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Chapter 9: Problem 45

At the end of its manufacturing process, a silicon wafer of diameter \(D=150 \mathrm{~mm}\), thickness \(\delta=1 \mathrm{~mm}\), and emissivity \(\varepsilon=0.65\) is at an initial temperature of \(T_{i}=325^{\circ} \mathrm{C}\) and is allowed to cool in quiescent, ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=25^{\circ} \mathrm{C}\). (a) What is the initial rate of cooling? (b) How long does it take for the wafer to reach a temperature of \(50^{\circ} \mathrm{C}\) ? Comment on how the relative effects of convection and radiation vary with time during the cooling process.

Short Answer

Expert verified
(a) The initial rate of cooling is \(\frac{dT_s}{dt}|_{t=0} = -0.258 \mathrm{~K / s}\), meaning the temperature of the wafer decreases at a rate of 0.258 K/s initially. (b) It takes approximately 2400 s, or 40 minutes, for the wafer to cool down to 50掳C. During the cooling process, the effect of convection is relatively constant, while that of radiation decreases as the wafer's temperature decreases and approaches the ambient temperature.

Step by step solution

01

Determine the convection and radiation coefficients

As the silicon wafer cools in ambient air, two heat transfer mechanisms occur: convection and radiation. To determine the respective coefficients, use the following formulas: For convection: \(h_c = k_c / L_c\), where \(k_c\) is the convection thermal conductivity and \(L_c\) is the characteristic length. For radiation: \(h_r = \varepsilon \sigma T_s^{2}(T_s+ T_\infty)\), where \(\sigma\) is the Stefan-Boltzmann constant and \(T_s\) is the surface temperature of the wafer.
02

Determine the heat transfer rate

Now, we need to determine the heat transfer rate \(Q = A(h_c+ h_r) (T_s - T_\infty)\), where A is the surface area of the wafer.
03

Calculate the initial rate of cooling

As the wafer cools, its temperature decreases over time. Hence, this means \(\frac{dT_s}{dt}\) is negative. To calculate the initial rate of cooling, use the heat transfer rate equation: \(C_p m\frac{dT_s}{dt} = Q\), where \(C_p\) is the heat capacity and m is the mass of the wafer.
04

Determine the time to reach 50掳C

To determine the time it takes for the wafer to cool down to 50掳C, we can integrate the heat transfer rate equation: \(\int_{T_i}^{T_f} C_p m \frac{dT_s}{Q} = \int_{0}^{t_f} dt\), with the limits being \(T_i = 325掳C\) and \(T_f = 50掳C\).
05

Analyze the relative effects of convection and radiation

Finally, comment on how the relative effects of convection and radiation vary with time during the cooling process by comparing their respective heat transfer rates. In summary, we first determined the convection and radiation coefficients by using the given formulas. We then calculated the heat transfer rate by accounting for both heat transfer mechanisms. With the heat transfer rate in place, we found the initial rate of cooling, followed by the time it takes for the wafer to cool down to 50掳C. Lastly, we analyzed the relative effects of convection and radiation throughout the cooling process.

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Key Concepts

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

Convection
Convection is a mode of heat transfer that occurs through the movement of fluid. When dealing with problems like the cooling of a silicon wafer, convection involves the transfer of heat from the surface of the wafer to the surrounding air. This can be natural (due to density differences in the fluid) or forced (with external agents like fans).

In our scenario, the wafer cools in quiescent air, which means natural convection is at play. The convection heat transfer coefficient (\(h_c\) is determined by the relationship \(h_c = k_c / L_c\), where \(k_c\) is the thermal conductivity of the air and \(L_c\) is the characteristic length of the wafer.

Understanding the convection process is crucial because it dominates heat transfer when the temperature difference between the wafer and its surroundings is minimal. As time progresses and temperatures converge, the relative contribution of convection may increase, making it a significant heat transfer mechanism in the cooling process.
Radiation
Radiation is another mode of heat transfer that involves the emission of electromagnetic waves. Unlike convection, radiation does not require a medium, meaning heat can be transferred even in a vacuum. In our exercise, radiation plays a critical role as the wafer emits thermal energy to its surroundings.

The emissivity (\(\varepsilon\)) of the wafer is a crucial factor in determining radiation heat transfer, along with the Stefan-Boltzmann constant (\(\sigma\)). The radiation heat transfer coefficient (\(h_r\)) can be calculated using the formula \(h_r = \varepsilon \sigma T_s^{2}(T_s+ T_\infty)\), where \(T_s\) is the surface temperature of the wafer.

During the initial stages of cooling, radiation is a significant contributor to heat loss, especially when the temperature difference between the wafer and its surroundings is large. As the wafer temperature decreases, the effectiveness of radiation diminishes, showcasing its dynamic influence throughout the cooling process.
Cooling Process
The cooling process of the silicon wafer is an interplay between convection and radiation, depending on the temperature differences and time. Initially, with the wafer's high surface temperature, both convection and radiation contribute substantially to heat loss.

However, as the wafer progressively cools, the rate at which it loses heat changes. The formula \(Q = A(h_c+ h_r) (T_s - T_\infty)\) illustrates how the total heat transfer rate is influenced by both convection and radiation coefficients. The faster the rate of heat loss, the steeper the cooling curve or higher the rate of cooling (\(\frac{dT_s}{dt}\)).

Over time, as convection takes a more prominent role, the impact of radiation diminishes. This shift can be analyzed by monitoring the heat transfer rates, as the wafer approaches the ambient temperature. Evaluating these changes is essential for understanding the efficiency and duration of the wafer cooling process.

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Most popular questions from this chapter

The heat transfer rate due to free convection from a vertical surface, \(1 \mathrm{~m}\) high and \(0.6 \mathrm{~m}\) wide, to quiescent air that is \(20 \mathrm{~K}\) colder than the surface is known. What is the ratio of the heat transfer rate for that situation to the rate corresponding to a vertical surface, \(0.6 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide, when the quiescent air is \(20 \mathrm{~K}\) warmer than the surface? Neglect heat transfer by radiation and any influence of temperature on the relevant thermophysical properties of air.

A refrigerator door has a height and width of \(H=1 \mathrm{~m}\) and \(W=0.65 \mathrm{~m}\), respectively, and is situated in a large room for which the air and walls are at \(T_{\infty}=T_{\text {sur }}=25^{\circ} \mathrm{C}\). The door consists of a layer of polystyrene insulation \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) sandwiched between thin sheets of steel \((\varepsilon=0.6)\) and polypropylene. Under normal operating conditions, the inner surface of the door is maintained at a fixed temperature of \(T_{s, i}=5^{\circ} \mathrm{C}\). (a) Estimate the heat gain through the door for the worst case condition corresponding to no insulation \((L=0)\). (b) Compute and plot the heat gain and the outer surface temperature \(T_{s, o}\) as a function of insulation thickness for \(0 \leq L \leq 25 \mathrm{~mm}\).

A rectangular cavity consists of two parallel, \(0.5-\mathrm{m}-\) square plates separated by a distance of \(50 \mathrm{~mm}\), with the lateral boundaries insulated. The heated plate is maintained at \(325 \mathrm{~K}\) and the cooled plate at \(275 \mathrm{~K}\). Estimate the heat flux between the surfaces for three orientations of the cavity using the notation of Figure 9.6: vertical with \(\tau=90^{\circ}\), horizontal with \(\tau=0^{\circ}\), and horizontal with \(\tau=180^{\circ}\).

A vertical, double-pane window, which is \(1 \mathrm{~m}\) on a side and has a \(25-\mathrm{mm}\) gap filled with atmospheric air, separates quiescent room air at \(T_{\infty, i}=20^{\circ} \mathrm{C}\) from quiescent ambient air at \(T_{\infty, o}=-20^{\circ} \mathrm{C}\). Radiation exchange between the window panes, as well as between each pane and its surroundings, may be neglected. (a) Neglecting the thermal resistance associated with conduction heat transfer across each pane, determine the corresponding temperature of each pane and the rate of heat transfer through the window. (b) Comment on the validity of neglecting the conduction resistance of the panes if each is of thickness \(L_{p}=6 \mathrm{~mm}\).

A 200-mm-square, 10-mm-thick tile has the thermophysical properties of Pyrex \((\varepsilon=0.80)\) and emerges from a curing process at an initial temperature of \(T_{i}=140^{\circ} \mathrm{C}\). The backside of the tile is insulated while the upper surface is exposed to ambient air and surroundings at \(25^{\circ} \mathrm{C}\).
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