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Chapter 1: Problem 31

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~mm}\) and emissivity \(\varepsilon=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the temperature range \(40 \leq T \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by considering values of \(0.20\) and \(0.30\).

Short Answer

Expert verified
The acceptable power dissipation range for the instrumentation package can be determined by utilizing the Stefan-Boltzmann law, accounting for the given temperature and emissivity values. For an emissivity of 0.25, calculate the net power radiated (Q_rad) at the lower (313.15 K) and upper (358.15 K) temperature limits, and determine the acceptable power dissipation range as Q_lower to Q_upper. Repeat these calculations for emissivity values of 0.20 and 0.30 to observe the effect of variations in emissivity. Graph the results to visually display the range of acceptable power dissipation with respect to temperature for each emissivity value.

Step by step solution

01

Calculate the Surface Area of the Sphere

The surface area of a sphere (A) can be calculated using the formula A = 4蟺r^2, where r is the radius of the sphere. We are given the diameter D = 100mm, so the radius r = D/2 = 50mm. Calculate the surface area (A) of the sphere: \(A = 4\pi(50\times 10^{-3})^2\)
02

Convert Temperature Ranges to Kelvin

Given that Celsius and Kelvin are related by the formula K = C + 273.15, we can convert the given temperature range (40掳C to 85掳C) and the chamber walls' temperature (77K) to Kelvin. 40掳C to 85掳C in Kelvin: Lower limit = 40 + 273.15 = 313.15 K Upper limit = 85 + 273.15 = 358.15 K
03

Determine the Theoretical Range of Net Power Radiated by the Sphere

We can use the Stefan-Boltzmann Law to calculate the theoretical range of net power radiated (Q_rad) by the sphere, where Q_rad = 蔚A蟽(T^4 - T_amb^4). Here 蔚 (emissivity) is given as 0.25, 蟽 is the Stefan-Boltzmann constant (5.67 x 10-8 W m^-2 K^-4), T is the temperature of the sphere, and T_amb is the ambient temperature (77K) in the chamber. Calculate Q_rad at the lower and upper limits of the temperature range. Q_rad at the lower limit (313.15 K): Q_lower = 0.25 * A * 蟽 * (313.15^4 - 77^4) Q_rad at the upper limit (358.15 K): Q_upper = 0.25 * A * 蟽 * (358.15^4 - 77^4)
04

Determine the Acceptable Power Dissipation Range

Now that we have the theoretical range of net power radiated, we need to determine the acceptable power dissipation range for the instrumentation package. Since the problem states that the operation is restricted by the mentioned temperature range, we can assume that the range of power dissipation is the same as the power radiated. Thus, the acceptable power dissipation range will be Q_lower to Q_upper.
05

Repeat Steps 3 and 4 for Different Emissivity Values

Now, we have to calculate the net power radiated and the acceptable power dissipation range for the given emissivity values of 0.20 and 0.30. The process is the same as in Steps 3 and 4, but using the new emissivity values: Perform the calculations for 蔚 = 0.20 and 蔚 = 0.30: Q_lower_0.20, Q_upper_0.20, Q_lower_0.30, and Q_upper_0.30.
06

Display the Results Graphically

To display the results graphically, plot the power dissipation (Y-axis) vs. temperature (X-axis) for each of the emissivity values, creating three separate curves. The X-axis represents the acceptable temperature range (313.15 K to 358.15 K), and the Y-axis represents the acceptable power dissipation range obtained in Steps 4 and 5. Now you have a graphical representation of the range of acceptable power dissipation for the instrumentation package.

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Key Concepts

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

Emissivity
Emissivity is a key property of materials that measures how effectively a surface emits thermal radiation compared to a perfect black body, which has an emissivity of 1. In this problem, we deal with an emissivity value of 0.25 for a spherical surface. This means that the surface emits only 25% of the thermal radiation that a perfect black body would emit at the same temperature.

The role of emissivity becomes evident when calculating the heat energy radiated by an object. A lower emissivity indicates that the object radiates less energy, which becomes crucial when determining the range of acceptable power dissipation for the device placed in a space simulation chamber. It affects how much heat is lost through radiation and thus affects the operational temperature of the device. Additionally, the exercise asks us to see how variations in emissivity (values of 0.20 and 0.30) alter these acceptable power levels, highlighting its practical importance in thermal management.
  • Low emissivity: The surface retains more heat.
  • High emissivity: The surface loses more heat.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in heat transfer that describes the power radiated from a black body in terms of its temperature. The law is given by the formula: \[ Q_{ ext{rad}} = ext{蔚} A ext{蟽} (T^4 - T_{ ext{amb}}^4) \]where:
  • \(Q_{\text{rad}}\) is the radiated power.
  • \(\text{蔚}\) is the emissivity of the material.
  • \(A\) is the surface area of the object.
  • \(\text{蟽}\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \,\text{W m}^{-2} \text{K}^{-4})\).
  • \(T\) and \(T_{\text{amb}}\) are the temperatures of the object and the surroundings in Kelvin, respectively.
This equation is essential when calculating how much energy the package in question can safely push out via radiation once it reaches its operating temperature range of 40 to 85掳C. The law elucidates why changing either the emissivity or temperature alters the radiation levels, offering a way to predict heat loss under different thermal conditions.
Spherical Surface Area
A sphere's surface area plays a crucial role in heat transfer calculations for objects with a spherical shape, like the instrumentation package in the exercise. It is calculated using the formula:\[ A = 4\pi r^2 \]where \(r\) is the radius of the sphere. Given a diameter of 100 mm, we find the radius to be 50 mm, or 0.05 meters. Consequently, the surface area, \(A\), of this sphere is \(4\pi(0.05)^2\).

Knowing the surface area is vital for accurately using the Stefan-Boltzmann Law, as the amount of radiated heat is directly proportional to the entire surface area that radiates energy. A larger surface area results in more heat being radiated, thus impacting the sphere's overall thermal balance when exposed to specific temperature conditions.

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Most popular questions from this chapter

A wall has inner and outer surface temperatures of 16 and \(6^{\circ} \mathrm{C}\), respectively. The interior and exterior air temperatures are 20 and \(5^{\circ} \mathrm{C}\), respectively. The inner and outer convection heat transfer coefficients are 5 and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

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.

In the thermal processing of semiconductor materials, annealing is accomplished by heating a silicon wafer according to a temperature-time recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follows, the wafer is in an evacuated chamber whose walls are maintained at \(27^{\circ} \mathrm{C}\) and within which heating lamps maintain a radiant flux \(q_{s}^{\prime \prime}\) at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and an emissivity that equals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{l}=0.65\right.\) ). For \(q_{s}^{\prime \prime}=3.0 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\), the temperature on its lower surface is measured by a radiation thermometer and found to have a value of \(T_{w, l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this condition being met?

The heat flux that is applied to one face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The opposite face is exposed to air at temperature \(30^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the wall exposed to air is measured and found to be \(50^{\circ} \mathrm{C}\). Do steady-state conditions exist? If not, is the temperature of the wall increasing or decreasing with time?

For a boiling process such as shown in Figure \(1.5 c\), the ambient temperature \(T_{\infty}\) in Newton's law of cooling is replaced by the saturation temperature of the fluid \(T_{\text {sat }}\). Consider a situation where the heat flux from the hot plate is \(q^{\prime \prime}=20 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). If the fluid is water at atmospheric pressure and the convection heat transfer coefficient is \(h_{w}=20 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the upper surface temperature of the plate, \(T_{s, w^{\circ}}\). In an effort to minimize the surface temperature, a technician proposes replacing the water with a dielectric fluid whose saturation temperature is \(T_{\text {sat,d }}=52^{\circ} \mathrm{C}\). If the heat transfer coefficient associated with the dielectric fluid is \(h_{d}=3 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), will the technician's plan work?
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