Controlled Thermonuclear Fusion — Question 1

Section 43-7 · Physics pages 7272–7273

Question 1 Both ions and electrons contribute to a fusion plasma energy inventory A uniform fully ionized D-T plasma has total ion density n = 1020 m−3, volume V = 100 m3, and equal electron and ion temperatures with kT=10.0 keV. Charge neutrality gives ne = n. Use 1 keV = 1.602 × 10−16 J. plasma Ph Ploss Schematic; labels, not drawing lengths, specify the data. (a) Calculate total thermal energy for classical nonrelativistic particles. (b) If net heating above losses is 20.0 MW, estimate the time needed to add this thermal energy and state the approximation.
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Question 1 - Solution Both ions and electrons contribute to a fusion plasma energy inventory plasma Ph Ploss (a) Method and calculation W = 3 2niV kT + 3 2neV kT ⇒W = 3nV kT . W = 3(1020)(100)(10)(1.602 × 10−16) = 48.06 MJ . (b) Interpretation and further calculation ˙W = Pheat −Ploss = 20.0 MW ⇒∆t = 48.06/20.0 = 2.403 s . ∆W = ∫ (Pheat −Ploss)dt if net heating is not constant. Physical check Counting ions alone would miss half the thermal energy. The time estimate assumes fixed volume, density and net heating, with negligible initial thermal energy.

Question 1

Both ions and electrons contribute to a fusion plasma energy inventory

A uniform fully ionized D-T plasma has total ion density n=1020m−3n=10^{20}\,\mathrm{m^{-3}}, volume V=100m3V=100\,\mathrm{m^3}, and equal electron and ion temperatures with kT=10.0 keV. Charge neutrality gives ne=nn_e=n. Use 1keV=1.602×10−161\,\mathrm{keV}=1.602\times10^{-16} J.

Diagram shown in the original SVG worksheet.

  1. Calculate total thermal energy for classical nonrelativistic particles.

  2. If net heating above losses is 20.0 MW, estimate the time needed to add this thermal energy and state the approximation.

Question 1 - Solution

Both ions and electrons contribute to a fusion plasma energy inventory

Diagram shown in the original SVG worksheet.

(a) Method and calculation

W=32niVkT+32neVkT⇒W=3nVkT.W=\tfrac32n_iVkT+\tfrac32n_eVkT\Rightarrow\boxed{W=3nVkT}.W=3(1020)(100)(10)(1.602×10−16)=48.06MJ.\boxed{W=3(10^{20})(100)(10)(1.602\times10^{-16})=48.06\,\mathrm{MJ}}.

(b) Interpretation and further calculation

Ẇ=Pheat−Ploss=20.0MW⇒Δt=48.06/20.0=2.403s.\dot W=P_{\mathrm{heat}}-P_{\mathrm{loss}}=20.0\,\mathrm{MW}\Rightarrow\boxed{\Delta t=48.06/20.0=2.403\,\mathrm s}.ΔW=∫(Pheat−Ploss)dtif net heating is not constant.\Delta W=\int(P_{\mathrm{heat}}-P_{\mathrm{loss}})dt\quad\text{if net heating is not constant}.

Physical check

Counting ions alone would miss half the thermal energy. The time estimate assumes fixed volume, density and net heating, with negligible initial thermal energy.