A small sphere (emissivity = 0.746, radius = r1) is located at the center of a s

Question

A small sphere (emissivity = 0.746, radius = r1) is located at the center of a spherical asbestos shell (thickness = 1.17 cm, outer radius = r2; thermal conductivity of asbestos is 0.090 J/(s m Co)). The thickness of the shell is small compared to the inner and outer radii of the shell. The temperature of the small sphere is 818 °C, while the temperature of the inner surface of the shell is 621 °C, both temperatures remaining constant. Assuming that r2/r1 = 8.79 and ignoring any air inside the shell, find the temperature in degrees Celsius of the outer surface of the shell.

Explanation / Answer

  I assume that ALL of the heat supplied to this system comes from the inner sphere. All other bodies are passive. This is a crucial fact which you've failed to give me.

Here is the most important step in the strategy: draw a thermal resistance diagram (analogous to electric circuits)

Visualizing this one, it seems as there is a radiation heat transfer resistance from the inner sphere to the inner shell surface.

In series with the radiation resistance, there is a conduction resistance.

Remember: current is analogous to heat flow rate, and voltage is analogous to temperature.

Caution: the equation you have written for heat conduction (first of all has errors), second of all, will not help you.

You will want to use a model for thermal resistance to conduction through a spherical shell. I will give your the formula.

Temperatures:
T1: inner sphere
T2: inner surface of spherical shell
T3: outer surface of spherical shell

Radii (RE-DEFINED as per my convenience):
r1: inner sphere
r2: inner surface of spherical shell
r3: outer surface of spherical shell

Areas:
A1: associated with r1. A1 = 4*Pi*r1^2
A2: associated with r2. A2 = 4*Pi*r2^2

Because this is a series circuit, the Q_dot is the same for all resistances.

Equation for radiation heat transfer (assuming inner surface of shell behaves as a black body):
Q_dot = epsilon*sigma*A1*(T1^4 - T2^4)

Equation for conduction heat transfer in terms of thermal resistance R:
Q_dot = (T2 - T3)/R

Thermal resistance of a spherical shell:
R = (r3 - r2)/(4*Pi*k*r2*r3)

Thus:
epsilon*sigma*A1*(T1^4 - T2^4) = (T2 - T3)*(4*Pi*k*r2*r3)/(r3 - r2)

Substitute A1:
epsilon*sigma*4*Pi*r1^2*(T1^4 - T2^4) = (T2 - T3)*(4*Pi*k*r2*r3)/(r3 - r2)

Solve for T3:
T3 = T2 - epsilon*sigma*r1^2*(T1^4-T2^4)*(r3-r2)/(...

In terms of the radius ratio I choose to name j for no good reason, and thickness of spherical shell t,
r3 = r1*j
r2 = r3 - t
r2 = r1*j - t


T3 = T2 - epsilon*sigma*r1^2*(T1^4-T2^4)*t/(k*(r1*... - t)*r1*j)

And it simplifies:
T3 = T2 - epsilon*sigma*r1*(T1^4-T2^4)*t/(k*j*(r1*... - t))

Assuming that r1*j is MUCH larger than t, our equation simplifies:
T3 = T2 - epsilon*sigma*(T1^4 - T2^4)*t/(k*j^2)

And the result doesn't depend on r1.

Data: (MUST use Kelvin for temperatures)
T1:=1091.15 K; T2:=894.15 K; epsilon:=0.746; sigma:=5.67e-8 W/m^2-K^4; k:=0.09 W/m-K; j:=8.79; t:=0.0117 m;

Result:
T3 = 838.75 Kelvin

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.