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Chapter 9: Problem 41

Many laptop computers are equipped with thermal management systems that involve liquid cooling of the central processing unit (CPU), transfer of the heated liquid to the back of the laptop screen assembly, and dissipation of heat from the back of the screen assembly by way of a flat, isothermal heat spreader. The cooled liquid is recirculated to the CPU and the process continues. Consider an aluminum heat spreader that is of width \(w=275 \mathrm{~mm}\) and height \(L=175 \mathrm{~mm}\). The screen assembly is oriented at an angle \(\theta=30^{\circ}\) from the vertical direction, and the heat spreader is attached to the \(t=3\)-mm-thick plastic housing with a thermally conducting adhesive. The plastic housing has a thermal conductivity of \(k=0.21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and emissivity of \(\varepsilon=0.85\). The contact resistance associated with the heat spreaderhousing interface is \(R_{t, c}^{\prime \prime}=2.0 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). If the CPU generates, on average, \(15 \mathrm{~W}\) of thermal energy, what is the temperature of the heat spreader when \(T_{\infty}=T_{\text {sur }}=23^{\circ} \mathrm{C}\) ? Which thermal resistance (contact, conduction, radiation, or free convection) is the largest?

Short Answer

Expert verified
The temperature of the heat spreader can be calculated by finding the sum of the individual thermal resistances (contact, conduction, radiation, and free convection) and using the heat transfer equation. The heat spreader temperature, \(T_{hs}\), is found as: \[T_{hs} = QR_t + T_{\infty}\] After calculating the individual thermal resistances, we can identify the largest one by comparing their magnitudes. Note that the radiation and free convection thermal resistances depend on the final calculated heat spreader temperature, \(T_{hs}\), and the properties of air at the film temperature.

Step by step solution

01

Calculate the contact thermal resistance.

To calculate the contact thermal resistance, we will use the given contact resistance \(R_{t, c}^{\prime \prime}\): \[R_{t, c} = R_{t, c}^{\prime \prime} \times A\] where \(A = w \times L\) is the surface area of the heat spreader.
02

Calculate the conduction thermal resistance.

To calculate the conduction thermal resistance, we will use the formula: \[R_{t, k} = \frac{t}{k \times A}\] where \(t\) is the thickness of the plastic housing and \(k\) is its thermal conductivity.
03

Calculate the radiation thermal resistance.

To calculate the radiation thermal resistance, we will use the formula: \[R_{t, rad} = \frac{1}{\varepsilon \sigma A (T_{\infty}+T_{hs})}\] where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(T_{\infty}\) is the surrounding temperature and \(T_{hs}\) is the heat spreader temperature. Since we don't know \(T_{hs}\) yet, we will leave this expression in terms of \(T_{hs}\).
04

Calculate the free convection thermal resistance.

To calculate the free convection thermal resistance, we will use the formula: \[R_{t, conv} = \frac{1}{h_{c} \times A}\] where \(h_{c}\) is the convection heat transfer coefficient. We will find \(h_c\) from the correlating Nusselt number for natural convection of a flat plate inclined at angle \(\theta\) (this expression can be found in heat transfer literature): \[Nu = 0.56 Ra_L^{\frac{1}{4}}\] where \(Ra_L = \frac{g \beta (T_{hs} - T_{\infty}) L^3}{\nu \alpha}\) is the Rayleigh number, \(g\) is the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(\nu\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity. We will use the properties of air at the film temperature \({(T_{hs} + T_{\infty})}/2\). Once we have the Nusselt number, we can find \(h_c = Nu \times k_f / L\), where \(k_f\) is the thermal conductivity of the fluid (air). Since we don't know \(T_{hs}\) yet, we will leave this expression in terms of \(T_{hs}\).
05

Calculate the total thermal resistance, \(R_t\).

We can calculate the total thermal resistance by summing the individual thermal resistances: \[R_t = R_{t, c} + R_{t, k} + R_{t, rad} + R_{t, conv}\]
06

Calculate the heat spreader temperature, \(T_{hs}\).

We will use the heat transfer equation, which states that the heat generation (\(Q\)) is equal to the temperature difference across the total thermal resistance: \[Q = (T_{hs} - T_{\infty}) R_t\] Now, we will solve for \(T_{hs}\): \[T_{hs} = Q R_t + T_{\infty}\] Once we have found the heat spreader temperature \(T_{hs}\), we can evaluate the thermal resistances and identify the largest one.

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Key Concepts

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

Thermal Conduction
Thermal conduction is a fundamental process where heat energy is transferred through a material without the physical movement of the substance itself. In a scenario like the one with the laptop's aluminum heat spreader, thermal conduction is significant. However, it's important to recognize that the conduction happens not just within the aluminum, but predominantly through the attached plastic housing.
This occurs because the plastic housing acts as a bridge for thermal energy transfer between the aluminum and external environments, having its own thermal conductivity denoted by the letter 'k'.
  • The formula used for calculating thermal resistance due to conduction is: \[R_{t, k} = \frac{t}{k \times A}\]
    Where:
    • R_{t, k} is the conduction resistance
    • t is the thickness of the material
    • k is the thermal conductivity, which is key for efficient thermal conduction
    • A is the area over which the heat is conducted, in this instance, the area of the heat spreader
The lower the thermal conductivity, the higher the thermal resistance, resulting in poorer heat transfer. Understanding these concepts can help in selecting materials with better thermal properties for improved thermal management.
Thermal Radiation
Thermal radiation is a mode of heat transfer that doesn't require contact or a medium. Instead, it relies on electromagnetic waves. This type of heat transfer occurs as energy emitted by bodies due to their temperature.
In the context of the aluminum heat spreader dispersing heat from a laptop, thermal radiation plays a crucial role. It functions in conjunction with the heat spreader's properties, such as emissivity \((\varepsilon)\), to determine how effective the radiation process is.
  • Formula for thermal radiation resistance:\[R_{t, rad} = \frac{1}{\varepsilon \cdot \sigma \cdot A (T_{\infty}+T_{hs})}\]
  • \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \text{W/m}^2\cdot\text{K}^4\)), and\(A\) represents the surface area
  • The temperatures involved include the surrounding \(T_{\infty}\) and the heat spreader \(T_{hs}\)
Emissivity is essential here because a high emissivity value means the material is more effective at radiating heat.
By understanding and manipulating these parameters, efficiency in heat dispersal can be maximized, crucial for electronic devices to prevent overheating.
Free Convection
Free convection refers to a type of heat transfer that occurs when a fluid (such as air or liquid) moves due to temperature differences, without being induced by external forces like fans or pumps. It is an essential mechanism for dissipating heat in electronic devices.
In our aluminum heat spreader scenario, free convection helps remove heat from the system into the air.
  • In the calculation of free convection resistance, the convection heat transfer coefficient, denoted as \(h_c\), comes into play:\[R_{t, conv} = \frac{1}{h_{c} \times A}\]
  • This is heavily dependent on the orientation of the surface, described by the angle \(\theta\), as well as temperature differences.
  • The Nusselt number \(Nu\) relates to this process for an inclined surface and is calculated using:\[Nu = 0.56 Ra_L^{\frac{1}{4}}\]Where \(Ra_L\) is the Rayleigh number, a dimensionless value that considers the thermal expansion, viscosity, and thermal properties at the film temperature.
Understanding and optimizing these variables enables effective heat management, consequently lifting the performance and longevity of devices such as laptops.
Thermal Management Systems
Thermal management systems are critical components in electronic devices, especially in maintaining operational efficiency and longevity. Such systems are designed to control and dissipate heat effectively, which is why they are implemented in laptops for cooling the CPU.
The goal is to prevent overheating by ensuring the produced heat is efficiently transferred away from critical components like CPUs. The aluminum heat spreader, in this scenario, is a core part of the thermal management system, effectively distributing and dissipating heat.
  • Key elements of thermal management systems include:
    • Heat conduction components like heat spreaders and heat sinks.
    • Liquid cooling mechanisms, enhancing thermal conductivity and providing uniform temperature distribution.
    • Passive heat dissipation via free convection and radiation, using materials with high emissivity.
The efficiency of these systems relies on a synergistic approach where conduction, radiation, and convection work in parallel. High-performing systems also include sensors and feedback loops to adapt to changes in thermal loads, ensuring optimal cooling strategies in real-time. Understanding and designing these systems effectively can greatly improve device durability and user experience, making them indispensable in modern electronic design.

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Most popular questions from this chapter

The maximum surface temperature of the \(20-\mathrm{mm}-\) diameter shaft of a motor operating in ambient air at \(27^{\circ} \mathrm{C}\) should not exceed \(87^{\circ} \mathrm{C}\). Because of power dissipation within the motor housing, it is desirable to reject as much heat as possible through the shaft to the ambient air. In this problem, we will investigate several methods for heat removal. (a) For rotating cylinders, a suitable correlation for estimating the convection coefficient is of the form $$ \begin{gathered} \overline{N u}_{D}=0.133 \operatorname{Re}_{D}^{2 / 3} \operatorname{Pr}^{1 / 3} \\ \left(\operatorname{Re}_{D}<4.3 \times 10^{5}, \quad 0.7<\operatorname{Pr}<670\right) \end{gathered} $$ where \(R e_{D} \equiv \Omega D^{2} / \nu\) and \(\Omega\) is the rotational velocity (rad/s). Determine the convection coefficient and the maximum heat rate per unit length as a function of rotational speed in the range from 5000 to \(15,000 \mathrm{rpm}\). (b) Estimate the free convection coefficient and the maximum heat rate per unit length for the stationary shaft. Mixed free and forced convection effects may become significant for \(R e_{D}<4.7\left(G r_{D}^{3} / P r\right)^{0.137}\). Are free convection effects important for the range of rotational speeds designated in part (a)? (c) Assuming the emissivity of the shaft is \(0.8\) and the surroundings are at the ambient air temperature, is radiation exchange important? (d) If ambient air is in cross flow over the shaft, what air velocities are required to remove the heat rates determined in part (a)?

Air at \(3 \mathrm{~atm}\) and \(100^{\circ} \mathrm{C}\) is discharged from a compressor into a vertical receiver of \(2.5-\mathrm{m}\) height and \(0.75-\mathrm{m}\) diameter. Assume that the receiver wall has negligible thermal resistance, is at a uniform temperature, and that heat transfer at its inner and outer surfaces is by free convection from a vertical plate. Neglect radiation exchange and any losses from the top. (a) Estimate the receiver wall temperature and the heat transfer to the ambient air at \(25^{\circ} \mathrm{C}\). To facilitate use of the free convection correlations with appropriate film temperatures, assume that the receiver wall temperature is \(60^{\circ} \mathrm{C}\). (b) Were the assumed film temperatures of part (a) reasonable? If not, use an iteration procedure to find consistent values. (c) Now consider two features of the receiver neglected in the previous analysis: (i) radiation exchange from the exterior surface of emissivity \(0.85\) to large surroundings, also at \(25^{\circ} \mathrm{C}\); and (ii) the thermal resistance of a 20 -mm-thick wall with a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Represent the system by a thermal circuit and estimate the wall temperatures and the heat transfer rate.

Consider an experiment to investigate the transition to turbulent flow in a free convection boundary layer that develops along a vertical plate suspended in a large room. The plate is constructed of a thin heater that is sandwiched between two aluminum plates and may be assumed to be isothermal. The heated plate is \(1 \mathrm{~m}\) high and \(2 \mathrm{~m}\) wide. The quiescent air and the surroundings are both at \(25^{\circ} \mathrm{C}\). (a) The exposed surfaces of the aluminum plate are painted with a very thin coating of high emissivity \((\varepsilon=0.95)\) paint. Determine the electrical power that must be supplied to the heater to sustain the plate at a temperature of \(T_{s}=35^{\circ} \mathrm{C}\). How much of the plate is exposed to turbulent conditions in the free convection boundary layer? (b) The experimentalist speculates that the roughness of the paint is affecting the transition to turbulence in the boundary layer and decides to remove the paint and polish the aluminum surface ( \(\varepsilon=0.05\) ). If the same power is supplied to the plate as in part (a), what is the steady- state plate temperature? How much of the plate is exposed to turbulent conditions in the free convection boundary layer?

In the central receiver concept of a solar power plant, many heliostats at ground level are used to direct a concentrated solar flux \(q_{s}^{\prime \prime}\) to the receiver, which is positioned at the top of a tower. However, even with absorption of all the solar flux by the outer surface of the receiver, losses due to free convection and radiation reduce the collection efficiency below the maximum possible value of \(100 \%\). Consider a cylindrical receiver of diameter \(D=7 \mathrm{~m}\), length \(L=12 \mathrm{~m}\), and emissivity \(\varepsilon=0.20\). (a) If all of the solar flux is absorbed by the receiver and a surface temperature of \(T_{s}=800 \mathrm{~K}\) is maintained, what is the rate of heat loss from the receiver? The ambient air is quiescent at a temperature of \(T_{\infty}=300 \mathrm{~K}\), and irradiation from the surroundings may be neglected. If the corresponding value of the solar flux is \(q_{S}^{\prime \prime}=10^{5} \mathrm{~W} / \mathrm{m}^{2}\), what is the collector efficiency? (b) The surface temperature of the receiver is affected by design and operating conditions within the power plant. Over the range from 600 to \(1000 \mathrm{~K}\), plot the variation of the convection, radiation, and total heat rates as a function of \(T_{s}\). For a fixed value of \(q_{S}^{\prime \prime}=10^{5} \mathrm{~W} / \mathrm{m}^{2}\), plot the corresponding variation of the receiver efficiency.

A probe, used to measure the velocity of air in a lowspeed wind tunnel, is fabricated of an \(L=100 \mathrm{~mm}\) long, \(D=8\)-mm outside diameter horizontal aluminum tube. Power resistors are inserted into the stationary tube and dissipate \(P=1.5 \mathrm{~W}\). The surface temperature of the tube is determined experimentally by measuring the emitted radiation from the exterior of the tube. To maximize surface emission, the exterior of the tube is painted with flat black paint having an emissivity of \(\varepsilon=0.95\). (a) For air at a temperature and cross flow velocity of \(T_{\infty}=25^{\circ} \mathrm{C}, V=0.1 \mathrm{~m} / \mathrm{s}\), respectively, determine the surface temperature of the tube. The surroundings temperature is \(T_{\text {sur }}=25^{\circ} \mathrm{C}\). (b) For the conditions of part (a), plot the tube surface temperature versus the cross flow velocity over the range \(0.05 \mathrm{~m} / \mathrm{s} \leq V \leq 1 \mathrm{~m} / \mathrm{s}\).
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