University Physics V · Nuclear Fission and Fusion · 16.4
Point Kinetics, Inhour Roots & Feedback
Reactivity alone says nothing about how fast a reactor moves. The period is an eigenvalue of a seven-state system, the roots of the inhour equation, and it is violently asymmetric: upward you can reach milliseconds, downward you can never beat −80.6 seconds.
Build the model
Connect the measurement to the mechanism.
Point kinetics buys tractability with one assumption: the flux keeps a fixed spatial shape, so the core collapses to an amplitude n(t) and six precursor concentrations. What is left is ṅ = [(ρ − β)/Λ]n + Σλᵢ Cᵢ with Ċᵢ = (βᵢ/Λ)n − λᵢCᵢ — at fixed reactivity a constant 7 × 7 matrix, solved exactly by eigendecomposition. Its seven eigenvalues are the roots of the inhour equation ρ = ωΛ + Σβᵢω/(ω + λᵢ), which has a pole at each −λᵢ and therefore one root in every gap, plus a largest root ω₀ that is all that survives after a few seconds.
Almost everything a reactor operator says is a statement about where ω₀ sits. Because ω₀ can never fall below −λ₁ = −0.0124 s⁻¹, no scram of any worth gives an asymptotic period shorter than −80.6 s. Because the delayed sum saturates at β once ω ≫ λ₆ = 3.0 s⁻¹, ρ = βeff is a cliff: below it the period is set by the 13 s mean precursor life, above it by Λ — 10⁻⁴ s in a thermal core, 4 × 10⁻⁷ s in a fast one.
That asymmetry is why reactivity is priced in dollars, ρ/βeff. The cost is what the fixed shape hides: no space-time tilt, no local void, no rod shadowing. And the feedback coefficients that close the loop — Doppler, void, xenon — are computed or measured elsewhere and inserted by hand; point kinetics never derives one.
- Simple definition
- Point kinetics is the approximation that the flux keeps one fixed spatial shape, so a whole core is a scalar amplitude n(t) plus six delayed-precursor concentrations, obeying a linear system whose eigenvalues — the inhour roots — set the reactor period.
- Example
- Insert ρ = +100 pcm into a U-235 thermal core (βeff = 0.0065, Λ = 1.0 × 10⁻⁴ s): power jumps by β/(β − ρ) = 1.18 within 18 ms, then climbs on a 71.8 s period — not the 0.1 s that Λ/ρ alone predicts.
One flux shape assumed, so an entire core becomes seven numbers — and at fixed ρ a constant 7 × 7 matrix that eigendecomposition solves exactly.
n ∝ power; Cᵢ in the same units; Λ in s; λᵢ in s⁻¹; βᵢ dimensionless with Σβᵢ = βeff ≈ 0.0065 for thermal U-235.
Prompt critical is 1.00 $ in every core, while 650 pcm means different things in different fuel, so dollars are the only portable unit.
ρ dimensionless, quoted in pcm (10⁻⁵) or dollars. βeff: 0.0065 U-235, 0.0021 Pu-239, ≈0.005 in a burnt LWR.
The characteristic polynomial of that matrix, rewritten. ω₀ is all that survives after seconds, and its lower bound is the −80.6 s scram floor.
ω in s⁻¹, period T = 1/ω. Poles at −λᵢ put one root in each gap; the largest obeys ω₀ > −λ₁ = −0.0124 s⁻¹.
Period and step size without an eigensolve: at 20 cents ×1.25 then 52 s, at 50 cents ×2.00 then 13 s, at 90 cents ×10 then 1.5 s.
λ̄ = 0.0767 s⁻¹ from 1/λ̄ = Σ(βᵢ/λᵢ)/β = 13.0 s. Holds for 0 < ρ < β with Λ ≪ 1/λ̄.
Above prompt critical the delayed sum saturates at β and only Λ is left, so the period drops from tens of seconds to milliseconds.
At 1.10 $: 6.5 s⁻¹ thermal (β 0.0065, Λ 10⁻⁴ s) against 875 s⁻¹ fast (β 0.0035, Λ 4 × 10⁻⁷ s).
Makes the system nonlinear and self-limiting: a prompt burst is stopped by fuel temperature, long before a rod can move.
α in K⁻¹ but quoted in pcm K⁻¹; C in J K⁻¹; E in J. Adiabatic, Doppler-only, ρ₀ > β.
Two clocks, and why the slow one wins
A prompt fission neutron in a thermal core is born, slows and is absorbed in a mean generation time Λ ≈ 1.0 × 10⁻⁴ s — 2 × 10⁻⁵ s in an LWR lattice, 10⁻³ s in graphite, 4 × 10⁻⁷ s in a fast core. About 0.65% of U-235 fission neutrons instead wait inside a precursor nucleus, Br-87 and I-137 and their relatives, sorted into six groups running from λ₁ = 0.0124 s⁻¹ to λ₆ = 3.01 s⁻¹. Weight the delays by yield and the picture inverts: 1/λ̄ = Σ(βᵢ/λᵢ)/β = 13.0 s, so the mean generation time over all neutrons is ℓ̄ = Λ + Σβᵢ/λᵢ = 0.085 s, some 850 times Λ. The same ratio appears as an inventory — at steady state the core holds Σβᵢ/(Λλᵢ) ≈ 850 precursor nuclei for every prompt neutron in flight. A 0.65% minority owns the clock, and every number in this topic follows from that.
Seven states, so diagonalise them
Stack the state as x = (n, C₁, …, C₆)ᵀ and the model is ẋ = Ax, with A₀₀ = (ρ − β)/Λ, A₀ᵢ = λᵢ, Aᵢ₀ = βᵢ/Λ and Aᵢᵢ = −λᵢ. At fixed ρ the matrix is constant, so x(t) = Σⱼ aⱼ vⱼ exp(ωⱼt) is exact and np.linalg.eig(A) returns the seven ωⱼ without the inhour equation ever being written down. Substituting the ansatz by hand instead gives ρ = ωΛ + Σβᵢω/(ω + λᵢ), whose poles at ω = −λᵢ force exactly one root into each interval between them, one below −λ₆ and one above −λ₁. That last one is ω₀. The spread of the roots is also the numerical warning: at ρ = 0 they run from −0.0124 s⁻¹ to about −β/Λ = −65 s⁻¹, a stiffness ratio near 5000. An explicit RK45 step is stability-limited to a few hundredths of a second, so a day-long xenon transient would need millions of steps; integrate with BDF or Radau, or exponentiate A directly.
The prompt jump is a level change, not a new equilibrium
Because Λ is 10⁻⁴ s while the precursors move on seconds, eliminate the fast variable: set ṅ ≈ 0 to get n ≈ ΛΣλᵢCᵢ/(β − ρ). A step in ρ cannot move the precursor inventory in 10⁻⁴ s, so the numerator is unchanged and the power jumps by β/(β − ρ) = 1/(1 − ρ$) on the time constant Λ/(β − ρ). At 20 cents that is ×1.25 in 19 ms; at 50 cents, ×2.00; at 90 cents, ×10. None of this is a new equilibrium. Reactivity is still positive, so ΣλᵢCᵢ keeps growing and the power tracks it with the asymptotic period T = (β − ρ)/(λ̄ρ) + Λ/ρ — 52 s at 20 cents, 13 s at 50 cents, 1.5 s at 90 cents. The prompt jump is a change of level; the inhour root is the change of rate, and only returning ρ to zero stops it.
A cliff at one dollar, a floor at −80.6 seconds
Drive ρ up toward β and ω₀ must outrun every λᵢ. Once ω ≫ λ₆ = 3.01 s⁻¹ the delayed sum saturates at β and the inhour equation collapses to ρ − β = ω₀Λ. At exactly ρ = β with Λ = 1.0 × 10⁻⁴ s the root is ω₀ = 4.6 s⁻¹, a 0.22 s period; at 1.10 $ it is 6.5 s⁻¹, T = 0.15 s. Do the same in a Pu-fuelled fast core, βeff = 0.0035 and Λ = 4 × 10⁻⁷ s, and 1.10 $ gives 875 s⁻¹, T = 1.1 ms. Downward there is no matching cliff, because every root lies above −λ₁ = −0.0124 s⁻¹: a scram worth ten dollars and one worth a hundred leave the same asymptotic period, −80.6 s, after which decay heat takes over. That asymmetry is the whole argument for quoting reactivity in dollars. Prompt critical is 1.00 $ in every core; in pcm it is 650 in fresh U-235, about 500 in a burnt LWR, 350 in a fast oxide core and 210 in pure Pu-239.
Feedback: which coefficient is prompt, and which waits
Point kinetics is linear only while ρ is externally fixed. Close the loop with ρ(t) = ρₑₓₜ + αf ΔTf + αc ΔTc and it becomes nonlinear. The fuel term is Doppler: hotter fuel broadens the U-238 capture resonances, the Doppler width ΓD = √(4E₀kT/A) growing from 54 meV at 300 K to 98 meV at 1000 K for the 6.67 eV resonance. Broadening conserves ∫σ dE but spoils self-shielding, so the resonance escape probability p falls, giving αf ≈ −2.5 pcm K⁻¹ and scaling as −K/√T. It counts as prompt because it tracks fuel temperature, which tracks power within milliseconds. Coolant and void terms wait for the heat to cross the gap, seconds later, and their sign is a lattice choice: an undermoderated LWR runs near −150 pcm per percent void, while graphite-moderated water-cooled channels ran strongly positive, and a large sodium-cooled core is positive in its interior. Not one of these coefficients falls out of point kinetics — each is computed from transport or fitted to a test and inserted by hand.
Xenon, and the table that sorts reactor types
The slowest feedback of all is a nuclide. I-135 (6.57 h) decays to Xe-135 (9.14 h), whose thermal absorption cross-section is 2.6 × 10⁶ b — worth about −2700 pcm at full power in an LWR. Its running level follows X∞ = (γI + γXe)Σf φ/(λXe + σₐ φ), so after a scram the burnout term σₐ φ vanishes while the iodine keeps feeding: xenon peaks roughly 11 h later, well above its operating value, and the core can be locked out for a day. That transient is a thermal-spectrum phenomenon only; at 100 keV the xenon cross-section is millibarns and a fast reactor never meets it. Line the discriminators up and the reactor types sort themselves: Λ (4 × 10⁻⁷ s fast, 2 × 10⁻⁵ s LWR, 10⁻³ s graphite), βeff (0.0065 fresh U-235, drifting toward 0.005 as Pu-239 builds in, 0.0021 for pure Pu-239), the sign of the void coefficient, and whether xenon exists at all.
Change one variable at a time
Make the relationship visible.
Hold ρ at 100 pcm and drag βeff from 650 down to 200: the same physical insertion goes from 15 cents to half a dollar, the jump grows from ×1.18 to ×2.00, and the period falls from 71.8 s to 13.1 s. Nothing about the core's reactivity changed — only what one dollar is worth.
REACTIVITY0.15 $
PROMPT JUMP n(0+)/n(0-)1.18
STABLE PERIOD71.8 s
DOUBLING TIME49.8 s
Live interpretationREACTIVITY: 0.15 $. PROMPT JUMP n(0+)/n(0-): 1.18. STABLE PERIOD: 71.8 s. DOUBLING TIME: 49.8 s
Catch the common trap
Explain before calculating.
A U-235 thermal core with βeff = 0.0065 and Λ = 1.0 × 10⁻⁴ s sits at steady power when a step of ρ = +0.0013 (20 cents) is inserted and held. Which description of what follows is right?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA U-235 thermal core has βeff = 0.0065, Λ = 1.0 × 10⁻⁴ s and a yield-weighted precursor constant λ̄ = 0.0767 s⁻¹. A step of ρ = +100 pcm is inserted and held. Find the reactivity in dollars, the prompt jump, the stable period and the power 10 s later — then repeat the period with the delayed neutrons deleted.
- Price it. ρ$ = ρ/βeff = 0.00100/0.00650 = 0.154 $, or 15.4 cents. Prompt critical is 1.00 $ by definition, so this insertion sits far inside the delayed-controlled regime.
- Prompt jump. Eliminating the fast variable gives n(0⁺)/n(0⁻) = β/(β − ρ) = 0.00650/0.00550 = 1.18, completed on the prompt time constant Λ/(β − ρ) = 1.0 × 10⁻⁴/0.00550 = 0.018 s.
- Stable period. T = (β − ρ)/(λ̄ρ) + Λ/ρ = 0.00550/(0.0767 × 0.00100) + 1.0 × 10⁻⁴/0.00100 = 71.7 + 0.1 = 71.8 s. The Λ/ρ term contributes 0.1% of the answer.
- Power at 10 s. n/n₀ = 1.18 × exp(10/71.8) = 1.18 × 1.149 = 1.36. A rod pull of this size is something an operator watches over a minute, not a millisecond.
- Delete the delayed neutrons and the same step gives ω = ρ/Λ = 10 s⁻¹, a 0.1 s period, and n/n₀ = e¹⁰⁰ ≈ 3 × 10⁴³ after 10 s. The 0.65% minority is the entire difference between a machine you can steer and one you cannot.
Answerρ = 0.154 $; prompt jump ×1.18 in 18 ms; T = 71.8 s; power ×1.36 after 10 s. Dropping the delayed groups predicts a 0.1 s period and ×3 × 10⁴³ — wrong by forty-three orders of magnitude.
MediumThe same 400 pcm control-rod error is made in two cores: a U-235 thermal core (βeff = 0.0065, Λ = 1.0 × 10⁻⁴ s) and a Pu-fuelled fast core (βeff = 0.0035, Λ = 4.0 × 10⁻⁷ s). Price the insertion in each, find the asymptotic period, and say what has to stop the transient.
- Thermal core: ρ$ = 0.00400/0.00650 = 0.615 $, still below prompt critical, so the delayed branch applies. Prompt jump β/(β − ρ) = 0.00650/0.00250 = 2.60.
- Its period: T = (β − ρ)/(λ̄ρ) + Λ/ρ = 0.00250/(0.0767 × 0.00400) + 0.025 = 8.15 + 0.03 = 8.17 s, so n/n₀ = 2.60 × exp(10/8.17) = 2.60 × 3.40 = 8.8 after 10 s. Scram rods insert in about a second; this is inside their reach.
- Fast core: ρ$ = 0.00400/0.00350 = 1.14 $. The insertion is 50 pcm above prompt critical, so the precursors are frozen on the transient timescale and ρ − β = ω₀Λ applies.
- ω₀ = (ρ − β)/Λ = 0.00050/(4.0 × 10⁻⁷) = 1250 s⁻¹, a period of 0.80 ms. In 10 ms the power multiplies by e¹²·⁵ = 2.7 × 10⁵; in 20 ms by 7 × 10¹⁰.
- So one number, 400 pcm, is a routine manoeuvre in one core and a prompt excursion in the other. Nothing mechanical acts in 0.8 ms: only fuel-temperature feedback — Doppler in oxide fuel, axial expansion in metal — can terminate it, which is why a fast core's individual reactivity worths are engineered to stay well under βeff.
AnswerThermal: 0.615 $, jump ×2.60, T = 8.17 s, ×8.8 in 10 s — controllable. Fast: 1.14 $, prompt supercritical, T = 0.80 ms, ×2.7 × 10⁵ in 10 ms — only Doppler feedback can stop it.
HardA pulsed core has βeff = 0.0065, Λ = 1.0 × 10⁻⁴ s, fuel heat capacity C = 3.0 × 10⁵ J K⁻¹ and a Doppler coefficient αf = −2.5 pcm K⁻¹, with no other feedback and no heat removal. A step of ρ₀ = 1.20 $ is inserted. Using the Nordheim–Fuchs model, find the reactivity removed per joule, the energy at peak power, the peak power, the total burst energy, the temperature rise and the burst width.
- Above prompt critical the precursors are frozen, so drop them: ṅ = [(ρ(t) − β)/Λ]n with ρ(t) = ρ₀ − γE and E(t) = ∫n dt. Here ρ₀ = 1.20 × 0.0065 = 0.00780, so ρ₀ − β = 0.00130, and γ = |αf|/C = 2.5 × 10⁻⁵/(3.0 × 10⁵) = 8.33 × 10⁻¹¹ J⁻¹.
- Change the independent variable from t to E: dn/dE = (ρ₀ − β − γE)/Λ, which integrates to n(E) = n₀ + [(ρ₀ − β)E − ½γE²]/Λ. Time has disappeared, which is why the burst has a closed form.
- Peak power is where ρ returns to β: Eₚ = (ρ₀ − β)/γ = 0.00130/(8.33 × 10⁻¹¹) = 1.56 × 10⁷ J = 15.6 MJ. Substituting back, nₘₐₓ = (ρ₀ − β)²/(2Λγ) = (0.00130)²/(2 × 1.0 × 10⁻⁴ × 8.33 × 10⁻¹¹) = 1.01 × 10⁸ W = 101 MW.
- The burst ends where n returns to zero, at Eₜₒₜ = 2Eₚ = 3.12 × 10⁷ J = 31.2 MJ. Check it against the feedback: ΔT = Eₜₒₜ/C = 104 K and |αf|ΔT = 2.5 × 10⁻⁵ × 104 = 2.60 × 10⁻³ = 2(ρ₀ − β), exactly as the model requires.
- Shape and width: n(t) = nₘₐₓ sech²[(ρ₀ − β)(t − tₚ)/(2Λ)], and sech² halves at argument 0.881, so FWHM = 3.526 Λ/(ρ₀ − β) = 3.526 × 1.0 × 10⁻⁴/0.00130 = 0.27 s.
- Read the engineering off the scaling. The excursion is stopped by fuel temperature, not by control: detection plus rod insertion takes of order a second and the burst is over in a quarter of that. Double ρ₀ − β and Eₜₒₜ doubles while nₘₐₓ quadruples — energy is linear in the prompt overshoot, peak power quadratic.
Answerγ = 8.33 × 10⁻¹¹ J⁻¹; Eₚ = 15.6 MJ; nₘₐₓ = 101 MW; Eₜₒₜ = 31.2 MJ; ΔT = 104 K; FWHM = 0.27 s. Doppler feedback alone terminates the burst, about four times faster than any scram could.