/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Approximately \(10^{6}\) discret... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 27

Approximately \(10^{6}\) discrete electrical components can be placed on a single integrated circuit (chip), with electrical heat dissipation as high as \(30,000 \mathrm{~W} / \mathrm{m}^{2}\). The chip, which is very thin, is exposed to a dielectric liquid at its outer surface, with \(h_{o}=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, 0}=20^{\circ} \mathrm{C}\), and is joined to a circuit board at its inner surface. The thermal contact resistance between the chip and the board is \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the board thickness and thermal conductivity are \(L_{b}=5 \mathrm{~mm}\) and \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The other surface of the board is exposed to ambient air for which \(h_{i}=40\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, i}=20^{\circ} \mathrm{C}\). (a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. In variable form, label appropriate resistances, temperatures, and heat fluxes. (b) Under steady-state conditions for which the chip heat dissipation is \(q_{c}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), what is the chip temperature? (c) The maximum allowable heat flux, \(q_{c, m}^{\prime \prime}\), is determined by the constraint that the chip temperature must not exceed \(85^{\circ} \mathrm{C}\). Determine \(q_{c, m}^{\prime \prime}\) for the foregoing conditions. If air is used in lieu of the dielectric liquid, the convection coefficient is reduced by approximately an order of magnitude. What is the value of \(q_{c, m}^{\prime \prime}\) for \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? With air cooling, can significant improvements be realized by using an aluminum oxide circuit board and/or by using a conductive paste at the chip/board interface for which \(R_{t, c}^{n}=10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) ?

Short Answer

Expert verified
The chip temperature under steady-state conditions, with a heat dissipation of \(30,000 \frac{W}{m^2}\), is \(953^{\circ} C\). The maximum allowable heat flux, \(q_{c, m}''\), ensuring the chip temperature does not exceed \(85^{\circ} C\) is \(2089.71 \frac{W}{m^2}\). To evaluate alternative cooling methods with air, aluminum oxide circuit boards, or conductive paste at the chip/board interface, adjust the corresponding thermal resistances and recalculate the maximum allowable heat flux to determine if significant improvements can be achieved.

Step by step solution

01

(a) Sketch the equivalent thermal circuit

Here, we are asked to sketch the equivalent thermal circuit. The circuit should include the following thermal resistances: 1. convection between the dielectric liquid and the chip, labeled as \(R_{conv, o}\) 2. contact resistance between the chip and the board, labeled as \(R_{t,c}\) 3. conduction through the board, labeled as \(R_{cond}\) 4. convection between the board and the ambient air, labeled as \(R_{conv, i}\) The heat flux across the entire circuit (\(q\)) is the same across each resistance. The temperature nodes we need to consider are \(T_{\infty, 0}\) (temperature of the dielectric liquid), \(T_{c}\) (chip temperature), \(T_{c, b}\) (chip/board interface temperature), \(T_{b, i}\) (board/air interface temperature) and \(T_{\infty, i}\) (temperature of the ambient air).
02

(b) Calculation of chip temperature

Now, we are asked to calculate the chip temperature (\(T_c\)) for a given chip heat dissipation (\(q_c'' = 30,000 \frac{W}{m^{2}}\)). For this, we need to find the total equivalent thermal resistance from the dielectric liquid to the ambient air. The total thermal resistance is: \[R_{total} = R_{conv,o} + R_{t, c} + R_{cond} + R_{conv, i}\] We have: 1. \(R_{conv,o} = \frac{1}{h_o}= \frac{1}{1000} \frac{m^{2}K}{W}\) 2. \(R_{t, c} = 10^{-4} \frac{m^{2}K}{W}\) 3. \(R_{cond} = \frac{L_{b}}{k_{b}} = \frac{0.005}{1} \frac{m^{2}K}{W}\) 4. \(R_{conv, i} = \frac{1}{h_i} = \frac{1}{40} \frac{m^{2}K}{W}\) Now calculate the total resistance: \[R_{total} = 0.001 + 0.0001 + 0.005 + 0.025 = 0.0311 \frac{m^{2}K}{W}\] Next, we can find the temperature difference between the dielectric liquid and the ambient air: \[\Delta T = T_{\infty,0} - T_{\infty,i} = 20 - 20 = 0 \] The heat flux across the thermal circuit: \[q = q_{c}'' = 30,000 \frac{W}{m^{2}}\] Now, you can use the total thermal resistance to find the difference between the chip's temperature and the dielectric liquid's temperature: \[\Delta T_{T_c} = q \cdot R_{total} = 30,000 \cdot 0.0311 = 933K \] Finally, we can find the chip temperature: \[T_c = T_{\infty,0} + \Delta T_{T_c} = 20 + 933 = 953^{\circ} C\]
03

(c) Calculate the maximum allowable heat flux

The maximum allowable heat flux (\(q_{c, m}''\)) is determined by the constraint that the chip temperature must not exceed \(85^{\circ} C\). Let's calculate the maximum heat flux for the given conditions. From the previous part, we obtained the total thermal resistance: \[R_{total} = 0.0311 \frac{m^{2}K}{W}\] Now, we can find the maximum temperature difference for the constraint: \[\Delta T_{T_c, max} = T_{c, max} - T_{\infty,0} = 85 - 20 = 65 K\] We can now calculate the maximum allowable heat flux: \[q_{c, m}'' = \frac{\Delta T_{T_c, max}}{R_{total}} = \frac{65}{0.0311} = 2089.71 \frac{W}{m^2}\] For an alternative cooling method using air, the convection coefficient will be \(h_o = 100 \frac{W}{m^2} K\). To calculate the new maximum allowable heat flux with the different cooling methods, replace the value of \(h_o\) in the total thermal resistance calculation and calculate the maximum allowable heat flux as previously done. As for the other alternative cooling solutions like using an aluminum oxide circuit board and/or using a conductive paste at the chip/board interface, you can repeat the previous calculation after adjusting the thermal resistances and find out the maximum allowable heat flux in each case. By comparing the heat flux values, we can determine if significant improvements can be realized in the cooling process.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer is a process where heat is transferred between a solid and a fluid. In the context of chip cooling, it occurs at the interface where the microchip comes in contact with a cooling fluid, such as a dielectric liquid or ambient air. The effectiveness of this process is largely determined by the convection heat transfer coefficient, denoted as \( h \). This coefficient represents how efficiently heat can be transferred from the chip to the fluid. A high value of \( h \) indicates that heat is efficiently removed, which is crucial for maintaining the chip at an optimal temperature.
  • In the given problem, the outer surface of the chip is exposed to a dielectric liquid with \( h_o = 1000 \, \frac{W}{m^2 \cdot K} \), which signifies a high heat transfer rate.
  • When the outer surface is cooled by air instead, the coefficient decreases to \( h_o = 100 \, \frac{W}{m^2 \cdot K} \), indicating a less efficient heat transfer process.
This difference in \( h \) values shows why convection heat transfer is a vital consideration in thermal circuit design, especially for high-performance electronics where cooling is paramount.
Thermal Resistance
Thermal resistance is akin to electrical resistance but in the context of heat flow. It measures how easily or difficult it is for heat to pass through a material. Every segment of a path that heat follows from a hot to a cold region can be seen as having a thermal resistance.
  • The overall resistance of a thermal circuit is the sum of individual resistances, including conduction through layers like circuit boards and convection from surfaces to fluids.
  • For example, in the exercise, we have distinct resistances for different processes: \( R_{conv,o} \) for convection at the outer chip surface, \( R_{t,c} \) for the thermal contact resistance, \( R_{cond} \) for conduction through the board, and \( R_{conv,i} \) for convection at the board's outer surface.
Understanding these resistances helps in determining how heat will traverse from the inside of a chip to its cooling medium, affecting the chip鈥檚 operating temperature and efficiency.
Chip Cooling
Chip cooling refers to methods and processes used to remove excess heat from electronic components to maintain functionality and avoid overheating. Efficient chip cooling is essential as modern chips can generate significant amounts of heat due to a high density of electronic components.
Effective cooling strategies often involve:
  • Enhancing convection processes by choosing fluids with high heat transfer coefficients
  • Minimizing thermal contact resistance through materials or interfaces that allow better heat flow from chip to board
  • Utilizing materials with high thermal conductivity for circuit boards to facilitate heat dissipation
In the given scenario, switching from a dielectric fluid to air as a cooling medium can significantly alter the heat removal efficiency due to changes in the convection heat transfer coefficient. Additionally, employing materials with improved conductive properties can aid in keeping the chip temperature within permissible limits, thus ensuring the maximum allowable heat flux remains manageable.
Steady-State Heat Transfer
Steady-state heat transfer occurs when the temperature distribution in a medium no longer changes with time. This is a common assumption in evaluating the thermal performance of electronic devices, like chips and circuit boards, reducing the complexity of the analysis.
  • In steady-state conditions, all temperatures, including that of the chip, the surrounding medium, and the interfaces, remain constant over time.
  • This assumption allows engineers to calculate the expected operating temperature of a chip for a given set of thermal resistances and power dissipation rates.
In the exercise, the chip's temperature was calculated assuming steady-state conditions, using the heat dissipation rate \( q'' = 30,000 \, W/m^2 \) and previously determined thermal resistances. The challenge is then to ensure that under these conditions, the temperature remains below critical limits like the maximum allowable 85掳C. When cooling solutions are modified, such as switching from liquid to air cooling, the impact on steady-state temperatures can be analyzed to ensure continuous safe operation.

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Most popular questions from this chapter

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.

A scheme for concurrently heating separate water and air streams involves passing them through and over an array of tubes, respectively, while the tube wall is heated electrically. To enhance gas-side heat transfer, annular fins of rectangular profile are attached to the outer tube surface. Attachment is facilitated with a dielectric adhesive that electrically isolates the fins from the current-carrying tube wall. (a) Assuming uniform volumetric heat generation within the tube wall, obtain expressions for the heat rate per unit tube length \((\mathrm{W} / \mathrm{m})\) at the inner \(\left(r_{i}\right)\) and outer \(\left(r_{o}\right)\) surfaces of the wall. Express your results in terms of the tube inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, e}\), and other pertinent parameters. (b) Obtain expressions that could be used to determine \(T_{s, i}\) and \(T_{s, o}\) in terms of parameters associated with the water- and air-side conditions. (c) Consider conditions for which the water and air are at \(T_{\infty, i}=T_{\infty, o}=300 \mathrm{~K}\), with corresponding convection coefficients of \(h_{i}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Heat is uniformly dissipated in a stainless steel tube \(\left(k_{w}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), having inner and outer radii of \(r_{i}=25 \mathrm{~mm}\) and \(r_{o}=30\) \(\mathrm{mm}\), and aluminum fins \(\left(t=\delta=2 \mathrm{~mm}, r_{t}=55\right.\) \(\mathrm{mm}\) ) are attached to the outer surface, with \(R_{t, c}^{\prime \prime}=\) \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Determine the heat rates and temperatures at the inner and outer surfaces as a function of the rate of volumetric heating \(\dot{q}\). The upper limit to \(\dot{q}\) will be determined by the constraints that \(T_{s, i}\) not exceed the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\) and \(T_{s, o}\) not exceed the decomposition temperature of the adhesive \(\left(250^{\circ} \mathrm{C}\right)\).

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of \(\dot{q}=\) \(2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The rod is concentric with a hollow ceramic cylinder, creating an enclosure that is filled with air. The thermal resistance per unit length due to radiation between the enclosure surfaces is \(R_{\mathrm{rad}}^{\prime}=0.30 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\), and the coefficient associated with free convection in the enclosure is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Construct a thermal circuit that can be used to calculate the surface temperature of the rod, \(T_{r}\). Label all temperatures, heat rates, and thermal resistances, and evaluate each thermal resistance. (b) Calculate the surface temperature of the rod for the prescribed conditions.

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.
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