Heat and Mass Transfer: Heat Out

  • Thread starter jdawg
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Homework Statement


The temperature distribution across a wall 0.3m thick at a certain instant of time is T(x)=a+bx+cx2, where T is in degrees Celsius and x is in meters, a = 200°C, b = -200°C/m, and c=30°C/m2. The wall has a thermal conductivity of 1W/m2K.
(a) On a unit surface area basis, determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall.
(b) If the cold surface is exposed to a fluid at 100°C, what is the convection coefficient?

Homework Equations


q'' conduction= -k[∂T/∂x]
q'' convection= h[Twall-T]

The Attempt at a Solution


So part A made perfect sense. Assuming 1D conduction, I got the heat transfer in to be 200 W/m2 and the heat transfer out to be 182 W/m2.

Part B is where I get a bit confused. So I found the temperature at the cold surface of the wall (at distance L) by just plugging 0.3 m into the given temperature distribution. Then you use the heat transfer equation for convection:

q'' convection= h[Twall-T]

So you know the temperature difference, but how to you know the heat transfer out? Why would it be acceptable to use the heat transfer out due to conduction from part A when we are considering convection in part B? Is it because if you do a surface balance at that surface of the wall, then the heat transfer due to conduction will equal the heat transfer due to convection? But then I suppose your conduction heat transfer would have to be heat in and not heat out in that case... I would greatly appreciate some clarification :)
 

Answers and Replies

  • #2
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You assessed this correctly. You calculated the heat transfer rate out as 182 W/m^2. This is in series with the heat transfer rate through the convection boundary layer from the wall to the fluid at 100 C. The 182 is the rate of heat transfer conducted through the wall into the interface.
 

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