2
$\begingroup$

Why is the criteria for fusion represented as a triple product nτT instead of the n,τ,T individually? i.e saying the density has to be above this value, temperature above this value? What are the advantage we gain from representing it this way?

$\endgroup$
1
  • $\begingroup$ Can you remember the formula $ pV=NkT $ from chemistry? The equation of the ideal gas. It's probably more complicated but it is something to look at. $\endgroup$
    – WalyKu
    Commented Aug 19, 2014 at 15:46

1 Answer 1

1
$\begingroup$

We need to have: $$P_{fusion}>P_{loss}$$
To do so, we begin by defining a new quantity $\tau$ , called the confinement time, which measures the rate at which a system loses energy to its environment. It is the energy density W (energy content per unit volume) divided by the power loss density $P_{loss}$ (rate of energy loss per unit volume):
$$\tau=\frac{W}{P_{loss}}$$
A well-known property of gases, which applies equally well to plasmas, is that their energy density W is equal to
$$W=3 k_b n T$$ This formula is a standard result in statistical mechanics and holds whenever there are no impurities in the plasma, an assumption that is justified for fusion reactors because they are typically well insulated. $$P_{fusion}>\frac{W}{\tau}=\frac{3 k_b n T}{\tau}$$

Now we can write: $$n_A n_B \langle {\sigma v}_{A,B} \rangle E_{fusion} >\frac{3 k_b n T}{\tau}$$

Putting $ \frac{3 n^2}{n_A n_B}=12$

$$n T \tau > \frac{12 k_b T^2}{\langle {\sigma v}_{A,B} \rangle E_{fusion} }$$

$\langle {\sigma v}_{A,B} \rangle$ is the averaged cross-section with the Maxwell–Boltzmann distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.