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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

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

Consider a carton of milk that is refrigerated at a temperature of \(T_{m \mathrm{r}}=5^{\circ} \mathrm{C}\). The kitchen temperature on a hot summer day is \(T_{\infty}=30^{\circ} \mathrm{C}\). If the four sides of the carton are of height and width \(L=200 \mathrm{~mm}\) and \(w=100 \mathrm{~mm}\), respectively, determine the heat transferred to the milk carton as it sits on the kitchen counter for durations of \(t=10 \mathrm{~s}, 60 \mathrm{~s}\), and \(300 \mathrm{~s}\) before it is returned to the refrigerator. The convection coefficient associated with natural convection on the sides of the carton is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(0.90\). Assume the milk carton temperature remains at \(5^{\circ} \mathrm{C}\) during the process. Your parents have taught you the importance of refrigerating certain foods from the food safety perspective. Comment on the importance of quickly returning the milk carton to the refrigerator from an energy conservation point of view.

An inexpensive food and beverage container is fabricated from 25 -mm-thick polystyrene \((k=0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and has interior dimensions of \(0.8 \mathrm{~m} \times 0.6 \mathrm{~m} \times 0.6 \mathrm{~m}\). Under conditions for which an inner surface temperature of approximately \(2^{\circ} \mathrm{C}\) is maintained by an ice-water mixture and an outer surface temperature of \(20^{\circ} \mathrm{C}\) is maintained by the ambient, what is the heat flux through the container wall? Assuming negligible heat gain through the \(0.8 \mathrm{~m} \times\) \(0.6 \mathrm{~m}\) base of the cooler, what is the total heat load for the prescribed conditions?

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.
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