/*! 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 113 A novel scheme for dissipating h... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 113

A novel scheme for dissipating heat from the chips of a multichip array involves machining coolant channels in the ceramic substrate to which the chips are attached. The square chips \(\left(L_{c}=5 \mathrm{~mm}\right)\) are aligned above each of the channels, with longitudinal and transverse pitches of \(S_{L}=S_{T}=20 \mathrm{~mm}\). Water flows through the square cross section \((W=5 \mathrm{~mm}\) ) of each channel with a mean velocity of \(u_{m}=1 \mathrm{~m} / \mathrm{s}\), and its properties may be approximated as \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=855 \times 10^{-6} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}, k=0.610\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=5.8\). Symmetry in the transverse direction dictates the existence of equivalent conditions for each substrate section of length \(L_{s}\) and width \(S_{T}\). (a) Consider a substrate whose length in the flow direction is \(L_{s}=200 \mathrm{~mm}\), thereby providing a total of \(N_{L}=10\) chips attached in-line above each flow channel. To a good approximation, all the heat dissipated by the chips above a channel may be assumed to be transferred to the water flowing through the channel. If each chip dissipates \(5 \mathrm{~W}\), what is the temperature rise of the water passing through the channel? (b) The chip-substrate contact resistance is \(R_{\mathrm{t}, c}^{\mathrm{r}}=\) \(0.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the three-dimensional conduction resistance for the \(L_{s} \times S_{T}\) substrate section is \(R_{\text {cond }}=0.120 \mathrm{~K} / \mathrm{W}\). If water enters the substrate at \(25^{\circ} \mathrm{C}\) and is in fully developed flow, estimate the temperature \(T_{c}\) of the chips and the temperature \(T_{s}\) of the substrate channel surface.

Short Answer

Expert verified
The temperature rise of the water passing through the channel is approximately 0.478 K. The temperature of the chips (T_c) is approximately 25.48掳C, and the temperature of the substrate channel surface (T_s) is approximately 31.48掳C.

Step by step solution

01

(Step 1: Calculate the volume flow rate of water)

(We're given the mean velocity of the water, u_m = 1 m/s. To find the volume flow rate Q, we have to multiply the mean velocity by the cross-sectional area of the channel, which is given by W^2 (5 mm x 5 mm).) First, convert the channel width to meters: \[W = 5 \mathrm{~mm} = 5 \times 10^{-3} \mathrm{~m}\] Next, calculate the cross-sectional area of the channel: \[A = W^2 = (5 \times 10^{-3} \mathrm{~m})^2 = 25 \times 10^{-6} \mathrm{~m}^2\] Now, calculate the volume flow rate Q: \[Q = u_m \times A = 1 \mathrm{~m/s} \times 25 \times 10^{-6} \mathrm{~m}^2 = 25 \times 10^{-6} \mathrm{~m^3/s}\]
02

(Step 2: Calculate the temperature rise of the water)

(To find the temperature rise of water, 螖T, we need to use the heat transfer equation: \(Q = \dot{m} c_p \Delta T\), where \(\dot{m}\) is the mass flow rate, c_p is the specific heat capacity of water, and 螖T is the temperature rise of the water. We also know that each chip dissipates 5 W, and there are 10 chips in the array, so the total heat transfer rate is 50 W.) First, calculate the mass flow rate, \(\dot{m}\), using the density of water (蟻 = 1000 kg/m鲁): \[\dot{m} = \rho Q = 1000 \mathrm{~kg/m^3} \times 25 \times 10^{-6} \mathrm{~m^3/s} = 0.025 \mathrm{~kg/s}\] Next, calculate the total heat transfer rate, Q_total: \[Q_{\text{total}} = N_L \times P_{\text{chip}} = 10 \times 5 \mathrm{~W} = 50 \mathrm{~W}\] Now, find the temperature rise of the water, 螖T, using the heat transfer equation: \[\Delta T = \frac{Q_{\text{total}}}{\dot{m} c_p} = \frac{50 \mathrm{~W}}{0.025 \mathrm{~kg/s} \times 4180 \mathrm{~J/kg\cdot K}} = 0.478 \mathrm{~K}\]
03

(Step 3: Estimate the temperature T_c of the chips and the temperature T_s of the substrate channel surface)

(We can use the given chip-substrate contact resistance and 3D conduction resistance to calculate the temperature of the chips and the substrate channel surface. We are given the following parameters: chip-substrate contact resistance (R_t,c^r = 0.5 x 10^-4 m虏K/W) and 3D conduction resistance (R_cond = 0.120 K/W). We also know that the water enters the substrate at 25掳C.) First, calculate the temperature at the water-substrate channel surface, T_s_0: \[T_{s_0} = 25^{\circ} \mathrm{C} + \Delta T = 25 + 0.478 = 25.478 ^{\circ} \mathrm{C}\] Next, calculate the heat transfer through the chip-substrate contact resistance, 螖T_c: \[\Delta T_c = Q_{\text{total}} \times R_{\text{t,c}^{\text{r}}} = 50 \mathrm{~W} \times 0.5 \times 10^{-4} \mathrm{~m}^2 \mathrm{K/W} = 0.0025 \mathrm{~K}\] Now, calculate the temperature T_c of the chips: \[T_c = T_{s_0} + \Delta T_c = 25.478 + 0.0025 = 25.4805^{\circ} \mathrm{C}\] Lastly, calculate the heat transfer through the 3D conduction resistance, 螖T_s: \[\Delta T_s = Q_{\text{total}} \times R_{\text{cond}} = 50 \mathrm{~W} \times 0.120 \mathrm{~K/W} = 6 \mathrm{~K}\] And find the temperature T_s of the substrate channel surface: \[T_s = T_{s_0} + \Delta T_s = 25.478 + 6 = 31.478^{\circ} \mathrm{C}\] So the temperature of the chips (T_c) is approximately 25.48掳C, and the temperature of the substrate channel surface (T_s) is approximately 31.48掳C.

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Key Concepts

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

Coolant Channel Design
When it comes to managing the temperature of multichip arrays, the design of coolant channels is a key factor. The aim is to efficiently remove the heat generated by the chips to prevent overheating and maintain optimal functioning. In our exercise, channels are machined into a ceramic substrate, with water as the coolant. The channel's cross-sectional dimensions are crucial, as they determine the flow rate 鈥� in our case, a 5mm square, which ensures sufficient water flow and heat transfer.

The velocity of the coolant and its properties, such as density, specific heat capacity, viscosity, and thermal conductivity, influence how effectively heat can be dissipated. The water's high specific heat capacity is beneficial, as it can absorb a considerable amount of heat with minimal temperature increase. The novel scheme in the exercise is designed to optimize these parameters, ensuring the chips鈥� heat is promptly absorbed by the water passing through the coolant channels.
Heat Dissipation in Electronics
Heat dissipation in electronics is vital for preventing thermal-related failures and ensures the longevity and reliability of electronic components. In our exercise, we see that the system is designed to transfer all the heat from the chips to the coolant. The multichip array setup has a total of 10 chips, each adding to the total heat load.

By ensuring that each chip's dissipation capacity is within the coolant鈥檚 ability to absorb heat (5W per chip in this scenario), designers can prevent overheating. The significance of proper heat dissipation is highlighted by the temperature rise calculation鈥攁n increase of just 0.478K indicates that the system is effectively managing the heat load from the chips. The positioning of chips above the flow channel allows for direct heat transfer into the coolant, showcasing the practical application of thermal management principles in electronic device design.
Thermal Resistance
Thermal resistance is a measure of the temperature difference across a material or a system when heat is transferred through it. Lower thermal resistance means higher heat transfer efficiency. In the context of our exercise, there are two types of thermal resistance to consider: chip-substrate contact resistance and three-dimensional conduction resistance of the substrate section.

The contact resistance, given as 0.5 x 10^-4 m虏K/W, represents the opposition to heat flow at the interface between the chip and the substrate. Meanwhile, the 3D conduction resistance, at 0.120 K/W, characterizes the hindrance to heat transfer within the ceramic substrate itself. By including these resistances in the calculation, we can determine precise temperatures for both the chips and the substrate channel surface after heat transfer occurs. Notably, the relatively low resistances confirm that the design effectively minimizes thermal bottlenecks, ensuring heat is efficiently conveyed away from critical components.

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Most popular questions from this chapter

Water flowing at \(2 \mathrm{~kg} / \mathrm{s}\) through a \(40-\mathrm{mm}\)-diameter tube is to be heated from 25 to \(75^{\circ} \mathrm{C}\) by maintaining the tube surface temperature at \(100^{\circ} \mathrm{C}\). (a) What is the required tube length for these conditions? (b) To design a water heating system, we wish to consider using tube diameters in the range from 30 to \(50 \mathrm{~mm}\). What are the required tube lengths for water flow rates of 1,2 , and \(3 \mathrm{~kg} / \mathrm{s}\) ? Represent this design information graphically. (c) Plot the pressure gradient as a function of tube diameter for the three flow rates. Assume the tube wall is smooth.

A circular tube of diameter \(D=0.2 \mathrm{~mm}\) and length \(L=\) \(100 \mathrm{~mm}\) imposes a constant heat flux of \(q^{\prime \prime}=20 \times 10^{3}\) \(\mathrm{W} / \mathrm{m}^{2}\) on a fluid with a mass flow rate of \(\dot{m}=0.1 \mathrm{~g} / \mathrm{s}\). For an inlet temperature of \(T_{m, i}=29^{\circ} \mathrm{C}\), determine the tube wall temperature at \(x=L\) for pure water. Evaluate fluid properties at \(\bar{T}=300 \mathrm{~K}\). For the same conditions, determine the tube wall temperature at \(x=L\) for the nanofluid of Example \(2.2\).

Engine oil is heated by flowing through a circular tube of diameter \(D=50 \mathrm{~mm}\) and length \(L=25 \mathrm{~m}\) and whose surface is maintained at \(150^{\circ} \mathrm{C}\). (a) If the flow rate and inlet temperature of the oil are \(0.5 \mathrm{~kg} / \mathrm{s}\) and \(20^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{m, o}\) ? What is the total heat transfer rate \(q\) for the tube? (b) For flow rates in the range \(0.5 \leq \dot{m} \leq 2.0 \mathrm{~kg} / \mathrm{s}\), compute and plot the variations of \(T_{m, o}\) and \(q\) with \(\dot{m}\). For what flow rate(s) are \(q\) and \(T_{m, \rho}\) maximized? Explain your results.

A liquid food product is processed in a continuousflow sterilizer. The liquid enters the sterilizer at a temperature and flow rate of \(T_{m, i, h}=20^{\circ} \mathrm{C}, \dot{m}=1 \mathrm{~kg} / \mathrm{s}\), respectively. A time-at-temperature constraint requires that the product be held at a mean temperature of \(T_{m}=90^{\circ} \mathrm{C}\) for \(10 \mathrm{~s}\) to kill bacteria, while a second constraint is that the local product temperature cannot exceed \(T_{\max }=230^{\circ} \mathrm{C}\) in order to preserve a pleasing taste. The sterilizer consists of an upstream, \(L_{k}=5 \mathrm{~m}\) heating section characterized by a uniform heat flux, an intermediate insulated sterilizing section, and a downstream cooling section of length \(L_{c}=10 \mathrm{~m}\). The cooling section is composed of an uninsulated tube exposed to a quiescent environment at \(T_{\infty}=20^{\circ} \mathrm{C}\). The thin-walled tubing is of diameter \(D=40 \mathrm{~mm}\). Food properties are similar to those of liquid water at \(T=330 \mathrm{~K}\). (a) What heat flux is required in the heating section to ensure a maximum mean product temperature of \(T_{m}=90^{\circ} \mathrm{C}\) ? (b) Determine the location and value of the maximum local product temperature. Is the second constraint satisfied? (c) Determine the minimum length of the sterilizing section needed to satisfy the time-at-temperature constraint. (d) Sketch the axial distribution of the mean, surface, and centerline temperatures from the inlet of the heating section to the outlet of the cooling section.

Atmospheric air enters the heated section of a circular tube at a flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(20^{\circ} \mathrm{C}\). The tube is of diameter \(D=50 \mathrm{~mm}\), and fully developed conditions with \(h=25 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) exist over the entire length of \(L=3 \mathrm{~m}\). (a) For the case of uniform surface heat flux at \(q_{s}^{\prime \prime}=1000 \mathrm{~W} / \mathrm{m}^{2}\), determine the total heat transfer rate \(q\) and the mean temperature of the air leaving the tube \(T_{m \rho^{-}}\)What is the value of the surface temperature at the tube inlet \(T_{s, i}\) and outlet \(T_{s, \rho}\) ? Sketch the axial variation of \(T_{s}\) and \(T_{m}\). On the same figure, also sketch (qualitatively) the axial variation of \(T_{s}\) and \(T_{m}\) for the more realistic case in which the local convection coefficient varies with \(x\). (b) If the surface heat flux varies linearly with \(x\), such that \(q_{s}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)=500 x(\mathrm{~m})\), what are the values of \(q, T_{m, o}, T_{s, j}\), and \(T_{s, o}\) ? Sketch the axial variation of \(T_{s}\) and \(T_{m-}\) On the same figure, also sketch (qualitatively) the axial variation of \(T_{s}\) and \(T_{m}\) for the more realistic case in which the local convection coefficient varies with \(x\). (c) For the two heating conditions of parts (a) and (b), plot the mean fluid and surface temperatures, \(T_{m}(x)\) and \(T_{s}(x)\), respectively, as functions of distance along the tube. What effect will a fourfold increase in the convection coefficient have on the temperature distributions? (d) For each type of heating process, what heat fluxes are required to achieve an air outlet temperature of \(125^{\circ} \mathrm{C}\) ? Plot the temperature distributions.
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