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Question : A 15- cm × 20- cm PCB has electronic components on one side...

A 15- cm \times 20- cm  PCB has electronic components on one side, dissipating a total of 7 W, as shown in Fig. (15-42). The PCB is mounted in a rack vertically together with other PCBs. If the surface temperature of the components is notto exceed 100^{\circ} C, determine the maximum temperature of the environment in which this PCB can operate safely at sea level. What would your answer be if this rack is located at a location at 4000 m altitude where the atmospheric pressure is 61.66 kPa ?

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SOLUTION The surface temperature of a PCB is not to exceed 100^{\circ} C. The maximum environment temperatures for safe operaton at sea level and at 4000 m altitude are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer is negligible since the PCB is surrounded by other PCBs at about the same temperature. 3 Heat transfer from the back surface of the PCB will be very small and thus negligible.

Analysis The entire heat load of the PCB is dissipated to the ambient air by natural convection from its front surface, which can be treated as a vertical flat plate. Using the simplified relation for a vertical surface from Table (15-1), the natural convection heat transfer coefficient for this PCB can be determined from

h_{\text {conv }}=1.42\left(\frac{\Delta T}{L}\right)^{0.25}=1.42\left(\frac{T_{s}-T_{\text {fuid }}}{L}\right)^{0.25}

The characteristic length in this case is the height (L=0.15 m ) of the PCB, which is the length in the path of heat flow. We cannot determine h_{\text {conv }}  yet, since we do not know the ambient temperature and thus \Delta T. But we can substitute this relation into the heat transfer relation to get

\dot{Q}_{\text {conv }} =h_{\text {conv }} A_{S}\left(T_{s}-T_{\text {fluid }}\right)=1.42\left(\frac{T_{s}-T_{\text {fluid }}}{L}\right)^{0.25} A_{s}\left(T_{s}-T_{\text {fluid }}\right)

 

=1.42 A_{s} \frac{\left(T_{s}-T_{\text {fluid }}\right) .25}{L^{0.25}}

 

The heat transfer surface area of the PCB is

A_{s}=(\text { Width })(\text { Height })=(0.2 m )(0.15 m )=0.03 m ^{2}

 

Substituting this and other known quantities in proper units \left( W\right.  for \dot{Q},{ }^{\circ} C  for T, m ^{2}  for A_{s}, and m for L  into this equation and solving for T_{\text {fluid }}  yields

7=1.42(0.03) \frac{\left(100-T_{\text {fluid }}\right)^{1.25}}{0.15^{0.25}} \longrightarrow T_{\text {fluid }}=59.5^{\circ} C

Therefore, the PCB will operate safely in environments with temperatures up to 59.5^{\circ} C  by relying solely on natural convection.At an altitude of 4000 m, the atmospheric pressure is 61.66 kPa, which is equivalent to

P=(61.66 kPa ) \frac{1 atm }{101.325 kPa }=0.609 atm

The heat transfer coefficient in this case is obtained by multiplying the value at sea level by \sqrt{P}, where P is in atm. Substituting

7=1.42(0.03) \frac{\left(100-T_{\text {fluid }}\right)^{1.25}}{0.15^{0.25}} \sqrt{0.609} \longrightarrow T_{\text {fluid }}=50.6^{\circ} C

which is about 10^{\circ} C  lower than the value obtained at 1 atm pressure. Therefore, the effect of altitude on convection should be considered in high-altitude applications.

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