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Chapter 1: Problem 16

A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If \(4 \mathrm{~W}\) are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

Short Answer

Expert verified
The steady-state temperature difference between the back and front surfaces of the silicon chip is approximately 1.067 K.

Step by step solution

01

Identify Relevant Formula

In this problem, we are dealing with heat conduction, so we will use Fourier's law, which is given by: q = -k * A * (dT/dx) Where: - q is the heat transfer rate (W) - k is the thermal conductivity (W/m路K) - A is the cross-sectional area of the chip (m^2) - dT/dx is the temperature gradient (K/m)
02

Calculate Cross-sectional Area

The chip has a square shape on one side with a width (w) of 5 mm. We first convert width to meters: w = 5 mm = 5 * 10^{-3} m Then, we can find the cross-sectional area (A) of the chip as: A = w^2 = (5 * 10^{-3})^2 = 25 * 10^{-6} m^2
03

Calculate Temperature Gradient

We are given the heat transfer rate (q) as 4 W. Now, we can rearrange the Fourier's law equation to solve for the temperature gradient (dT/dx): dT/dx = -q / (k * A)
04

Calculate the Steady-State Temperature Difference

Now, substitute the given values into the equation and find the temperature gradient (dT/dx): dT/dx = -(-4 W) / (150 W/m路K * 25 * 10^{-6} m^2) = 4 / (150 * 25 * 10^{-6}) = 1066.67 K/m Finally, to find the temperature difference (螖T) between the back and front surfaces, we multiply the temperature gradient by the thickness (t) of the chip and convert the thickness to meters: 螖T = dT/dx * t = 1066.67 K/m * 1 * 10^{-3} m = 1.067 K The steady-state temperature difference between the back and front surfaces of the chip is approximately 1.067 K.

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Key Concepts

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

Fourier's Law
Fourier's law is fundamental in understanding heat conduction. It describes how heat energy transfers through materials, highlighting that this transfer is proportional to the negative temperature gradient. This law is mathematically expressed as:
  • \( q = -k \cdot A \cdot \frac{dT}{dx} \)
Here:
  • \( q \) is the rate of heat transfer, measured in watts (W).
  • \( k \) represents the material's thermal conductivity, given in watts per meter-kelvin (W/m路K).
  • \( A \) stands for the cross-sectional area through which the heat transfers, measured in square meters (m虏).
  • \( \frac{dT}{dx} \) is the temperature gradient, showing how temperature changes across the material per unit distance.
By using Fourier's law, you can determine how much heat flows in a system, which is crucial for designing effective thermal management in devices like silicon chips.
This law emphasizes the linear relationship between heat transfer, material properties, and temperature difference, serving as a guiding principle for solving heat conduction problems in various engineering fields.
Thermal Conductivity
Thermal conductivity \( (k) \) is a key property of materials that indicates how well they conduct heat. A higher \( k \) value means better thermal conductivity. It's intrinsic to the materials used and is measured in watts per meter-kelvin (W/m路K). This property is crucial when evaluating how efficient a material is in transferring heat.

In our exercise, the silicon chip has a thermal conductivity of \( 150 \text{ W/m}\cdot \text{K} \). This relatively high value implies that silicon can effectively transfer heat, making it suitable for electronic applications where dissipating heat is important. Silicon's high thermal conductivity also allows for efficient management of heat produced by electronic circuits. Here are some other common materials and their typical thermal conductivities:
  • Copper: \( 400 \text{ W/m}\cdot \text{K} \)
  • Aluminum: \( 235 \text{ W/m}\cdot \text{K} \)
  • Glass: \( 1.4 \text{ W/m}\cdot \text{K} \)
A material's thermal conductivity determines its role in applications requiring thermal management, as it affects energy efficiency and stability.
Temperature Gradient
The temperature gradient \( \left( \frac{dT}{dx} \right) \) measures how temperature changes across a material's length. It's a decisive factor in heat conduction, representing the rate of temperature change per unit distance. In the context of Fourier's law, it's expressed in kelvins per meter (K/m). This concept indicates where heat is transferring: from hotter to cooler areas, as energy moves to even out temperatures.

In our example of a silicon chip, with a given heat rate \( (q) \) and thermal conductivity \( (k) \), the temperature gradient was calculated using the rearranged Fourier's law:
  • \( \frac{dT}{dx} = -\frac{q}{k \cdot A} \)
This means that knowing the heat flow and the material's properties allows us to determine how rapidly temperature shifts within the object.

The calculated temperature gradient of \( 1066.67 \, \text{K/m} \) shows the changes in temperature from the back to the front of the chip, crucial for understanding the heat dissipation behavior in micro-electronic settings. This gradient guides engineers in setting up adequate cooling measures to maintain device performance.

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Most popular questions from this chapter

Three electric resistance heaters of length \(L=250 \mathrm{~mm}\) and diameter \(D=25 \mathrm{~mm}\) are submerged in a 10 -gal tank of water, which is initially at \(295 \mathrm{~K}\). The water may be assumed to have a density and specific heat of \(\rho=990 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If the heaters are activated, each dissipating \(q_{1}=500 \mathrm{~W}\), estimate the time required to bring the water to a temperature of \(335 \mathrm{~K}\). (b) If the natural convection coefficient is given by an expression of the form \(h=370\left(T_{s}-T\right)^{1 / 3}\), where \(T_{s}\) and \(T\) are temperatures of the heater surface and water, respectively, what is the temperature of each heater shortly after activation and just before deactivation? Units of \(h\) and \(\left(T_{s}-T\right)\) are \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{K}\), respectively. (c) If the heaters are inadvertently activated when the tank is empty, the natural convection coefficient associated with heat transfer to the ambient air at \(T_{\infty}=300 \mathrm{~K}\) may be approximated as \(h=0.70\) \(\left(T_{s}-T_{\infty}\right)^{1 / 3}\). If the temperature of the tank walls is also \(300 \mathrm{~K}\) and the emissivity of the heater surface is \(\varepsilon=0.85\), what is the surface temperature of each heater under steady-state conditions?

A wall is made from an inhomogeneous (nonuniform) material for which the thermal conductivity varies through the thickness according to \(k=a x+b\), where \(a\) and \(b\) are constants. The heat flux is known to be constant. Determine expressions for the temperature gradient and the temperature distribution when the surface at \(x=0\) is at temperature \(T_{1}\).

An overhead 25-m-long, uninsulated industrial steam pipe of \(100-\mathrm{mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{\circ} \mathrm{C}\), and the coefficient associated with natural convection is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steam is generated in a gas-fired boiler operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02\) per \(\mathrm{MJ}\), what is the annual cost of heat loss from the line?

An internally reversible refrigerator has a modified coefficient of performance accounting for realistic heat transfer processes of $$ \mathrm{COP}_{m}=\frac{q_{\text {in }}}{\dot{W}}=\frac{q_{\text {in }}}{q_{\text {out }}-q_{\text {in }}}=\frac{T_{c, i}}{T_{h, i}-T_{c, i}} $$ where \(q_{\text {in }}\) is the refrigerator cooling rate, \(q_{\text {out }}\) is the heat rejection rate, and \(\dot{W}\) is the power input. Show that \(\mathrm{COP}_{m}\) can be expressed in terms of the reservoir temperatures \(T_{c}\) and \(T_{h}\), the cold and hot thermal resistances \(R_{L, c}\) and \(R_{t, h}\), and \(q_{\text {in }}\), as $$ \mathrm{COP}_{m}=\frac{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}}{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{tot}}} $$ where \(R_{\mathrm{tot}}=R_{t, c}+R_{t, h}\). Also, show that the power input may be expressed as $$ \dot{W}=q_{\mathrm{in}} \frac{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{id \textrm {t }}}}{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}} $$

A thin electrical heating element provides a uniform heat flux \(q_{o}^{\prime \prime}\) to the outer surface of a duct through which airflows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What heat flux \(q_{o}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of part (a), what is the temperature \(\left(T_{o}\right)\) of the duct surface next to the heater? (c) With \(T_{i}=85^{\circ} \mathrm{C}\), compute and plot \(q_{o}^{\prime \prime}\) and \(T_{o}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.
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