Worked Example: SWU Calculation

Problem: Calculate the mass of natural uranium feed and the Separative Work Units (SWU) needed to produce 1 kg of 3.5% enriched uranium with a tails assay of 0.3%.

Given:

P = 1 kg (product mass)
xₚ = 0.035 (3.5% enriched product)
x(f) = 0.00711 (natural uranium feed, 0.711%)
xₜ = 0.003 (0.3% tails)

Step 1: Calculate the feed mass (F) and tails mass (T)

Using the mass balance equations:

$F = P + T \quad \text{...(i)}$ $F \cdot x_f = P \cdot x_p + T \cdot x_t \quad \text{...(ii)}$

From (i): T = F - P = F - 1

Substituting into (ii):

$F \times 0.00711 = 1 \times 0.035 + (F - 1) \times 0.003$

$0.00711F = 0.035 + 0.003F - 0.003$

$0.00711F - 0.003F = 0.032$

$0.00411F = 0.032$

$\boxed{F = \frac{0.032}{0.00411} = 7.785 \text{ kg}}$

Therefore:

$T = F - P = 7.785 - 1 = 6.785 \text{ kg}$

Check: F x(f) = 7.785 x 0.00711 = 0.05531; P xₚ + T xₜ = 1 x 0.035 + 6.785 x 0.003 = 0.035 + 0.02036 = 0.05536. These agree (small rounding difference).

Result: To produce 1 kg of 3.5% enriched uranium with 0.3% tails, you need approximately 7.8 kg of natural uranium feed, generating approximately 6.8 kg of depleted uranium tails.

Step 2: Calculate the value functions V(x)

The value function is: $V(x) = (2x - 1) \cdot \ln\left(\frac{x}{1-x}\right)$

V(xₚ) — Product (xₚ = 0.035):

$V(0.035) = (2 \times 0.035 - 1) \cdot \ln\left(\frac{0.035}{1 - 0.035}\right)$

$= (0.07 - 1) \cdot \ln\left(\frac{0.035}{0.965}\right)$

$= (-0.93) \cdot \ln(0.03627)$

$= (-0.93) \cdot (-3.3168)$

$= 3.085$

V(xₜ) — Tails (xₜ = 0.003):

$V(0.003) = (2 \times 0.003 - 1) \cdot \ln\left(\frac{0.003}{1 - 0.003}\right)$

$= (0.006 - 1) \cdot \ln\left(\frac{0.003}{0.997}\right)$

$= (-0.994) \cdot \ln(0.003009)$

$= (-0.994) \cdot (-5.806)$

$= 5.771$

V(x(f)) — Feed (x(f) = 0.00711):

$V(0.00711) = (2 \times 0.00711 - 1) \cdot \ln\left(\frac{0.00711}{1 - 0.00711}\right)$

$= (0.01422 - 1) \cdot \ln\left(\frac{0.00711}{0.99289}\right)$

$= (-0.98578) \cdot \ln(0.007161)$

$= (-0.98578) \cdot (-4.9388)$

$= 4.868$

Step 3: Calculate the Separative Work (SW)

$SW = P \cdot V(x_p) + T \cdot V(x_t) - F \cdot V(x_f)$

$SW = (1 \times 3.085) + (6.785 \times 5.771) - (7.785 \times 4.868)$

$SW = 3.085 + 39.156 - 37.897$

$\boxed{SW \approx 4.34 \text{ kg-SWU}}$

Summary of Results:

Product mass (P): 1.000 kg
Product enrichment (xₚ): 3.5%
Feed mass (F): 7.785 kg natural U
Feed enrichment (x(f)): 0.711% (natural)
Tails mass (T): 6.785 kg
Tails assay (xₜ): 0.3%
Separative Work (SW): 4.34 kg-SWU

Interpretation: To produce 1 kg of 3.5% enriched uranium (with 0.3% tails), you need approximately 7.8 kg of natural uranium and 4.3 SWU of enrichment work. The cost of enrichment is then 4.3 multiplied by the price per SWU (typically $100-160/SWU in recent years).

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