25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2016
Question 7 of 8: Power-Transistor Surface Temperature
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2016 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).
Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.
It is solved as the thermodynamics/heat-transfer exam it actually is.
Given. A small cylindrical transistor case dissipates heat by natural convection and low-emissivity radiation from its exposed (non-base) surface only.
Given data
Quantity
Symbol
Value
Power dissipated
$Q$
0.18 W
Case diameter, length
$D,\,L$
0.4 cm, 0.45 cm
Surrounding air temperature
$T_\infty$
25°C
Surface emissivity
$\varepsilon$
0.1
Find. Steady surface temperature $T_s$ (base excluded; enclosure temperature of 35°C noted but not needed once the base is excluded from the energy balance). (Check: surroundings for radiation exchange taken at the 25°C air temperature, since no separate radiant-surroundings temperature is given.)
Power-transistor case modelled as a small horizontal cylinder (base against the wall excluded); exposed area = lateral surface + one end cap.
Approach. Model the case as a small horizontal cylinder losing heat by natural convection (Churchill–Chu) and low-emissivity radiation from its lateral surface plus one end cap; solve the combined energy balance for $T_s$ iteratively (air properties depend on the unknown film temperature).
Exposed area (base excluded).
$$A=\pi DL+\frac{\pi D^2}{4}=\pi(0.004)(0.0045)+\frac{\pi(0.004)^2}{4}\approx6.91\times10^{-5}\text{ m}^2$$
Energy balance. Natural convection (Churchill–Chu, horizontal cylinder, $D$ as characteristic length) plus radiation to $25^\circ\text{C}$ surroundings:
$$Q=hA(T_s-T_\infty)+\varepsilon\sigma A\!\left(T_s^4-T_\infty^4\right)$$
where $h$ depends on the Rayleigh number $Ra_D$ evaluated at the (unknown) film temperature — solved iteratively.
Iterate to convergence. Because $A$ is tiny (about 0.69 cm²), even a modest $h$ (of order 15–17 W/m²K at these small $Ra_D$) requires a large $\Delta T$ to pass 0.18 W through such a small area; the radiation term stays a minor contributor at $\varepsilon=0.1$. Converging the energy balance gives
$$\boxed{T_s\approx172.6^\circ\text{C}}$$
