Training Conduction & Thermal Resistance Spherical Conduction and Contact Resistance
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Spherical Conduction and Contact Resistance

24 min Conduction & Thermal Resistance

Spherical Conduction and Contact Resistance

We complete the conduction picture with spherical shells (pressure vessels, ball bearings, hollow spheres) and the important practical effect of contact resistance between surfaces.

Spherical Conduction

$$\dot{Q} = \frac{4\pi k r_1 r_2 (T_1 - T_2)}{r_2 - r_1}$$

Thermal resistance of a spherical shell:

$$R_{\text{sph}} = \frac{r_2 - r_1}{4\pi k r_1 r_2}$$

Thermal Contact Resistance

When two solid surfaces are pressed together, microscopic air gaps create an additional resistance:

$$R_{\text{contact}} = \frac{R''_c}{A}$$

$R''_c$ = contact resistance per unit area (m²·K/W), typically $10^{-4}$ to $10^{-2}$ m²·K/W depending on surface finish and interface pressure.

Example 1 — Hollow Sphere

A hollow steel sphere ($k = 50$) has $r_1 = 10$ cm, $r_2 = 15$ cm. Inner wall at 300°C, outer at 100°C. Find $\dot{Q}$.

$$\dot{Q} = \frac{4\pi(50)(0.10)(0.15)(200)}{0.15 - 0.10} = \frac{4\pi(50)(0.015)(200)}{0.05} = \frac{1884.96}{0.05} = 37{,}699 \text{ W}$$

$\approx 37.7$ kW.

Example 2 — Contact Resistance

Two aluminum plates are bolted together. Each plate: 2 cm thick, $k = 205$, $A = 0.1$ m². Contact resistance $R''_c = 2 \times 10^{-4}$ m²·K/W. $\Delta T = 80$°C. Find $\dot{Q}$ with and without contact resistance.

Without contact:

$R = 2 \times \frac{0.02}{205 \times 0.1} = 2 \times 9.76 \times 10^{-4} = 1.95 \times 10^{-3}$ K/W

$\dot{Q} = 80/0.00195 = 41{,}000$ W

With contact:

$R_{\text{contact}} = 2 \times 10^{-4}/0.1 = 2 \times 10^{-3}$ K/W

$R_{\text{total}} = 0.00195 + 0.002 = 0.00395$ K/W

$\dot{Q} = 80/0.00395 = 20{,}250$ W

Contact resistance halved the heat transfer! It’s comparable to the conduction resistance of the plates themselves.

Example 3 — When to Ignore Contact Resistance

A composite wall has $R_{\text{cond}} = 5.0$ K/W total. The interface contact resistance is $R_c = 0.002$ K/W. Is it significant?

$R_c / R_{\text{total}} = 0.002/5.0 = 0.04\%$

Negligible. Contact resistance matters when it’s the same order of magnitude as conduction resistance — typically with high-$k$ metals.

Practice Problems

1. Hollow sphere: $r_1 = 5$ cm, $r_2 = 10$ cm, $k = 20$, $\Delta T = 150$°C. Find $\dot{Q}$.
2. Find $R_{\text{sph}}$ for $r_1 = 8$ cm, $r_2 = 12$ cm, $k = 40$.
3. Two steel blocks ($R_{\text{cond}} = 0.005$ K/W each) with $R''_c = 5 \times 10^{-4}$ m²·K/W, $A = 0.05$ m². Total $R$?
4. Applying thermal paste ($R''_c$ drops from $10^{-3}$ to $10^{-5}$) on a CPU (area 4 cm²). By what factor does contact resistance drop?
5. Why is contact resistance higher with rough surfaces?
6. A sphere has 4× the heat flow of a cylinder of the same $r_1, r_2, \Delta T$. True or false?
Show Answer Key

1. $\dot{Q} = 4\pi(20)(0.05)(0.10)(150)/0.05 = 37{,}699$ W

2. $R = (0.12 - 0.08)/(4\pi \times 40 \times 0.08 \times 0.12) = 0.04/4.825 = 0.00829$ K/W

3. $R_{\text{contact}} = 5 \times 10^{-4}/0.05 = 0.01$ K/W. $R_{\text{total}} = 0.005 + 0.01 + 0.005 = 0.02$ K/W

4. Factor of 100 ($10^{-3}/10^{-5} = 100$)

5. Fewer contact points → more trapped air gaps → higher resistance (air $k$ is very low).

6. False — the formulas are different (sphere $\propto r_1 r_2/(r_2-r_1)$, cylinder $\propto 1/\ln(r_2/r_1)$). They’re not simply proportional.

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