Controlled Thermonuclear Fusion — Question 9

Section 43-7 · Physics pages 7288–7289

Question 9 Pulse energy gain must be combined with repetition and electrical costs A hypothetical pulsed experiment releases 1.00 GJ of fusion energy per pulse and consumes .200 GJ of electrical driving energy per pulse. Pulses repeat every 20.0 s. A converter captures only fusion energy at efficiency .400; continuous auxiliary loads are 5.00 MW. t ∆t fusion pulses Schematic; labels, not drawing lengths, specify the data. (a) Calculate average fusion power and average gross electrical output. (b) Find average net electrical output and the longest pulse interval allowing positive net output if per-pulse energies remain fixed.
Show solutionHide solution
Question 9 - Solution Pulse energy gain must be combined with repetition and electrical costs t ∆t fusion pulses (a) Method and calculation P f = 1.00 GJ/20.0 s = 50.0 MW, P gross = 20.0 MW . P drive = .200 GJ/20.0 s = 10.0 MW . (b) Interpretation and further calculation P net = 20.0 −10.0 −5.00 = 5.00 MW . .400(1.00) −.200 ∆t GJ > 5.00 MW ⇒∆t < 40.0 s . Physical check The driving energy is already electrical input; no extra driver-efficiency factor belongs here. A pulse energy ratio is insufficient without repetition rate and continuous loads.

Question 9

Pulse energy gain must be combined with repetition and electrical costs

A hypothetical pulsed experiment releases 1.00 GJ of fusion energy per pulse and consumes .200 GJ of electrical driving energy per pulse. Pulses repeat every 20.0 s. A converter captures only fusion energy at efficiency .400; continuous auxiliary loads are 5.00 MW.

Diagram shown in the original SVG worksheet.

  1. Calculate average fusion power and average gross electrical output.

  2. Find average net electrical output and the longest pulse interval allowing positive net output if per-pulse energies remain fixed.

Question 9 - Solution

Pulse energy gain must be combined with repetition and electrical costs

Diagram shown in the original SVG worksheet.

(a) Method and calculation

P¯f=1.00GJ/20.0s=50.0MW,P¯gross=20.0MW.\boxed{\overline P_f=1.00\,\mathrm{GJ}/20.0\,\mathrm s=50.0\,\mathrm{MW},\quad\overline P_{\mathrm{gross}}=20.0\,\mathrm{MW}}.P¯drive=.200GJ/20.0s=10.0MW.\boxed{\overline P_{\mathrm{drive}}=.200\,\mathrm{GJ}/20.0\,\mathrm s=10.0\,\mathrm{MW}}.

(b) Interpretation and further calculation

P¯net=20.0−10.0−5.00=5.00MW.\boxed{\overline P_{\mathrm{net}}=20.0-10.0-5.00=5.00\,\mathrm{MW}}..400(1.00)−.200ΔtGJ>5.00MW⇒Δt<40.0s.\frac{.400(1.00)-.200}{\Delta t}\,\mathrm{GJ}>5.00\,\mathrm{MW}\Rightarrow\boxed{\Delta t<40.0\,\mathrm s}.

Physical check

The driving energy is already electrical input; no extra driver-efficiency factor belongs here. A pulse energy ratio is insufficient without repetition rate and continuous loads.