Chain Reactions and Reactor Criticality
Chain Reactions and Reactor Criticality
A nuclear chain reaction occurs when a fission event produces neutrons that go on to cause additional fission events, each of which releases more neutrons, in a self-sustaining cycle. This is the fundamental principle behind both nuclear reactors and nuclear weapons, and understanding the mathematics of chain reactions is the core discipline of reactor physics.
When a uranium-235 or plutonium-239 nucleus absorbs a neutron, it fissions into two smaller fragments and releases an average of about 2.5 additional neutrons. Whether these neutrons go on to cause more fissions depends on many factors: how many escape the reactor, how many are absorbed by non-fissile materials, and how many slow down to the right energy to cause fission.
The effective multiplication factor, k-effective, is the ratio of neutrons in one generation to neutrons in the previous generation. If k equals exactly 1, the reactor is critical — the chain reaction is self-sustaining at a steady power level. If k is greater than 1, the reactor is supercritical and power increases exponentially. If k is less than 1, the reactor is subcritical and the chain reaction dies out. The entire art of reactor control is maintaining k at precisely 1.
The four-factor formula gives k-infinity — the multiplication factor for an infinitely large reactor — as the product of four factors: eta (the reproduction factor), epsilon (the fast fission factor), p (the resonance escape probability), and f (the thermal utilization factor). For a finite reactor, k-effective equals k-infinity times the non-leakage probability. Each of these factors captures a different physical process, and optimizing reactor design means carefully balancing all of them.
$$k_{\text{eff}} = \frac{\text{neutrons in generation } n+1}{\text{neutrons in generation } n}$$
$k_{\text{eff}} = 1$: Critical (steady state). $k_{\text{eff}} > 1$: Supercritical (power rising). $k_{\text{eff}} < 1$: Subcritical (power falling).
$$k_\infty = \eta \, \varepsilon \, p \, f$$
where:
$\eta$ = reproduction factor (neutrons produced per absorption in fuel)
$\varepsilon$ = fast fission factor ($\approx 1.03$)
$p$ = resonance escape probability (fraction that avoid resonance capture)
$f$ = thermal utilization factor (fraction of thermal absorptions in fuel)
For a finite reactor: $k_{\text{eff}} = k_\infty \cdot P_{NL}$ where $P_{NL}$ is the non-leakage probability.
Reactivity $\rho$ measures the departure from criticality:
$$\rho = \frac{k_{\text{eff}} - 1}{k_{\text{eff}}}$$
$\rho = 0$: critical. $\rho > 0$: supercritical. $\rho < 0$: subcritical. Often expressed in pcm (1 pcm = $10^{-5}$).
A reactor has $\eta = 2.04$, $\varepsilon = 1.03$, $p = 0.87$, $f = 0.92$, and $P_{NL} = 0.96$. Is it critical?
Step 1: Compute $k_\infty$.
$$k_\infty = 2.04 \times 1.03 \times 0.87 \times 0.92 = 1.682$$
Step 2: Apply non-leakage.
$$k_{\text{eff}} = 1.682 \times 0.96 = 1.615$$
Step 3: Since $k_{\text{eff}} = 1.615 > 1$, the reactor is supercritical. Reactivity $\rho = (1.615-1)/1.615 = 0.381 = 38{,}100$ pcm. This is far too supercritical — control rods would need to be inserted.
Starting with 1000 neutrons and $k_{\text{eff}} = 1.001$, how many neutrons exist after 100 generations?
Step 1: After $n$ generations: $N_n = N_0 \cdot k^n$
$$N_{100} = 1000 \times 1.001^{100}$$
Step 2: $1.001^{100} = e^{100 \ln 1.001} = e^{0.09995} \approx 1.105$
$$N_{100} \approx 1105 \text{ neutrons}$$
Even a tiny supercriticality of 0.1% causes a 10.5% increase in 100 generations.
A reactor must reduce $k_{\text{eff}}$ from 1.05 to 1.00 by adding boron (a neutron absorber). If boron reduces $f$ by 5%, and the original $f = 0.92$, verify this achieves criticality.
Step 1: New $f = 0.92 \times 0.95 = 0.874$.
Step 2: Since $k_{\text{eff}} \propto f$, new $k_{\text{eff}} = 1.05 \times (0.874/0.92) = 1.05 \times 0.9500 = 0.9975$.
This is slightly subcritical ($k = 0.9975$), so the boron insertion is slightly too much. Fine-tuning the boron concentration would achieve exactly $k = 1$.
Practice Problems
Show Answer Key
1. Subcritical ($k < 1$).
2. $k_\infty = 1.8 \times 1.02 \times 0.90 \times 0.95 = 1.571$.
3. $\rho = (1.003-1)/1.003 = 0.002991 = 299.1$ pcm.
4. $N = 10000 \times 0.99^{200} = 10000 \times e^{-2.01} \approx 10000 \times 0.134 = 1340$.
5. $k_{\text{eff}} = 1.25 \times 0.82 = 1.025$. Supercritical.
6. $k_\infty \propto \eta$. New $k_\infty = 1.2 \times (2.1/2.0) = 1.26$.
7. A small fraction of neutrons cause fission in U-238 before slowing down (fast fission). This adds slightly more neutrons than thermal fission alone, making $\varepsilon \gtrsim 1$.
8. $k_{\text{eff}} \propto f$. New $k = 1.02 \times (0.88/0.92) = 0.9757$. Now subcritical.
9. $2 = 1.01^n \Rightarrow n = \ln 2/\ln 1.01 = 69.7 \approx 70$ generations.
10. Larger reactors have a smaller surface-to-volume ratio, so fewer neutrons reach the surface and escape. Thus $P_{NL}$ increases with size, approaching 1 for very large reactors.