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Join Date: Jun 2008
Posts: 2

Radiation heat loss

06/19/2008 8:55 AM

Hi,

I have a hot pipe at 500degF surrounded by a metal screen for protection. I want to find how hot the screen will become. I have found that the radiation heat loss is the main factor and calculated the heat load. If I consider 50% view favtor and then equate it to heat loss due to air cooling effect, is that the right approach? Also for this overall heat transfer coeff varies from 2-50 Btu/hrft2F. How can I calculate the tmperature of the metal screen?

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#1

Re: Radiation heat loss

06/19/2008 9:21 AM

It's not possible to calculate it. If it already exists, why not measure it?

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Associate

Join Date: Mar 2008
Location: Southern California
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#2

Re: Radiation heat loss

06/19/2008 11:06 PM

If this is an OSHA issue, then it may be better to just insulate the pipe. I would think that the amount of heat loss from the bare pipe itself is not energy efficient (not to mention unsafe).

If you wanted to provide a theoretical calculation you would have to:

- calculate how hot the ambient air around the pipe would get

- calculate how much heat energy the air would transfer to the metal screen

Though, I agree with the previous post. Take an actual temperature reading, and call it good (or bad if this is a burn issue) and insulate the pipe instead.

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Active Contributor

Join Date: Sep 2007
Location: Florence, Tuscany, Italy
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#3

Re: Radiation heat loss

06/20/2008 4:08 AM

Without entering into detail, if you want to estimate the temperature of the screen you must calculate the heat imbalance of the item, which shall be zero in staedy state conditions. By reasonably assuming that heat flows from hot pipe to colder environment, this means that the heat entering the screen, coming from hot pipe, will equal the heat leaving the same screen towards external environment.

I understood that the main heat vector from pipe to screen is radiation, whilst, on the external side, the screen exchanges heat with air by convection. Are you sure that convection is negligible inside the cavity between pipe and screen? This is actually true when vacuum is provided in the cavity, but in presence of air I suppose that a large part of the heat transfer will occur by convection, relying upon common experience.

The calculation of the theoretical radiative exchange is not so hard, as the geometry of the system is simple (I suppose two concentric cylindrical surfaces). On literature you may find many a relationship for the calculation of the radiant heat exchange, provided that you may know the thermo-optical properties (namely, the infra-red emissivity) of the facing surfaces.

What I find more difficult is the calculation of the convective heat transfer. It is true that you can find so many empirical correlations in literature or on the web, but these relationships are generally obtained with specific configurations of the systems. Besides, the calculation requires in general the knowledge of physical parameters (air density, dynamic viscosity, etc.), which can only be estimated, on the basis of well-established databases, which can however provide statistical figures (besids depending on temperature). In conclusion, the calculated parameters are affected with a certain error. This, on turn, will involve an uncertainty margin to be considered in your evaluations.

I agree with other guys that, if you can measure the temperature, this is the fastest solution. Unless you don't want to find a theoretical basis to the thermal performance, and in this case you could exploit the physical measurements of the temperature (pipe, screen, still air) to correlate your own thermal model.

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Anonymous Poster
#4

Re: Radiation heat loss

06/20/2008 6:22 AM

gtjon windstream com

found radio shack $29. IR temp reader

as well as $125. field piece

some surfaces require satin-white or black

or

small piece of electrical or masking tape

where acceptable to leave ( or hi-temp tape)

to pick up for instance, truer differential temps.

convection readings seem hard also

probably due to some laminar-fluid/air dynamics

of measurable , but hard to separate,'layers'

(i believe)

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Anonymous Poster
#5

Re: Radiation heat loss

06/20/2008 6:39 AM

note: 2

gtjon windstream net (not com)

how hot ?

[YMK] You May Know the following:

Temperature- 'hot' does not really tell the

heat load, unless referenced to material knowns;

if we do not know the work/hr for instance

we would need a measurement of differential(s)

of 'cooling' air and screen material and, and ,

? can the source of heat be measured

entering versus leaving at the pipe end ?

Source of dropping in known capacity = BtuH lost

if that is the 'HEAT' equated to 'LOAD'

-as questioned-

typical of "how loaded is the chiller" q's heard...

later showing surface temperatures, alone, of objects

don't depict heat as a work load

until heat capacity and/or contents are known

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Power-User

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#6

Re: Radiation heat loss

06/20/2008 9:37 AM

Calculate?? Why not measure? Put a thermostat on the screen and read the temperature.

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Guru

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#7
In reply to #6

Re: Radiation heat loss

06/20/2008 11:46 AM

A thermostat is normally used to control the temperature. A thermometer will more accurately measure the temperature.

Just a quibble from a troglodyte!

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Power-User

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#8
In reply to #7

Re: Radiation heat loss

06/20/2008 3:52 PM

Damn!! You got me. I am surprised. Usually I am the one being picky like that. You got me.

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