I'll try to answer my own question, trying to be as generous as possible to the polywell, within the assumption that the ion speed is much larger than the speed of the electrons.
I published
a similar calculation already for Rostocker's CBFR, where I concluded that the ratio P_fusion/P_fric cannot be more than 0.12, even taking credit for goodies like the 580 keV (CM energy) resonance and spin polarization. P_fric was power loss due to direct collisions between the protons and boron ions, but I also said "The power balance would be at least another factor of three less favorable than this estimate because the coupling of the ions through the electrons would be stronger than the direct coupling if Te < E0/15 = 40 keV." This would be the regime we are discussing here. In fact, we have to deduct another factor of 1.6 because we cannot polarize the fuel ions in a polywell. With P_fusion/P_loss = 0.025, assuming the power lost from the ions can be harvested and recycled with the same efficiency eta as the fusion power, breakeven would require
eta = P_loss / (P_fusion+P_loss) = (1+P_fusion/P_loss)^-1 = 97.5%
Significant net power production would require eta to be at least 99%.
I did not publish the details of this calculation (I wonder if I have notes somewhere?), but it shouldn't be hard to reproduce. According to the
NRL Plasma Formulary, p. 32, in this limit the slowing down rate is given by
nu_s[s^-1] = 1.7e-4 mu^0.5 * n_e[cm^-3] * Z^2 * lambda_ie * epsilon[eV]^-1.5
nu_s = (1.7e-4 s^-1 cm^3 eV^1.5) mu^0.5 * n_e * Z^2 * lambda_ie * epsilon^-1.5
for p: mu = 1 and Z =1
for B11: mu = 11 and Z = 5
I'll just take lambda_ie = 10. (Anyone who wants to refine the decimal places is welcome to do so.)
The energy loss rate is twice as fast (due to the v^2 dependence).
The most generous assumption I can make is a colliding beam configuration like the CBFR, although the core of the polywell will have an approximately isotropic distribution (even if it is monoenergetic), so not all the collisions can be at the resonance energy. In that case, the protons have to have 11/12 of the CM energy and the boron ions the other 1/12.
epsilon_p = (11/12)*epsilon_CM
epsilon_B = (1/12)*epsilon_CM
The power lost to collisions per unit volume will be
P_loss = n_p*epsilon_p*2*nu_s(p) + n_B*epsilon_B*2*nu_s(B)
= (3.4e-4 s^-1 cm^3 eV^1.5) * n_e * Z^2 * lambda_ie * [
n_p * mu_p^0.5 * epsilon_p^-0.5 +
n_B * mu_B^0.5 * epsilon_B^-0.5 ]
= (3.4e-4 s^-1 cm^3 eV^1.5) * n_e * lambda_ie * [
Z_p^2 * n_p * mu_p^0.5 * epsilon_p^-0.5 +
Z_B^2 * n_B * mu_B^0.5 * epsilon_B^-0.5 ]
= (3.4e-4 s^-1 cm^3 eV^1.5) * n_e * lambda_ie * epsilon_CM^-0.5 * [ 1.04 * n_p + 275 * n_B ]
Obviously, the power loss from the boron ions will dominate for any reasonable fuel mixture. (This suggests running boron lean.) Dropping the n_p term, setting n_e = n_p, lambda = 10, and epsilon_CM = 580 keV, we arrive at
P_loss = (3.4e-4 s^-1 cm^3 eV^1.5) * n_p * n_B * 10 * (5.8e5 eV)^-0.5 * [ 275 ]
P_loss = (1.23e-3 s^-1 cm^3 eV) * n_p * n_B
That looks good. The next step is to compare this to the fusion power. sigma*v at resonance is (8.1e-16 cm^3 s^-1) (no bonus for spin polarization here), and the energy gain per fusion reaction is 8.7 MeV. That gives
P_fusion = n_p * n_B * (8.7e6 eV) * (8.1e-16 cm^3 s^-1)
= n_p * n_B * (8.7e6 eV) * (8.1e-16 cm^3 s^-1)
= (7.0e-9 s^-1 cm^3 eV) * n_p * n_B
and
P_fusion/P_loss = 5.7e-6
(Do I believe this result? It wouldn't surprise me if I lost a factor of lambda or mu here or there, but the drag on cold electrons is really pretty drastic, so, yeah, I think you will be dead before you hit the ground if you try to use cold electrons to make p-B11 fusion.)