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ypchen January 3, 2012 10:02

heat flux problem
 
Hi, everybody, my model is a cylinder tank and can be assumed 2-D axisymmetric(the axis is centerline). The tank wall (parellel the axis) has a heat flux 70W, so I have to calculate w/m2 to correspond the unit in FLUENT, but I don't know how to determine the area, if the wall length is H, how do I get the area ?

Thank you very much.

sunflower January 3, 2012 21:48

If the diameter of cylinder is D, the area should be A=Pi*D*H, and heat flux is 70/A.




Quote:

Originally Posted by ypchen (Post 337668)
Hi, everybody, my model is a cylinder tank and can be assumed 2-D axisymmetric(the axis is centerline). The tank wall (parellel the axis) has a heat flux 70W, so I have to calculate w/m2 to correspond the unit in FLUENT, but I don't know how to determine the area, if the wall length is H, how do I get the area ?

Thank you very much.


ypchen January 3, 2012 22:12

Thank for your reply, Sunflower.
I know the surface area is pi*D*H, and I had set the value as 70/A, but the simulation result is not good, so I am not sure whether the relation is what I want, maybe there is another problem in my case. thank you !

sunflower January 3, 2012 22:20

What do you mean by saying that simulation result is not good. Is the temperature too low or high? Can you describe the problems?


Quote:

Originally Posted by ypchen (Post 337761)
Thank for your reply, Sunflower.
I know the surface area is pi*D*H, and I had set the value as 70/A, but the simulation result is not good, so I am not sure whether the relation is what I want, maybe there is another problem in my case. thank you !


ypchen January 3, 2012 22:41

Quote:

Originally Posted by sunflower (Post 337762)
What do you mean by saying that simulation result is not good. Is the temperature too low or high? Can you describe the problems?

I calculate the surface area as 0.000971 m2 (D = 6mm, H = 50 mm), the surface area include side and top without bottom. so I can get the heat flux = Q/A = 70/0.000971 = 72090.63 W/m2. because of the wall is cooling the inside fluid, so I set the flux as -72090.63. after calculation, the temperature near the wall is too low, below 293 K, but the fluid inside the tank should not below 293 K. I think the flux is too large, so maybe I need to use UDF to control the heat flux,for example, if the heat transfer rate reach some value, the flux = 0.

sunflower January 3, 2012 22:55

You define the negative heat flux on the cylinder wall, so temperature of fluid inside the tank will become lower and lower if there is no heat source inside the fluid. So the current simulation result is reasonable. It is not what you expected but it is correct.

To maintain the temperature of fluid inside the tank to a certain level, there should be a heat source to provide the energy. My suggestion is to go back to physical model to check if something is missing.



Quote:

Originally Posted by ypchen (Post 337764)
I calculate the surface area as 0.000971 m2 (D = 6mm, H = 50 mm), the surface area include side and top without bottom. so I can get the heat flux = Q/A = 70/0.000971 = 72090.63 W/m2. because of the wall is cooling the inside fluid, so I set the flux as -72090.63. after calculation, the temperature near the wall is too low, below 293 K, but the fluid inside the tank should not below 293 K. I think the flux is too large, so maybe I need to use UDF to control the heat flux,for example, if the heat transfer rate reach some value, the flux = 0.


ypchen January 3, 2012 23:02

Quote:

Originally Posted by sunflower (Post 337767)
You define the negative heat flux on the cylinder wall, so temperature of fluid inside the tank will become lower and lower if there is no heat source inside the fluid. So the current simulation result is reasonable. It is not what you expected but it is correct.

To maintain the temperature of fluid inside the tank to a certain level, there should be a heat source to provide the energy. My suggestion is to go back to physical model to check if something is missing.

Thank for your suggestion. actually, there is a heat source in the tank, but the heat remove from the tank is higher than the system generate, so I think I should set a stop point.

sunflower January 3, 2012 23:06

Are you running a transient case or steady state one? What is the meaning of stop point? I am just curious.


Quote:

Originally Posted by ypchen (Post 337768)
Thank for your suggestion. actually, there is a heat source in the tank, but the heat remove from the tank is higher than the system generate, so I think I should set a stop point.


ypchen January 3, 2012 23:32

Quote:

Originally Posted by sunflower (Post 337769)
Are you running a transient case or steady state one? What is the meaning of stop point? I am just curious.

The case is transient sate. My cylinder tank is a heat pipe, the heat pipe is put in a hydride tank, and the tank has a heat source so I use the heat pipe to remove heat from the hydride tank. The heat pipe has a maximum heat transfer rate as 70W. When the heat transfer rate across side and top surface reach 70W, the heat pipe will dryout, it means the heat pipe would not working so the heat flux equal to zero. I think the reason why I get the wrong temperature is the heat pipe still working while it reach the maximum value.


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