EPRI - Reactor Theory| Engineering Systems
Training| 77 QUESTIONS| WITH COMPLETE
SOLUTIONS
course
Electric Power Research Institute EPRI
1. Four-factor formula and multiplication factor
Q: A thermal reactor has ε = 1.03, η = 2.05, p = 0.96, f = 0.70.
(a) Calculate k∞.
(b) If non-leakage probability = 0.995, compute keff and state whether the reactor is critical,
subcritical, or supercritical.
Solution:
k∞ = ε × p × f × η = 1.03 × 0.96 × 0.70 × 2.05 = 1.41893.
keff = k∞ × PNL = 1.41893 × 0.995 = 1.41183.
Since keff > 1 → supercritical.
2. Reactor point kinetics
Q: Λ = 1.0×10⁻⁴ s, βeff = 0.0065, step reactivity insertion ρ = 0.003.
Calculate the prompt jump and describe behavior.
Solution:
Prompt jump factor = β / (β − ρ) = 0..0035 = 1.857.
Neutron population jumps to 1.857 times initial. Since ρ < β, power rise is delayed-neutron
dominated, exponential with period of seconds.
3. Diffusion length and slab thickness
Q: Bare slab reactor, diffusion length L = 60 cm. Find half-thickness a for criticality.
Solution:
Critical condition: (π/2a) = 1/L → a = πL/2 = 94.25 cm.
Full thickness ≈ 188.5 cm.
Flux shape: φ(x) = φ₀ cos(πx/2a).
4. Doppler broadening
Q: Why does Doppler broadening provide negative reactivity?
Solution:
Fuel temperature rise broadens absorption resonances (mainly U-238), lowering resonance
, escape probability p. Since k∞ ∝ p, reactivity decreases. This is immediate negative feedback,
stabilizing the reactor.
5. Control rod worth
Q: k changes from 1.005 to 0.995. Find rod worth in pcm and Δk/k.
Solution:
ρi = (1.005−1)/1.005 = 0.004975.
ρf = (0.995−1)/0.995 = −0.005025.
Δρ = −0.01 ≈ −1000 pcm.
Δk/k = (0.995−1.005)/1.005 = −0.00995 ≈ −0.995%.
Rod worth = −1000 pcm (negative).
6. Reactor period
Q: Λ = 2×10⁻⁵ s, β = 0.007, ρ = 0.001, λ = 0.08 s⁻¹. Estimate period.
Solution:
Approx formula: T ≈ (β − ρ)/(λρ) = 0.006 / (0.08×0.001) = 75 s.
So period ≈ 75 s (slow rise due to delayed neutrons).
7. Collisions in hydrogen
Q: ξ = 1.0. Neutrons from 2 MeV to 0.025 eV. Find collisions to thermalize.
Solution:
n = ln(E0/Eth)/ξ = ln(2×10⁶ / 0.025) ≈ ln(8×10⁷) ≈ 18.2.
So ~18 collisions.
8. Absorption rate in fuel pin
Q: Fuel pin radius = 0.5 cm, flux = 1.0×10¹⁴ n/cm²·s, Σa = 0.12 cm⁻¹. Find absorptions per cm
length.
Solution:
Rate density = Σa φ = 0.12 × 1.0×10¹⁴ = 1.2×10¹³ cm⁻³·s⁻¹.
Volume = πr²×1 = π×0.25 = 0.785 cm³.
Rate = 1.2×10¹³ × 0.785 = 9.42×10¹² reactions/s per cm.
Training| 77 QUESTIONS| WITH COMPLETE
SOLUTIONS
course
Electric Power Research Institute EPRI
1. Four-factor formula and multiplication factor
Q: A thermal reactor has ε = 1.03, η = 2.05, p = 0.96, f = 0.70.
(a) Calculate k∞.
(b) If non-leakage probability = 0.995, compute keff and state whether the reactor is critical,
subcritical, or supercritical.
Solution:
k∞ = ε × p × f × η = 1.03 × 0.96 × 0.70 × 2.05 = 1.41893.
keff = k∞ × PNL = 1.41893 × 0.995 = 1.41183.
Since keff > 1 → supercritical.
2. Reactor point kinetics
Q: Λ = 1.0×10⁻⁴ s, βeff = 0.0065, step reactivity insertion ρ = 0.003.
Calculate the prompt jump and describe behavior.
Solution:
Prompt jump factor = β / (β − ρ) = 0..0035 = 1.857.
Neutron population jumps to 1.857 times initial. Since ρ < β, power rise is delayed-neutron
dominated, exponential with period of seconds.
3. Diffusion length and slab thickness
Q: Bare slab reactor, diffusion length L = 60 cm. Find half-thickness a for criticality.
Solution:
Critical condition: (π/2a) = 1/L → a = πL/2 = 94.25 cm.
Full thickness ≈ 188.5 cm.
Flux shape: φ(x) = φ₀ cos(πx/2a).
4. Doppler broadening
Q: Why does Doppler broadening provide negative reactivity?
Solution:
Fuel temperature rise broadens absorption resonances (mainly U-238), lowering resonance
, escape probability p. Since k∞ ∝ p, reactivity decreases. This is immediate negative feedback,
stabilizing the reactor.
5. Control rod worth
Q: k changes from 1.005 to 0.995. Find rod worth in pcm and Δk/k.
Solution:
ρi = (1.005−1)/1.005 = 0.004975.
ρf = (0.995−1)/0.995 = −0.005025.
Δρ = −0.01 ≈ −1000 pcm.
Δk/k = (0.995−1.005)/1.005 = −0.00995 ≈ −0.995%.
Rod worth = −1000 pcm (negative).
6. Reactor period
Q: Λ = 2×10⁻⁵ s, β = 0.007, ρ = 0.001, λ = 0.08 s⁻¹. Estimate period.
Solution:
Approx formula: T ≈ (β − ρ)/(λρ) = 0.006 / (0.08×0.001) = 75 s.
So period ≈ 75 s (slow rise due to delayed neutrons).
7. Collisions in hydrogen
Q: ξ = 1.0. Neutrons from 2 MeV to 0.025 eV. Find collisions to thermalize.
Solution:
n = ln(E0/Eth)/ξ = ln(2×10⁶ / 0.025) ≈ ln(8×10⁷) ≈ 18.2.
So ~18 collisions.
8. Absorption rate in fuel pin
Q: Fuel pin radius = 0.5 cm, flux = 1.0×10¹⁴ n/cm²·s, Σa = 0.12 cm⁻¹. Find absorptions per cm
length.
Solution:
Rate density = Σa φ = 0.12 × 1.0×10¹⁴ = 1.2×10¹³ cm⁻³·s⁻¹.
Volume = πr²×1 = π×0.25 = 0.785 cm³.
Rate = 1.2×10¹³ × 0.785 = 9.42×10¹² reactions/s per cm.
