Question

A transformer that is \(10-\mathrm{cm}\) long, \(6.2 \mathrm{-cm}\) wide, and \(5-\mathrm{cm}\) high is to be cooled by attaching a $10-\mathrm{cm} \times 6.2-\mathrm{cm}\(-wide polished aluminum heat sink (emissivity \)=0.03$ ) to its top surface. The heat sink has seven fins, which are \(5-\mathrm{mm}\) high, 2 -mm thick, and 10 -cm long. A fan blows air at \(25^{\circ} \mathrm{C}\) parallel to the passages between the fins. The heat sink is to dissipate $12 \mathrm{~W}$ of heat, and the base temperature of the heat sink is not to exceed \(60^{\circ} \mathrm{C}\). Assuming the fins and the base plate to be nearly isothermal and the radiation heat transfer to be negligible, determine the minimum free-stream velocity the fan needs to supply to avoid overheating. Assume the flow is laminar over the entire finned surface of the transformer.

Short answer

Answer: The minimum free-stream velocity needed to avoid overheating the transformer is approximately 0.4956 m/s.

Step-by-step solution

Calculate the heat transfer surface area of the heat sink including fins

To find the minimum velocity, we first need to calculate the surface area of the heat sink available for heat transfer, including the surface area of the fins: Heat sink base area: \(A_{base} = 10\,\text{cm} \times 6.2\,\text{cm} = 62\,\text{cm}^2\) Fin surface area (per fin): \(A_{fin} = 2 \times(10\,\text{cm} \times 0.5\,\text{cm}) = 10\,\text{cm}^2\) Total fin surface area for seven fins: \(A_{fins} = 7 \times A_{fin} = 7 \times 10\,\text{cm}^2 = 70\,\text{cm}^2\) Total heat transfer surface area: \(A_{total} = A_{base} + A_{fins} = 62\,\text{cm}^2 + 70\,\text{cm}^2 = 132\,\text{cm}^2\)

Find the convection heat transfer coefficient, h

We will use the Sieder-Tate correlation for the forced convection heat transfer coefficient, assuming the flow is laminar: \(h = \frac{k_f}{L_{char}}Nu\) Where \(k_f\) is the fluid thermal conductivity, \(L_{char}\) is the characteristic length, and \(Nu\) is the Nusselt number. For air flow at 25°C, the thermal conductivity of air (\(k_f\)) is approximately \(0.0262\,\text{W/(m}\cdot\text{K)}\). From Sieder-Tate correlation for flow parallel to the plates, \(Nu = 0.664Re^{0.5}Pr^{1/3}\) Where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number of air at \(25^{\circ}\mathrm{C}\), approximately equal to \(0.7\).

Calculate the Reynolds number based on free-stream velocity

To calculate the Reynolds number based on free-stream velocity, we need to find the hydraulic diameter, which is defined as: \(D_h = \frac{4A_c}{P}\) Where \(A_c\) is the flow cross-sectional area and \(P\) is the wetted perimeter of the flow channel. Since we have seven fins and the air flows between them, we have six passages each with a height of \(0.5\,\text{cm}\) (as given) and a width of \(6.2\,\text{cm}\), which results in: \(A_c = 0.5\,\text{cm} \times 6.2\,\text{cm} = 3.1\,\text{cm}^2\) \(P = 2\times(0.5\,\text{cm} + 6.2\,\text{cm}) = 13.4\,\text{cm}\) Now we can calculate the hydraulic diameter: \(D_h = \frac{4(3.1\,\text{cm}^2)}{13.4\,\text{cm}} = 0.924\,\text{cm}\) The Reynolds number is defined as: \(Re = \frac{u_\infty D_h}{\nu}\) Where \(u_\infty\) is the free-stream velocity and \(\nu\) is the kinematic viscosity of air at \(25^{\circ}\mathrm{C}\), approximately equal to \(1.568 \times 10^{-5}\,\text{m}^2/\text{s}\).

Calculate the minimum free-stream velocity

To find the minimum free-stream velocity, we use the known heat transfer rate, \(Q = 12\,\text{W}\), and the convective heat transfer relation: \(Q = hA_{total}(T_s - T_\infty)\) Where \(T_s = 60^{\circ}\mathrm{C}\) is the base temperature of the heat sink and \(T_\infty = 25^{\circ}\mathrm{C}\) is the free-stream temperature. Using the heat transfer coefficient, we rewrite the equation as: \(12\,\text{W} = (\frac{k_f}{L_{char}}0.664Re^{0.5}Pr^{1/3})(132\,\text{cm}^2)(60^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C})\) From this equation, we can find \(Re\) and then calculate the minimum free-stream velocity. After solving for \(Re\), we get a value of 292.88. Now, we can find the free-stream velocity: \(u_\infty = \frac{Re \cdot \nu}{D_h} = \frac{292.88 \times 1.568 \times 10^{-5}\,\text{m}^2/\text{s}}{0.924 \times 10^{-2}\,\text{m}} = 0.4956\,\text{m/s}\) Hence, the minimum free-stream velocity needed to avoid overheating the transformer is approximately \(0.4956\,\text{m/s}\).