Convective and conductive heat transfer

In summary, the conversation is about a cylindrical copper block being cooled by convection heat transfer. The first part of the question asks to determine the final temperature of the block, assuming it is uniform throughout. The second part asks to determine the time it takes for the block to cool from 60 degrees to the final temperature, using the average temperature of the block during the cooling process. The equation Qdotin=Qdotout is used, but there is difficulty in finding an expression for Tsurface. The second part of the question is approached using the formula m*c*deltaT/t = hA(Ts-Tf), but the answer is not correct.
  • #1
DarkBlitz
22
0
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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  • #2
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.
 
  • #3
ah ok, but then what do i do with the initial 60 degree temperature?
 
  • #4
oh, that might be for the second part of the question, ill give it a go thanks!
 
  • #5
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?
 
  • #6
DarkBlitz said:
oh, that might be for the second part of the question

Yup, that's right. Give it a go...
 

1. What is the difference between convective and conductive heat transfer?

Convective heat transfer is the transfer of heat through the movement of fluids, while conductive heat transfer is the transfer of heat through a solid material or between two solids in direct contact.

2. What are some examples of convective and conductive heat transfer in everyday life?

Convective heat transfer can be seen in a pot of boiling water, where the heat from the stove is transferred through the movement of the water molecules. Conductive heat transfer can be seen when you touch a hot pan and feel the heat transfer from the pan to your hand.

3. How is heat transfer affected by the material of an object?

The material of an object can affect how quickly heat is transferred through it. Materials with high thermal conductivity, such as metals, will transfer heat more quickly than materials with low thermal conductivity, such as wood or plastic.

4. What are some factors that impact convective and conductive heat transfer?

The rate of convective and conductive heat transfer can be affected by factors such as the temperature difference between the objects, the surface area of contact, the type of materials involved, and the presence of insulating materials.

5. How is convective and conductive heat transfer used in engineering and technology?

Convective and conductive heat transfer play a crucial role in many engineering and technological applications, such as in the design of heat exchangers, refrigeration systems, and thermal insulation. Understanding these heat transfer processes is essential for developing efficient and effective systems in various industries.

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