \int_{A_i}\int_{A_j} \frac{\theta_i \theta_j}{R^2} dA_i dA_j[/itex]
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
isabella said:
this is the diagram of the cylinder and the smaller surface
Is the problem looking for what a person sees - or does the problem ask for the total heat radiated to the person? There is a big difference.

The problem is complicated by the tilting. One then has to include the angle of tilt in order to generate a general (non-specific) solution.

Also if any part of the person 'sees' the top, then that is another view factor to consider.

Let me see if I can find some discussion of tilted cylinders.
 
The problem is about how much radiation is received by the person. It is assumed that the amount of heat recieved by the person only comes from the area that is visible to that person. It is also assumed that the person is not able to see the top part of the cylinder. I have the area visible to that person. the equation is

Area=integration(lim phi1 to phi2)integration(lim zmin to zmax)|V1xV2|dzd(phi)

|V1xV2|=(D/2cos(theta))*(SQRT(cos^2(theta)-cos^2(theta)*cos^2(phi)+cos^2)(phi))

phi1 and phi2 are the angles that determines the visible area(phi1 is the smaller angle, phi2 is the larger angle)
zmin=0
zmax=max. length of the cylinder.
theta=tilt angle
D=diameter of the cylinder
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
OK, I think I see what you are doing.

To simplify the problem, one can approximate the person as a rectangle of some height, h, and some width, w.

Then one has to determine the solid angle that the person (h x w) represents to each dA of the cylinder.

In addition to the viewing angles described above, one had the geometrical angle of the slant of the cylinder.

Basically because this is a double area integral, one will have 4 integration limits.

I would like to get arildno to look at this problem. He has very good insights into these problems.
 

Astronuc

Staff Emeritus
Science Advisor
18,542
1,659
Two references which might help.

Guelzim, A., Souil, J.M., and Vantelon, J.P., 1993, "Suitable configuration factors for radiation calculation concerning tilted flames," J. Heat Transfer, vol. 115, no. 2, pp. 489-492, May.

Factors are given in closed form between differential elements in various configurations to tilted cylinders with faces parallel to the base plane.


Rein, R.G., Jr., Sliepcevich, C.M., and Welker, J.R., 1970, "Radiation view factors for tilted cylinders," J. Fire Flammability, vol. 1, pp. 140-153.

Factors given from a vertical differential element normal to line through the base of a tilted circular cylinder for various cylinder length/radius ratios, and for various distances of the element from the cylinder. Discusses use of configuration factor algebra for cases when the element is above or below the cylinder base plane, and effects of cases when element lies under tilted cylinder or far from the cylinder.

This figure should go with the tex equations above.
 

Attachments

Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top