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Chapter 8 • Theory & Derivations

Reactor Kinetics, Reactivity Feedback, Control Rods & Poisoning

Time-dependent dynamics, kinetics, control, and transient stability of nuclear reactors: derivation of the Point Reactor Kinetics Equations (PRKE) with six delayed neutron precursor groups; prompt neutron lifetime l versus generation time Lambda; prompt criticality threshold rho = beta and explosive power runaway; the Inhour (Inverse Hour) equation relating static reactivity rho to stable asymptotic reactor period T; analytical solution of step reactivity insertions and the prompt jump approximation; control rod physics, perturbation theory, differential and integral control rod worth curves, and burnable absorbers; inherent negative reactivity feedback coefficients (fuel Doppler broadening alpha_D, moderator temperature alpha_M, and void alpha_V); fission product poisoning dynamics, coupled non-linear differential equations for the Iodine-135 / Xenon-135 chain, equilibrium xenon poisoning, the post-shutdown 'iodine pit' xenon peak, and Samarium-149 equilibrium poisoning.

§8.1 Time-Dependent Neutron Diffusion & Point Reactor Kinetics Equations (PRKE)

1. The Time-Dependent Diffusion Equation

When neutron populations vary with time, the net time rate of change of neutron density is governed by the time-dependent diffusion equation: $$\frac{1}{v} \frac{\partial \phi(\vec{r}, t)}{\partial t} = D \nabla^2 \phi(\vec{r}, t) - \Sigma_a \phi(\vec{r}, t) + S(\vec{r}, t)$$ In a multiplying medium, the source $S(\vec{r}, t)$ consists of two distinct components:

  1. Prompt Neutrons: Fraction $1 - \beta$ of all fission neutrons, emitted promptly: $(1 - \beta) k_\infty \Sigma_a \phi$.
  2. Delayed Neutrons: Produced via radioactive decay of six delayed precursor groups: $\sum_{i=1}^6 \lambda_i C_i(\vec{r}, t)$.

2. The Point Reactor Kinetics Equations (PRKE)

Assuming the spatial flux shape remains fixed during transients (the point reactor approximation $\phi(\vec{r}, t) \approx \psi(\vec{r}) P(t)$), integrating over core volume yields the classical Point Reactor Kinetics Equations for total neutron population $n(t)$ (or thermal power $P(t)$): $$\mathbf{\frac{dn(t)}{dt} = \frac{\rho(t) - \beta}{\Lambda} n(t) + \sum_{i=1}^6 \lambda_i C_i(t)}$$ $$\mathbf{\frac{dC_i(t)}{dt} = \frac{\beta_i}{\Lambda} n(t) - \lambda_i C_i(t) \quad (i = 1, 2, \dots, 6)}$$ where:

  • $\rho(t) = \frac{k(t) - 1}{k(t)}$ is the net reactivity.
  • $\beta = \sum_{i=1}^6 \beta_i$ is the total delayed neutron fraction ($\beta \approx 0.0065$ for $^{235}\text{U}$).
  • $\Lambda \equiv \frac{l}{k} \approx l$ is the prompt neutron generation time ($\Lambda \sim 10^{-4}\text{ s}$ in PWRs, $\sim 10^{-7}\text{ s}$ in fast reactors).
  • $C_i(t)$ is the precursor population of group $i$, decaying with decay constant $\lambda_i$.

§8.2 Prompt Neutron Generation Time & The Prompt Criticality Barrier (rho = beta)

1. The Purely Prompt Regime ($\rho \ge \beta$)

If a reactor receives a reactivity insertion equal to or exceeding the delayed neutron fraction: $$\rho \ge \beta \implies \rho - \beta \ge 0$$ The prompt production term $\frac{\rho - \beta}{\Lambda} n(t)$ becomes positive on its own! In this terrifying condition—known as Prompt Criticality ($\rho = 1\$$)—the chain reaction sustains itself entirely on prompt neutrons, completely bypassing the slow decay of delayed precursors! The neutron population explodes as: $$n(t) = n_0 \exp\left( \frac{\rho - \beta}{\Lambda} t \right)$$ For $\rho = 1.1 \beta$ ($\rho - \beta = 0.1 \beta \approx 6.5 \times 10^{-4}$) in a thermal reactor with $\Lambda = 10^{-4}\text{ s}$: $$\frac{\rho - \beta}{\Lambda} = \frac{6.5 \times 10^{-4}}{10^{-4}\text{ s}} = 6.5\text{ s}^{-1} \implies n(t) = n_0 e^{6.5 t}$$ In just one second, power multiplies by $e^{6.5} \approx 665$! In a fast reactor ($\Lambda = 10^{-7}\text{ s}$), $\frac{\rho - \beta}{\Lambda} = 6500\text{ s}^{-1}$, multiplying power by $10^{28}$ in one-hundredth of a second (a catastrophic explosive disassembly).

2. The Delayed Critical Buffer

For all normal reactor operations, reactivity is strictly confined to: $$\mathbf{0 \le \rho < \beta \quad (0 \le \rho < 1\$)}$$ In this delayed critical regime, the prompt term is negative ($\rho - \beta < 0$), forcing the system to wait for delayed precursors to decay before each new generation can be sustained.

§8.3 The Inhour (Inverse Hour) Equation & Stable Reactor Period

1. Derivation of the Characteristic Equation

Assuming exponential solutions of the form $n(t) = n_0 e^{\omega t}$ and $C_i(t) = C_{i, 0} e^{\omega t}$ for constant reactivity $\rho$: Substitute into the precursor equation: $$\omega C_{i, 0} = \frac{\beta_i}{\Lambda} n_0 - \lambda_i C_{i, 0} \implies C_{i, 0} = \frac{\beta_i}{\Lambda (\omega + \lambda_i)} n_0$$ Substitute $C_{i, 0}$ into the neutron equation: $$\omega n_0 = \frac{\rho - \beta}{\Lambda} n_0 + \sum_{i=1}^6 \lambda_i \left[ \frac{\beta_i}{\Lambda (\omega + \lambda_i)} n_0 \right]$$ Dividing through by $n_0$ and multiplying by $\Lambda$: $$\omega \Lambda = \rho - \beta + \sum_{i=1}^6 \frac{\lambda_i \beta_i}{\omega + \lambda_i}$$ Since $\beta = \sum \beta_i$, we write $\rho = \omega \Lambda + \sum \beta_i - \sum \frac{\lambda_i \beta_i}{\omega + \lambda_i} = \omega \Lambda + \sum \beta_i \left( 1 - \frac{\lambda_i}{\omega + \lambda_i} \right) = \omega \Lambda + \sum \frac{\omega \beta_i}{\omega + \lambda_i}$. Setting reactor period $T \equiv 1/\omega$: $$\mathbf{\rho = \frac{\Lambda}{T} + \sum_{i=1}^6 \frac{\beta_i}{1 + \lambda_i T}}$$ This is the foundational Inhour Equation (named from historic units of inverse hours, $\text{hr}^{-1}$).

2. Roots and Asymptotic Stable Period

For any step reactivity $\rho$, the inhour equation possesses seven real roots: $\omega_0, \omega_1, \dots, \omega_6$. For $\rho > 0$, the largest root $\omega_0 > 0$ defines the stable reactor period $T = 1/\omega_0$. The remaining six roots $\omega_1, \dots, \omega_6$ are negative and die out within tens of seconds. After these transients decay, reactor power escalates cleanly as: $$\mathbf{P(t) = P_0 e^{t / T}}$$ The doubling time $T_d$ (time for power to double) is: $$\mathbf{T_d = T \ln 2 \approx 0.69315 \, T}$$

§8.4 Transient Response to Step Reactivity: The Prompt Jump Approximation

1. The Prompt Jump Phenomenon

When a positive step reactivity $\rho < \beta$ is inserted into a critical reactor at $t = 0$, the neutron population exhibits a two-stage response:

  1. The Prompt Jump (0 to 10 ms): Because delayed precursors cannot change instantaneously ($C_i(0^+) = C_i(0^-) = \frac{\beta_i}{\Lambda} n_0$), the prompt neutron population jumps almost instantaneously until the prompt sink balances precursor emission: $$\frac{dn}{dt} \approx 0 \implies \frac{\rho - \beta}{\Lambda} n(0^+) + \sum \lambda_i C_i(0) = 0 \implies \frac{\beta - \rho}{\Lambda} n(0^+) = \frac{\beta}{\Lambda} n_0$$ $$\mathbf{n(0^+) = n_0 \frac{\beta}{\beta - \rho} = n_0 \frac{1}{1 - \rho/\beta} = n_0 \frac{1}{1 - \$}}$$
  2. Precursor-Controlled Growth ($t > 1\text{ s}$): Following the prompt jump, power increases on the slow stable period $T \approx \frac{\beta - \rho}{\lambda_{\text{eff}} \rho}$: $$\mathbf{n(t) \approx n_0 \left[ \frac{\beta}{\beta - \rho} \exp\left( \frac{\lambda_{\text{eff}} \rho}{\beta - \rho} t \right) - \frac{\rho}{\beta - \rho} \exp\left( - \frac{\beta - \rho}{\Lambda} t \right) \right]}$$

§8.5 Control Rod Physics: Perturbation Theory, Rod Worth & Burnable Absorbers

1. Control Rod Worth via First-Order Perturbation Theory

Control rods contain strong neutron-absorbing materials (such as Boron Carbide $\text{B}_4\text{C}$, Cadmium, Silver-Indium-Cadmium Ag-In-Cd, or Hafnium). When a control rod is partially inserted to depth $z$ into a cylindrical reactor of active height $H$, first-order perturbation theory shows that the differential reactivity worth $d\rho/dz$ is proportional to the square of the unperturbed axial flux: $$\frac{d\rho}{dz} \propto \phi^2(z) = \phi_0^2 \sin^2\left( \frac{\pi z}{H} \right)$$ Integrating from $0$ to $z$ yields the cumulative Integral Control Rod Worth $\rho(z)$: $$\rho(z) = \rho_{\text{tot}} \frac{\int_0^z \sin^2(\pi z'/H) dz'}{\int_0^H \sin^2(\pi z'/H) dz'} = \mathbf{\rho_{\text{tot}} \left[ \frac{z}{H} - \frac{1}{2\pi} \sin\left( \frac{2\pi z}{H} \right) \right]}$$ This exhibits the characteristic S-curve:

  • Near the top ($z \approx 0$) and bottom ($z \approx H$), flux is low, so rod movement produces tiny reactivity change.
  • Near the core midplane ($z = H/2$), flux is maximal, so rod movement produces maximum differential worth: $\left.\frac{d\rho}{dz}\right|_{\max} = \frac{2 \rho_{\text{tot}}}{H}$.

2. Burnable Absorbers

To compensate for the large excess reactivity of freshly loaded fuel without requiring dozens of physical control rods, reactors incorporate burnable poisons (such as Gadolinium Oxide $\text{Gd}_2\text{O}_3$ or Boron in fuel cladding). These high cross-section isotopes burn away at the same rate as fuel fissile inventory decreases, keeping net core reactivity nearly flat throughout a 24-month operating cycle.

§8.6 Inherent Reactivity Feedback: Fuel Doppler, Moderator & Void Coefficients

1. Reactivity Coefficients of Temperature

Any temperature increase in a reactor core alters material densities, thermal cross sections, and resonance absorption, producing net reactivity feedback: $$\frac{d\rho}{dt} = \alpha_F \frac{dT_{\text{fuel}}}{dt} + \alpha_M \frac{dT_{\text{mod}}}{dt} + \alpha_V \frac{d\alpha_{\text{void}}}{dt}$$

  • Fuel Doppler Coefficient ($\alpha_F$): Promptly negative ($\alpha_F \sim -2\text{ to }-4\text{ pcm/}^\circ\text{C}$). Operates with zero thermal delay because heat is born directly inside the $\text{UO}_2$ crystals, instantly broadening ${}^{238}\text{U}$ resonance capture.
  • Moderator Temperature Coefficient ($\alpha_M$): Governed by coolant thermal expansion. In an under-moderated PWR, heating decreases water density, reducing moderation and lowering $k_{\text{eff}}$ ($\alpha_M \sim -10\text{ to }-40\text{ pcm/}^\circ\text{C} < 0$).
  • Void Coefficient of Reactivity ($\alpha_V$): In BWRs and PWRs, steam bubble formation displaces water. Because the core is under-moderated, voiding decreases reactivity ($\alpha_V < 0$). Conversely, in the Chernobyl RBMK reactor, the over-moderated graphite design caused a positive void coefficient ($\alpha_V > 0$), which contributed directly to the runaway power explosion in 1986.
Modern nuclear regulatory safety mandates an unconditionally negative power coefficient of reactivity under all operating states.

§8.7 Fission Product Poisoning: Xenon-135 Dynamics, The Iodine Pit & Samarium-149

1. Xenon-135: The Super-Poison

Xenon-135 (${}^{135}\text{Xe}$) is the most potent neutron poison known to science, with a microscopic thermal absorption cross section of: $$\mathbf{\sigma_a({}^{135}\text{Xe}) \approx 2.65 \times 10^6\text{ barns} = 2.65 \times 10^{-18}\text{ cm}^2}$$ (Over $3{,}800$ times greater than ${}^{235}\text{U}$ fission!).

2. The Iodine-Xenon Dynamic System

${}^{135}\text{Xe}$ is formed through two pathways:

  1. Direct fission yield: $\gamma_{\text{Xe}} \approx 0.003$ ($0.3\%$).
  2. Radioactive decay of Iodine-135 (${}^{135}\text{I}$, yield $\gamma_I \approx 0.061 = 6.1\%$): $$\text{Fission} \longrightarrow {}^{135}_{53}\text{I} \xrightarrow[\beta^-, \, T_{1/2}=6.57\text{ h}]{} {}^{135}_{54}\text{Xe} \xrightarrow[\beta^-, \, T_{1/2}=9.14\text{ h}]{} {}^{135}_{55}\text{Cs}$$
The coupled non-linear differential equations are: $$\mathbf{\frac{d I(t)}{dt} = \gamma_I \Sigma_f \phi(t) - \lambda_I I(t)}$$ $$\mathbf{\frac{d X(t)}{dt} = \gamma_{\text{Xe}} \Sigma_f \phi(t) + \lambda_I I(t) - \lambda_{\text{Xe}} X(t) - \sigma_a^{\text{Xe}} X(t) \phi(t)}$$ where: $$\lambda_I = \frac{\ln 2}{6.57 \times 3600\text{ s}} \approx 2.93 \times 10^{-5}\text{ s}^{-1}, \qquad \lambda_{\text{Xe}} = \frac{\ln 2}{9.14 \times 3600\text{ s}} \approx 2.10 \times 10^{-5}\text{ s}^{-1}$$ At steady-state full power: $$I_0 = \frac{\gamma_I \Sigma_f \phi_0}{\lambda_I}, \qquad X_0 = \frac{(\gamma_I + \gamma_{\text{Xe}}) \Sigma_f \phi_0}{\lambda_{\text{Xe}} + \sigma_a^{\text{Xe}} \phi_0}$$

3. Post-Shutdown Xenon Peak ("The Iodine Pit")

When a reactor trips or is shut down from high power ($\phi \to 0$):

  • Xenon destruction by neutron burnup ($\sigma_a^{\text{Xe}} X \phi$) abruptly drops to zero!
  • Meanwhile, the large pre-existing inventory of Iodine-135 continues to decay into Xenon-135 at rate $\lambda_I I(t)$.
  • Because $\lambda_I > \lambda_{\text{Xe}}$ ($T_{1/2}^I = 6.6\text{ h} < T_{1/2}^{\text{Xe}} = 9.1\text{ h}$), Xenon production initially outpaces its decay!
Xenon concentration swells to a colossal peak approximately $10\text{ to }11\text{ hours}$ after shutdown: $$t_{\text{peak}} = \frac{1}{\lambda_{\text{Xe}} - \lambda_I} \ln\left( \frac{\lambda_{\text{Xe}}}{\lambda_I} \left[ 1 + \frac{\lambda_{\text{Xe}} - \lambda_I}{\lambda_I} \frac{X_0}{I_0} \right] \right) \approx \mathbf{10.5\text{ hours}}$$ In high-flux reactors, this peak introduces a massive negative reactivity deficit ($\Delta\rho \sim -3000\text{ to }-5000\text{ pcm}$) that completely exceeds control rod withdrawal worth—locking the reactor in a temporary shutdown state known as Xenon Deadtime!

ADVANCED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, quantitative calculations, and step-by-step examination solutions for Unit 1.

Solved Problem Example 8.1: Prompt Jump and Stable Reactor Period Calculation for Step Reactivity Insertions

A critical Uranium-235 fueled thermal research reactor operates at steady thermal power $P_0 = 100\text{ kW}$. The kinetics parameters are:

  • Total delayed neutron fraction $\beta = 0.0065$
  • Prompt neutron generation time $\Lambda = 5.0 \times 10^{-5}\text{ seconds}$
  • Effective one-group precursor decay constant $\bar{\lambda} = 0.080\text{ s}^{-1}$

At $t = 0$, an operator accidentally withdraws a control rod, inserting a positive step reactivity of $\rho = +0.0020$ ($+200\text{ pcm} = 0.308$$). (a) Calculate the instantaneous power $P(0^+)$ immediately following the prompt jump. (b) Using the one delayed group inhour equation $\rho = \frac{\Lambda}{T} + \frac{\beta}{1 + \bar{\lambda} T}$, determine the asymptotic stable reactor period $T$ in seconds. (c) Calculate the reactor doubling time $T_d$, and compute the reactor power at $t = 30\text{ seconds}$ after insertion.

(a) Power Immediately Following Prompt Jump: Using the prompt jump formula:

$$P(0^+) = P_0 \frac{\beta}{\beta - \rho} = P_0 \frac{0.0065}{0.0065 - 0.0020} = P_0 \frac{0.0065}{0.0045} = P_0 \left( \frac{13}{9} \right) \approx 1.4444 \, P_0$$
$$P(0^+) = 1.4444 \times 100\text{ kW} \approx \mathbf{144.4\text{ kW}}$$

Within a fraction of a millisecond, the power jumps by $+44.4\%$ before delayed precursors have had time to decay!

(b) Stable Reactor Period $T$: Since $T \gg \Lambda/\beta$, the prompt term $\Lambda/T$ is negligible compared to precursor terms:

$$\rho \approx \frac{\beta}{1 + \bar{\lambda} T} \implies 1 + \bar{\lambda} T = \frac{\beta}{\rho}$$
$$\bar{\lambda} T = \frac{\beta}{\rho} - 1 = \frac{\beta - \rho}{\rho}$$
$$T \approx \frac{\beta - \rho}{\bar{\lambda} \rho}$$

Substituting numerical values:

$$T \approx \frac{0.0065 - 0.0020}{(0.080\text{ s}^{-1})(0.0020)} = \frac{0.0045}{0.000160} = \mathbf{28.125\text{ seconds}}$$

Checking the prompt term correction:

$$\rho = \frac{5.0 \times 10^{-5}}{28.125} + \frac{0.0065}{1 + 0.080 \times 28.125} = 1.78 \times 10^{-6} + \frac{0.0065}{3.25} = 0.0000018 + 0.0020000 \approx 0.002002$$

The stable asymptotic reactor period is $T = 28.1\text{ seconds}$.

(c) Doubling Time and Power at $t = 30\text{ s}$: Doubling time:

$$T_d = T \ln 2 = 28.125 \times 0.69315 \approx \mathbf{19.5\text{ seconds}}$$

After the prompt jump, power increases as $P(t) = P(0^+) e^{t/T}$:

$$P(30\text{ s}) = 144.4\text{ kW} \times \exp\left( \frac{30}{28.125} \right) = 144.4 \times e^{1.0667} = 144.4 \times 2.9056 \approx \mathbf{419.7\text{ kW}}$$

At $30\text{ seconds}$, reactor power has risen to $420\text{ kW}$.

Solved Problem Example 8.2: Equilibrium Xenon-135 Poisoning and Maximum Post-Shutdown Iodine Pit

A high-power commercial PWR operates at steady thermal neutron flux $\phi_0 = 1.20 \times 10^{14}\text{ neutrons/cm}^2\cdot\text{s}$. The nuclear parameters are:

  • Macroscopic thermal fission cross section $\Sigma_f = 0.115\text{ cm}^{-1}$
  • Iodine-135 fission yield $\gamma_I = 0.061$, decay constant $\lambda_I = 2.87 \times 10^{-5}\text{ s}^{-1}$ ($T_{1/2} = 6.70\text{ h}$)
  • Xenon-135 fission yield $\gamma_{\text{Xe}} = 0.003$, decay constant $\lambda_{\text{Xe}} = 2.09 \times 10^{-5}\text{ s}^{-1}$ ($T_{1/2} = 9.20\text{ h}$)
  • Xenon microscopic absorption cross section $\sigma_a^{\text{Xe}} = 2.65 \times 10^6\text{ b} = 2.65 \times 10^{-18}\text{ cm}^2$
  • Core macroscopic absorption cross section $\Sigma_a = 0.160\text{ cm}^{-1}$

(a) Calculate the steady-state equilibrium concentrations of Iodine-135 ($I_0$) and Xenon-135 ($X_0$) in $\text{atoms/cm}^3$. (b) Calculate the equilibrium xenon reactivity worth $\rho_{\text{Xe}} = -\frac{\Sigma_a^{\text{Xe}}}{\Sigma_a} = -\frac{X_0 \sigma_a^{\text{Xe}}}{\Sigma_a}$ in pcm. (c) The reactor is suddenly scrammed (shutdown to zero flux $\phi = 0$). Calculate the time $t_{\text{peak}}$ at which the post-shutdown Xenon concentration reaches its maximum, and determine the peak xenon reactivity penalty in pcm.

(a) Equilibrium Concentrations $I_0$ and $X_0$: Fission rate density:

$$R_f = \Sigma_f \phi_0 = (0.115\text{ cm}^{-1})(1.20 \times 10^{14}\text{ cm}^{-2}\text{s}^{-1}) = 1.380 \times 10^{13}\text{ fissions/cm}^3\cdot\text{s}$$

Equilibrium Iodine-135:

$$I_0 = \frac{\gamma_I \Sigma_f \phi_0}{\lambda_I} = \frac{0.061 \times 1.380 \times 10^{13}}{2.87 \times 10^{-5}\text{ s}^{-1}} = \frac{8.418 \times 10^{11}}{2.87 \times 10^{-5}} \approx \mathbf{2.933 \times 10^{16}\text{ atoms/cm}^3}$$

Xenon destruction rate constant at power:

$$\lambda_{\text{Xe}} + \sigma_a^{\text{Xe}} \phi_0 = 2.09 \times 10^{-5} + (2.65 \times 10^{-18} \times 1.20 \times 10^{14}) = 2.09 \times 10^{-5} + 3.18 \times 10^{-4} \approx 3.389 \times 10^{-4}\text{ s}^{-1}$$

Note that neutron burnup dominates Xenon radioactive decay by a factor of $15$! Total direct and decay yield:

$$\gamma_{\text{tot}} R_f = (0.061 + 0.003) \times 1.380 \times 10^{13} = 0.064 \times 1.380 \times 10^{13} = 8.832 \times 10^{11}\text{ atoms/cm}^3\cdot\text{s}$$

Equilibrium Xenon-135:

$$X_0 = \frac{8.832 \times 10^{11}}{3.389 \times 10^{-4}} \approx \mathbf{2.606 \times 10^{15}\text{ atoms/cm}^3}$$

(b) Equilibrium Xenon Reactivity Worth:

$$\Sigma_a^{\text{Xe}} = X_0 \cdot \sigma_a^{\text{Xe}} = (2.606 \times 10^{15}\text{ cm}^{-3})(2.65 \times 10^{-18}\text{ cm}^2) \approx 0.006906\text{ cm}^{-1}$$

Reactivity worth:

$$\rho_{\text{Xe}} = -\frac{\Sigma_a^{\text{Xe}}}{\Sigma_a} = -\frac{0.006906}{0.160} \approx -0.04316 = \mathbf{-4316\text{ pcm}}$$

At full power, equilibrium xenon absorbs over $4.3\%$ of core reactivity!

(c) Post-Shutdown Xenon Peak: Following shutdown ($\phi = 0$), the Xenon concentration as a function of time $t$ is:

$$X(t) = X_0 e^{-\lambda_{\text{Xe}} t} + \frac{\lambda_I I_0}{\lambda_I - \lambda_{\text{Xe}}} \left( e^{-\lambda_{\text{Xe}} t} - e^{-\lambda_I t} \right)$$

Setting $dX/dt = 0$:

$$t_{\text{peak}} = \frac{1}{\lambda_I - \lambda_{\text{Xe}}} \ln\left( \frac{\lambda_I}{\lambda_{\text{Xe}}} \left[ 1 - \frac{\lambda_I - \lambda_{\text{Xe}}}{\lambda_I} \frac{X_0}{I_0} \right]^{-1} \right)$$

Here $\lambda_I - \lambda_{\text{Xe}} = (2.87 - 2.09) \times 10^{-5} = 0.78 \times 10^{-5}\text{ s}^{-1}$:

$$\frac{\lambda_I - \lambda_{\text{Xe}}}{\lambda_I} \frac{X_0}{I_0} = \frac{0.78}{2.87} \times \frac{2.606 \times 10^{15}}{2.933 \times 10^{16}} = 0.2718 \times 0.08885 \approx 0.02415$$
$$1 - 0.02415 = 0.97585 \implies \frac{\lambda_I}{\lambda_{\text{Xe}} \times 0.97585} = \frac{2.87}{2.09 \times 0.97585} \approx 1.4072$$
$$t_{\text{peak}} = \frac{\ln(1.4072)}{0.78 \times 10^{-5}\text{ s}^{-1}} = \frac{0.3416}{0.78 \times 10^{-5}} \approx 43{,}790\text{ seconds} \approx \mathbf{10.2\text{ hours}}$$

Evaluating $X(t_{\text{peak}})$ at $t = 43{,}790\text{ s}$:

$$e^{-\lambda_{\text{Xe}} t} = \exp(-2.09 \times 10^{-5} \times 43790) = e^{-0.9152} \approx 0.4004$$
$$e^{-\lambda_I t} = \exp(-2.87 \times 10^{-5} \times 43790) = e^{-1.2568} \approx 0.2846$$
$$X_{\text{peak}} = (2.606 \times 10^{15})(0.4004) + \frac{2.87 \times 10^{-5} \times 2.933 \times 10^{16}}{0.78 \times 10^{-5}} (0.4004 - 0.2846)$$
$$X_{\text{peak}} = 1.043 \times 10^{15} + (1.0792 \times 10^{17})(0.1158) = 1.043 \times 10^{15} + 1.250 \times 10^{16} \approx \mathbf{1.354 \times 10^{16}\text{ atoms/cm}^3}$$

Peak Xenon reactivity penalty:

$$\rho_{\text{Xe, peak}} = -\frac{(1.354 \times 10^{16})(2.65 \times 10^{-18})}{0.160} = -\frac{0.03588}{0.160} \approx \mathbf{-22{,}425\text{ pcm}} = \mathbf{-22.4\%\ \Delta k/k}$$

Ten hours after shutdown, the xenon reactivity pit deepens by an extra $-18{,}100\text{ pcm}$, completely poisoning the reactor core!

Solved Problem Example 8.3: Cylindrical Control Rod Worth and Reactivity Depth Profile via Perturbation Theory

A central control rod in a critical cylindrical reactor of core height $H = 3.60\text{ m}$ has a total shutdown worth of $\rho_{\text{tot}} = -3200\text{ pcm}$. The axial thermal neutron flux is $\phi(z) = \phi_0 \sin\left( \frac{\pi z}{H} \right)$, where $z = 0$ is the top of the core and $z = H$ is the bottom. The rod is inserted from the top ($z = 0$) downward. (a) Using first-order perturbation theory, write the expression for the differential rod worth $w(z) = \frac{d\rho}{dz}$ and find the position of maximum differential worth. (b) Calculate the maximum differential worth in $\text{pcm/cm}$. (c) Calculate the integral reactivity inserted when the control rod is inserted to depths $z = 0.90\text{ m}$ (quarter insertion), $z = 1.80\text{ m}$ (half insertion), and $z = 2.70\text{ m}$ (three-quarter insertion).

(a) Differential Rod Worth Expression: By first-order perturbation theory:

$$\frac{d\rho}{dz} = C \sin^2\left( \frac{\pi z}{H} \right)$$

Normalizing such that $\int_0^H \frac{d\rho}{dz} dz = \rho_{\text{tot}}$:

$$\int_0^H \sin^2\left( \frac{\pi z}{H} \right) dz = \frac{H}{2} \implies C \frac{H}{2} = \rho_{\text{tot}} \implies C = \frac{2 \rho_{\text{tot}}}{H}$$

Thus:

$$w(z) = \frac{d\rho}{dz} = \mathbf{\frac{2 \rho_{\text{tot}}}{H} \sin^2\left( \frac{\pi z}{H} \right)}$$

Maximum differential worth occurs at the core midplane where $\sin(\pi z/H) = 1$:

$$z_{\max} = \frac{H}{2} = \frac{3.60\text{ m}}{2} = \mathbf{1.80\text{ m}}$$

(b) Maximum Differential Worth Value: With $\rho_{\text{tot}} = -3200\text{ pcm}$ and $H = 360\text{ cm}$:

$$w_{\max} = \frac{2 \times (-3200\text{ pcm})}{360\text{ cm}} \approx \mathbf{-17.78\text{ pcm/cm}}$$

At the core center, moving the control rod by just $1.0\text{ cm}$ introduces $-17.8\text{ pcm}$ of reactivity.

(c) Integral Worth at Selected Depths: Using the integral S-curve formula $\rho(z) = \rho_{\text{tot}} \left[ \frac{z}{H} - \frac{1}{2\pi} \sin\left( \frac{2\pi z}{H} \right) \right]$:

  • At quarter insertion ($z = 0.90\text{ m} = H/4$):
$$\frac{z}{H} = 0.25, \qquad \frac{2\pi z}{H} = \frac{\pi}{2} \implies \sin\left(\frac{\pi}{2}\right) = 1$$
$$\rho(H/4) = \rho_{\text{tot}} \left[ 0.25 - \frac{1}{2\pi} \right] = (-3200) [0.25 - 0.15915] = (-3200)(0.09085) \approx \mathbf{-290.7\text{ pcm}}$$

Quarter rod travel produces only $9.1\%$ of total worth!

  • At half insertion ($z = 1.80\text{ m} = H/2$):
$$\frac{z}{H} = 0.50, \qquad \frac{2\pi z}{H} = \pi \implies \sin(\pi) = 0$$
$$\rho(H/2) = \rho_{\text{tot}} [0.50 - 0] = 0.50 \times (-3200) = \mathbf{-1600.0\text{ pcm}}$$

Exactly half the worth ($50\%$) is inserted at the core midplane.

  • At three-quarter insertion ($z = 2.70\text{ m} = 3H/4$):
$$\frac{z}{H} = 0.75, \qquad \frac{2\pi z}{H} = \frac{3\pi}{2} \implies \sin\left(\frac{3\pi}{2}\right) = -1$$
$$\rho(3H/4) = \rho_{\text{tot}} \left[ 0.75 - \left(-\frac{1}{2\pi}\right) \right] = (-3200) [0.75 + 0.15915] = (-3200)(0.90915) \approx \mathbf{-2909.3\text{ pcm}}$$

Three-quarter insertion yields $90.9\%$ of total worth. Moving through the middle half of the core ($H/4$ to $3H/4$) accounts for over $81.8\%$ of the entire control rod reactivity!