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Note: This is not homework, and I am not at the age to receive such homework. This is simply something that I am interested about.

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Note: This is not homework, and I am not at the age to receive such homework. This is simply something that I am interested about.

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mgb_phys

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1, The rate of heat input - just saying you supply heat at 60deg is not enough. Suppose you connect a very thin wire at 60deg to one face it will heat much more slowly than connecting a huge electrical heater even if that was also at 60deg.

2, Then there is the rate of heat loss. There are 3 main types:

Radiation will depend on the reflectivity of the surface and the temperature of the surrounding walls - if the block was pointing at outer space (cold black) there would be more heat loss than if it was inside a reflective sphere.

Convection will depend on the rate of air flow and moisture content of the air.

Conduction doesn't matter if we assume the block is not touching anything (I know this is impossible but we are allowed to cheat in thought experiments!)

3, The rate of heat transfer in the block. This is the easiest to work out since it only depends on the properties of aluminium and the temperature difference. Paradoxically since the rate of heat transfer slows as the temperature difference is reduced the two sides of the block will never reach the same temperature ( in theory) - and your coffee will never actually get cold!

I hope I hit the right technical level - please post back if you want anything explained.

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Hootenanny

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Hootenanny

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The heat equation is as follows,

[tex]\frac{\partial u}{\partial t} = k\nabla^2 u[/tex]

Where [itex]u=u(x,y,z,t)[/itex] is the temperature at a given point and time in the cube. I as said previously (and mgb_physics has alluded to), you could potentially end up with some nasty boundary conditions. You could simplify the boundary conditions considerably by assuming that the face of the cube is uniformally heated and the boundaries of the other faces remain at a constant temperature, but perhaps this would be an over-simplification.

Since your query is more practical that theoretical, perhaps it would be better addressed by the engineers in their forum, I could get the mentor's attention to move it there if you like?

[tex]\frac{\partial u}{\partial t} = k\nabla^2 u[/tex]

Where [itex]u=u(x,y,z,t)[/itex] is the temperature at a given point and time in the cube. I as said previously (and mgb_physics has alluded to), you could potentially end up with some nasty boundary conditions. You could simplify the boundary conditions considerably by assuming that the face of the cube is uniformally heated and the boundaries of the other faces remain at a constant temperature, but perhaps this would be an over-simplification.

Since your query is more practical that theoretical, perhaps it would be better addressed by the engineers in their forum, I could get the mentor's attention to move it there if you like?

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Hootenanny

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This is what I was talking about when I said that the boundary conditions could get a little ugly. Your boundary conditions will be temperature dependent, my best guess would be to solve the heat equation subject to the heat lost determined by Fourier's law. However, this would only be a guess, I haven't attempted anything so complicated myself.

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Could you please demostrate it mathematically?

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Hootenanny

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Aftert thinking about the problem a little more this morning it is clear that no matter how many assumptions you make, your boundary conditions are either going to be really ugly or completely unrealistic. In any case, the PDE which you obtain is unlikely to have any analytical solutions, you're probably going to have to use numerical analysis to determine the temperature. I've managed to find a few decent resources that deal with the numerical analysis of heat transfers,

http://icb.olin.edu/fall_03/mc/reading/conduction.pdf [Broken]

http://www.buildingphysics.com/manuals/avh_TB.pdf

http://icb.olin.edu/fall_03/mc/reading/conduction.pdf [Broken]

http://www.buildingphysics.com/manuals/avh_TB.pdf

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Hootenanny

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Indeed there is, I hadn't spotted that! Equation (1.6) should read,

[tex]\frac{dT_i}{dt} = \alpha\frac{T_{i+1}+T_{i-1}}{L_i^2}}[/tex]

And the final equation (1.7) should read,

[tex]\frac{dT_i}{dt} = N^2\frac{T_{i+1}+T_{i-1}}{\tau_\text{cond}}[/tex]

I apologise for referencing incorrect material. The second document is far more comprehensive and will probably tell you what you need to know.

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[tex]\frac{\partial^2 T}{\partial z^2}=0\iff T(z)=az+b .[/tex]

By Fourier's Law, [tex]-\kappa\frac{\partial T}{\partial z}=h(T-S),[/tex]

where S is the room temperature, and [tex]\kappa[/tex] and h are constants. Letting [tex]T(0)=T_0[/tex], then the temperature at any point may be obtained. Is the Fourier's Law correctly written? Thanks in advance.

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Hootenanny

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Yes, your steady state solution is correct, as is your expression of Fourier's Law, provided that the aluminum block is undergoing forced convection (i.e. a constant flow of air at room temperature is passing over it's surface). As an aside Fourier's Law in this form is commonly referred to as

[tex]\frac{\partial^2 T}{\partial z^2}=0\iff T(z)=az+b .[/tex]

By Fourier's Law, [tex]-\kappa\frac{\partial T}{\partial z}=h(T-S),[/tex]

where S is the room temperature, and [tex]\kappa[/tex] and h are constants. Letting [tex]T(0)=T_0[/tex], then the temperature at any point may be obtained. Is the Fourier's Law correctly written? Thanks in advance.

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the free convection heat transfer coefficient is dependent on the temperature of the body, and since the temperature of the body is changing as it gets hotter , the coefficient is also changing all the time so you have to solve PDEs numerically for the exact solution.

for most of the free convection problems the value of the convection coefficient is between about 5-10 W/(M^2*Deg C)- it doesn't change much so if you don't need high precision you may assume it as constant.

Finally, the most easy, accurate and fast way to solve this problem is with a software, such as COMSOL. you just sketch the model, all the PDFs are already in the software (with the material properties) and you solve it numerically, it doesn't take more then 10-15min if you have some basic background on heat transfer.

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