A 0.2-W small cylindrical resistor mounted on a PCB is 1 cm long and has a diameter of 0.3 cm, as shown in Fig. 15-41. The view of the resistor is largely blocked by the PCB facing it, and the heat transfer from the connecting wires is negligible. The air is free to flow through the parallel flow passages between the PCBs. If the air temperature at the vicinity of the resistor is 50^{\circ} C, determine the surface temperature of the resistor.
SOLUTION A small cylindrical resistor mounted on a PCB is being cooled by natural convection and radiation. The surface temperature of the resistor is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The device is located at sea level so that the local atmospheric pressure is 1 atm. 3 Radiation is negligible in this case since the resistor is surrounded by surfaces that are at about the same temperature, and the net radiation heat transfer between two surfaces at the same temperature is zero. This leaves natural convection as the only mechanism of heat transfer from the resistor.
Analysis Using the relation for components on a circuit board from Table (15-1), the natural convection heat transfer coefficient for this cylindrical component can be determined from
h_{\text {conv }}=2.44\left(\frac{T_{s}-T_{\text {fluid }}}{D}\right)^{0.25}where the diameter D=0.003 m, which is the length in the heat flow path, is the characteristic length. We cannot determine h_{\text {conv }} yet since we do not know the surface temperature of the component and thus \Delta T. But we can substitute this relation into the heat transfer relation to get
Q_{\text {conv }} =h_{\text {conv }} A_{s}\left(T_{s}-T_{\text {fluid }}\right)=2.44\left(\frac{T_{s}-T_{\text {fuid }}}{D}\right)^{0.25} A_{s}\left(T_{s}-T_{\text {fuid }}\right)=2.44 A_{s}\left(\frac{T_{s}-T_{\text {fluid }}}{D}\right)^{1.25}
The heat transfer surface area of the component is
A_{s}=2 \times \frac{1}{4} \pi D^{2}+\pi D L=2 \times \frac{1}{4} \pi(0.3 cm )^{2}+\pi(0.3 cm )(1 cm )=1.084 cm ^{2}Substituting this and other known quantities in proper units (W for \dot{Q},{ }^{\circ} C for T, m ^{2} for A, and m for D ) into this equation and solving for T_{s} yields
0.2=2.44\left(1.084 \times 10^{-4}\right) \frac{\left(T_{s}-50\right)^{1.25}}{0.003^{0.25}}-\rightarrow T_{s}=113^{\circ} CTherefore, the surface temperature of the resistor on the PCB will be 113^{\circ} C, which is considered to be a safe operating temperature for the resistors. Note that blowing air to the circuit board will lower this temperature considerably as a result of increasing the convection heat transfer coefficient and decreasing the air temperature at the vicinity of the components due to the larger flow rate of air.
