Combined Free Convection and Radiation from a Power Transistor: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2016, Q7

Question 7: Combined Free Convection and Radiation from a Power Transistor

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Power dissipation 0.18 W; ambient air $25\,{}^{\circ}\text{C}$; surrounding enclosure (radiation sink) $35\,{}^{\circ}\text{C}$; surface emissivity 0.1; case 0.4 cm × 0.4 cm × 0.45 cm, mounted base-in (base excluded from heat transfer).

[Figure not reproduced: Power transistor case dimensions. See the official exam paper or the cited reference text.]

Fig. 6 — power transistor case as printed: 0.4 cm square face, 0.45 cm deep.

Given data
Quantity	Symbol	Value
Power dissipated	$\dot{Q}$	0.18 W
Ambient air temperature	$T_{\text{air}}$	$25\,{}^{\circ}\text{C}$
Enclosure (radiation) temperature	$T_{\text{surr}}$	$35\,{}^{\circ}\text{C}$
Surface emissivity	$\varepsilon$	0.1
Case dimensions	$H\times W\times D$	0.4 × 0.4 × 0.45 cm

Find. The steady surface temperature $T_s$ of the transistor case.

Approach. Sum natural-convection and radiation losses from the five exposed faces (base excluded), set equal to the dissipated power, and solve implicitly for $T_s$ — convection exchanges with the AIR temperature while radiation exchanges with the (different) ENCLOSURE temperature, since they are physically distinct surroundings.

Exposed area (base excluded). One far face ($H\times W$) plus four side faces ($H\times D$ and $W\times D$, two each): $$A_s=(0.004)(0.004)+2(0.004)(0.0045)+2(0.004)(0.0045)=8.80\times10^{-5}\text{ m}^2$$
Natural-convection coefficient (Cengel's simplified free-convection relation for a small vertical surface in air, characteristic length = case height $L_c=H=0.004$ m): $$h=1.42\left(\frac{\Delta T}{L_c}\right)^{1/4}\quad[\text{W/m}^2\text{K, }\Delta T\text{ in K, }L_c\text{ in m}]$$
Energy balance (implicit in $T_s$). $$\dot{Q}=h(T_s)\,A_s\,(T_s-T_{\text{air}}) + \varepsilon\sigma A_s\!\left(T_s^4-T_{\text{surr}}^4\right)$$ Solved numerically (the natural-convection $h$ itself depends on $T_s$ through $\Delta T$): $$\boxed{T_s\approx 132.0\,{}^{\circ}\text{C}},\qquad h\approx18.2\text{ W/m}^2\text{K}$$
Check the energy split. At the solution, convection carries $h A_s(T_s-T_{\text{air}})\approx0.166$ W and radiation carries the remaining $\approx0.014$ W — consistent with the low emissivity (0.1) making radiation a minor contributor for this surface.

A tiny (88 mm²) exposed area, a low emissivity, and no forced airflow together make this a poor passive heat-rejection path, so a $107\,{}^{\circ}\text{C}$ rise above ambient air for only 0.18 W is physically reasonable — the same reason small unheat-sinked transistor packages are commonly derated or fitted with a small clip-on heat sink in practice.

Question 7 — results
Quantity	Value
Exposed surface area, $A_s$	88.0 mm²
Natural-convection coefficient, $h$	18.2 W/m²K
Surface temperature, $T_s$	$132.0\,{}^{\circ}\text{C}$

Topic. Combined natural convection and surface radiation from a small electronic-component case with two DIFFERENT surrounding temperatures.

Key relations. Simplified free-convection relation for a small vertical surface in air, $h=1.42(\Delta T/L_c)^{1/4}$; net radiation to large isothermal surroundings $\dot{Q}_{\text{rad}}=\varepsilon\sigma A_s(T_s^4-T_{\text{surr}}^4)$; steady-state energy balance $\dot{Q}=\dot{Q}_{\text{conv}}+\dot{Q}_{\text{rad}}$.

Why this works. The problem deliberately supplies two DIFFERENT temperatures (air 25°C, enclosure 35°C) because convection and radiation genuinely exchange heat with two different reservoirs in a real enclosure — the surrounding air for convection, and the solid enclosure walls (which need not be at the air temperature) for radiation view-factor exchange; folding both onto a single "ambient" temperature is a common oversimplification.

Common pitfall. Including the wall-mounted "base" face in the exposed area directly contradicts the question's own "disregarding any heat transfer from the base" instruction and understates $T_s$; conversely, using the cylindrical-package Churchill–Chu correlation here would be geometrically wrong once the illustration shows a rectangular, not cylindrical, case.

Source. Çengel & Ghajar, Heat and Mass Transfer (simplified natural-convection relations for surfaces in air); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 9 (free convection), Ch. 13 (radiation exchange with surroundings).

Question 7 of 8: Combined Free Convection and Radiation from a Power Transistor

Question 7: Combined Free Convection and Radiation from a Power Transistor