thermal conductivity question/confusion

[Plasmech]

I have a machine that blows hot air (350F) into a tube that is 7 inches ID by 18.5 inches long. The tube is currently made of 1020 CRS, 18 gauge (.048" thick). The idea is that I want minimal heat loss in this tube, ideally having 350F air at the exit end. Here are thermal conductivity ratings of three different materials:

1020 CRS: 51.9 W/m*k

304SS: 16.2 W/m*k

glass: 1.1 W/m*k

So one would think that if I made the tube out of 304SS I would get a higher air outlet temperature (less heat loss), and if I made it out of glass the outlet temp would be higher still. However, this is not the case, in fact there was almost no measurable difference when I tried it. What am I missing in my thinking? Is there another property I need to be looking at? Is the tube so thin that it doesn't matter what it is made of? Thanks for any help thermal guys and gals!

 

[prex]

For such a short tube the heat losses through the wall are largely negligible with respect to the heat content of the air flow. This means that you'll have the outlet temperature substantially equal to the inlet one irrespective of pipe material.
As you mention temperature measurements, did you measure the inlet temperature together with the oulet one? And how much is the flow rate?

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[Plasmech]

I would think so too, that with such a short tube the outlet temp should almost equal the inlet temp. However, using an RTD (very accurate) at the inlet and the outlet shows a 30 degree (!) heat loss. I am pumping 50 SCFM of hot air through this tube.

It's not an instrument error, I have used several diffferent RTD's and thermocouples and they all read the same.

 

[dvd]

You probably can't accurately measure the transient response of the heated air while the different tubes are heating up. The conductivity of the tubes is more important in this phase. The dominant effects after the tubes have warmed up will be convection of the ambient air on the outside of the tube and radiation. You need to look at natural convection on a (horizontal?) tube and radiative heat transfer equations to get a better understanding.

 

[IRstuff]

Your flow , while it sounds large, results in something like 5 ft/s when corrected for temperature. With such a large ID, there may be a considerable amount of mixing with respect to the external air, so you might need to sample several points in the flow to determine if that's the case. You should also look at different locations radially from the center of the tube. Is your measurement right against the wall of the tube? You need to accurately characterize the temperature distribution in the air stream. Then, you can determine the rate of heat loss, which will give you some clues about where the heat loss is occuring.

You should have two heat loss mechanisms, convection and radiation, assuming your tube is conductively isolated, otherwise, you have conduction as well. The largest, because of temperature, is radiation.

You didn't describe the external environment. Is it at room temp? How much air flow is there across the tube, either from air conditioning or just convection from the tube?

A simple solution might be to place a highlt reflective tube with an ID 1 inch larger than the OD of your tube. This will minimize radiated loss as well as convection.



TTFN

FAQ731-376

 

[Plasmech]

Unsure what you mean by mixing with external air? This is a completely sealed tube.

Temperature around the tube is ambient (say 70F). No airflow across the tube, it's stagnant inside the cabinet of a machine. The temperature measurements are actually done in 2" OD tubes at the inlet and outlet of the big tube (it necks down on either end with flat plates, no taper/transition). Any heat that's being lost is through the walls of the tube, I'm sure of that. My question is really about thermal conductivity, and whether or not I understand it properly. Why would a glass tube with such a low thermal conductivity now show and difference from a carbon steel tube?

 

[IRstuff]

Again, you need to do the numbers for your configuration. A BOTE calculation puts the radiated component as much as 30 times higher than convection. Look at your glass, it can support the estimated radiated emission with only a 4ºC temperature drop. The CRS would have less than 0.2ºC drop.

If my numbers are correct, you're losing about 1 kW, mostly from radiation. Convection might only account for about 200 W, tops.

However, I would disagree that your cabinet temperature is only 70ºF if there's no air flow, since you're clearly dumping a fairly substantial heat load into the cabinet. It would need to have forced air cooling fans to remove the heat from the cabinet, therefore, it would have to have air flow. Something with external walls at nearly 350ºF is NOT going to have 70ºF air around it.

TTFN

FAQ731-376

 

[Plasmech]

IR, first of all, thanks a lot for your help here, appreciate it!

You're right, the inside cabinet temp is higher than 70F while the unit is running, like you said due to the heat loss that we're talking about in the first place. It might be closer to 120F inside that cabinet.

So you're saying that what's really killing me is radiation? Is there a readily available reflective foil I can put around that thing to keep the radiation down to a mimimun?

 

[davefitz]

when comparing different tube materials, the ultimate comparison is the " resistance to heat transfer", or theinverse of teh overall heat transfer coeficient, Roa= 1/ Uoa.

The sum of the resistances of inside convective film, inside fouling, tube metal conductivity, outside fouling, and outside convectivew film are calculated in a conventional manner. The effect of tube metal is so small, you can neglect it in this case, just as it is nearly negligible in tubular condensers.

 

[Plasmech]

I'm a bit confused as to why the word "convection" keeps coming up. I figured that most of the energy was being transfered via _conduction_ through the metal wall? Or Are we talking about the fact that once the energy gets through the wall, it must be convected away via the outside air?

 

[IRstuff]

The foil would be similar to what you'd find surrounding a catalytic converter.

Oops, had a slight glitch, I was trying to do two different things. The radiated component is only 1/2 of convected component, so they're both contributing to your heat loss.

You need to accept that you might not practically have zero heat loss. How much can you tolerate? How much are you willing to spend to achieve whatever you can achieve?

A simple, 1-in thick foam insulation with aluminum reflective foil might reduce the total heat loss by a factor of 5.

TTFN

FAQ731-376

 

[prex]

Plasmech, you are correct with the last sentence of your last post: convection+radiation from the outer tube wall to the surroundings is the limiting heat transport phenomenon in your case, and the change of tube material will anyway have a negligible impact.
As also recalled above, the heat loss through the tube wall is the result of various thermal resistances in series, and, as for electrical resistances, the largest one dominates. With your conditions (thin wall, sustained flow at inside) the controlling resistance is on the outside.
To evaluate the losses through the tube wall you can account for a heat transfer coefficient of 10 W/sq.m/degC on the outer surface: this includes convection and radiation losses, each contributing for some 50% to the total.
To crunch some numbers, 50 SCFM of air means some 0.03 kg/s of air, or some 30 W per degC of heat content. In other words a loss of 30 W would cause a temperature drop of 1 degC.
The heat loss through the tube wall can be estimated, with the coefficient above, to some 400 W, say 500 W to include some losses by conduction to the tube supports.
As a consequence we would expect a temperature drop of 17 degC or 30 degF (!).
By suitably insulating the tube wall (and a portion of the connected pipes and equipment), you can definitely reduce that loss by a large amount, say a factor of 10.

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[Plasmech]

So it looks like, going back to my original post, that I was right in the sense that the wall is so thin that it simply doesn't matter what the thermal properties of the tube itself are. If I add a decent insulation with a reflective foil, the heat loss should be reduced dramatically.

 

[dvd]

The point of using foil is that it won't radiate the heat as well. If you use a good pipe insulation the outside temperature of the insulation won't be much warmer than ambient and thus there won't be much radiation anyway. A more durable pipe insulation covering material may be a better choice.

 

[Plasmech]

We tried 1/4" of fiberglass with a silicone cloth on top...that barely makes a 5 degree F difference. Probably need something closer to 1 inch with a foil covering.

 

[IRstuff]

Speaking of which, how is this tube supported?

TTFN

FAQ731-376

 

[Plasmech]

(4) 3/8" diameter trunions, two of which bear the weight, the other two just to loacte it. I don't imagine a whole ton of heat going through the supports. The contact area of all 4 is theoretically infinitely small, a round on a flat.

 

[IRstuff]

It might be worthwhile measuring the temperature along those supports, just for thoroughness.

Are they aluminum or steel? If aluminum, they could be contributing to the heat loss.

TTFN

FAQ731-376

 

[Serrand]

If you look at the resistivity, which is going to be the important factor in this case, you will see that the differences in the thermal conductivities don't make a large difference. ln(rout/rin)/(2*pi*L*k)
Cheers

 

[zerosum]

Plasmech,

I think davefitz and Serrand are on the right track. very often in gas-to-gas heat transfer, almost all the resistance comes from the boundry layers (inside convective film and outside convective film) compared to conductance of the tube