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Considering Radiation from a Stream to a Pipe Wall
- Thread starter rvivac
- Start date
racookpe1978
Nuclear
"Duct" means you have gas flowing at your 515 C, not a liquid in a pipe, right?
The radiation will depend on a few things:
> The temperature difference between Ti and T2. If they are sufficiently close, then T2 will radiate back to Ti, and the net radiated heat is nil
> The emissivity of the gas, which is not close to 1, and is directly affected by the density of the gas in the duct
> The optical thickness of the gas, which for a small duct, is negligible, and radiative effects would be ignored.
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faq731-376
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> The temperature difference between Ti and T2. If they are sufficiently close, then T2 will radiate back to Ti, and the net radiated heat is nil
> The emissivity of the gas, which is not close to 1, and is directly affected by the density of the gas in the duct
> The optical thickness of the gas, which for a small duct, is negligible, and radiative effects would be ignored.
TTFN
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Need help writing a question or understanding a reply? forum1529
- Thread starter
- #4
racookpe1978, yes! The duct carries hot air at 515°C (788,15K).
IRstuff, the radiation heat exchange occurs from hot ~800K air to surface inside the duct. I think it is just a matter of the gas itself: hot air and its emissivity. I know that, for CO2 it is around 0.15 @ ~800K & 1 atm. Using this value may be very conservative, since CO2 is less than 1% of the air or just ignore, like you said. Thanks.
IRstuff, the radiation heat exchange occurs from hot ~800K air to surface inside the duct. I think it is just a matter of the gas itself: hot air and its emissivity. I know that, for CO2 it is around 0.15 @ ~800K & 1 atm. Using this value may be very conservative, since CO2 is less than 1% of the air or just ignore, like you said. Thanks.
- Thread starter
- #6
Emissivity for a gas, even carbon dioxide, is a tricky thing. CO2 does not behave like a blackbody, so it only emits where it absorbs. In the 3-5 um regime, that primarily occurs between ~4.18 um to 4.38 um. However, for a 1-m path length, the integrated absorption still results in >97% transmission over the 3-5 um band, meaning that it, and all the other gases are emitting at less than 3%
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racookpe1978
Nuclear
Now, be very careful about where you are assuming each heat transfer mechanism occurs.
1) Inside the duct, Hot Air @ 515 C to duct wall
2) Outside the duct, Hot duct wall (through duct wall insulation) to room air
At the gas temperature you use, radiation will always be a substantial part of the heat transfer problem, but it will always only be a part of the heat losses. Make sense?
10 Inside the duct, gas is flowing. That gas is (partially) transparent to radiation.
Therefore, every molecule of that hot gas will be trying to radiate its energy to something else - andpart of that emitted energy si going to hit the wall and be lost. And 9almost all of the rest of it) will hit other gas molecules and do nothing (won't be "lost" to the gas stream). At the same time, the inside of the wall will be trying to radiate its energy - the old (Thot^4-Tcolder^4) factor, but much less radiant energy will be lost from the wall than will be gained. Still, include it.
Forced Convection losses from the hot gas to the cold wall will be very large, perhaps 50% of the total energy lost from the hot gasses to the duct wall.
All of the energy lost from the gasses to the wall will be transmitted through the wall (little temperature change probably) to the insulation to the insulation shroud to the room at large. Those outside losses will be also both radiant and natural convection losses to the room. Area of the shroud (wrapping) arounfd the insulation will make the outside area much larger than the inside area.
But, because the outside shroud wall is much, much cooler than the inside gas 515 temperature, the outside radiation losses are much smaller compared to the outside natural convection losses to ambient. Probably 15% of the total losses outside will be radiation.
1) Inside the duct, Hot Air @ 515 C to duct wall
2) Outside the duct, Hot duct wall (through duct wall insulation) to room air
At the gas temperature you use, radiation will always be a substantial part of the heat transfer problem, but it will always only be a part of the heat losses. Make sense?
10 Inside the duct, gas is flowing. That gas is (partially) transparent to radiation.
Therefore, every molecule of that hot gas will be trying to radiate its energy to something else - andpart of that emitted energy si going to hit the wall and be lost. And 9almost all of the rest of it) will hit other gas molecules and do nothing (won't be "lost" to the gas stream). At the same time, the inside of the wall will be trying to radiate its energy - the old (Thot^4-Tcolder^4) factor, but much less radiant energy will be lost from the wall than will be gained. Still, include it.
Forced Convection losses from the hot gas to the cold wall will be very large, perhaps 50% of the total energy lost from the hot gasses to the duct wall.
All of the energy lost from the gasses to the wall will be transmitted through the wall (little temperature change probably) to the insulation to the insulation shroud to the room at large. Those outside losses will be also both radiant and natural convection losses to the room. Area of the shroud (wrapping) arounfd the insulation will make the outside area much larger than the inside area.
But, because the outside shroud wall is much, much cooler than the inside gas 515 temperature, the outside radiation losses are much smaller compared to the outside natural convection losses to ambient. Probably 15% of the total losses outside will be radiation.
- Thread starter
- #9
Thanks racookpe1978.
You described the phenomenon just perfectly. Thanks.
We are just finishing the construction of a Reformer Furnace for a Hydrogen Generating Unit. The designer / supplier used ASTM A36 on an expansion joint on a breeching duct that is subjected to a 515°C stream, but the item 12.5.4 of the API 560 says that on temperatures above 425°C, stainless steel (SS) or alloy steels (AS) shall be used, instead of carbon steel due to graphitization and loss of ductility that occurs on this type of material at temperatures above 425°C.
However, since we have no insulation: it is an expansion joint, we found:
INSIDE:
ε=0,15 -> radiation: 30%.q" - I believe this value may be conservative since it is for CO2 @ 800K, 1 atm.
h=27 W/m2.K -> forced convection: 70%.q"[W/m2]
OUSIDE:
ε=0,9 -> radiation: 80%.q"
h=5 W/m2.K - natural convection: 20%.q"[W/m2]
The most important here is the metal wall temperature that we found to be adequate for the material and way below 515°C, way below the 425°C limit.
You described the phenomenon just perfectly. Thanks.
We are just finishing the construction of a Reformer Furnace for a Hydrogen Generating Unit. The designer / supplier used ASTM A36 on an expansion joint on a breeching duct that is subjected to a 515°C stream, but the item 12.5.4 of the API 560 says that on temperatures above 425°C, stainless steel (SS) or alloy steels (AS) shall be used, instead of carbon steel due to graphitization and loss of ductility that occurs on this type of material at temperatures above 425°C.
However, since we have no insulation: it is an expansion joint, we found:
INSIDE:
ε=0,15 -> radiation: 30%.q" - I believe this value may be conservative since it is for CO2 @ 800K, 1 atm.
h=27 W/m2.K -> forced convection: 70%.q"[W/m2]
OUSIDE:
ε=0,9 -> radiation: 80%.q"
h=5 W/m2.K - natural convection: 20%.q"[W/m2]
The most important here is the metal wall temperature that we found to be adequate for the material and way below 515°C, way below the 425°C limit.
racookpe1978
Nuclear
Now, be careful of your overall efficiency and total product cost.
1) Remember, if your gas stream needs to be 515 dec C, then you spent a LOT of money heating it up to 515 deg C and you don't want to cool it off and waste your money. Is it worth the saved money between a A36 bellows and a stainless bellows to lose all of that expensive hot gas you need? (Or is the hot gas truly a "waste product" that you can't run into a heat exchanger somehow? )
2) If your gas stream is 515 deg C, but because there is "no insulation" on this particular expansion joint, the joint metal itself is less than 425 deg C and is therefore "safe" because the metal temperature is less than 425 deg C, then "plant safety" is only maintained IF you can keep the metal temperature that low ALL of the time. So now you have to go the other way, and verify that on the hottest day of the year with the greatest humidty and highest (dirtiest) reasonable surface and least wind flow at all, the metal surface will still be less than 425 deg C.
3) In part, I'd question this assumption anyway of 425 degrees steel temperature, because that inside gas temp of 515 is going to be hitting the bellows' wall, so even if the outside is 420 deg C, the inside will be hotter.
Make sense? Even a "partially insulated expansion joint" now becomes like a plugged relief valve: It will be getting too hot and violate your material spec about maximum temperature.
1) Remember, if your gas stream needs to be 515 dec C, then you spent a LOT of money heating it up to 515 deg C and you don't want to cool it off and waste your money. Is it worth the saved money between a A36 bellows and a stainless bellows to lose all of that expensive hot gas you need? (Or is the hot gas truly a "waste product" that you can't run into a heat exchanger somehow? )
2) If your gas stream is 515 deg C, but because there is "no insulation" on this particular expansion joint, the joint metal itself is less than 425 deg C and is therefore "safe" because the metal temperature is less than 425 deg C, then "plant safety" is only maintained IF you can keep the metal temperature that low ALL of the time. So now you have to go the other way, and verify that on the hottest day of the year with the greatest humidty and highest (dirtiest) reasonable surface and least wind flow at all, the metal surface will still be less than 425 deg C.
3) In part, I'd question this assumption anyway of 425 degrees steel temperature, because that inside gas temp of 515 is going to be hitting the bellows' wall, so even if the outside is 420 deg C, the inside will be hotter.
Make sense? Even a "partially insulated expansion joint" now becomes like a plugged relief valve: It will be getting too hot and violate your material spec about maximum temperature.
- Thread starter
- #11
racookpe1978. Thanks for your help.
1) Remember, if your gas stream needs to be 515 dec C, then you spent a LOT of money heating it up to 515 deg C and you don't want to cool it off and waste your money. Is it worth the saved money between a A36 bellows and a stainless bellows to lose all of that expensive hot gas you need? (Or is the hot gas truly a "waste product" that you can't run into a heat exchanger somehow? )
I agree with you about the energy lost, but I think the designer thought that it is not an important issue since the feed gas is heated by the downstream combustion gas in many sets of coils. We have plenty of energy on this case and it would be much more expensive to try to use this heat for other process outside the unit plant.
2) If your gas stream is 515 deg C, but because there is "no insulation" on this particular expansion joint, the joint metal itself is less than 425 deg C and is therefore "safe" because the metal temperature is less than 425 deg C, then "plant safety" is only maintained IF you can keep the metal temperature that low ALL of the time. So now you have to go the other way, and verify that on the hottest day of the year with the greatest humidty and highest (dirtiest) reasonable surface and least wind flow at all, the metal surface will still be less than 425 deg C.
I agree with you. I really do not know how much cost the designer saved on this, using A36 instead of some alloy steel. But if you boil the world at 100°C, the metal temperature is going to be ok. I made that calculation.
3) In part, I'd question this assumption anyway of 425 degrees steel temperature, because that inside gas temp of 515 is going to be hitting the bellows' wall, so even if the outside is 420 deg C, the inside will be hotter.
That was our question at first. That's why we needed to calculate the heat flux through the system. That is all a schedule matter and if we stop the plant just because of that, it's going to be nuts.
The outside ambient temperature is expected to be bellow 41°C on that height. But if goes to 60°C, the maths says the inside metal temperature will not achieve 350°C.
1) Remember, if your gas stream needs to be 515 dec C, then you spent a LOT of money heating it up to 515 deg C and you don't want to cool it off and waste your money. Is it worth the saved money between a A36 bellows and a stainless bellows to lose all of that expensive hot gas you need? (Or is the hot gas truly a "waste product" that you can't run into a heat exchanger somehow? )
I agree with you about the energy lost, but I think the designer thought that it is not an important issue since the feed gas is heated by the downstream combustion gas in many sets of coils. We have plenty of energy on this case and it would be much more expensive to try to use this heat for other process outside the unit plant.
2) If your gas stream is 515 deg C, but because there is "no insulation" on this particular expansion joint, the joint metal itself is less than 425 deg C and is therefore "safe" because the metal temperature is less than 425 deg C, then "plant safety" is only maintained IF you can keep the metal temperature that low ALL of the time. So now you have to go the other way, and verify that on the hottest day of the year with the greatest humidty and highest (dirtiest) reasonable surface and least wind flow at all, the metal surface will still be less than 425 deg C.
I agree with you. I really do not know how much cost the designer saved on this, using A36 instead of some alloy steel. But if you boil the world at 100°C, the metal temperature is going to be ok. I made that calculation.
3) In part, I'd question this assumption anyway of 425 degrees steel temperature, because that inside gas temp of 515 is going to be hitting the bellows' wall, so even if the outside is 420 deg C, the inside will be hotter.
That was our question at first. That's why we needed to calculate the heat flux through the system. That is all a schedule matter and if we stop the plant just because of that, it's going to be nuts.
The outside ambient temperature is expected to be bellow 41°C on that height. But if goes to 60°C, the maths says the inside metal temperature will not achieve 350°C.
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