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Chapter 12: Problem 23

The extremely high temperatures needed to trigger nuclear fusion are proposed to be generated by laserirradiating a spherical pellet of deuterium and tritium fuel of diameter \(D_{p}=1.8 \mathrm{~mm}\). (a) Determine the maximum fuel temperature that can be achieved by irradiating the pellet with 200 lasers, each producing a power of \(P=500 \mathrm{~W}\). The pellet has an absorptivity \(\alpha=0.3\) and emissivity \(\varepsilon=0.8\). (b) The pellet is placed inside a cylindrical enclosure. Two laser entrance holes are located at either end of the enclosure and have a diameter of \(D_{\mathrm{LEH}}=2 \mathrm{~mm}\). Determine the maximum temperature that can be generated within the enclosure.

Short Answer

Expert verified
The maximum fuel temperature that can be achieved by irradiating the pellet with 200 lasers, each producing a power of 500 W, is approximately 49,900 K. However, without knowing the size, shape, or emissivity values of the walls of the cylindrical enclosure, it is impossible to calculate the maximum temperature within the enclosure. More information is needed to determine the maximum temperature within the enclosure.

Step by step solution

01

Calculate the total laser power

The total power produced by the 200 lasers is given by \(P_\text{total} = N \times P\), where \(N\) represents the number of lasers, and \(P\) represents the power of each laser. In this case, \(N=200\) and \(P=500 \mathrm{~W}\). The total laser power is: \[P_\text{total} = 200 \times 500 \mathrm{~W} = 100000 \mathrm{~W}\]
02

Calculate the power absorbed by the pellet

The power absorbed by the pellet is determined by its absorptivity, \(\alpha\). The power absorbed, \(P_\text{abs}\), can be calculated as: \(P_\text{abs} = \alpha \times P_\text{total} \). In this case, \(\alpha=0.3\). The power absorbed is: \[P_\text{abs} = 0.3 \times 100000 \mathrm{~W} = 30000 \mathrm{~W}\]
03

Calculate the power emitted by the pellet

The power emitted by the pellet is determined by its emissivity, \(\varepsilon\), and temperature, \(T\). The power emitted, \(P_\text{emit}\), is given by the equation: \(P_\text{emit} = \varepsilon \sigma A T^4\), where \(\varepsilon\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\) is the Stefan-Boltzmann constant, \(A\) is the surface area of the pellet, and \(T\) is the temperature in Kelvin. First, calculate the surface area of the pellet. The pellet is a sphere with diameter \(D_p = 1.8 \mathrm{~mm}\). The radius \(r_p = \frac{1}{2}D_p = 0.9 \mathrm{~mm} = 9 \times 10^{-4} \mathrm{~m}\). The surface area, \(A\), is given by: \[A = 4\pi r_p^2 = 4\pi (9 \times 10^{-4})^2 \mathrm{m^2} \]
04

Find the maximum fuel temperature

We know that the power absorbed by the pellet is equal to the power emitted by the pellet in equilibrium, i.e., \(P_\text{abs} = P_\text{emit}\). Therefore, we can write: \[30000 \mathrm{~W} = 0.8 \times 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}} \times 4\pi (9 \times 10^{-4})^2 \mathrm{m^2} \times T^4\] From this equation, we can solve for the maximum fuel temperature, \(T\): \[T = \sqrt[4]{\frac{30000 \mathrm{~W}}{0.8 \times 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}} \times 4\pi (9 \times 10^{-4})^2 \mathrm{m^2}}}\] After calculating, we get the maximum fuel temperature: \[T \approx 49900 \mathrm{~K}\] (b) Determine the maximum temperature that can be generated within the enclosure.
05

Calculate the laser power into the enclosure

For the pellet placed inside the cylindrical enclosure, there are two laser entrance holes each with diameter \(D_\text{LEH} = 2 \mathrm{~mm}\). We need to determine the total laser power entering the enclosure. Since we already calculated the total laser power in (a), the laser power entering the enclosure is also \(P_\text{total} = 100000 \mathrm{~W}\).
06

Calculate the power absorbed by the walls of the enclosure

Using the same method we used in Step 2(a) to find the power absorbed by the pellet, in this case, let's assume that the absorptivity of the walls of the enclosure is also \(\alpha = 0.3\). Therefore, the power absorbed by the walls of the enclosure is also given by \(P_\text{abs} = \alpha \times P_\text{total}\). In this case: \[P_\text{abs} = 0.3 \times 100000 \mathrm{~W} = 30000 \mathrm{~W}\]
07

Determine the maximum temperature within the enclosure

Since we're considering equilibrium conditions, the power absorbed by the walls of the enclosure is equal to the power emitted by them, \(P_\text{abs} = P_\text{emit}\). However, without knowing the size, shape or emissivity values of the walls it is impossible to calculate the temperature within the enclosure. More information is needed to calculate the maximum temperature within the enclosure.

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Key Concepts

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

Laser Irradiation of Fusion Fuel
Imagine trying to light the sun in a laboratory. This is similar to what scientists aim to achieve when they use lasers to induce nuclear fusion, a process that could potentially provide abundant, clean energy. Fusion fuel, typically made of deuterium and tritium, requires extreme temperatures to overcome repulsive forces between nuclei and facilitate fusion. Laser irradiation is one of the methods to achieve these temperatures.

When a fusion fuel pellet is bombarded by high-energy laser beams, the energy is absorbed by the fuel, increasing its temperature. This methodology involves precisely controlling the lasers to deliver uniform energy to avoid instabilities. The challenge is to reach temperatures high enough for fusion to occur while maintaining the integrity of the containment apparatus.

Through multiple lasers focusing on a small pellet, scientists can create the necessary conditions for fusion 鈥� high temperature and pressure 鈥� converting the fuel into a hot plasma. The temperature calculation involves understanding the power delivered by the lasers and the rate at which the fuel absorbs this energy.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in thermodynamics, describing how a black body in thermal equilibrium radiates energy. It states that the power emitted per unit area of the surface of a black body is directly proportional to the fourth power of the body's thermodynamic temperature. Mathematically, the law is expressed as: \[ P = \sigma T^4 \]where \( P \) is the power radiated per unit area, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature in Kelvin.

In the context of nuclear fusion, the Stefan-Boltzmann law helps us understand the relationship between the temperature of the fuel pellet and the power it emits once it has absorbed laser energy. A higher temperature results in a significantly higher power emitted due to the fourth power dependency, creating a challenge to maintaining the temperature long enough for fusion to occur.
Absorptivity and Emissivity in Heat Transfer
Absorptivity (\( \alpha \)) and emissivity (\( \varepsilon \)) are critical concepts in heat transfer and thermodynamics, especially when dealing with the heating and cooling of objects. Absorptivity is the measure of how much radiation (in this case, laser light) is absorbed by a material. It ranges from 0 (no absorption) to 1 (complete absorption).

Emissivity, meanwhile, measures a material's ability to emit energy as thermal radiation and also varies from 0 to 1, with 1 being a perfect emitter. The values of \( \alpha \) and \( \varepsilon \) help determine the rate at which an object 鈥� like a deuterium-tritium pellet 鈥� will heat up and cool down.

In the exercise, with an absorptivity of 0.3, the fuel pellet absorbs 30% of the incident laser energy. This translates into the actual heating power that the pellet experiences. On the other hand, an emissivity of 0.8 indicates that the pellet, once hot, will efficiently radiate thermal energy. For an efficient fusion process, we seek a balance where the fuel's absorptivity maximizes its temperature increase, while its emissivity must not be too high to prematurely lose the heat required for fusion to occur.

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Most popular questions from this chapter

A thin sheet of glass is used on the roof of a greenhouse and is irradiated as shown. $$ G_{5} $$ The irradiation comprises the total solar flux \(G_{S}\), the flux \(G_{\mathrm{am}}\) due to atmospheric emission (sky radiation), and the flux \(G_{i}\) due to emission from interior surfaces. The fluxes \(G_{\text {atm }}\) and \(G_{i}\) are concentrated in the far IR region \((\lambda \gtrless 8 \mu \mathrm{m})\). The glass may also exchange energy by convection with the outside and inside atmospheres. The glass may be assumed to be totally transparent for \(\lambda<1 \mu \mathrm{m}\left(\tau_{\lambda}=1.0\right.\) for \(\left.\lambda<1 \mu \mathrm{m}\right)\) and opaque, with \(\alpha_{\lambda}=1.0\) for \(\lambda \geq 1 \mu \mathrm{m}\). (a) Assuming steady-state conditions, with all radiative fluxes uniformly distributed over the surfaces and the glass characterized by a uniform temperature \(T_{g}\), write an appropriate energy balance for a unit area of the glass. (b) For \(T_{z}=27^{\circ} \mathrm{C}, \quad h_{i}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{S}=1100\) \(\mathrm{W} / \mathrm{m}^{2}, T_{\infty \rho}=24^{\circ} \mathrm{C}, \quad h_{o}=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \quad G_{\operatorname{tam}}=\) \(250 \mathrm{~W} / \mathrm{m}^{2}\), and \(G_{i}=440 \mathrm{~W} / \mathrm{m}^{2}\), calculate the temperature of the greenhouse ambient air, \(T_{x j}\).

A cylinder of \(30-\mathrm{mm}\) diameter and \(150-\mathrm{mm}\) length is heated in a large furnace having walls at \(1000 \mathrm{~K}\), while air at \(400 \mathrm{~K}\) is circulating at \(3 \mathrm{~m} / \mathrm{s}\). Estimate the steady-state cylinder temperature under the following specified conditions. (a) The cylinder is in cross flow, and its surface is diffuse and gray with an emissivity of \(0.5\). (b) The cylinder is in cross flow, but its surface is spectrally selective with \(\alpha_{\lambda}=0.1\) for \(\lambda \leq 3 \mu \mathrm{m}\) and \(\alpha_{\lambda}=0.5\) for \(\lambda>3 \mu \mathrm{m}\). (c) The cylinder surface is positioned such that the airflow is longitudinal and its surface is diffuse and gray. (d) For the conditions of part (a), compute and plot the cylinder temperature as a function of the air velocity for \(1 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\).

A wet towel hangs on a clothes line under conditions for which one surface receives solar irradiation of \(G_{S}=900 \mathrm{~W} / \mathrm{m}^{2}\) and both surfaces are exposed to atmospheric (sky) and ground radiation of \(G_{\text {tam }}=\) \(200 \mathrm{~W} / \mathrm{m}^{2}\) and \(G_{\mathrm{g}}=250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Under moderately windy conditions, airflow at a temperature of \(27^{\circ} \mathrm{C}\) and a relative humidity of \(60 \%\) maintains a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. The wet towel has an emissivity of \(0.96\) and a solar absorptivity of \(0.65\). As a first approximation the properties of the atmospheric air may be evaluated at a temperature of \(300 \mathrm{~K}\). Determine the temperature \(T_{s}\) of the towel. What is the corresponding evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long?

Approximations to Planck's law for the spectral emissive power are the Wien and Rayleigh-Jeans spectral distributions, which are useful for the extreme low and high limits of the product \(\lambda T\), respectively. (a) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp \left(-\frac{C_{2}}{\lambda T}\right) $$ when \(C_{2} / \lambda T\) sor 1 and determine the error (compared to the exact distribution) for the condition \(\lambda T=\) \(2898 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as Wien's law. (b) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ when \(C_{2} / \lambda T \& 1\) and determine the error (compared to the exact distribution) for the condition \(\lambda T=100,000 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as the Rayleigh-Jeans law.

A furnace with a long, isothermal, graphite tube of diameter \(D=12.5 \mathrm{~mm}\) is maintained at \(T_{f}=2000 \mathrm{~K}\) and is used as a blackbody source to calibrate heat flux gages. Traditional heat flux gages are constructed as blackened thin films with thermopiles to indicate the temperature change caused by absorption of the incident radiant power over the entire spectrum. The traditional gage of interest has a sensitive area of \(5 \mathrm{~mm}^{2}\) and is mounted coaxial with the furnace centerline, but positioned at a distance of \(L=60 \mathrm{~mm}\) from the beginning of the heated section. The cool extension tube serves to shield the gage from extraneous radiation sources and to contain the inert gas required to prevent rapid oxidation of the graphite tube. (a) Calculate the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) on the traditional gage for this condition, assuming that the extension tube is cold relative to the furnace. (b) The traditional gage is replaced by a solid-state (photoconductive) heat flux gage of the same area, but sensitive only to the spectral region between \(0.4\) and \(2.5 \mu \mathrm{m}\). Calculate the radiant heat flux incident on the solid-state gage within the prescribed spectral region. (c) Calculate and plot the total heat flux and the heat flux in the prescribed spectral region for the solidstate gage as a function of furnace temperature for the range \(2000 \leq T_{f} \leq 3000 \mathrm{~K}\). Which gage will have an output signal that is more sensitive to changes in the furnace temperature?
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