Find Temp - Radiative Heat Transfer 2

[joedvo]

I am trying to calculate the temperature of the closest side of a structural steel frame (10"x10"x.5", 304ss) to a 304ss vessel wall that is at 216 degF and 10" away.
|Outer vessel T |
\ 216 degF / <--- Rudimentary vessel
\__T1__A1__/
air | 40 degF Ambient
gap |10"
_______________|________________
T2 A2
10"x10"x.5" 304SS HSS
________________________________
Top View

I have used every resource in my possession to try and find a way to calculate T2 on the inner side of the frame. I have already calculated the configuration factor for two parallel plates in one direction (F1-2), used the view factor reciprocity rule to find F2-1. I keep getting a T2 that is equal to T1 which cannot be correct. I cannot figure out what I am doing wrong, can someone please give me some assistance. Thank you.

Relevant Equations:
Qnet=(emissivity)(s-bc)(A1)(F1-2)(((T1+460)^4)-(T2+460)^4))
Qnet1-2=-Qnet2-1
(F1-2)(A1)=(F2-1)(A2)

 

[zekeman]

You ignored convective heat transfer and you left out the view factor and heat flow to the space outside the ship, (1-F2-1)*emmisivity*A2(......-.....)
plus the opposite side of the frame convective + radiative heat transfer.

 

[joedvo]

Could you please elaborate? Are you saying that my Qnet should be:

Qconvection1 + Qrad(F1-2) + Qrad (1-F1-2) = Qrad(1-F2-1)+ Qrad(F2-1) + Qcovection2

And solve for the temperature of the structural steel that way?

 

[hacksaw]

1. you've assumed equal emissivities
2. ignored heat flows in the bottom plate to a thermal reservoir at some temperature
3. for example, include the emissivity of your plate to a non-radiative ambient i.e. free space. with heat flows your see temperature differences. What you have calculated is two objects in a well insulated container and at equilibrium

 

[joedvo]

Hacksaw,
I have assumed equal emissivities, I am not sure if you talking about the structural steel beam when you wrote bottom plate. The beam is a square tube (10x10), so I think what you are saying is that I have ignored any heat flows already existing in the beam which is also true. I am assuming that the beam is at ambient temperature (40 deg). #3 could you put that statement into equation form, I am not sure I understand. What is a non-radiative ambient? Thank you.

 

[hacksaw]

joedvo,

in other words if the tube (or the plate in your initial modeling) radiated to free space on its far side, it will have an equilibium lower than the source of heat (your vessel). You've already written the equation needed but an iterative or graphical calculation will be required.

 

[corus]

Another way of explaining what others have said, is that you have heat flow to a temperature of 216C, but no heat loss to anywhere else. That'll mean your surface temperature will equal 216C too. If you add in an extra term to your radiation to that of the ambient at 40c then you'll have a heat loss. The view factor to the ambient will be one less than the view factor to the underside of the vessel.

ex-corus (semi-detached)

 

[joedvo]

Corus,

That is a bit clearer, thank you. Do you just determine the area of the ambient using the view factor reciprocity rule and then iterate the beam Temp that zeros out the Qnet balance?

 

[joedvo]

Okay, I have used the following formula to find T2. I have tried every temperature between 216 degF and 40 degF and the two sides do not equal out. Still not sure what I am doing wrong here. I have written below exactly what I have been using to try and solve for T2. Please review and let me know if this equation is correct/incorrect. Thank you.

S-Bc = Stefan-Boltzman Constant
Ta = Ambient Temperature
Aa = Ambient Area (calculated using the view factor reciprocity rule, see below)

Qnet=-Qnet

Qnet=(h*A1*(T1-Ta))+(emissivity*(S-Bc)*F1-2*A1*(((T1+460)^4)-(T2+460)^4))+(emissivity*(S-Bc)*(1-(F2-1))*Aa1*(((T1+460)^4)-((Ta+460)^4)))

-Qnet=-(h*A2*(T2-Ta))+(emissivity*(S-Bc)*F2-1*A2*(((T2+460)^4)-(T1+460)^4))+(emissivity*(S-Bc)*(1-(F1-2))*Aa2*(((T2+460)^4)-((Ta+460)^4)))

when

Aa1 = (F1-2/(1-(F2-1)))*A1

Aa2 = (F2-1/(1-(F1-2)))*A2

F1-2 and F2-1 are both calculated and verified to graph

h and emissivity are assumed to be the same for both sides

 

[ione]

joedvo,

At first sight the conversion between °F and K is wrong.

K = (°F + 459,67) / 1.8

 

[joedvo]

The +460 is to convert to degrees Rankine to accomadate the form of the Stefan-Boltzman constant I used. This conversion from degrees Fahrenheit to degrees Rankine is accurate.

 

[hacksaw]

just for discussion, ignoring convection

Qout=emiss*SB*A2*((T2+460)^4)-((Ta+460)^4)

Qin =emiss*SB*F12*A1*((T1+460)^4-(T2+460)^4)

or something to that effect

so that given a T1 and Tamb you should be able to find a non-trivial T2 with Qin=Qout.