Automotive Exhaust Pipe Radiation

[uglyfatjoe]

Heat transfer was not one of my strongest classes and I've been out of school long enough that my text books didn't make it through the 15 moves. Anyway, I seem to be lost on the fundamentals as I search the web for information but here is what I am trying to do:

I have a cylinder (automotive exhaust pipe). I want to know what the temperature of a body would be at various distances from the cylinder and various steady state temperatures of the pipe from radiation only.

I find the simple basic equation q=??AT^4. Applying it for the above problem has be at a loss. So I can solve for q knowing the emissivity of the body, the area of the cylinder, and the surface temperature of the cylinder. Beyond that what formula do I need to plug q unto to find out the temperature of the body?

 

[Zapster]

You need to include convection to get to an estimated temperature that is meaningful. Look at projected surfaces, both sides of objects, convection, conduction, and radiation. Write equations for all of the surfaces. Then iterate temperatures such that there is an energy balance.

 

[kchan711]

Try the wikipedia article.

http://en.wikipedia.org/wiki/Thermal_radiation

Also, see the extra links at the end. It's a good refresher on heat transfer:radiation.

Like Zapster said, you'll need to include conduction and convection to achieve an appropriate answer. With that said, setting up a simple resistor-network problem solution (the nodes are your points of interest and the resistors are your convective heat laws through different mediums) will make the problem much simpler.

I would give that a try.

Kyle Chandler

http://www.chiefengineering.net

 

[uglyfatjoe]

Thanks.

I found an online text through some links elsewhere on this site. The more I read I realize I'll have to consider everything as suggested as feedback here.

I'm sure I'll have many more questions once I re-educate myself. I was looking for an easy way out but I guess that's not going to happen.

 

[chicopee]

You have not defined the size of the body ,whether it is insulated, whether it is infinetly long, etc. for which you are trying to calculate its temperature. Then would it be its surface temparature or bulk temperature you are trying to calculate. Is this body also exposed to cooling medium on the heaty trasfer size? One thing you should always do is a sketch of the problem with all variables identified and then set up initial conditions to simplify the problem. The problem may be a little more complex than you think.

 

[uglyfatjoe]

I'm still trudging through the theory and trying to get up to speed. However, the attached picture defines (well at least I hope it does) the problem I am trying to solve.

I guess I could simplify it by neglecting gravity and assuming the pipe is a black body. Anyway I'll be back with many more questions.

 

[kchan711]

If you're working with theory, I would start within a 2D plane with two point bodies. You want to start with the simplest model possible to understand the concepts. I know this sounds horrible, but a good heat transfer book is always a perfect reference. The steps are outlined well for solving problems, and it becomes quite handy when addressing such problems on a regular basis.

I would check online for a discounted book. Just something to think about...

Kyle Chandler

http://www.chiefengineering.net

 

[chicopee]

The equation q=??AT^4 is for the entire circumference of the pipe. The actual amount of radiation hitting the entire width(W) of the plate would be reduced by a factor of 2*arctan(W/2D) where D is the distance between plate and pipe center.
As far as determining the temperature distribution along an elemental strip parallel to the width edge this is what I would do:
1) Assume plate is thin ,therefore no temperature gradient;air between plate and pipe does not absorb radiation heat; pipe has uniform temperature distribution around its circumference and along its length; grey bodies;and steady state heat transfer.
2) Equally divide an elemental strip so as to get rectangular elemental areas(and don't make them too small initially). The elemental areas starting from the projected center lines of the pipe unto the plate(hopefully,the pipe is centered) will be labelled as 1,2,3....
3) Do a nodal analysis at each elemental area. Each elemental area will have a different s.s.temperature such as T1,T2,T3 etc..
4) The nodal analysis will consist of conduction heat transfer between nodes, convection heat along the faces of the elemental areas to ambient, and radiation heat input from the pipe. At the free edge of the elemental strip you'll convection heat loss to ambient.
Don't forget do these analysis for only half of the nodes if the pipe is centered over the plate.
5)Radiation heat input on each elemental area will not be the same and gets to be less as from the center to the outer edge of the plate. So first take the total heat input as calculated in the first paragraph and divide that value by the total number of nodes in that elemental strip. The actual radiation absorbed by each elemental area is reduced by the cosine of the angle between the radiation ray and the surface of the elemntal area. You have to figure out the cosine values for the elemental areas based on the geometry of pipe and plate.
5) Once you have developed s.s.heat balance at each node develop residual equation for each node and some textbooks may refer to these residual equations as "Equation for Q', you will need to do the relaxation operations to determine the temperature at each node and I am sure that your textbook has some info on that subject.

 

[IRstuff]

Again, you cannot ignore either convective, nor secondary radiation, nor even conduction, from the target material. Additionally, the material's thickness and thermal conductivity will also affect the eventual answer.

At distances on the order of <13 mm separation, the air behaves like a thermal conductor and the effective heat transfer coefficient is quite high.

This plate is presumably attached to something else in the car, so there may be conducted losses in the supports. If the material is thermally conductive, then there'll be heat movement away from the closest proximity to the pipe as well. Since the target is a graybody, any substantial increase in its temperature results in emission into anything in its field of view that's cooler.

All and all, a fairly complicated scenario. I think it almost screams for an FEA solution.

TTFN

FAQ731-376

 

[uglyfatjoe]

I agree that for a complete solution all paths/sources of radiation, convection, and conduction need to be considered and it does scream for an FEA solution.

However, what I am trying to do is generate a 'simple' equation based on radiation heat transfer which can be used as a rule of thumb for a starting point.

This way I can say hmmm well I have a 2" pipe at 450C and if I travel away from it by X mm (based on the rule of thumb) anything outside this zone would not require shielding if I am looking to keep the surface at 80C or less (assuming blackbody thermal radiation, 1 dimension, steady state, etc...) This could help establish the base shield neccessary so when the FEA is ran there is already a reasonable baseline in place.

What I can't get my hands around, for some reason - possibly because the beer I have consumed over the years has killed many brain cells, is the complete equation to factor out q to solve for X.

I guess where I started going down the path of 'simple' equation came from the wikipedia page on the Stephan-Boltzmann law. About 2/3rds of the way down the page the effective temperature of the earth is caluclated using the approximation Te=Ts*(Rs/2a)^1/2 where Rs is the radius of the sun and a is the distance between the sun and the earth. So I though wow sweet...a=(Rp/2)*(Tp/T)^2 where Rp is the radius of the pipe, Tp is the temperature of the pipe, and a is the distance at which T=80C. But I have no clue how the equation was derived and I assume that since it's a sphere radiating on a sphere and the distance is 1x10^12 larger than I am considering that maybe this equation is not applicable.

 

[IRstuff]

You did apparently lose too many gray cells ;-)

The example you cite is derived from the heat balance at the Earth, i.e., it radiates all its received solar energy as well as any internal energy that makes it to the surface, otherwise, the Earth would be worse that Venus.

The captured energy at about 1300 W/m^2 at the Earth's orbit is radiated into ~4pi steradians of absolute zero, neglecting the sun. That's why I told you that you couldn't ignore secondary radiation, since that's what balances the energy flow.

TTFN

FAQ731-376

 

[uglyfatjoe]

I would like to thank everyone who has given me their input thus far. I'm going to have you losing your willingness to help through my stubborn ignorance however:

Instead of the cylinder/plate example what if I consider two concentric cylinders of infinite length with open ends with the outer surface if the inner cylinder being the hot surface at 450C and the inner surface of the outer cylinder being the surface I want at 80C. If the inner pipe is a black body it will radiate ?AT^4 energy or it's energy flux (q/A) will be ?T^4. Is it correct to assume that as I move away from the inner pipe the amount of energy remains constant but the energy flux will become smaller because the surface area becomes larger? So thus at some radius the energy flux would equate to a lower temperature if a cylindrical surface located at that radius was also a blackbody and absorbed all of that energy?

Thefore q1=q2 and A1?T1^4=A2?T2^4. Since the surface area of a capless cylinder is 2PIrL I get r2=r1T1^4/T2^4. For an example of a pipe surface at 450C(723.15K) a pipe radius of 25.4 mm and a desired radius at which the resultant surface temperature would be 80C(353.15K) I calculated r2 = 446.6 mm or a distance between the surface of the pipe of 446.6 - 25.4 = 421.2 mm.

This number seems too large for a rule of thumb but if the theory above is correct then I should be able to factor in the emissivity of the pipe (steel or stainless steel), the shape factor between the inner cylinder and the outer cylinder, and the emissivity of the receiving medium (black HDPE or white HDPE) and reduce that number to something more realistic...No?

 

[IRstuff]

NO.

Again, this ONLY has meaning in the context of an external heat loss from the outer cylinder.

Just consider if the outer cylinder had ZERO heat loss. What happens with a constant heat flux entering, but nothing leaving? In a hypothetical situation, the outer cylinder will head to infinite temperature. In real life, the radiation back to the inner cylinder will equalize, resulting in ZERO temperature difference between the two cylinders.

This is no different than taking a charged capacitor and connecting it to another, uncharged capacitor through a resistor. You can make it 10 ohms, ot 10 megohms, but eventually, with no other losses, the two capacitors must be at the same voltage.

TTFN

FAQ731-376