Nuclear Fuel Cycle

(a) A single irradiated fuel assembly produces a gamma dose rate of 5 Gy/h at 3 metres (unshielded). Treating the assembly as a point source, calculate the dose rate at:

(i) 1 metre from the assembly (unshielded)

(ii) 10 metres from the assembly (unshielded)

(b) The assembly is placed in a cooling pond with 3 metres of water above it. If the half-value layer (HVL) of water for the average gamma energy from fission products is approximately 18 cm, estimate the dose rate at the water surface directly above the assembly.

(c) Comment on the implications for pond operations if the water level were to drop to only 1 metre above the fuel.

Solution

(a) Inverse square law calculations:

For a point source, the dose rate varies with the inverse square of the distance:

$\dot{D}_1 r_1^2 = \dot{D}_2 r_2^2$

(i) Dose rate at 1 metre:

Step 1: Apply the inverse square law.

$\dot{D}_{1\text{m}} = \dot{D}_{3\text{m}} \times \left(\frac{r_{3\text{m}}}{r_{1\text{m}}}\right)^2 = 5 \times \left(\frac{3}{1}\right)^2 = 5 \times 9$

$\boxed{\dot{D}_{1\text{m}} = 45 \text{ Gy/h}}$

(ii) Dose rate at 10 metres:

Step 1: Apply the inverse square law.

$\dot{D}_{10\text{m}} = \dot{D}_{3\text{m}} \times \left(\frac{3}{10}\right)^2 = 5 \times 0.09$

$\boxed{\dot{D}_{10\text{m}} = 0.45 \text{ Gy/h} = 450 \text{ mGy/h}}$

These are lethal dose rates. Even at 10 m, a worker would receive 450 mGy in one hour, which is well above any permissible limit.

(b) Dose rate at the water surface (3 m of water shielding):

Step 1: Calculate the number of HVLs in 3 m of water.

$\text{Number of HVLs} = \frac{300 \text{ cm}}{18 \text{ cm}} = 16.67 \text{ HVLs}$

Step 2: Calculate the attenuation factor.

Each HVL reduces the dose rate by a factor of 2, so:

$\text{Attenuation factor} = \left(\frac{1}{2}\right)^{16.67} = 2^{-16.67}$

$2^{16.67} = 2^{16} \times 2^{0.67} = 65536 \times 1.59 = 1.04 \times 10^{5}$

Step 3: Calculate the shielded dose rate.

First, account for the inverse square law. With the fuel at the bottom and 3 m of water above, the distance from the fuel to the surface is approximately 3 m (same as our reference distance), so the unshielded dose rate at the surface would be ~5 Gy/h (same as the reference). With attenuation:

$\dot{D}_{\text{surface}} = \frac{5}{1.04 \times 10^5} \approx 4.8 \times 10^{-5} \text{ Gy/h} = 0.048 \text{ mGy/h}$

$\boxed{\dot{D}_{\text{surface}} \approx 0.05 \text{ mGy/h} = 50 \text{ } \mu\text{Gy/h}}$

This is a manageable dose rate, well below alarm levels, demonstrating the effectiveness of the water shielding.

Note: This is a simplified calculation treating the source as a point and using narrow-beam attenuation. In practice, build-up factors would increase the dose rate somewhat due to scattered radiation, but the order of magnitude is correct.

(c) Implications of water level dropping to 1 m:

Step 1: Calculate the number of HVLs in 1 m of water.

$\text{Number of HVLs} = \frac{100}{18} = 5.56$

Step 2: Calculate the new attenuation factor.

$2^{5.56} = 2^5 \times 2^{0.56} = 32 \times 1.47 = 47.2$

Step 3: Estimate the dose rate at the new surface.

With the fuel now at 1 m below the surface, using inverse square law from the 3 m reference:

$\dot{D}_{\text{unshielded at 1m}} = 5 \times (3/1)^2 = 45 \text{ Gy/h}$

With 1 m of water attenuation:

$\dot{D}_{\text{surface}} = \frac{45}{47.2} \approx 0.95 \text{ Gy/h} = 950 \text{ mGy/h}$

$\boxed{\dot{D}_{\text{surface}} \approx 950 \text{ mGy/h}}$

Comment: This is nearly 1 Gy/h at the water surface --- an immediately life-threatening dose rate. This is approximately 20,000 times higher than with the normal 3 m of water cover. It would trigger the evacuation alarm (set at ~100 mSv/h) by a factor of nearly 10 and would deliver a potentially lethal dose to anyone remaining near the pond for even a short time.

This dramatically illustrates why:

• Maintaining the pond water level is critical to safety
• Gamma monitors with appropriate alarm levels must be installed
• Loss-of-coolant scenarios in cooling ponds are taken extremely seriously

Common Mistakes --- Question 5

Mistake	How to Avoid It
Forgetting the inverse square law when the distance changes	Always check whether the source-to-point distance has changed as well as the shielding thickness
Using HVL without converting units (m vs cm)	Ensure the shielding thickness and HVL are in the same units before dividing
Confusing Gy with Sv	For gamma radiation, 1 Gy $\approx$ 1 Sv (radiation weighting factor = 1), but state your units clearly