Heat Transfer Conduction 2

[joedvo]

When an insulated tank of 306 degrees F has an uninsulated lug sitting on an uninsulated frame. I am trying to determine how far, inch by inch, from the lug to the frame and down, will the temperature decrease. I have been studying the heat transfer equations for conduction, convection, and radiation. I feel that I understand those well, but what I cannot determine is the starting point, exactly how many Btu's the tank would generate through the lug at the vessel wall. The tank is 53-1/4" diameter, a height is 24", and the dome top of the tank is 21-3/8". The vessel wall is 1-1/2". The cross sectional area of the lug one inch from the vessel wall is .63 ft^2. The tank should remain at 306 degrees F and the ambient temperature of the air outside the vessel is 59 degrees F. Both the lug and the vessel are stainless steel with a thermal conductivity value of 9.4 (btu in / ft^2 hr deg F). How do I calculate the maximum Q leaving the vessel wall? An EE will be greatful for your help. Thank you!

 

[IRstuff]

You have two possible limiters, the lug thermal conductivity, or the convection from the frame. Whichever is limiting, will determine the amount of heat that can flow. Once you know that, you can then map out the rough temperature differentials. Given that it's not a simple structure, an FEA might be required.

TTFN

FAQ731-376

 

[Unotec]

Chapter 3, sec 6, subsection 2 of the incropera will help you. Basically you can consider you have a rectalgular fin and ignore tht radiation from the insulated part of the tank. This is for the fin, though. The tank, insulated, will also have losses.
Is the process so sensitive to temperature that you have to calculate this? The heat will be disipated very quickly. Only close, very close to the tank it will be hot.

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[joedvo]

I just have to mathmatically prove that the frame will not exceed 140 degrees F. The cross sectional area and surface area change several times as the distance increases, mostly due to the frame the lug is resting on. The average cross sectional area of the first fifteen inches from the vessel is 1.05 ft^2. There is some play in the temperature, I am only concerned with approximate temperature values.

 

[IRstuff]

thermal_conductivity*cross_sec_area*(Thigh-Tlug)/len_lug = conv_area*conv_coeff*(Tlug-Tamb)

find respective areas and solve for Tlug

TTFN

FAQ731-376

 

[joedvo]

I didn't think the rate of Btu's generated by conduction was equal to the rate of Btu's generated by convection. I was under the impression that Q convection was much less... Otherwise this method and gave a very realistic lug temperature. Please confirm my question about Q conduction = Q convection. Also, does Q radiation play a role in setting these equal? Thank you!

 

[IRstuff]

It's not, in general. The point of the exercise was to determine the temperature of the lug and frame. Therefore, what heat that goes into the lug from the tank must leave, by any combination of convection, conduction, emission. The balance of the entering and exiting heat determines the temperature at any given point.

TTFN

FAQ731-376

 

[chicopee]

Check out the steady state relaxation method to determine the temperature gradient along the lug and part of the frame. Once the analysis stabilized then you can figure out the heat transfer at the junction of the tank and lug.

 

[sailoday28]

Unotec (Chemical) has the key word "fin". Does the cross sectional area of the lug vary with length?

 

[Unotec]

Sailoday, a fin does not necessarily vary in crossectional area. They can have a uniform crossectional area and still be considered fins. The definition comes more from the heat transfer mechanism (convection) than from the shape.
What I was trying to express is that at these temperature gradients, it is very likely that the lug will dissipate most of the heat, as a fin, before it reaches the frame and causes any significant temperature increase in it.
I have not done the calculations, so I could be proven wrong.

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>