Controlled Thermonuclear Fusion — Question 4

Section 43-7 · Physics pages 7278–7279

Question 4 Positive plasma gain does not by itself establish net electrical output A hypothetical plant produces Pf = 500 MW of fusion power with Q=10. External heating has electrical-to-plasma efficiency ηh = .500. All fusion and deposited heating power enters a thermal converter of efficiency ηt = .400. Other electrical loads total 30.0 MW. plasma heat converter gross recirculated electricity Schematic; labels, not drawing lengths, specify the data. (a) Calculate gross and net electrical power. (b) At fixed fusion power and efficiencies, derive the minimum Q for positive net electrical power.
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Question 4 - Solution Positive plasma gain does not by itself establish net electrical output plasma heat converter gross recirculated electricity (a) Method and calculation Ph = Pf/Q = 50 MW, Pgross = ηt(Pf + Ph) = 220 MW . Pnet = 220 −50/.500 −30 = 90 MW . (b) Interpretation and further calculation Pnet = ηtPf −Paux −Pf Q (1/ηh −ηt) > 0. Q > Pf(1/ηh −ηt) ηtPf −Paux = 800 170 = 4.70588 . Physical check The accounting explicitly recovers deposited heating as heat. Converter output is offset by heating electricity and auxiliary loads; plasma Q alone omits these costs.

Question 4

Positive plasma gain does not by itself establish net electrical output

A hypothetical plant produces Pf=500P_f=500 MW of fusion power with Q=10. External heating has electrical-to-plasma efficiency ηh=.500\eta_h=.500. All fusion and deposited heating power enters a thermal converter of efficiency ηt=.400\eta_t=.400. Other electrical loads total 30.0 MW.

Diagram shown in the original SVG worksheet.

  1. Calculate gross and net electrical power.

  2. At fixed fusion power and efficiencies, derive the minimum Q for positive net electrical power.

Question 4 - Solution

Positive plasma gain does not by itself establish net electrical output

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ph=Pf/Q=50MW,Pgross=ηt(Pf+Ph)=220MW.P_h=P_f/Q=50\,\mathrm{MW},\quad\boxed{P_{\mathrm{gross}}=\eta_t(P_f+P_h)=220\,\mathrm{MW}}.Pnet=220−50/.500−30=90MW.\boxed{P_{\mathrm{net}}=220-50/.500-30=90\,\mathrm{MW}}.

(b) Interpretation and further calculation

Pnet=ηtPf−Paux−PfQ(1/ηh−ηt)>0.P_{\mathrm{net}}=\eta_tP_f-P_{\mathrm{aux}}-\frac{P_f}{Q}(1/\eta_h-\eta_t)>0.Q>Pf(1/ηh−ηt)ηtPf−Paux=800170=4.70588.\boxed{Q>\frac{P_f(1/\eta_h-\eta_t)}{\eta_tP_f-P_{\mathrm{aux}}}=\frac{800}{170}=4.70588}.

Physical check

The accounting explicitly recovers deposited heating as heat. Converter output is offset by heating electricity and auxiliary loads; plasma Q alone omits these costs.