Heat Transfer - Radiation Question

[macmet]

I hope this isn't too academic for this forum, but I have been trying to find the skin temperature of a wall today and am stumped. I am using Steam - Ed. 41 by Babcock and Wilcox as reference, specifically example 2 on pages 4-29/4-30.

Can anyone tell me where equation 79a comes from? Where did the 4 come from? I cannot find this forumla anywhere else and I no longer have my heat transfer text from school.

For those without access to the book the formula is,

(rad heat trans coeff) = 4.0 * boltzmann's constant * emissivity * F * ((T1+T2)/2)^3

The example states taht both emissivity and F can be assumed to be 1.0

 

[hacksaw]

it uses the Boltzman equation to define the radiant heat transfer,based on average T, then divides by the mean delta T, more or less

that is easy enough, but that is also where the really tough part of the calculation starts, as the simple estimate ignores reflected heat, the actual emissivity, radiation of the media, etc., etc....

the theory then sums the conduction, convection, and radiation terms to estimate the total heat transfer coefficient...under the assumption that there is no change in phase...

 

[IRstuff]

F is the "view factor" which is an estimate of the "visibility" of the emitter to the surroundings and vice-versa.

I think the "4" may be a typo and should be "A" to represent the area of the emitter.

I don't get the bit about the temperatures. The factored Boltzman equation should be (T1+T2)(T1^2+T2^2), leaving you the (T1-T2) term to multiply. Given today's calculators, there's really no reason not to use the T1^4-T2^4 expression directly, .i.e.:

area*sigma*emissivity*view*(T1^4-T2^4)

TTFN

FAQ731-376

 

[corus]

.... where T1 and T2 are in degrees Kelvin, or whatever units you're using.

corus

 

[macmet]

IRstuff,

The 4 may be a typo, but the area for the question is 600ft2, so that alone would be a large change to the final result. And the question says that the gases and the surface of the inner wall is small. I take that as saying I can assume they're the same so that whole term would cancel out.

When analyzing our steym I went through it blindly (eventhough I don't understand them and that bothers me) and I get a reasonable answer. But I suppose that could just be coincidence really. Does anyone know if B&W will answer questions about the book?

 

[Powerman81]

I think the equation is calculating the radiation heat transfer coefficent "h_r" between the radiating gas and the surface, NOT the heat transfer rate "q_r".

where, h_r=q/A(T1-T2), and q_r= A* boltzmann's constant * emissivity * F * T_s^4;

Since, you are looking at the average temp, substitute T1-T2 with skin temp T_s which is equal to average temp between gas and surface (Tg+Ts)/2. Now work the math and solve for h, you will get

h_r= (A*BC*emm*F*T_s^4)/(A*T_s) = BC*emm*F*T_s^3

The units work out well, but there is no 4, huh.

My other guess is that the 4 comes from differntiating the heat rate q with respect to Temp to obtain the radiation heat transfer coefficent "h_r", hence,

dq/dT = A*BC*emm*F*4*T_s^3.

Moe

 

[IRstuff]

The coefficient, as I understand it, is to take
sigma*emiss*f*(T1^4-T2^4)

and transform it into:

htc*(T1-T2), where htc = sigma*emiss*f*(T1^4-T2^4)/(T1-T2)

The objective, in the end, is to have an equation for heat transfer looking like the convective equation:

Q = area*htc*(T1-T2)

But, since you already have all the numbers, and you have sufficient processing horsepower, there's no reason to use an approximation formula, when you can use the real thing.

TTFN

FAQ731-376

 

[macmet]

IRStuff,

Do you have access to the question I referred to?

 

[IRstuff]

Sorry, no.

TTFN

FAQ731-376

 

[SomptingGuy]

Factors of 4 are normally associated with spheres, since the number appears in the expressions for volume and surface area. But 4 normally brings pi with it somewhere.

- Steve

 

[macmet]

Well, this is a flat surface. I am going to try to get this question into a pic format so I can post it.

I went through some of the older work that we outsourced to consultants years ago by setting a up spreadsheet and using the previous setup. With the B&W method the skin temp is about 50 deg F cooler. But, if I change the 4 in the formula to a 1 (or remove it entirely) I get the same skin temp. The other junction temperatures are also very close to what the consultants got in the past.

The problem with that, is I have no way of knowing if that's just a coincident. I just don't feel comfortable using formula that make no sense to me, nor do I like just taking a formula and changing it b/c I don't understand where a constant came from.

 

[macmet]

Here is a file...

 

[vpl]

macmet

The last file worked, but now I'm confused (no comments about that being my normal state)

The example is clearly labeled as "Heat flow in a composite wall with convection[\b]" and the discussion states that it demonstrates the procedure for combining thermal resistance of [multiple materials.] It also states that the formula 79a is an approximation[\u] that can be used when the temperature difference between the radiating gas and the surface temperature is small. Finally, at the end of the example it states that, if the temperature differential is large, than the full equation for the radiation resistance from Table 4 may be required.

Since it's an approximation, you probably need to go back to the material immediately preceding the "Applications" and read through it to see if there is an explanation of how the approximation was derived.

Besides not having any idea where the "4" comes from, I am totally stymied by the temperature being raised to the third power, because everything I read in my heat transfer books says the radiation heat transfer coefficient is to the fourth power.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[macmet]

This example is very much similar to the what I am trying to figure out. I think the assumption between the radiating gas and surface temperature is valid in my application.

I'm with you about the temperature being raised to the third power. I was really hoping someone might have looked at this example before and figured it out or actually contacted B&W for explanation.

 

[MechEng2005]

Perhaps a derivative somewhere?

The derivative of T^4 would be 4*T^3...

This may explain the factor of 4 and the power of 3. Haven't done heat transfer for awhile, and don't have time to review now, but I'd check into that...

-- MechEng2005

 

[IRstuff]

Sure, for small changes, as alluded to in the text:

dQ = F' * dT ~ 4t^3 * delta_T

By the way, while a bit long in the tooth, B&W's 35th ed, c.1913 is available for download from GoogleBooks:

http://books.google.com/books?id=nq...babcock+&+wilcox"+"steam:+its+generation"&lr=

TTFN

FAQ731-376

 

[prex]

Something is not convincing in that book page.
The first approximation implied in the formula
T_g⁴-T_s⁴[≅]4(T_g-T_s)((T_g+T_s)/2)³
is quite correct: with the figures of the book, T_g=1540R and T_s=540R the approximated formula introduces an error of the order of 20%, quite acceptable for a heat transfer calculation.
However when it comes to the second approximation,
[≈]4(T_g-T_s)T_g³
things are different, the error is higher than 100% with the same figures: the two temperatures should be much closer for that approximation to work.
In conclusion the formula quoted by macmet is correct for relatively large temperature differences, though is still unclear why one shouldn't use the exact formula.

prex

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[IRstuff]

Don't forget that the basis for all of these calculations is the assumption of:
>> doing things by hand, with maybe a 4-banger
>> using a htc*delta_T paradigm

Neither is relevant to modern calculations, calculation tools, and modern users.

TTFN

FAQ731-376