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Chapter 38: Problem 63

In the dangerous situation of prompt criticality in a fission reactor, the generation time drops to \(100 \mu\) s as prompt neutrons alone sustain the chain reaction. If a reactor goes prompt critical with \(k=1.001,\) how long does it take for a 100 -fold increase in reactor power?

Short Answer

Expert verified
Therefore, in case of prompt criticality, it would take approximately 0.46 seconds for a 100-fold increase in reactor power.

Step by step solution

01

Understand the model

Given that the reactor goes prompt critical, it means the system reaction is sustained by the prompt neutrons alone. The system is described by the equation \(N(t) = N_0e^{(k-1)t/T}\), where \(N(t)\) is the number of neutrons at time \(t\), \(N_0\) is the initial number of neutrons, \(k\) is the multiplication factor, and \(T\) is the generation time.
02

Substitute the Values

We are looking for \(t\) when \(N(t) = 100N_0\). Thus, this gives us: \(100N_0 = N_0e^{(1.001-1)t/100*10^{-6}}\). Note the generation time (100 microseconds) has been converted to seconds.
03

Complete the calculation

Rearrange the equation to solve for \(t\). Cancel \(N_0\) from both sides to give \(100 = e^{0.001t/100*10^{-6}}\). The logarithm of both sides then gives \(ln(100) = 0.001t / 100 * 10^{-6}\) Thus, \(t = ln(100) / (0.001 / 100 * 10^{-6})\).
04

Evaluate the expression

Calculating the right-hand side of the final equation in the previous step, we get \(t = ln(100) / (0.001 / 100 * 10^{-6}) = 0.46 \) seconds approximately.

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