BTUH Per Ft² Formula for Stove Surface Area Vs. Temperature

How many BTUs does a surface radiate that is 72 degrees, or how about 98.6?

This will be interesting.

I think the average skin temp. of the human body is about 91 degrees. 21 degrees above room temp.

21 + 459.67 = 480.67 degrees Rankine

The human body has a surface area of about 18 sq-ft.

BTUH = FT²*°Ra⁴ /583,235,639

BTUH = 18 x 480.67⁴ / 583,235,639

BTUH = 1,647

That would be 483 Watt Hours. But based on calorie consumption I believe 100 Watt Hours is closer to the truth here. Hmmm ????

I would think that most of those BTUs are reflected back onto our skin by clothes.

OK, when I calculate for 18 sq-ft at 70 degrees straight up I get:

BTUH = 134.95/sq-ft x 18 sq-ft = 2,429

And when I calculate for 18 sq-ft at 91 degrees straight up I get:

BTUH = 157.66/sq-ft x 18 sq-ft = 2,838

2,838 BTUH - 2,429 BTUH = 409 BTUH

409 BTUH / 3.412 BTUH/Watt-Hour = 120 watt hours for the human body (much closer)

The BTUH's calculated by this formula are likely therefore for objects (bodies) radiating energy into the coldest depths of outer space. Thus the need to reference them to absolute zero via their Rankine temperature. Hmmm ?????

It's either that or the average nude human body radiates about 4.8 times more energy away than does the average fully clothed human body. Hmmm again. ????

Well ain't that sexy. Must be why we're hotter without any clothes on.

Anyways, lol ... two objects at the same temperature next to each other gain and loose BTU's to each other at the same rate, I mean unless it's nude bodies hahaha. That would be more about chemistry instead of thermodynamics. Lol

The formula given on the reference page you've cited is a significant simplification of the equations you need to use to solve what you are trying to do.

dj

dlj wrote:The formula given on the reference page you've cited is a significant simplification of the equations you need to use to solve what you are trying to do.

dj

I gave thought to that potentially being the case. But the output appears to closely match that which I've seen going back years ago on this forum whereby others have sited data for surface temperature vs. BTU's for stoves. The difference being that wherein they listed the radiating BTU's at each temperature, they did not to my knowledge ever site a formula by which to derive the BTU's. Until something better comes along I'm willing to work with this formula and its output.

An excellent book on this subject is "Transport Phenomena" by Bird.

http://www.amazon.com/Transport-Phenomena-Revised-2nd-Edition/dp/0470115394

A book aimed more at practical applications of Bird's work could be "Transport Phenomena in Metallurgy" by Geiger and Poirier.

http://www.amazon.com/Transport-Phenomena-Metallurgy-G-Geiger/dp/0201023520

These are not light reads... But cover this subject in detail.

dj

The formula that I converted for use with square feet and Fahrenheit degrees is derived from what is known as the "Stefan-Boltzmann Law". It is highly regarded as accurate. My presentation of the formula seems to at least fairly well match the chart that I have linked here.
(chart as attached was sourced from this website): http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
The results as can be seen (and as I had suspected above) are referenced against absolute zero.

Lightning wrote:ain't that sexy. Must be why we're hotter without any clothes on.

Well, some's Hotter than others...

Scroll down on this website until you hit the 'Radiation Calculation' section.
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3

To match my rendition of the Stefan-Boltzmann formulas output you must input the following:

1) Enter the numeral 1 in the Emissivity box.
2) On the left and below the image of the sun enter 588.7 degrees kelvin in the box just left of the letter 'K' (NOTE: 588.7K = 600 degrees F.)
3) On the right enter 20 in the square feet box.

Then you must click directly onto the letter P as seen in the ugly pink P=Q/T formula below these two and you will get an output of 43,187 BTU's.

Then go to my first post in this thread and see where I derived 43,240 BTU's

Close enough for government work as they say. Any difference will likely vanish if the conversion of 600 degrees F. to 588.7 degrees K is carried out to a few more significant decimal places.

I am quite familiar with the Stefan-Boltzmann equation. And yes, it is considered an excellent equation when applied to the region of thermal transport that it works best in. In your difficulty above, where you have taken the equation and applied it first to a stove at temperatures around 600*F and then applied it to the human body at some lower temperature you have not taken into consideration a couple major aspects: first the Stefan-Boltzmann equation is best applied when talking about heat transfer by radiation, and second, you have to know the emissivity of the different materials that you compared. As you can see from the text below, there are a number of assumptions using this equation, especially when using as you did to the two very disparate examples.

You should also be aware, that the use of strictly radiation heat transfer calculation (e.g. use of the Stefan-Boltzmann equation) is limited to high temperature applications. In the realm of our stoves, you would have to use equations that utilize both radiation and convection heat transfer equations to achieve accuracy. But it's all fun to play with isn't it?

dj

I'll agree fully that the Stefan-Boltzmann equation as I have presented it is for the case of a perfect "black body" radiator with an ideal (or perfect) emmisivity of 1. The real world clearly does not function with such perfection, and stoves are not by any means perfect radiators. But sticking with the case for perfection makes it relatively easy to use, as well as fun to play with as you stated. Plus I was bored today.

The complexities involved are why I have always preferred to equate BTUH (both as input and output) to coal consumption. There is only so much energy in the coal to begin with, and you simply can't get out of it and deliver into the home all that it has in it. That fact alone establishes the upper limit for BTUH quite nicely.

I give you "Brainiacks" a lot of credit.........My head hurts!

This website gives a similar formula for convective heat transfer. I believe it may be as much as the radiant heat transfer. I ran some basic numbers on both on a couple of my stoves.

http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html