I have a relatively simple problem on heat flow, but my knowledge in this department is limited at best. I have the following situation: I need to heat up an 8" sch. 80 304H SS pipe to 1900°F and do some light bending. (We will use a 4" wide low voltage heater band typically designed for PWHT.) I would like to find out what the temperature of the pipe will be 12" away from the heated band. I need to know the equation(s) used for this.
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Basic Thermodynamics Help 2
- Thread starter Montana1
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I assume a PWHT heater is a post weld heat treater, which will attach around the circumference of the pipe. I am assuming you need the 1900 F at a distance of 12" from your heater.
Your equation you need is:
Qdot=-kA*(t2-t1)/L
Qdot is the rate of heat transfer, which you will know from your heater, in watts.
k is thermal conductivity, which is 16.2 W/mK at 100C, 21.5 at 500C. So at 1030C(which is 1900 F), your k value will be 28.6 W/mK assuming a linear relation between thermal conductivity and temperature.
L is 12 inches, which is .3048 m.
t2 is 1030 C
A is your cross sectional area. For 8" schedule 80 pipe, your A is pi*.25*(8.625^2-7.625^2)=12.762 in^2, which is .008234 m^2
So your temperature in degrees C will be:
1037-(Qdot*.7726)
This neglects any convection cooling from the surface, but it should get you in the ballpark.
If you need convection information, just reply to this post saying so.
Reidh
Your equation you need is:
Qdot=-kA*(t2-t1)/L
Qdot is the rate of heat transfer, which you will know from your heater, in watts.
k is thermal conductivity, which is 16.2 W/mK at 100C, 21.5 at 500C. So at 1030C(which is 1900 F), your k value will be 28.6 W/mK assuming a linear relation between thermal conductivity and temperature.
L is 12 inches, which is .3048 m.
t2 is 1030 C
A is your cross sectional area. For 8" schedule 80 pipe, your A is pi*.25*(8.625^2-7.625^2)=12.762 in^2, which is .008234 m^2
So your temperature in degrees C will be:
1037-(Qdot*.7726)
This neglects any convection cooling from the surface, but it should get you in the ballpark.
If you need convection information, just reply to this post saying so.
Reidh
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I don't think there's going to be a good shortcut beyond trying to solve the problem in more detail.
The solution above assumes that ALL of the heat going into that pipe is transferred by conduction down the length of that pipe to points unknown. But in order to show that is a reasonable approximation, you'd just about have to solve the more general problem.
Example: Suppose the pipe is only 1' long beyond the heater. Then conduction out the end is very low, and the assumption made above is off. If the pipe is short and convection/radiation losses are low, then in fact, the entire pipe ought to be about the same temperature.
Or, if convection and radiation losses are very very large, then the temperature a foot away ought to be near ambient.
The real case will surely be somewhere in between. You should be able to include convection, radiation, and transient effects and still be able to solve either analytically or numerically by spreadsheet. (Perhaps a mathcad problem once you set up the differentical equations)(and even then, the solution is fairly approximate due to uncertainties in convection rates, assuming 2-d flow when it's 3-d, etc). Don't forget to check temperature differential within the heated portion- to get the ends of the heated part that hot, the center will be hotter.
I'm assuming you have checked into the metallurgy end of this first. Also that you know your heater can handle that temperature (should be glowing a good red at that temp if I recall).
The solution above assumes that ALL of the heat going into that pipe is transferred by conduction down the length of that pipe to points unknown. But in order to show that is a reasonable approximation, you'd just about have to solve the more general problem.
Example: Suppose the pipe is only 1' long beyond the heater. Then conduction out the end is very low, and the assumption made above is off. If the pipe is short and convection/radiation losses are low, then in fact, the entire pipe ought to be about the same temperature.
Or, if convection and radiation losses are very very large, then the temperature a foot away ought to be near ambient.
The real case will surely be somewhere in between. You should be able to include convection, radiation, and transient effects and still be able to solve either analytically or numerically by spreadsheet. (Perhaps a mathcad problem once you set up the differentical equations)(and even then, the solution is fairly approximate due to uncertainties in convection rates, assuming 2-d flow when it's 3-d, etc). Don't forget to check temperature differential within the heated portion- to get the ends of the heated part that hot, the center will be hotter.
I'm assuming you have checked into the metallurgy end of this first. Also that you know your heater can handle that temperature (should be glowing a good red at that temp if I recall).
If you are concerned about getting the exact temperature 12 inches from your heat source, instrument your pipe with a thermocouple.
If you want to accurately calculate the temperature, much more details are needed.
Maybe I should have added that the above equation will get you in the ballpark, which is probably good enough for bending pipe.
Reidh
If you want to accurately calculate the temperature, much more details are needed.
Maybe I should have added that the above equation will get you in the ballpark, which is probably good enough for bending pipe.
Reidh
Star for JStephen for correctly characterising the problem.
At 1900F radiation is definately not negligible. Convection either. You will probably also induce a draft inside the pipe, so you will need to include convection on the inside as well.
You should also check with a metalurgist about the bad things that might happen to the material at those temperatures.
At 1900F radiation is definately not negligible. Convection either. You will probably also induce a draft inside the pipe, so you will need to include convection on the inside as well.
You should also check with a metalurgist about the bad things that might happen to the material at those temperatures.
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- #7
Thanks to all for the valuable information. The ends of the pipe will be covered up so drafting would be minimal. I really just needed the equation for a ballpark estimate since there is an abrasion resistant refractory lining that I'd like to stay away from. The attachment of a T/C is definitely a must to monitor accurate temperature and check my math. Thanks again for the valuable advice.
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