22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2016
Question 7 of 8: Surface Temperature of a Power Transistor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, steam tables, vapour and gas power cycles, and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — cylindrical composite-wall conduction, internal-flow and cross-flow convection correlations, natural convection with radiation, and the ε–NTU heat-exchanger method. Freon-12 property data are taken from the appendix supplied with the exam; steam, air and engine-oil data from standard tables.
Paper format: National Examination 07-Mec-A1, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three from one part, two from the other). All eight questions are solved in full below.
Question 7: Surface Temperature of a Power Transistor (equal value)
Figure 7 — Energy balance on the small horizontal cylindrical can: the 0.18 W is removed by natural convection to the 25 °C air plus (weak, $\varepsilon=0.1$) radiation to the 35 °C enclosure walls, over the lateral plus one end face (base excluded).
Given. A small cylindrical power transistor (0.4 cm dia × 0.45 cm) dissipating 0.18 W, in an enclosure with 35 °C walls and 25 °C air, surface emissivity 0.1, base insulated. Find. The transistor surface temperature.
Approach. Set the dissipated power equal to natural convection plus radiation from the exposed surface, and iterate on the surface temperature (the convection coefficient itself depends on $T_s$).
Exposed area (base excluded). $A=\pi DL+\tfrac{\pi}{4}D^2=\pi(0.004)(0.0045)+\tfrac{\pi}{4}(0.004)^2=6.91\times10^{-5}\ \text{m}^2.$
Energy balance. $$Q=hA(T_s-T_\text{air})+\varepsilon\sigma A(T_s^4-T_\text{wall}^4)=0.18\ \text{W}.$$
Natural-convection coefficient. Treating the can as a small horizontal cylinder at a film temperature ~100 °C, $\text{Ra}_D\approx3.3\times10^{2}$ and the Churchill–Chu correlation gives $\text{Nu}_D\approx2.0$, so $h\approx16\ \text{W/m}^2\text{K}.$
Iterate for the surface temperature. Converging the balance gives $$T_s\approx\boxed{173\,{}^\circ\text{C}}\ (446\ \text{K}),$$ with convection removing 0.168 W and radiation only 0.012 W.
Interpretation. With so little area and a low emissivity, the tiny can must run very hot to shed even 0.18 W — radiation contributes under 7 %, so the transistor relies almost entirely on natural convection.
Check — small-body correlation.
Air properties are evaluated at the film temperature and updated each iteration. The horizontal-cylinder Churchill–Chu correlation is applied to the whole exposed area; for such a small object the Nusselt number sits near its low-Rayleigh floor, so $h$ (and hence $T_s$) carries a modest correlation uncertainty of order ±10 °C, but the result — a transistor running well above 150 °C — is firm.
