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45 lines
1.7 KiB
45 lines
1.7 KiB
id: phys-ex-03a
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type: calculation
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difficulty: hard
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points: 20
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related_lesson: phys-03
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question: |
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A burst-mode coil has ε = 60 J/m. To reach L = 1.5 m in a 200 μs pulse,
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what power is required? Is this realistic for a burst-mode Tesla coil?
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hints:
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- "Use growth rate equation: dL/dt = P/ε"
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- "Rearrange: P = ε × dL/dt"
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- "Calculate dL/dt = L/T for the pulse duration"
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- "Consider typical DRSSTC power levels"
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solution:
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steps:
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- "Calculate growth rate needed:"
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- "dL/dt = L / T = 1.5 m / (200×10⁻⁶ s) = 7,500 m/s"
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- "Calculate required power:"
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- "P = ε × dL/dt"
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- "P = 60 J/m × 7,500 m/s"
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- "P = 450,000 W = 450 kW"
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- "Analysis of realism:"
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- "Energy per pulse: E = P × T = 450 kW × 200 μs = 90 J"
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- "For primary: C = 0.5 μF, need V² = 2E/C = 360,000, so V ≈ 600 V"
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- "Peak power: 450 kW is high but achievable for large DRSSTC"
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- "Conclusion: Challenging but realistic for large coil"
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answer: "450"
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unit: "kW"
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energy_per_pulse: "90"
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realistic: "yes, but requires large DRSSTC"
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tolerance: 5.0
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explanation: |
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Growing 1.5 m in just 200 μs requires extremely high instantaneous power
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(450 kW). However, the total energy per pulse is only 90 J, which is achievable
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with a 0.5 μF primary capacitor charged to 600 V. This high power/short duration
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trade-off is characteristic of burst mode operation. The high ε = 60 J/m reflects
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inefficiency (branching, radiation) in burst mode. A QCW coil with ε = 10 J/m
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would need only 75 kW for the same growth rate, or could grow the same length
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with less power over a longer time.
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related_concepts: ["energy-per-meter", "growth-rate", "burst-mode", "power-requirements"]
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