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1. What the question is really asking

The problem gives a specific energy stored in irradiated graphite because of the Wigner effect:

[ E_{\text{W}} = 2.7\;\text{MJ kg}^{-1}=2.7\times10^{6}\;\text{J kg}^{-1} ]

The student wants to know the power density – i.e. how many watts (J s⁻¹) are produced per unit mass (or per unit volume) when that stored energy is released.
Power density can be written as

[ \dot{q}= \frac{E_{\text{W}}}{\Delta t} ]

where (\Delta t) is the time over which the energy is liberated.
Since the release time is not given, we must treat it as an unknown and express the answer as a function of (\Delta t). We can also give a few realistic “order‑of‑magnitude” estimates based on what actually happened at Windscale.

2. Step‑by‑step calculation

Step 1 – Write the basic relationship

[ \boxed{\dot{q}_{\text{mass}} = \frac{2.7\times10^{6}\;\text{J kg}^{-1}}{\Delta t\;(\text{s})}\;\text{W kg}^{-1}} ]

If you prefer volumetric power density, multiply by the bulk density of graphite.

Step 2 – Graphite density

Typical nuclear‑grade graphite has a bulk density

[ \rho \approx 1.8\;\text{g cm}^{-3}=1.8\times10^{3}\;\text{kg m}^{-3} ]

Step 3 – Convert to volumetric power density

[ \dot{q}{\text{vol}} = \dot{q}{\text{mass}}\;\rho = \frac{2.7\times10^{6}\;\text{J kg}^{-1}}{\Delta t}\;(1.8\times10^{3}\;\text{kg m}^{-3}) ]

[ \boxed{\dot{q}_{\text{vol}} = \frac{4.86\times10^{9}}{\Delta t\;(\text{s})}\;\text{W m}^{-3}} ]

Step 4 – Insert plausible release times

Assumed release time (\Delta t) (\dot{q}_{\text{mass}}) (W kg⁻¹) (\dot{q}_{\text{vol}}) (W m⁻³)
1 s (instantaneous) (2.7\times10^{6}) (4.9\times10^{9})
1 min (60 s) (4.5\times10^{4}) (8.1\times10^{7})
1 h (3600 s) (7.5\times10^{2}) (1.35\times10^{6})
10 h (3.6 × 10⁴ s) – approximate Windscale event (75) (1.35\times10^{5})
1 day (8.64 × 10⁴ s) (31) (5.6\times10^{4})

These numbers illustrate the huge range: the power density can be megawatts per kilogram if the energy is dumped in a fraction of a second, but only a few tens of watts per kilogram if the release stretches over many hours.

Step 5 – Relate to the Windscale accident

During the Windscale fire (October 1957) the graphite core was deliberately heated to ~250 °C to anneal stored Wigner energy. The temperature rise and the subsequent “energy release” lasted on the order of 10 h before the fire became uncontrolled. Using (\Delta t\approx 3.6\times10^{4}) s gives

[ \dot{q}{\text{mass}} \approx 75\;\text{W kg}^{-1} \qquad \dot{q}{\text{vol}} \approx 1.3\times10^{5}\;\text{W m}^{-3} ]

These values are consistent with the observed heat flux that ignited the surrounding fuel channels.

3. Final answer

The power density (mass basis) of Wigner‑energy release from graphite is

[ \boxed{\displaystyle \dot{q}_{\text{mass}} = \frac{2.7\times10^{6}\;\text{J kg}^{-1}}{\Delta t\;(\text{s})}\; \text{W kg}^{-1}} ]

and, using the bulk density (\rho = 1.8\times10^{3}\;\text{kg m}^{-3}),

[ \boxed{\displaystyle \dot{q}_{\text{vol}} = \frac{4.9\times10^{9}}{\Delta t\;(\text{s})}\; \text{W m}^{-3}} ]

For the Windscale incident, where the energy was released over roughly 10 h (≈ 3.6 × 10⁴ s), the power densities are

[ \dot{q}{\text{mass}} \approx 75\;\text{W kg}^{-1}, \qquad \dot{q}{\text{vol}} \approx 1.3\times10^{5}\;\text{W m}^{-3}. ]

4. Common Mistakes

Mistake Why it’s wrong How to avoid it
Treating the 2.7 MJ kg⁻¹ as a power instead of an energy Power already includes a time factor; the given number is purely an energy per mass. Remember that Power = Energy / Time; you must introduce a time scale (\Delta t).
Using the crystal density of graphite (≈ 2.2 g cm⁻³) instead of the bulk density of the reactor core Reactor graphite is porous and has a lower bulk density (~1.8 g cm⁻³). Using the higher value over‑estimates volumetric power density. Use the actual bulk density of the material in the specific configuration (often supplied in reactor design data).
Assuming the release is instantaneous In reality the Wigner energy anneals over minutes to hours; an instantaneous assumption gives absurdly high power densities (MW kg⁻¹) that are not physically realized. Cite experimental or historical release times (e.g., Windscale ~10 h) and present the result as a function of (\Delta t).
Neglecting the fact that only a fraction of the stored energy is released in a single event Not all stored Wigner energy may be liberated; some remains trapped after the first anneal. State the assumption “all 2.7 MJ kg⁻¹ is released” and, if needed, introduce a fraction (f) (0 < f ≤ 1) to scale the answer.
Mixing up mass‑ and volume‑based power densities Confusing units (W kg⁻¹ vs. W m⁻³) leads to errors in subsequent heat‑transfer calculations. Keep a clear conversion step: (\dot{q}{\text{vol}} = \dot{q}{\text{mass}} \times \rho).

By keeping the time factor explicit and using the correct material density, you obtain a reliable estimate of the power density associated with the Wigner effect in graphite.

Original question: What is the power density of graphite caused by the Wigner effect? on Chemistry Stack Exchange, licensed CC BY-SA.