Question

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

Short answer

Answer: Without the cooling system, the temperatures on the sides of the circuit board are 39.389°C and 39.222°C. With the aluminum plate and fins cooling system, the temperatures are 134.720°C and 93.501°C.

Step-by-step solution

Calculate temperature without cooling system

At first, we will determine the temperatures on both sides without the cooling system. We know the power dissipated as heat from the components is 15W and the area of the board is 10cm * 15cm. To find the temperature on the opposite side of the board, we will apply Fourier's Law of Conduction: \(q = -kA\frac{\Delta T}{\Delta x}\) Where \(q\) is the heat transfer, \(k\) is the thermal conductivity, \(A\) is the area, \(\Delta T\) is the temperature difference, and \(\Delta x\) is the thickness. We can rewrite the equation like this: \(\Delta T = \frac{q\Delta x}{kA}\). Now let's plug in the values and calculate the temperature difference: \(\Delta T = \frac{15W * (0.002m)}{12 \frac{W}{m\cdot K} * (0.1m * 0.15m)} = \frac{0.03}{0.18}K = 0.1667K\) Now we will apply Newton's Law of Cooling to find the temperature difference between the board's opposite side (where heat transfer occurs) and the medium: \(q = hA(\Delta T_{conv})\). Rewrite the equation to find the \(\Delta T_{conv}\): \(\Delta T_{conv} = \frac{q}{hA}\) Plugging in the values: \(\Delta T_{conv} = \frac{15W}{45\frac{W}{m^2 \cdot K} * (0.1m * 0.15m)} = \frac{15}{6.75}K = 2.2222K\) Now we have both temperature differences: \(\Delta T_{cond} = 0.1667K\) and \(\Delta T_{conv} = 2.2222K\). The temperature on the side with the components is: \(T_{comp} = 37^{\circ}C + \Delta T_{conv} + \Delta T_{cond} = 37 + 2.2222 + 0.1667 = 39.389 ^{\circ}C\) The temperature on the opposite side is: \(T_{opp} = 37^{\circ}C + \Delta T_{conv} = 37 + 2.2222 = 39.222 ^{\circ}C\)

Calculate temperature with cooling system

Now we have to calculate the new temperatures on both sides, including the aluminum plate and fins cooling system. To do that, we need to find the effective thermal conductivity of the combined layers: \(K_{eff} = \frac{\Delta x_1}{k_1} + \frac{\Delta x_2}{k_2} + \frac{\Delta x_3}{k_3} + \frac{\Delta x_4}{k_4}\) Where \(k_1\) is the conductivity of the circuit board (12 W/mK), \(\Delta x_1\) is its thickness (0.002 m), \(k_2\) is the epoxy (1.8 W/mK), \(\Delta x_2\) is its thickness (0.0003 m), \(k_3\) is the aluminum (237 W/mK), and \(\Delta x_3\) (0.001 m). Plugging in the values: \(K_{eff} = \frac{0.002}{12} + \frac{0.0003}{1.8} + \frac{0.001}{237} = 0.0001667 + 0.0001667 + 0.0000042 = 0.0003376\) Now we have an effective conductivity for the whole multi-layered board. We can now use the same temperature difference equation as in step 1: \(\Delta T = \frac{q\Delta x_{eff}}{k_{eff}A}\) \(\Delta T = \frac{15W * (0.002m + 0.0003m + 0.001m)}{(0.0003376 \frac{W}{m\cdot K}) * (0.1m * 0.15m)} = \frac{0.0495}{0.0005064}K = 97.7204K\) The temperature on the component side is: \(T_{comp} = 37^{\circ}C + \Delta T = 37 + 97.7204 = 134.720 ^{\circ}C\) The temperature on the opposite side is: \(T_{opp} = 37^{\circ}C + \Delta T - \Delta x_3\frac{q}{k_3A} = 37 + 97.7204 - \frac{(0.001m) * 15W}{(237\frac{W}{m\cdot K}) * (0.1m * 0.15m)} = 37 + 97.7204 - 4.2194 = 93.501 ^{\circ}C\) The new temperatures on the sides of the circuit board are \(T_{comp} = 134.720 ^{\circ}C\) and \(T_{opp} = 93.501 ^{\circ}C\).

Key concepts

Fourier's Law of Conduction

Understanding Fourier's Law of Conduction is essential for anyone interested in heat transfer in electronics. This law provides us with a mathematical explanation of how heat flows through materials. It's expressed with the formula:
\[q = -kA\frac{\Delta T}{\Delta x}\]
Here, \(q\) stands for the heat transfer rate, \(k\) is the material's thermal conductivity, \(A\) refers to the cross-sectional area through which heat is flowing, \(\Delta T\) is the temperature difference driving the heat transfer, and \(\Delta x\) represents the thickness of the material. The negative sign in the equation indicates that heat flows from a high-temperature region to a low-temperature one.
In the context of the exercise, application of Fourier's Law helps determine the temperature difference across the circuit board due to the electronic components dissipating heat. For students tackling similar problems, it's imperative to understand how to apply this law to calculate the internal heat flow within solid materials.

Newton’s Law of Cooling

In the realm of electronics cooling, Newton's Law of Cooling plays a pivotal role in describing how heat is expelled from surfaces to their surroundings. The law takes the form:
\[q = hA(\Delta T_{conv})\]
where \(q\) is still the heat transfer rate, \(h\) represents the heat transfer coefficient, \(A\) the area where convection is occurring, and \(\Delta T_{conv}\) the temperature difference between the surface and the surrounding fluid. This law is particularly useful when analyzing how effective a cooling system is in removing heat from a device.
Looking at the given exercise, we use Newton’s Law of Cooling to evaluate the heat rejected from the circuit board into the air. Getting comfortable with this concept allows students to handle a variety of thermal management questions, including those involving both passive and active electronic cooling strategies.

Thermal Conductivity

Thermal conductivity, symbolized as \(k\), is a fundamental property of materials that quantifies how well they conduct heat. It's measured in watts per meter-kelvin \(\left(\frac{W}{m\cdot K}\right)\). A high thermal conductivity indicates that the material is a good conductor of heat — metals like copper and aluminum are prime examples, often used in heat sinks due to their high thermal conductivity.
In our exercise, we are provided with the thermal conductivity of the circuit board's material. By understanding the role of thermal conductivity, students can predict how changes in material composition can influence the effectiveness of thermal management in electronic systems. Remember, the effectiveness of materials in conducting heat is vital for designing electronics that are reliable and prevent overheating.

Circuit Board Cooling

Circuit board cooling is a crucial aspect of electronic design. Without it, components may overheat, leading to failure or reduced lifespan. There are various methods of cooling, including passive heat sinks, active cooling systems like fans, or even more advanced techniques such as liquid cooling. Materials with high thermal conductivity, such as the metal fillings in the circuit board mentioned in the textbook problem, enhance conduction of heat away from components. But that's just half the battle. The heat must then be efficiently dissipated into the environment, which is where concepts like Fourier's Law of Conduction and Newton’s Law of Cooling come into play.
Our textbook exercise example addresses the significance of such a cooling system by adding an aluminum plate with fins to the board. It shows how this method decreases the surface temperature, benefiting the overall thermal management of the electronics. Students intending to solve problems related to electronic cooling should comprehend the mechanisms and calculations related to heat transfer and the efficiency of different cooling techniques.