Controlled Thermonuclear Fusion — Question 10

Section 43-7 · Physics pages 7290–7291

Question 10 Helium ash dilutes fuel even while charge neutrality is maintained Hold electron density ne and temperature fixed in a plasma containing D, T and fully ionized helium ash. Let x = nα/ne = .0500 and nD = nT . Charge neutrality requires nD + nT + 2nα = ne. Reactivity is unchanged. D + T He ash less fuel at fixed ne Schematic; labels, not drawing lengths, specify the data. (a) Find D and T densities relative to ne and fusion-rate reduction relative to an ash-free plasma. (b) Find total particle pressure relative to the ash-free pressure, including electrons and all ions.
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Question 10 - Solution Helium ash dilutes fuel even while charge neutrality is maintained D + T He ash less fuel at fixed ne (a) Method and calculation 2nD = ne −2nα ⇒nD = nT = 1 2ne(1 −2x) = .450ne . R/R0 = (1 −2x)2 = (.900)2 = .810 . (b) Interpretation and further calculation nions = nD + nT + nα = ne(1 −x), p = ne(2 −x)kT. p/p0 = (2 −x)/2 = .975 while fusion rate falls by 19 percent. Physical check A small pressure change can conceal a much larger fuel-dilution penalty. Fixed electron density is the constraint; fixed ion density would give a different comparison.

Question 10

Helium ash dilutes fuel even while charge neutrality is maintained

Hold electron density nen_e and temperature fixed in a plasma containing D, T and fully ionized helium ash. Let x=nα/ne=.0500x=n_\alpha/n_e=.0500 and nD=nTn_D=n_T. Charge neutrality requires nD+nT+2nα=nen_D+n_T+2n_\alpha=n_e. Reactivity is unchanged.

Diagram shown in the original SVG worksheet.

  1. Find D and T densities relative to nen_e and fusion-rate reduction relative to an ash-free plasma.

  2. Find total particle pressure relative to the ash-free pressure, including electrons and all ions.

Question 10 - Solution

Helium ash dilutes fuel even while charge neutrality is maintained

Diagram shown in the original SVG worksheet.

(a) Method and calculation

2nD=ne−2nα⇒nD=nT=12ne(1−2x)=.450ne.2n_D=n_e-2n_\alpha\Rightarrow\boxed{n_D=n_T=\tfrac12n_e(1-2x)=.450n_e}.R/R0=(1−2x)2=(.900)2=.810.\boxed{R/R_0=(1-2x)^2=(.900)^2=.810}.

(b) Interpretation and further calculation

nions=nD+nT+nα=ne(1−x),p=ne(2−x)kT.n_{\mathrm{ions}}=n_D+n_T+n_\alpha=n_e(1-x),\quad p=n_e(2-x)kT.p/p0=(2−x)/2=.975while fusion rate falls by 19 percent.\boxed{p/p_0=(2-x)/2=.975}\quad\text{while fusion rate falls by 19 percent}.

Physical check

A small pressure change can conceal a much larger fuel-dilution penalty. Fixed electron density is the constraint; fixed ion density would give a different comparison.