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?
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?
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
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.
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.
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 A_{i} and A_{j}. Then consider the elemental areas designated as dA_{i} and dA_{j}, 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 n_{i} and n_{i}, 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 dA_{i} and is intercepted by dA_{j}, 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 dA_{j} when viewed from dA_{i}.
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
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)
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
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.
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