Thermal Conduction Through a Sphere

In summary,The problem supplies the temperature difference between the center/surface and the radius of the sphere. My question is whether it would be correct to apply the conduction formula Q/t=kA(T2-T1)/L even though the area of contact between two adjacent surfaces will be changing at different radii. If applying this equation is incorrect, what method should be used for this problem? The best solution I could find is to approximate the heat flow equation using the surface area and radius.
  • #1
Brooks
3
0
I have come across a problem where I must find the rate of heat flow to the surface of a sphere from the center. The sphere has a constant coefficient of thermal conduction. The problem also supplies the temperature difference between the center/surface and the radius of the sphere. My question is whether it would be correct to simply apply the conduction formula Q/t=kA(T2-T1)/L even though the area of contact between two adjacent surfaces will be changing at different radii. If applying this equation is incorrect, what method should be used for this problem? Will calculus be required due to the varying area of contact? Thank you for your time.
 
Science news on Phys.org
  • #2
Q/t=kA(T2-T1)/L would apply in one dimension in rectilinear coordinate system.

See this for a spherical system, and yes, calculus is involved.

http://rpaulsingh.com/teaching/LecturesIFE/CondSphere/condsph.htm

or

http://www.rh.edu/~ernesto/C_S2002/CHT/notes/s08/s08.html
 
Last edited by a moderator:
  • #3
Thanks for the reply, Though I can't seem to find any elementary techniques for finding the rate of heat transfer through a sphere with a constant thermal coefficient k.
 
  • #4
Oops! Sorry about that. Let me see what I can find.

Also, is this steady-state, i.e. does [tex]\frac{\partial{T(r)}}{\partial{t}}[/tex] = 0?
 
  • #5
Yes it is steady state
 
  • #6
To solve this you need to solve the heat equation wit h appropriate boundary conditions on the sphere. You may be able to use a simple Newtons law of cooling for the heat loss from the surface of the sphere. Another condition will be 0 heat loss from the center and perhaps a initial heat distribution. Due to the symmetries this problem can be reduced to a single variable, r, greatly simpling the solution.
 
  • #7
I don't quite feel that Newton's law of cooling would be any substitute for the treatment required by the OP. That is more appropriate when the inside temperature is almost constant, unlike here. But of course, without any calculus, this can't be solved at all. The best is http://rpaulsingh.com/teaching/LecturesIFE/CondSphere/condsph.htm , which Astronuc has already sited. It requires the minimum of calculus, and the diagram is good.

Could the OP tell us under what chapter he came across this?
 
Last edited by a moderator:
  • #8
For a sphere no formulas are available to my knowledge because of diverging problems at the center. They are available, as can be read in the given links, for a hollow sphere, not a complete solid one. One can check that the formula is indeed diverging for an inner radius going towards 0.

However, there might be an approximate solution to the problem. Consider therefore the formula for the heat flow derived in the link:

[tex]q=-4 \pi k r^2\frac{dT}{dr}[/tex]

This can be approximated in stating the following:

outer radius sphere: [tex]r=R[/tex]
temperature difference: [tex]dT=\Delta T[/tex], from the original post
"radius difference": [tex]dr=0-R=-R[/tex]

giving thus:

[tex]q=4 \pi k R \Delta T[/tex]

Can someone confirm this before using it...
 
  • #9
So, q/t = kA(T2-T1)/L becomes q/t = k(T2-T1)*4pi*r^2/r = k(T2-T1)*surface area/radius. I don't know...
 
Last edited:
  • #10
As has been pointed out, no exact solution exists even using calculus, because of the div at 0. Some sort of mean of the max area and min area (which is 0) has to be taken. The one given by coomast seems as good as any. It's useless to ponder on this any further without knowing at what level the problem was presented.
 
  • #11
Heat conduction equation in spherical coordinates and with transient surface temperature is not an easy problem to solve. There is no analytical solution but only approximations that some times are not accurate. If you are familiar with numerical methods and discretization have a look to my publication:

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TWM-4SBRTK1-1&_user=153063&_coverDate=10%2F31%2F2008&_alid=1290075656&_rdoc=2&_fmt=high&_orig=search&_cdi=5566&_sort=r&_docanchor=&view=c&_ct=8&_acct=C000012698&_version=1&_urlVersion=0&_userid=153063&md5=6d0c4eddd4872a90ae0a0d1c5a698946
 

What is thermal conduction?

Thermal conduction is the transfer of heat from one object to another through direct contact. It occurs when there is a temperature difference between the two objects.

What is a sphere?

A sphere is a three-dimensional geometric shape that is perfectly round and has a constant diameter at every point on its surface. It is often used to represent objects such as planets, marbles, and balls.

How does thermal conduction occur through a sphere?

Thermal conduction through a sphere occurs when there is a temperature difference between the inside and outside of the sphere. Heat will flow from the hotter side to the cooler side until the two temperatures equalize.

What factors affect thermal conduction through a sphere?

The rate of thermal conduction through a sphere is affected by several factors including the material of the sphere, its size and shape, the temperature difference between the inside and outside, and the surrounding environment.

How is thermal conduction through a sphere calculated?

The rate of thermal conduction through a sphere can be calculated using the equation q = kA(ΔT)/d, where q is the heat flow, k is the thermal conductivity of the material, A is the surface area of the sphere, ΔT is the temperature difference, and d is the thickness of the sphere.

Similar threads

  • Thermodynamics
Replies
2
Views
1K
Replies
13
Views
3K
  • Thermodynamics
Replies
5
Views
1K
Replies
16
Views
12K
Replies
10
Views
2K
Replies
26
Views
3K
Replies
1
Views
516
  • Thermodynamics
Replies
2
Views
708
  • Mechanical Engineering
Replies
4
Views
1K
Replies
1
Views
1K
Back
Top