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Chapter 4: Problem 35

An electronic device, in the form of a disk \(20 \mathrm{~mm}\) in diameter, dissipates \(100 \mathrm{~W}\) when mounted flush on a large aluminum alloy (2024) block whose temperature is maintained at \(27^{\circ} \mathrm{C}\). The mounting arrangement is such that a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-5}\) \(\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\) exists at the interface between the device and the block. (a) Calculate the temperature the device will reach, assuming that all the power generated by the device must be transferred by conduction to the block. (b) To operate the device at a higher power level, a circuit designer proposes to attach a finned heat sink to the top of the device. The pin fins and base material are fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and are exposed to an airstream at \(27^{\circ} \mathrm{C}\) for which the convection coefficient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the device temperature computed in part (a), what is the permissible operating power?

Short Answer

Expert verified
(a) The temperature the device will reach is approximately \(49.82^{\circ} \mathrm{C}\). (b) The permissible operating power with a pin-fin heat sink is approximately \(238.8 \mathrm{~W}\).

Step by step solution

01

Calculate the temperature of the device

First, we need to find the temperature at which the device operates. We will be using Fourier's Law of heat conduction. The formula is: \[q = kA\frac{\Delta T}{R_t}\] Where: - q is the heat transfer rate between the device and the block (\(100 W\)), - k is the thermal conductivity of the aluminum alloy block (\(2024\)), - A is the contact area between the device and the block (to be calculated), - 螖T is the temperature difference between the device and the block (T_device - T_block), - \(R_t\) is the total thermal resistance, which includes the contact resistance \(R_{t, c}^{\prime \prime} = 5 \times 10^{-5} \mathrm{~m}^2 \cdot \mathrm{K} / \mathrm{W}\). First, we need to calculate the contact area A. Since it's a disk with diameter \(20 mm\): \[A = \pi \cdot \left(\frac{d}{2}\right)^2 = \pi \cdot \left(\frac{20 \times 10^{-3}}{2}\right)^2 = 3.14 \times 10^{-4} \mathrm{~m}^2\] Now, we can rearrange the heat transfer equation to solve for the temperature difference 螖T: \[\Delta T = R_t \cdot \frac{q}{kA}\] From the given contact resistance \(R_{t, c}^{\prime \prime}\), we can find the total thermal resistance \(R_t\): \[R_t = \frac{R_{t, c}^{\prime \prime}}{A} = \frac{5 \times 10^{-5}}{3.14 \times 10^{-4}} = 0.159 \mathrm{~K / W}\] Now we can substitute the values into the 螖T equation: \[\Delta T = 0.159 \frac{\mathrm{K}}{\mathrm{W}} \cdot \frac{100 \mathrm{~W}}{2024(\mathrm{W}/\mathrm{mK})(3.14 \times 10^{-4} \mathrm{~m}^2)} = 22.82 \mathrm{~K}\] Finally, we can find the temperature of the device: \[T_{device} = T_{block} + \Delta T = 27^{\circ} \mathrm{C} + 22.82 \mathrm{~K} \approx 49.82^{\circ} \mathrm{C}\] (a) The temperature the device will reach is approximately \(49.82^{\circ} \mathrm{C}\).
02

Calculate the permissible operating power with a pin-fin heat sink

To find the permissible operating power with a heat sink, we will be using Newton's law of cooling: \[q = hA_s(T_{device} - T_{\infty})\] Where: - h is the convection coefficient (\(1000 \mathrm{~W/m}^2 \cdot \mathrm{K}\)), - \(A_s\) is the surface area of the heat sink (can be approximated as base area \(A\)), - \(T_{\infty}\) is the temperature of the airstream (\(27^{\circ} \mathrm{C}\)). We can rearrange the equation to find the heat transfer rate q: \[q = hA_s(T_{device} - T_{\infty})\] Plugging in the values: \[q = (1000 \mathrm{~W/m}^2 \cdot \mathrm{K})(3.14\times10^{-4} \mathrm{~m}^2)(49.82^{\circ} \mathrm{C} - 27^{\circ} \mathrm{C})\] \[q \approx 238.8 \mathrm{~W}\] (b) The permissible operating power with a pin-fin heat sink is approximately \(238.8 \mathrm{~W}\).

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Key Concepts

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

Thermal Resistance
Thermal resistance is the opposition to heat flow through a material. In many ways, it is like electrical resistance, but for heat energy. It is a crucial concept in understanding heat transfer. In our scenario, the electronic device uses a contact resistance at the interface with an aluminum alloy block.
If the material offers higher thermal resistance, it means that it will protect itself from heat more effectively, which can be a problem for devices in need of quick heat dissipation.
The equation for calculating thermal resistance is:
  • It essentially comes from the relation between the contact resistance and the contact area: \[ R_t = \frac{R_{t, c}^{\prime \prime}}{A} \]
  • In this problem, it helps us find how much temperature difference we need to maintain a steady state when managing heat flow.
Understanding thermal resistance helps us optimize our systems to become more efficient in handling heat.
Conduction
Conduction is a way of heat transfer where heat moves through a material without the movement of the material itself. Imagine heating one end of a metal rod: eventually, that heat travels to the other end. This is conduction in action.
When dealing with solids, conduction is often the main mode of heat transfer. Factors influencing conduction include the material's thermal conductivity, surface area, and temperature difference.
In our problem, conduction plays a vital role because the heat dissipated by the electronic device needs to move into an attached aluminum block.
The higher the thermal conductivity of the material, the faster the heat can travel through it. In this case, aluminum is used due to its good thermal conductivity.
Fourier's Law
Fourier's Law describes the relationship between heat transfer through conduction and the physical characteristics of the material through which heat is being transferred. It enables the calculation of heat flow through a material.
The law can be expressed as:
  • \[ q = kA \frac{\Delta T}{R_t} \]
  • Here, \( q \) is the heat transfer rate, \( k \) is the thermal conductivity, \( A \) is the contact area, \( \Delta T \) is the temperature difference, and \( R_t \) is the thermal resistance.
Applying Fourier's Law allows us to calculate the temperature the device will reach by determining how much heat is conducted from the electronic device through the mounting block.
This is particularly useful when trying to maintain specific thermal conditions in electronic equipment to prevent overheating.
Convection Coefficient
The convection coefficient is a key factor in determining how effectively a surface can lose heat to the surrounding air or fluid. It reflects the efficiency of heat transfer between a solid surface and a moving fluid (such as air).
In this scenario, the convection coefficient is used to calculate the heat dissipation with the help of the Newton's law of cooling, especially when additional cooling mechanisms like heat sinks are considered. It's depicted in the formula:
  • \[ q = hA_s(T_{device} - T_{\infty}) \]
  • Here, \( h \) is the convection coefficient, \( A_s \) is the surface area, \( T_{device} \) is the device temperature, and \( T_{\infty} \) is the fluid temperature.
This equation helps us determine the permissible operating power, leading us to understand how much heat can be dissipated naturally by the heat sink to prevent the device from overheating.

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Most popular questions from this chapter

Consider a one-dimensional fin of uniform crosssectional area, insulated at its tip, \(x=L\). (See Table \(3.4\), case B). The temperature at the base of the fin \(T_{b}\) and of the adjoining fluid \(T_{\infty}\), as well as the heat transfer coefficient \(h\) and the thermal conductivity \(k\), are known. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for a node \(n\) located at the insulated tip.

A long bar of rectangular cross section is \(60 \mathrm{~mm} \times\) \(90 \mathrm{~mm}\) on a side and has a thermal conductivity of \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). One surface is exposed to a convection process with air at \(100^{\circ} \mathrm{C}\) and a convection coeffient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the remaining surfaces are maintained at \(50^{\circ} \mathrm{C}\). (a) Using a grid spacing of \(30 \mathrm{~mm}\) and the Gauss-Seidel iteration method, determine the nodal temperatures and the heat rate per unit length normal to the page into the bar from the air. (b) Determine the effect of grid spacing on the temperature field and heat rate. Specifically, consider a grid spacing of \(15 \mathrm{~mm}\). For this grid, explore the effect of changes in \(h\) on the temperature field and the isotherms.

Hot water at \(85^{\circ} \mathrm{C}\) flws through a thin-walled copper tube of \(30-\mathrm{mm}\) diameter. The tube is enclosed by an eccentric cylindrical shell that is maintained at \(35^{\circ} \mathrm{C}\) and has a diameter of \(120 \mathrm{~mm}\). The eccentricity, defined as the separation between the centers of the tube and shell, is \(20 \mathrm{~mm}\). The space between the tube and shell is filled with an insulating material having a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Calculate the heat loss per unit length of the tube, and compare the result with the heat loss for a concentric arrangement.

Consider the nodal point 0 located on the boundary between materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\). Derive the finite-difference equation, assuming no internal generation.

A plate \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is stiffened by a series of longitudinal ribs having a rectangular cross section with length \(L=8 \mathrm{~mm}\) and width \(w=4 \mathrm{~mm}\). The base of the plate is maintained at a uniform temperature \(T_{b}=\) \(45^{\circ} \mathrm{C}\), while the rib surfaces are exposed to air at a temperature of \(T_{\infty}=25^{\circ} \mathrm{C}\) and a convection coeffient of \(h=600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using a finite-difference method with \(\Delta x=\Delta y=\) \(2 \mathrm{~mm}\) and a total of \(5 \times 3\) nodal points and regions, estimate the temperature distribution and the heat rate from the base. Compare these results with those obtained by assuming that heat transfer in the rib is one- dimensional, thereby approximating the behavior of a fin. (b) The grid spacing used in the foregoing finitedifference solution is coarse, resulting in poor precision for estimates of temperatures and the heat rate. Investigate the effect of grid refinement by reducing the nodal spacing to \(\Delta x=\Delta y=1 \mathrm{~mm}\) (a \(9 \times 3\) grid) considering symmetry of the center line. (c) Investigate the nature of two-dimensional conduction in the rib and determine a criterion for which the one-dimensional approximation is reasonable. Do so by extending your finite-difference analysis to determine the heat rate from the base as a function of the length of the rib for the range \(1.5 \leq L w \leq 10\), keeping the length \(L\) constant. Compare your results with those determined by approximating the rib as a fin.
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