Training Conduction & Thermal Resistance Radial Conduction — Pipes and Insulation
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Radial Conduction — Pipes and Insulation

24 min Conduction & Thermal Resistance

Radial Conduction — Pipes and Insulation

Cylindrical geometry is everywhere: pipes, tubes, wires, insulation wraps. The key difference from planar conduction: the area changes with radius, so the resistance formula uses a logarithm.

Radial Conduction (Cylindrical)

$$\dot{Q} = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}$$

$r_1$ = inner radius, $r_2$ = outer radius, $L$ = pipe length.

Cylindrical Thermal Resistance

$$R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L}$$

For convection on a cylindrical surface of radius $r$: $R_{\text{conv}} = \frac{1}{h(2\pi r L)}$.

Critical Radius of Insulation

$$r_{\text{cr}} = \frac{k_{\text{ins}}}{h_{\text{outer}}}$$

Adding insulation to a small pipe can increase heat loss until radius reaches $r_{\text{cr}}$, because the added surface area overcomes the added resistance.

Example 1 — Steam Pipe

A steel pipe ($k = 50$) carries steam. Inner radius = 5 cm, outer = 6 cm, length = 10 m. $T_{\text{inner}} = 200$°C, $T_{\text{outer}} = 190$°C. Find $\dot{Q}$.

$$\dot{Q} = \frac{2\pi(50)(10)(200 - 190)}{\ln(0.06/0.05)} = \frac{2\pi(50)(10)(10)}{\ln 1.2} = \frac{31{,}416}{0.1823} = 172{,}300 \text{ W}$$

That’s 172 kW — steel has very low resistance.

Example 2 — Insulated Pipe

The same pipe gets a 4 cm layer of fiberglass insulation ($k = 0.04$). Now $r_3 = 10$ cm. Overall outer convection $h = 10$ W/(m²·K), $T_\infty = 25$°C. Find $\dot{Q}$ (per 10 m).

$R_{\text{steel}} = \frac{\ln(0.06/0.05)}{2\pi(50)(10)} = \frac{0.1823}{3141.6} = 5.8 \times 10^{-5}$ K/W (negligible!)

$R_{\text{ins}} = \frac{\ln(0.10/0.06)}{2\pi(0.04)(10)} = \frac{0.5108}{2.513} = 0.2033$ K/W

$R_{\text{conv}} = \frac{1}{10 \times 2\pi(0.10)(10)} = \frac{1}{62.83} = 0.0159$ K/W

$R_{\text{total}} = 0 + 0.2033 + 0.0159 = 0.2192$ K/W

$\dot{Q} = (200 - 25)/0.2192 = 798$ W

Insulation reduced heat loss from 172 kW to 0.8 kW!

Example 3 — Critical Radius

An electrical wire has radius 2 mm. Insulation $k = 0.2$ W/(m·K), outer $h = 15$ W/(m²·K). Will adding insulation increase or decrease heat loss?

$$r_{\text{cr}} = \frac{k}{h} = \frac{0.2}{15} = 0.0133 \text{ m} = 13.3 \text{ mm}$$

Wire radius (2 mm) < $r_{\text{cr}}$ (13.3 mm), so adding insulation will increase heat loss until the insulation outer radius reaches 13.3 mm. Beyond that, it starts decreasing.

Practice Problems

1. Find $R_{\text{cyl}}$ for a copper pipe: $r_1 = 3$ cm, $r_2 = 4$ cm, $k = 385$, $L = 5$ m.
2. Heat loss through a 2 m pipe: $k = 1.0$, $r_1 = 5$ cm, $r_2 = 8$ cm, $\Delta T = 50$°C.
3. Critical radius: $k_{\text{ins}} = 0.05$, $h = 20$. Find $r_{\text{cr}}$.
4. A pipe has $r_1 = 2$ cm. Does 1 cm of insulation ($k = 0.05$, $h = 10$) help?
5. Convection resistance on a 10 cm radius pipe, $L = 3$ m, $h = 50$. Find $R_{\text{conv}}$.
6. Why does the cylindrical resistance use $\ln(r_2/r_1)$ instead of $(r_2 - r_1)$?
Show Answer Key

1. $R = \ln(4/3)/(2\pi \times 385 \times 5) = 0.2877/12{,}095 = 2.38 \times 10^{-5}$ K/W

2. $\dot{Q} = 2\pi(1.0)(2)(50)/\ln(8/5) = 628.3/0.470 = 1337$ W

3. $r_{\text{cr}} = 0.05/20 = 2.5$ mm

4. $r_{\text{cr}} = 0.05/10 = 5$ mm. Pipe $r_1 = 20$ mm > 5 mm, so yes, insulation helps (reduces $\dot{Q}$).

5. $R = 1/(50 \times 2\pi \times 0.10 \times 3) = 1/94.25 = 0.0106$ K/W

6. Because the area ($2\pi r L$) changes with $r$. Integrating Fourier’s law in cylindrical coordinates produces the logarithm.

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