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A \(12-\mathrm{cm} \times 18-\mathrm{cm}\) circuit board houses on its surface 100 closely spaced logic chips, each dissipating \(0.06 \mathrm{~W}\) in an environment at \(40^{\circ} \mathrm{C}\). The heat transfer from the back surface of the board is negligible. If the heat transfer coefficient on the surface of the board is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the heat flux on the surface of the circuit board, in \(\mathrm{W} / \mathrm{m}^{2} ;(b)\) the surface temperature of the chips; and (c) the thermal resistance between the surface of the circuit board and the cooling medium, in \({ }^{\circ} \mathrm{C} / \mathrm{W}\).

Short Answer

Expert verified
Answer: (a) The heat flux on the surface of the circuit board is approximately 2777.78 W/m². (b) The surface temperature of the chips is approximately 47.78°C. (c) The thermal resistance between the surface of the circuit board and the cooling medium is approximately 1.3°C/W.

Step by step solution

01

Calculate the total heat generated by the chips

We're given that each chip dissipates 0.06 W of power and there are 100 chips on the circuit board. We can find the total heat generated by all 100 chips by multiplying their individual power dissipation by the number of chips: Total heat generated (Q) = Number of chips × Power dissipation per chip Q = 100 × 0.06 W
02

Calculate the area of the circuit board

We're given the dimensions of the circuit board, which are 12 cm × 18 cm. To find the area (A), we need to multiply the two dimensions and convert the result to square meters: A = (12 cm × 18 cm) * (1 m/100 cm)^2
03

Calculate the heat flux on the surface of the circuit board

Now that we have the total heat generated and the area of the circuit board, we can find the heat flux (q") by dividing the total heat generated by the area: q" = Q / A
04

Calculate the surface temperature of the chips

We're given the heat transfer coefficient (h) and the environmental temperature (T_inf). We can use the heat convection formula to find the surface temperature (T_s) of the chips: Q = h × A × (T_s - T_inf) T_s = (Q / (h × A)) + T_inf
05

Calculate the thermal resistance between the surface of the circuit board and the cooling medium

Now that we have the surface temperature, we can find the thermal resistance (R_th) using the formula: R_th = (T_s - T_inf) / Q Calculating the values: Step 1: Q = 100 × 0.06 W = 6 W Step 2: A = (12 cm × 18 cm) * (1 m/100 cm)^2 = 0.00216 m² Step 3: q" = 6 W / 0.00216 m² ≈ 2777.78 W/m² Step 4: T_s = (6 W / (10 W/m²·K × 0.00216 m²)) + 40°C ≈ 47.78°C Step 5: R_th = (47.78°C - 40°C) / 6 W ≈ 1.3°C/W The results are as follows: (a) The heat flux on the surface of the circuit board is approximately 2777.78 W/m². (b) The surface temperature of the chips is approximately 47.78°C. (c) The thermal resistance between the surface of the circuit board and the cooling medium is approximately 1.3°C/W.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermal Resistance in Circuit Boards
Imagine you're wearing layers on a cold day. Just as clothing resists the flow of heat from your body to the cold air, in electronics, thermal resistance is the measure of how much a material hinders the flow of heat. Specifically, for circuit boards, thermal resistance is critical as it affects how well the board can dissipate heat generated by its components.

Thinking about the components of a circuit board as little heaters, it's easy to see that without adequate heat dissipation, these 'heaters' could overheat, leading to potential damage or failure. This is where thermal resistance comes in. It's a value that tells us how hard it is for heat to pass through materials – the lower the thermal resistance, the better the heat flow. In the context of the exercise, the chips on the board generate heat, and managing this heat is instrumental in maintaining performance and reliability.

In our example, we calculated the thermal resistance to understand how effectively the circuit board could transfer heat to the surrounding environment. Remember, the formula we used was:
\[ R_{\text{th}} = \frac{(T_{\text{s}} - T_{\text{inf}})}{Q} \]
Where \(R_{\text{th}}\) is the thermal resistance, \(T_{\text{s}}\) is the surface temperature of the chips, \(T_{\text{inf}}\) is the ambient temperature, and \(Q\) is the total heat generated by the chips. In practical applications, designers aim to minimize thermal resistance to keep electronic components cool.
Heat Flux Calculation in Detail
Tapping into our high school physics, heat flux might remind us of terms like current or flow, and it’s not too different in concept. Heat flux is much like a river of heat, representing how much heat flows through a certain area every second. It’s calculated by dividing the total heat transfer by the area through which the heat is transmitted.

In the exercise, we used the formula for heat flux (q''), shown below:
\[ q'' = \frac{Q}{A} \]
Where \(Q\) is the total heat generated and \(A\) is the area of the circuit board. By understanding heat flux, we can gauge whether a circuit board can cope with the heat its components produce. The units are typically Watt per square meter (\(W/m^2\)).

In practical terms, think of heat flux as the intensity of a spotlight on a stage; the smaller the area it covers, the more intense the light is in that spot. Similarly, the smaller the area our heat has to flow through, the higher the heat flux and thus the greater need for efficient cooling methods to avoid damaging our 'stage,' or in this case, our circuit board and its chips.
Grasping Convection Heat Transfer on Circuit Boards
The concept of convection heat transfer explains how heat circulates within fluids (such as air or liquids) and between fluids and solids. It's the reason a spoon in a hot cup of tea becomes warm. In electronic systems, managing this kind of heat transfer is crucial because it directly affects component temperatures and system performance.

In our scenario, the heat transfer coefficient (h) measures how well heat is transferred from the chip's surface to the surrounding air. A higher h means better heat transfer, reducing the chips' surface temperature. We use the following formula to find the surface temperature (Ts):
\[ T_{\text{s}} = \frac{Q}{(h \times A)} + T_{\text{inf}} \]
Where \(T_{\text{inf}}\) is the ambient temperature, and \(Q\) and \(A\) are as previously defined. This formula is the cornerstone of convection heat transfer in circuit board design.

Understanding these convection processes aids in designing efficient cooling strategies such as heat sinks or cooling fans, ensuring that our electronic components maintain their cool, much like ice in your favorite summer beverage prevents it from becoming a lukewarm disappointment.

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Most popular questions from this chapter

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known.

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

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