Nuclear Fuel Cycle

The following tutorial questions are adapted from Worksheet 1, with corrections, additional questions, and fully worked solutions provided.

Q: State the three main forms of uranium mining. Which form of uranium mining produces the greatest radiological hazards? How are these hazards controlled? State the environmental hazards of all three types of mining.

A:

The three main forms are:

1. Open pit mining
2. Underground mining
3. In situ leaching (ISL)

Greatest radiological hazard: Underground (deep pit) mining presents the greatest radiological hazard. The two main exposure routes are:

• External gamma exposure from gamma-emitting nuclides in the ore (especially Bi-214)
• Internal exposure from inhalation of radon gas (Rn-222) and its short-lived daughter products attached to airborne dust particles

Controls:

• Forced ventilation to maintain radon concentrations below regulatory limits
• Personal dosimetry for external gamma monitoring
• Air filtration and respiratory protective equipment for the internal (inhalation) hazard
• Dust suppression measures

Environmental hazards by mining type:

Q: Describe the process by which UF₆ is produced from uranium oxide.

A:

The conversion from uranium oxide to UF₆ is a two-step process:

Step 1: Uranium oxide (UO₂ or UO₃) is reacted with anhydrous hydrofluoric acid (HF) at approximately 450 degrees C in a rotary kiln to produce uranium tetrafluoride (UF₄), a green powder:

$\text{UO}_2 + 4\text{HF} \xrightarrow{450°\text{C}} \text{UF}_4 + 2\text{H}_2\text{O}$

Step 2: UF₄ powder is then reacted with elemental fluorine gas (F₂) in a fluidised bed reactor. This exothermic reaction produces gaseous UF₆:

$\text{UF}_4 + \text{F}_2 \rightarrow \text{UF}_6$

The gaseous UF₆ is then cooled and solidified into cylinders for transport.

The HF used in Step 1 is what likely “caused the Wicked Witch of the West to dissolve” — hex reacts with water to produce HF (hydrofluoric acid), which is extremely corrosive and would dissolve organic matter.

Q: Early uranium enrichment centrifuges suffered from chronic rotor bearing failures, which delayed the adoption of this technology. Which type of enrichment plant is this? What type of bearings are used in modern versions of this enrichment technology?

A:

This describes the gas centrifuge enrichment plant. Early centrifuge development was hampered by severe problems with rotor bearings and structural components.

Modern gas centrifuges use magnetic bearings to suspend the rotor, minimising mechanical contact and friction. The rotor spins at extremely high speeds (supersonic) on a small pivot point, with the magnetic bearing providing stability. The entire assembly is enclosed in a vacuum casing to eliminate aerodynamic drag.

Q: A nuclear power plant requires 10,000 kg of uranium enriched to 3.2% $^{235}$U. The enrichment plant uses natural uranium feed (0.711% $^{235}$U) and produces tails with an assay of 0.25% $^{235}$U.

(a) Calculate the mass of natural uranium feed required. (b) Calculate the mass of depleted uranium tails produced. (c) Calculate the number of SWU required.

A:

Given:

• P = 10,000 kg
• xₚ = 0.032 (3.2%)
• x(f) = 0.00711 (0.711%)
• xₜ = 0.0025 (0.25%)

(a) Feed mass (F):

Using the feed factor:

$\frac{F}{P} = \frac{x_p - x_t}{x_f - x_t} = \frac{0.032 - 0.0025}{0.00711 - 0.0025} = \frac{0.0295}{0.00461} = 6.399$

$F = 6.399 \times 10{,}000 = 63{,}990 \text{ kg}$

$\boxed{F \approx 64{,}000 \text{ kg of natural uranium}}$

(b) Tails mass (T):

$T = F - P = 63{,}990 - 10{,}000 = 53{,}990 \text{ kg}$

$\boxed{T \approx 54{,}000 \text{ kg of depleted uranium}}$

Check: F x(f) = 63,990 x 0.00711 = 454.97; P xₚ + T xₜ = 10,000 x 0.032 + 53,990 x 0.0025 = 320 + 134.98 = 454.98. Checks out.

(c) SWU required:

First, calculate the value functions:

V(xₚ) = V(0.032):

$V(0.032) = (2 \times 0.032 - 1) \cdot \ln\left(\frac{0.032}{0.968}\right) = (-0.936) \times \ln(0.03306) = (-0.936) \times (-3.409) = 3.191$

V(xₜ) = V(0.0025):

$V(0.0025) = (2 \times 0.0025 - 1) \cdot \ln\left(\frac{0.0025}{0.9975}\right) = (-0.995) \times \ln(0.002506) = (-0.995) \times (-5.989) = 5.959$

V(x(f)) = V(0.00711):

$V(0.00711) = (2 \times 0.00711 - 1) \cdot \ln\left(\frac{0.00711}{0.99289}\right) = (-0.98578) \times (-4.939) = 4.868$

Now calculate SW:

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

$SW = (10{,}000 \times 3.191) + (53{,}990 \times 5.959) - (63{,}990 \times 4.868)$

$SW = 31{,}910 + 321{,}724 - 311{,}503$

$\boxed{SW \approx 42{,}131 \text{ kg-SWU}}$

Interpretation: Producing 10 tonnes of 3.2% enriched uranium requires approximately 64 tonnes of natural uranium feed and 42,131 SWU of separative work. At a typical SWU cost of $130/SWU, the enrichment cost alone would be approximately $5.5 million.

Q: Using the data from Tutorial Question 4, recalculate the feed requirement and SWU if the tails assay is changed to 0.20% instead of 0.25%. Comment on the effect.

A:

Given (changed): xₜ = 0.002 (0.20%), all other values as before.

(a) Feed mass:

$\frac{F}{P} = \frac{0.032 - 0.002}{0.00711 - 0.002} = \frac{0.030}{0.00511} = 5.871$

$F = 5.871 \times 10{,}000 = 58{,}710 \text{ kg}$

$T = 58{,}710 - 10{,}000 = 48{,}710 \text{ kg}$

(b) SWU:

$V(0.002) = (2 \times 0.002 - 1) \cdot \ln\left(\frac{0.002}{0.998}\right) = (-0.996) \times (-6.214) = 6.189$

$SW = (10{,}000 \times 3.191) + (48{,}710 \times 6.189) - (58{,}710 \times 4.868)$

$SW = 31{,}910 + 301{,}564 - 285{,}800$

$SW \approx 47{,}674 \text{ kg-SWU}$

Comparison:

Comment: Reducing the tails assay means that less natural uranium feed is required (better utilisation of the feed), but more separative work is needed (higher enrichment cost). This trade-off is the basis for choosing the economically optimal tails assay: when natural uranium is expensive, a lower tails assay is preferred to reduce feed costs, even though enrichment costs increase. When uranium is cheap, a higher tails assay reduces enrichment costs at the expense of using more feed.

Q: Calculate the maximum theoretical separation factor for gaseous diffusion enrichment of uranium using UF₆. How many stages would be needed in principle if each stage achieved this maximum separation factor and you wanted to enrich from 0.711% to 3.5%?

A:

Separation factor:

The molecular masses are:

• $^{235}$UF₆: 235 + 6(19) = 235 + 114 = 349 amu
• $^{238}$UF₆: 238 + 6(19) = 238 + 114 = 352 amu

$\alpha_{\text{max}} = \sqrt{\frac{352}{349}} = \sqrt{1.008596} = 1.00429$

Number of stages (rough estimate):

For a simple cascade, the number of enriching stages n can be estimated from:

$\alpha^n \approx \frac{x_p/(1-x_p)}{x_f/(1-x_f)}$

$1.00429^n = \frac{0.035/0.965}{0.00711/0.99289} = \frac{0.03627}{0.007161} = 5.065$

$n \cdot \ln(1.00429) = \ln(5.065)$

$n = \frac{1.6224}{0.004281} = 379 \text{ enriching stages}$

A similar number of stripping stages (below the feed point) would be needed, giving a total cascade of roughly 700-1,000+ stages (accounting for the fact that real separation factors are lower than the theoretical maximum, and the stripping section is also needed).

This demonstrates why gaseous diffusion plants are so enormous and energy-intensive.

Q: Yellowcake is produced as a bright yellow solid during uranium milling. Consider a scenario in which workers must transit through a storage area where yellowcake drums have been opened and the product is exposed. Would workers incur significant radiation doses? Give two possible exposure routes and state suitable ALARP measures.

A:

Yellowcake is the common name for ammonium diuranate (ADU), the bright yellow product of uranium milling.

Radiation exposure assessment:

Yellowcake has a very low specific radioactivity because all the high-activity daughter products (radium, radon precursors) have been removed during milling. The two main exposure routes are:

1. External exposure (direct shine): Beta and gamma radiation from the yellowcake surface. Assuming the yellowcake is in solid “brick” form, contact beta dose rates could be about 1 mSv/h, but dose rates to the body (at ~1 m height) would be only a few microSv/h. Footwear and distance provide significant shielding.

2. Internal exposure (inhalation of resuspended dust): If the yellowcake surface erodes and dust becomes airborne, inhalation of uranium-bearing particles is a hazard. However, for consolidated “bricks” with pedestrian traffic, resuspension would be minimal (pedestrians cause little erosion).

ALARP measures:

• Wear footwear (shielding against contact beta dose)
• Minimise time spent on the road
• Walk at a steady pace (minimise dust generation)
• Avoid disturbing the surface
• Use respiratory protection in dusty conditions

The overall dose from walking on a yellowcake road would be small, of the order of a few microSv/h, well below levels of concern for short exposures.

Q: In the nuclear industry, uranium hexafluoride is commonly referred to as ‘hex’.

(i) What is hex and why is it used in enrichment? (ii) In what form is hex stored, and what are its hazards? (iii) Where in the UK would one find significant stockpiles of hex?

A:

(i) “Hex” is the informal name for uranium hexafluoride (UF₆). It is used in uranium enrichment because:

• At suitable temperatures and pressures, UF₆ is a gas, which is essential for both the gaseous diffusion and gas centrifuge enrichment processes.
• Fluorine has only one stable isotope (F-19), so the mass difference between molecules of UF₆ containing U-235 and those containing U-238 is entirely due to the uranium isotope. This maximises the efficiency of isotopic separation.

(ii) UF₆ is normally stored as a crystalline solid in large, thick-walled steel cylinders. At room temperature and atmospheric pressure UF₆ is a solid; it sublimes directly to gas at 56.5 °C.

Hazards of UF₆:

• Chemical toxicity: UF₆ reacts violently with water (moisture) to produce uranyl fluoride (UO₂F₂) and hydrofluoric acid (HF). HF is extremely corrosive and toxic, causing severe chemical burns and systemic fluoride poisoning.
• Radiological hazard: UF₆ contains uranium, which is an alpha emitter.
• Criticality risk: If enriched UF₆ comes into contact with water and moderating conditions arise, there is a potential criticality hazard (particularly for highly enriched material).

(iii) In the UK, significant stockpiles of UF₆ (hex) are found at Capenhurst, Cheshire, which is the site of the Urenco UK enrichment facility. There are also stockpiles of depleted UF₆ (tails) at the same site.

End of Chapter 2