# Heat Transfer through a Sphere

1. Dec 22, 2016

### Zulumike

Hello everyone.
I've been doing a bit of independent study for this topic without much background and so my thermodynamic knowledge is fairly limited. I came across this problem and I'd like some assistance with it! If anyone can help out, I would be very grateful. A lot of the equations came from my own research, so I'm not sure if my starting point is altogether correct.
1. The problem statement, all variables and given/known data
A 0.0985 m radius hollow sphere contains a spherical cavity with 0.0785 m radius. The sphere is made out of a material with a k-value of 0.003 W/m.K. Ambient temperature around the sphere is 25 C, and the initial temperature within the cavity is 5.5 C. Convection heat transfer coefficient of the air is 10.45 W/m2.K for both the internal and external air. Calculate the steady-state heat transfer rate at this given temperature.

2. Relevant equations
Conductive heat transfer through spherical wall.
$$Q_{sphere} = -kA \frac{dT} {dr}$$

Convective heat transfer on a wall
$$q_{wall} = hA (T_{ambient} - T_{wall})$$
3. The attempt at a solution
Looking at a few calculations online, I determined that the $Q_{sphere}$ can be estimated as
$$Q_{sphere} = 4 \pi k r_1 r_2 \frac {(T_{intwall} - T_{exwall})} {r_2-r_1}$$

Where $r_1$ is the interior radius and $T_1$ is the internal temperature.

Using this http://www.learnthermo.com/examples/ch04/p-4b-2.php link, I think I found a way to connect steady state calculations to combine convective and conductive heat transfer. Using a similar method to the one in the link, I assume that the internal convection heat transfer rate is equal to the heat conduction rate which is also equal to the external convection heat transfer rate.

Internal convection equations:
$$q_{internal} = 4 \pi h r_1^2 (T_1 - T_{intwall})$$
$$q_{external} = 4 \pi h r_2^2 (T_{exwall} - T_2)$$
Where $T_2$ is the external ambient air and $T_1$ is the internal ambient air.
Isolating these for the $T_{intwall}$ and $T_{exwall}$:
$$T_{intwall} = T_1 - \frac {q} {4 \pi h r_1^2}$$
$$T_{exwall} = T_2 + \frac {q} {4 \pi h r_2^2}$$

Plugging these into the $Q_{sphere}$ equation:
$$q = 4 \pi k r_1 r_2 \frac {(T_1 - \frac {q} {4 \pi h r_1^2} - (T_2 + \frac {q} {4 \pi h r_2^2}))} {r_2-r_1}$$

Plug this into wolframalpha to isolate $q$:
$$q = -\frac {(4 \pi h k r_1^2 r_2^2 (T_2 - T_1))} {(r_1^2 (k - h r_2) + h r_2^2 r_1 + k r_2^2)}$$

Plugging in values:
h=10.45
r1 = 0.0785
r2 = 0.0985
k = 0.003
T1 = 25
T2 = 5.5
$$q = -0.276 W$$

Is this correct? Is my method at all correct? I'm really not entirely sure on the equations I used or whether I connected them correctly. Any help or pointing in the right direction would be much appreciated.

Thank you!

Last edited: Dec 22, 2016
2. Dec 22, 2016

### TSny

It all looks good to me. I noted that you used k = .003 in the calculation whereas it stated k = .002 in the problem.

3. Dec 22, 2016

### Zulumike

Thanks for looking it over. That was just a typo on my part, my bad. It's fixed now.

4. Dec 22, 2016

### Staff: Mentor

This was not done quite correctly. For heat conduction through the shell, $$-k\frac{dT}{dr}=\frac{Q}{4\pi r^2}$$where r is the local radius within the shell. You need to integrate this between the inner and out radii of the shell.

5. Dec 23, 2016

### Zulumike

I just integrated it again according to your parameters, but I'm getting the same equation. Could you let me know what I'm doing wrong? Below is my calculation.

6. Dec 23, 2016

### Staff: Mentor

Oops. Sorry for my confusion. You, of course, did it corretly.