/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The neutron generation time \(t_... [FREE SOLUTION] | 91Ó°ÊÓ

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The neutron generation time \(t_{\mathrm{gen}}\) in a reactor is the average time needed for a fast neutron emitted in one fission event to be slowed to thermal energies by the moderator and then initiate another fission event. Suppose the power output of a reactor at time \(t=0\) is \(P_{0}\). Show that the power output a time \(t\) later is \(P(t)\), where \(P(t)=P_{0} k^{t t \text { tgen }}\) and \(k\) is the multiplication factor. For constant power output, \(k=1\).

Short Answer

Expert verified
Power output is given by \(P(t) = P_0 k^{t/t_{\text{gen}}}\). If \(k=1\), power is constant.

Step by step solution

01

Understanding the Problem

We need to express the power output of a reactor at time \(t\) given that, at time \(t=0\), the power output is \(P_0\). The power output changes over time as it depends on the multiplication factor \(k\) and the neutron generation time \(t_{\text{gen}}\).
02

Define the Exponential Growth Model

Reactors are modeled by exponential growth when there is a reproducible process like fission events. This is similar to compound interest in finance or population biology. So the power at time \(t\) can be described by an exponential function, \(P(t) = P_0 e^{rt}\), where \(r\) is the growth rate.
03

Relate the Multiplication Factor and Growth Rate

The multiplication factor \(k\) describes the average number of neutrons from each fission event that go on to cause additional fission events. Hence, each fission increases the neutron population by a factor of \(k\). For neutron generation, this relationship can be seen as \(k = e^{r t_{\text{gen}}}\), solving for \(r\), we get \(r = \frac{\ln(k)}{t_{\text{gen}}}\).
04

Substitute the Growth Rate into the Exponential Model

Replacing \(r\) in the power function with \(\frac{\ln(k)}{t_{\text{gen}}}\), we get: \[ P(t) = P_0 e^{t \left(\frac{\ln(k)}{t_{\text{gen}}}\right)} = P_0 \left(e^{\ln(k)}\right)^{\frac{t}{t_{\text{gen}}}} = P_0 \left(k\right)^{\frac{t}{t_{\text{gen}}}} \]This simplifies to the requested form.
05

Conclusion

Thus, we have shown that the power output at any time \(t\) is given by \(P(t) = P_0 k^{\frac{t}{t_{\text{gen}}}}\). If the power output remains constant, \(k = 1\).

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Key Concepts

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

Neutron Generation Time
Neutron generation time is a crucial concept in nuclear reactor physics. It represents the average time taken for a fast neutron generated from one fission event to slow down to a thermal neutron, which can then cause another fission. This slows down because the neutron collides with nuclei called moderators. These moderators, such as heavy water or graphite, are materials used to reduce the speed of the neutrons, making them more likely to cause further fission when they interact with fissile material.

In a practical sense, neutron generation time helps us understand how quickly a reactor can respond and adjust its power output. A shorter generation time implies quicker responses to changes in neutron flux, which is important for controlling the reactor's power level. All together, this makes neutron generation time a vital parameter for reactor kinetics and control.
Exponential Growth Model
The exponential growth model is a mathematical description used to represent processes where the rate of change is proportional to the size of the current state. In the context of a nuclear reactor, this model elegantly illustrates how the number of neutrons, and hence power output, grows over time. Each fission event produces neutrons, which can cause further fission, effectively leading to a multiplicative effect similar to compounding interest in finance.

Using the equation \(P(t) = P_0 e^{rt}\), where \(P(t)\) is the power output at time \(t\), \(P_0\) is the initial power output, and \(r\) is the growth rate, we can visualize how the process starts small and rapidly increases over time. This exponential relationship is vital for predicting the changes in power levels over time in a reactor. Understanding this growth model also helps in seeing how small changes in parameters like the multiplication factor or generation time can have large effects on the power output over time.
Multiplication Factor
The multiplication factor, denoted as \(k\), is another fundamental concept related to the operation of nuclear reactors. It represents the average number of neutrons from each fission event that goes on to cause subsequent fission events. It is a measure of the chain reaction's sustainability.
A multiplication factor greater than 1 indicates a supercritical state, where the reaction is growing over time, leading to an increase in neutron population and power. When \(k\) equals 1, the reaction is critical, meaning it is self-sustaining but not growing. If \(k\) is less than 1, the reactor is subcritical, and the reaction will slowly dwindle.

Understanding the multiplication factor is crucial for controlling reactor dynamics. By knowing and adjusting \(k\), operators can manage how quickly or slowly the reaction progresses, maintaining desired power levels and ensuring safety. Ultimately, this helps in achieving a balance, as sustained or self-sustaining operations are needed for efficient power output and safety.

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Most popular questions from this chapter

For overcoming the Coulomb barrier for fusion, methods other than heating the fusible material have been suggested. For example, if you were to use two particle accelerators to accelerate two beams of deuterons directly toward each other so as to collide head-on, (a) what voltage would each accelerator require in order for the colliding deuterons to overcome the Coulomb barrier? (b) Why do you suppose this method is not presently used?

Calculate the energy released in the fission reaction $$ { }^{235} \mathrm{U}+\mathrm{n} \rightarrow{ }^{141} \mathrm{Cs}+{ }^{93} \mathrm{Rb}+2 \mathrm{n} $$ Here are some atomic and particle masses. \(\begin{array}{cccc}{ }^{235} \mathrm{U} & 235.04392 \mathrm{u} & { }^{93} \mathrm{Rb} & 92.92157 \mathrm{u} \\\ { }^{141} \mathrm{Cs} & 140.91963 \mathrm{u} & \mathrm{n} & 1.00866 \mathrm{u}\end{array}\)

(a) A neutron of mass \(m_{\mathrm{n}}\) and kinetic energy \(K\) makes a head-on elastic collision with a stationary atom of mass \(m\). Show that the fractional kinetic energy loss of the neutron is given by $$ \frac{\Delta K}{K}=\frac{4 m_{\mathrm{n}} m}{\left(m+m_{\mathrm{n}}\right)^{2}} $$ Find \(\Delta K / K\) for each of the following acting as the stationary atom: (b) hydrogen, (c) deuterium, (d) carbon, and (e) lead. (f) If \(K=1.00 \mathrm{MeV}\) initially, how many such head-on collisions would it take to reduce the neutron's kinetic energy to a thermal value \((0.025 \mathrm{eV})\) if the stationary atoms it collides with are deuterium, a commonly used moderator? (In actual moderators, most collisions are not head-on.)

Assume that a plasma temperature of \(1 \times 10^{8} \mathrm{~K}\) is reached in a laser-fusion device. (a) What is the most probable speed of a deuteron at that temperature? (b) How far would such a deuteron move in a confinement time of \(1 \times 10^{-12}\) s?

A \({ }^{236} \mathrm{U}\) nucleus undergoes fission and breaks into two middle- mass fragments, \({ }^{140} \mathrm{Xe}\) and \({ }^{96}\) Sr. (a) By what percentage does the surface area of the fission products differ from that of the original \({ }^{236} \mathrm{U}\) nucleus? (b) By what percentage does the volume change? (c) By what percentage does the electric potential energy change? The electric potential energy of a uniformly charged sphere of radius \(r\) and charge \(Q\) is given by $$ U=\frac{3}{5}\left(\frac{Q^{2}}{4 \pi \varepsilon_{0} r}\right) $$

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