Heat transfer by conduction in a truncated cone

  • Thread starter Kincaid
  • Start date
  • #1
7
0
1. The problem statement
A truncated cone 30cm high is made of Aluminum. The dia at the top is 7.5cm, and 12.5cm at the bottom. The lower surface is maintained at 93 deg C, the upper surface at 540 deg C. the other surface is insulated. Assuming 1 dimentional heat flow, calculate the rate of heat transfer in watts.


2. Homework Equations
Conduction equation:
q/a=k(dt/dx)


3. The Attempt at a Solution

I have the k value from a textbook at 204 w/(m C)

Made the assumptions that k is not a function of x or t and that the rest of the cone is perfectly insulated besides the top and the bottom.

That being the case I simply plugged in the values using

q=k*(A1*T1-A2*T2)/x where T1 is the high temperature and T2 is the low temperature. x is the thickness.

I'm not sure if this involves some kind of integration though. I fear I may be oversimplifying the problem.

This question was asked previously in the forum but was unanswered.

Thanks for the help.

K

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
783
9
Perform a heat balance on the cone. Since the sides of the cone are perfectly insulated, the cone does not generate energy nor does it store energy, so the amount of heat entering the cone must balance the amount of heat leaving the cone.

Thanks
Matt
 
  • #3
7
0
So then can it be said that

q1=k1A1T1

and

q2=k2A2T2

and the overall heat transfer is equal to q1/x?

Or are you saying that q=0?

Or do I need to use constant specific heats to solve this question to get the change in enthalpy and use that to get q?

I'm afraid this has just confused me further.
 
  • #4
783
9
Or are you saying that q=0?
No. q is not equal to zero.

This is what I am saying,

Ein - Eout = 0 therfore Ein = Eout

Or do I need to use constant specific heats to solve this question to get the change in enthalpy and use that to get q?
No this is not needed.

Sorry to have confused you further. I will work on this later today.

Thanks
Matt
 
  • #5
783
9
Let me give you a push in the right direction.

First lets find an expession for the area of the cone as a function of x.

let r1 = radius of the small end
let r2 = radius of the large end
let h = height of the cone

Then the function will be

r(x) = r1 + ((r2 - r1) / h)*x

The area as a function of x is A(x) = PI*r(x)^2

Therefore

A(x) = PI*(r1 + (((r2-r1)/h)*x)^2

Now using Fouriers law for one-dimensional heat conduction

Q' = -A(x)*k*(dt/dx)

Rearanging yields

Q'*(dx/(A(x)) = -k*dt

Now integrate the above function between 0 and h on the left hand side and integrate the right hand side between Th and To where Th = temperature on upper surface and To the temperature on the lower surface.

Remeber Q and PI are constant so they stay outside of the integral on the left hand side.

Think you can take it from here?

PS - This problem is stated in a slightly misleading manner because it is really a 2-dimensional problem and to bring it into a 1-dimensional problem the equation for the area as a function of x is required.


Thanks
Matt
 
Last edited:
  • Like
Likes 1 person
  • #6
7
0
Thanks so much. This makes it really clear.

Because the area is constantly changing as we move through the cone we must transform the area into terms of one of the available variables, in this case the height. Once this is done we can take the integral with respect to that variable in order to determine the values of the area at each individual point. This allows us to state the rate of heat flow through the cone as a whole instead of as a cylinder as I was doing before.

I don't know if I've verbalized the mathematics correctly but I feel I understand it much better.

Thanks

K
 
  • #7
783
9
For the most part you are on the correct track. Again, sorry to have mislead you with the heat balance method.

I have the final answer, so when you complete the integrations and substitutions we can compare my result with yours.

Thanks
Matt
 
  • #8
7
0
I think I must have made a mistake somewhere because I end up with a negative heat transfer which doesn't make sense.

My integration of dx/A(x) yields (1/PI)*(h/(r2-r1))*(r1+(r2-r1)*x/h)^-1*(-1) between 0 and h. This yields (1/PI)*(h/(r2-r1))*(1/r2)*(-1).

Multiplying this to the opposite side where the integration of -k(dT) between T1 and T2 yields k(T1-T2) gives me an answer of -1.49 kW.

Am I missing a sign somewhere or have I made a graver mistake?
 
  • #9
7
0
I neglected the 0 part of the integration. I now have a value of 2.24 kW which makes more sense.
 
  • #10
783
9
You have the right hand side integrated correctly.

The mistake is on the left hand side.

You are not far off though. Here is what I have. (Do you see your mistake?)

1/PI * (h/(r2-r1)) * (r1 + *((r2-r1)/h)*x)^-1 evaluated between 0 and h

Think you can take it from here?

Thanks
Matt
 
  • #11
783
9
Looks like you have corrected yourself.

I have a final result of 2.216 kW.

The difference is simply the value of k that I used was 202 W/(m-C) as opposed to your value of 204 W/(m-C).

Glad to have helped you out.

Thanks
Matt
 
  • #12
7
0
I am so thankful for your help on this. I feel like I have a much better understanding of the material now as well.

Thanks again

K
 
  • #13
1
0
Where did you find this question? If it's possible, I'd like to know which book this comes from and the answer to it as well. Thank you!
 

Related Threads on Heat transfer by conduction in a truncated cone

  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
0
Views
5K
Replies
1
Views
8K
  • Last Post
Replies
9
Views
9K
Replies
12
Views
11K
Replies
3
Views
2K
  • Last Post
Replies
0
Views
8K
Replies
4
Views
4K
Replies
3
Views
688
Top