Convective and conductive heat transfer

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SUMMARY

The discussion focuses on calculating the final temperature and cooling time of a cylindrical copper block subjected to convective and conductive heat transfer. The block, with dimensions of 0.3m in diameter and 0.4m in length, is initially at 60°C and experiences a heat flux of 450 W/m² on half its curved surface while being cooled by convection to surrounding air at 20°C. The convective heat transfer coefficient is 20 W/m²K, and the specific heat capacity of copper is 400 J/kg. The final temperature of the block is determined using energy balance equations, leading to a cooling time of approximately 5.4 hours.

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DarkBlitz
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Hey, I was doing some revison for a test and i came across this question which I can't get my head around

a clyndrical copper block, 0.3m in diameter and 0.4m long is initially at a uniform temp of 60°C and half its curved surface is exposed to a heat flux of 450w/m^2. If the complete surface is being cooled by convection heat transfer to the surrounding air at 20°C, determine the final temperature of the block, assuming it is uniform throughout.

Determine the time to cool the block from 60 degrees to the final temperature. assuming the convective heat transfer is constant throughout the cooling process and can be calculated using the average temperature of the block during the cooling process.

For each case, identify an appropriate system and energy balance and use the following date: p(copper) =8900 kg/m^3
C(copper) =400j/kg
convective heat transfer coefficient = 20w/m^2k

I'm haven't tried the second half of the question as I can not get the first part.
So far, i have said that:
Qdotin=Qdotout
flux*Half curved surface= h* Atotal * (Tsurface - Tair)
But my problem with this equation is to find an expression for T surface in terms of the final and initial temperature.

Am I on the right track?

Thanks
 
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The first part says
determine the final temperature of the block, assuming it is uniform throughout.
That is telling you to assume Tsurface is equal the the uniform final temperature.

It is physically impossible for the temperature to be uniform throughout, because there would be no heat conduction in the block. But you have to use the assumptions the question tells you to use.
 
ah ok, but then what do i do with the initial 60 degree temperature?
 
oh, that might be for the second part of the question, ill give it a go thanks!
 
I got the first part, but for the second part of the question,
I used the formula:
m*c*deltaT/t = hA(Ts-Tf)
Where Ts is the average temperature of the copper during cooling and we are solving for T, however i do not get the right answer. The answer is 5.4hours.

Which part of the equation is wrong?
 
DarkBlitz said:
oh, that might be for the second part of the question

Yup, that's right. Give it a go...
 

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