Heat dissipation of an annular fin heatsink

P8.1.6Heat dissipation of an annular fin heatsink

Some electronic components is cooled by annular fins (heatsinks) which conduct heat away from the component and provide a larger surface area for that heat to dissipate to the surroundings.

The cooling efficiency of an annular fin of width $$2w$$ and inner and outer radii $$r_0$$ and $$r_1$$ may be written in terms of modified Bessel functions of the first and second kinds:

\eta = \frac{2r_0}{\beta(r_1^2-r_0^2)}\frac{K_1(u_0)I_1(u_1) - I_1(u_0)K_1(u_1)}{K_0(u_0)I_1(u_1)+I_0(u_0)K_1(u_1)},

where $u_0 = \beta r_0$ , $u_1 = \beta r_1$ and

\beta = \sqrt{\frac{h_c}{\kappa w}}.

$$h_c$$ is the heat transfer coefficient (which is taken to be constant over the fin's surface) and $\kappa$ is the thermal conductivity of the fin material.

What is the cooling efficiency of an aluminium annular fin with dimensions $r_0 = 5\;\mathrm{mm}$ , $r_1 = 10\;\mathrm{mm}$ , $w = 0.1\;\mathrm{mm}$ ? Take $h_c = 10\;\mathrm{W\,m^{-2}\,K^{-1}}$ and $\kappa = 200\;\mathrm{W\,m^{-1}\,K^{-1}}$ .

Calculate the heat dissipation, $\dot{Q}$ (the product of the efficiency, the fin area and the temperature difference) for a component temperature of $T_0 = 400\;\mathrm{K}$ and ambient temperature $T_\mathrm{e} = 300\;\mathrm{K}$ .

Show solution

Note that in the below code we use the scipy.special functions k0, k1, i0 and i1 which are faster and more accurate implementations of the more general kn and iv functions for calculating the modified Bessel functions of the first and second kinds.

heatsink

import numpy as np
from scipy.special import k0, k1, i0, i1

# Component temperature, ambient temperature (K)
T0, Te = 400, 300

# Fin width, internal radius, external radius (m)
w, r0, r1 = 0.1 * 1.0e-3, 0.005, 0.01
# Heat transfer coefficient (W.m-2.K-1), thermal conductivity (W.m-1.K-1)
hc, kappa = 10, 200

beta = np.sqrt(hc / kappa / w)
u0, u1 = beta * r0, beta * r1

# Cooling efficiency
eta = (
    2
    * r0
    / beta
    * (k1(u0) * i1(u1) - i1(u0) * k1(u1))
    / (r1**2 - r0**2)
    / (k0(u0) * i1(u1) + i0(u0) * k1(u1))
)

print(f"eta = {eta:5.3f}")

fin_area = 2 * np.pi * (r1**2 - r0**2)
# Heat dissipation (W)
Qdot = eta * hc * fin_area * (T0 - Te)
print(f"Qdot = {Qdot:.2f} W")

Output:

    eta = 0.994
    Qdot = 0.47 W