Basic Math Check for the Heating of 304 Stainless . . . 3

[AlbertG]

Hello folks.

Ran out of fingers and toes; and just needed a little help with some simple heating work on a small amount of 304 stainless (8000 kg/m³; 500J/kg.K (0-100°C) --

http://www.azom.com/details.asp?ArticleID=965).

Here's what I have.

First, I am working with what amounts to a cylindrical can; 5.25" diameter X 6" tall, with an .035" wall. The heat source is entirely from within; and is capable of uniformly raising the temperature of the body from 25ºC to 1200ºC in 35 seconds (still air).

Question: Is the basic Q = mc(?T) equation appropriate to call in consideration of the material specs/configuration?

If so, I seem to have come up with a value of about 13kW as the heating energy represented here. Correct?


Thanks again.

 

[chicopee]

Is that a seal can?

 

[AlbertG]

@chicopee:

For the sake of this discussion, please consider the cylinder closed on both ends; with, again, all of the heat radiating through the walls from within.

In this way, a "sealed can" it is

Thanks.

 

[dvd]

No, Q = mc(?T) does not work because you haven't used your time rate. Qdot = mc(?T/?t) would give you a result in watts. Your units are in joules with your present equation.

I think you've got a bull somewhere in your calcs to get 13KW to be required to heat something that light to 1200C in 35 seconds. Check your conversions for the can volume. It is only .035 wall, can't weigh more than a few ounces. Write out all of your units and make sure that everything cancels correctly. Don't use the volume of the can, use the volume of the skin of the can.

Another important note is that 304 won't stand up to that kind of temperature for long; if you are heating up and cooling down for very many cycles the material will degrade. You might want to consider 309 or 310.

 

[AlbertG]

@dvd:

Thanks for your post.

I agree with your contention concerning the materials selection. Off the cuff, Hastelloy X would be closer to my choice for anything like this; but alas, nobody asked me...

Here's my math:

Outer volume: 5.25" X 6" -- 129.82 ci
Inner volume: 5.18" X 5.93" -- 124.91 ci
Difference: 4.91 ci; or .0000804 m³

Q = mc(?T)

Total weight (.0000804 m³ * 8000 kg/m³): .6432 kg (m)
Specific heat: 500J/kg.K (c)
Temperature rise: 1175°C (?T)

Q = .6432 * 500 * 1175
Q = 377880 Joules

Watts (J/s) = 377880 / 35 (seconds)
Watts = 10,796

As you can see, my original ballpark math was off just a bit; but this is the methodology which I used anyway...

In consideration of the fact that the specific heat is specified in the literature only for the 0-100°C span; is there any cause for concern with this approach?

Is there anything else which I might be missing here?


Thanks again.

 

[dvd]

I think you are on the right track, but I would concur with you to get some high temperature specific heat data. I don't know if c for stainless steel goes up or down as a function of temperature.

Just curious, what is this for?

 

[AlbertG]

@dvd:

I'd love to be specific, but it's more of a thought problem for my (complicated) end of things right now; and I'd probably get a 5.25" diameter X 6" tall lump on my head from going into the details. I like to think of it as a rather hot little suppository

Anyway, specific heat data seems to be a real bugaboo for this; as I can't find anything beyond 100°C for 304 (huh?).

Any ideas???


Thanks a bunch again.

 

[IRstuff]

At 1200°C, and 0.5 emissivity, your can radiates more than 8 kW, so you should not ignore the radiated loss.

TTFN

FAQ731-376

 

[chicopee]

I agree with IRstuff about radiation losses which was not taken into account in your calculations. This radiative heat loss will increase as the can temperature increases; you may want to take an average value for radiative heat loss and then refine your calculations by with time step calculations as the temperature rises.
By the way be cautious about "sealed" cans, they can explode.

 

[AlbertG]

Thanks folks.

So, if I understand a more complete picture now (neglecting the explosion potential , we need to factor in gray body radiation,

q = ??(T_h⁴ - T_c⁴) A_c

where

q = heat transfer per unit time (W)
? = emissivity of the object (one for a black body)
? = Stefan-Boltzmann Constant: 5.6703 10^-8 (W/m²°K⁴)
T_h = hot body absolute temperature (°K)
T_c = cold surroundings absolute temperature (°K)
A_c = area of the object (m²)

q = .5 * .000000056703 * (1473.15⁴ - 298.15⁴) * .085084
q = .5 * .000000056703 * (4709641832864.5 - 7902040563.8) * .085084
q = .5 * .000000056703 * 4701739792300.7 * .085084
q = 11,341.8

Total power (W) = 11,341.8 (radiated power) + 10,796 (heating power; from previous calculation) = 22,137.8

Did I miss on one of my calculations, or is this value correct?

If so, is this all we need to solve for the total heating energy represented through the cylindrical vessel?


Thanks again for all of the help.

 

[chicopee]

In all likelyhood you are too high as you figured out the radiative heat loss based on a max body temperature when actually the body temperature is increasing to its max value.

 

[AlbertG]

Errata:

Error for A_c; should be corrected to .091729

q = .5 * .000000056703 * 4701739792300.7 * .091729
q = 12,227.6

Total power (W) = 12,227.6 (radiated power) + 10,796 (heating power; from previous calculation) = 23,023.6

 

[AlbertG]

@chicopee:

Our posts passed in transit...

Ironically, I seem to have been getting a larger value with my latest correction.

As to your contention: Doesn't the radiative heat loss remains constant after the body temperature has stopped increasing? Therefore, would the instantaneous collective value not be the sum of that which was demonstrated in the forgoing J/s rampup to peak temperature combined with the steady-state radiative losses?

In other words, at this point in time (full thermal stability; 35 seconds out from initiation) does the body not provide a total of 23,023.6 Watts of heating energy through the cylindrical vessel?


Thanks again for your help.

 

[IRstuff]

I think you are confusing the heat powers. One of your terms is an input power, the other is an output power. At steady state temperature, you should expect that input=output, otherwise, your temperature would be going up, or down.

Based on the steady state loss, though, it's clear that your heating power must be substantially higher than the radiated loss, otherwise, you'll reach steady state temperature much slower than you'd think.

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

Fair enough.

So, we should consider the two results not to be additive; and should also reckon the heating energy through the cylindrical vessel to be well in excess of 12,227.6 Watts (q).

Therefore, if I'm on the same page with everyone here, Q = mc(?T) was indeed not an appropriate choice for the situation at hand.

Is this correct?

 

[IRstuff]

It's the correct starting point. That's what you need to put in to get the Joule heat and temperature that you want.

You then have to take into account the attendant losses.

You would then create a time-phased heating curve to determine the desired heating and the amount of power required to achieve that.

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

OK.

To recap, we need to start with the steady-state gray body radiation equation,

q = ??(T_h⁴ - T_c⁴) A_c

instead of Q = mc(?T).

Then, we should take the value of q and factor in a set of losses to arrive at the steady-state heating energy available through the body. That being correct, what would these factors be for a body such as this in still air at 25°C?

Lastly, in consideration of the stabilized input/output nature of this setting, would the time-phased heating curve of which you speak still be required?

20 questions


Thanks again.

 

[IRstuff]

Again,
Q = mass*specific_heat*delta_T is the place to start.

That's what you need to put in to get the Joule heat and temperature that you want.

You then have to take into account the attendant losses.

You would then create a time-phased heating curve to determine the desired heating and the amount of power required to achieve that.

TTFN

FAQ731-376

 

[AlbertG]

@IRstuff:

OK. So...

We need to start with Q = mc(?T).

Then, we should take the value of Q and factor in a set of losses to arrive at the steady-state heating energy available through the body.

What would these factors be for a body such as this in still air at 25°C?

Lastly, in consideration of the stabilized input/output nature of this setting, would the time-phased heating curve of which you speak still be required?

I truly appreciate all of the help which you have provided: Thank you.

On that note, I have a proposal for anyone who may presently this way come:

Since we seem to have made it 'round and back to the matter of my original posting, is anyone interested in solving for my former query:

Here's my math:

Outer volume: 5.25" X 6" -- 129.82 ci
Inner volume: 5.18" X 5.93" -- 124.91 ci
Difference: 4.91 ci; or .0000804 m³

Q = mc(?T)

Total weight (.0000804 m³ * 8000 kg/m³): .6432 kg (m)
Specific heat: 500J/kg.K (c)
Temperature rise: 1175°C (?T)

Q = .6432 * 500 * 1175
Q = 377880 Joules

Watts (J/s) = 377880 / 35 (seconds)
Watts = 10,796

As you can see, my original ballpark math was off just a bit; but this is the methodology which I used anyway...

In consideration of the fact that the specific heat is specified in the literature only for the 0-100°C span; is there any cause for concern with this approach?

Is there anything else which I might be missing here?

Tangibly bringing all of this together with a solution would certainly settle the matter without further question; as looking at the work which a better hand might put to this example would probably be the best way for me to see just what must be done here...

A star for anyone interested?


Thanks again.

 

[dvd]

AlbertG:

I agree that radiation is a good thing to consider. You stated above that the unit is a sort of suppository. If that is the case then convection or conduction to the medium that this suppository is being stuck into may be more of an issue than radiation.