plasma-expert/spark-lessons/lessons/03-spark-physics/03-energy-per-meter.md

id	title	section	difficulty	estimated_time	prerequisites	objectives	tags
phys-03	Energy Per Meter Concept	Spark Growth Physics	intermediate	40	[phys-01 phys-02]	[Understand the concept of energy per meter (ε) for spark growth Apply the growth rate equation dL/dt = P/ε Calculate total energy and average power for target spark length Recognize the difference between theoretical minimum and practical ε values]	[energy-per-meter epsilon growth-rate power ionization]

Energy Per Meter Concept

Extending a spark requires energy. Surprisingly, the energy needed is approximately constant per unit length, regardless of how long the spark already is. This fundamental concept enables practical spark growth modeling.

The Energy Per Meter Parameter (ε)

Definition: ε (epsilon) is the energy required to extend a spark by one meter.

Energy to grow from L₁ to L₂:
ΔE ≈ ε × (L₂ - L₁)  [Joules]

where ε has units [J/m]

Key characteristics:

• Approximately constant for a given operating mode
• Independent of current spark length (first-order approximation)
• Depends strongly on operating regime (QCW vs burst)
• Empirical parameter that must be calibrated per coil

Why is this useful?

• Simple relationship: energy scales linearly with length
• Easy to calculate power requirements
• Enables growth rate predictions
• Separates voltage limit (field) from power limit (energy)

What Does ε Include?

The energy per meter is NOT just the ionization energy. It includes all energy processes:

1. Initial Ionization

Breaking molecular bonds to create ions and free electrons:

E_ionize ≈ 15 eV per molecule

2. Heating to Operating Temperature

Raising channel temperature from ambient to 5,000-20,000 K:

E_thermal = m × c_p × ΔT

3. Work Against Pressure

Expanding the channel against atmospheric pressure:

E_expansion = P × ΔV

4. Radiation Losses

Emitted light, UV, infrared, and RF:

E_radiation = ∫ σ T⁴ dA dt (blackbody + line emission)

5. Branching Losses

Energy wasted in short branches that don't contribute to main channel:

E_branching = ε × L_branches (failed growth attempts)

6. General Inefficiencies

Non-productive heating, turbulence, and other losses:

E_losses = various mechanisms

Result: Practical ε is 20-300× larger than theoretical ionization minimum!

Theoretical Minimum Energy

Let's estimate the absolute minimum energy needed for ionization alone:

Given:

• Ionization energy per molecule: ~15 eV
• Air density: n ≈ 2.5×10²⁵ molecules/m³
• Channel diameter: d = 1 mm (typical)
• Length increment: ΔL = 1 m

Calculation:

Volume of 1 m channel:
V = π(d/2)² × L = π(0.5×10⁻³)² × 1 = 7.85×10⁻⁷ m³

Number of molecules:
N = n × V = 2.5×10²⁵ × 7.85×10⁻⁷ = 1.96×10¹⁹ molecules

Energy to ionize:
E_min = N × 15 eV × (1.6×10⁻¹⁹ J/eV)
      = 1.96×10¹⁹ × 15 × 1.6×10⁻¹⁹
      = 0.47 J/m

Theoretical minimum: ε_theory ≈ 0.3-0.5 J/m

Why is practical ε so much higher?

Compare to real values:

• QCW: ε ≈ 5-15 J/m (10-30× theoretical)
• Burst mode: ε ≈ 30-100 J/m (60-200× theoretical)

The difference accounts for:

• Heating to high temperature (major contribution)
• Radiation losses (visible light alone is significant)
• Expansion work (pushing air aside)
• Branching inefficiency (many failed paths)
• Re-ionization (especially in pulsed modes)

The Growth Rate Equation

When the field threshold is met (E_tip > E_propagation), the growth rate is determined by power:

dL/dt = P_stream / ε  [m/s]

where:
  P_stream = power delivered to spark [W]
  ε = energy per meter [J/m]

Physical interpretation:

• More power → faster growth
• Higher ε (inefficiency) → slower growth for same power
• Linear relationship: double power → double growth rate

When growth stops:

If E_tip < E_propagation:
  dL/dt = 0 (stalled)

Cannot grow regardless of available power
(voltage-limited condition)

Predicting Growth Time

For constant power during ramp:

Growth rate: dL/dt = P_stream / ε

Integrating: L(t) = (P_stream / ε) × t

Time to reach target length:
T = ε × L_target / P_stream

More realistic scenario: Power changes as spark grows (loading changes):

T = ∫₀^L_target (ε / P_stream(L)) dL

Requires simulation or numerical integration

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WORKED EXAMPLE 3.2: Energy Budget

Given:

• Target spark length: L = 2 m
• Operating mode: QCW with ε = 10 J/m
• Growth time: T = 12 ms

Find: (a) Total energy required (b) Average power required (c) If 80 kW is available, what changes?

Solution

Part (a): Total energy

E_total = ε × L
        = 10 J/m × 2 m
        = 20 J

Remarkably modest! Only 20 J to create a 2 m spark.

Part (b): Average power

P_avg = E_total / T
      = 20 J / 0.012 s
      = 1,667 W
      ≈ 1.7 kW

For 12 ms growth, need ~1.7 kW average power.

Part (c): With 80 kW available

Available power is 80 kW, but only need 1.7 kW!

Power ratio: 80 kW / 1.7 kW = 47× more than needed

Option 1: Grow much faster

T_min = E_total / P_available
      = 20 J / 80,000 W
      = 0.00025 s
      = 0.25 ms (burst-like growth)

Option 2: Grow to longer length (in same 12 ms)

L_max_power = P_available × T / ε
            = 80,000 W × 0.012 s / 10 J/m
            = 960 J / 10 J/m
            = 96 m (!!)

Reality check: 96 m is absurd! What limits this?

Voltage limit kicks in first:

• Cannot maintain E_tip > E_propagation for 96 m
• Spark stalls at voltage-limited length
• Typical: L_max ≈ 2-4 m for practical topload voltages

Key insight: Tesla coils are almost always voltage-limited, not power-limited. Excess power goes into brightening, heating, and branching rather than length.

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WORKED EXAMPLE 3.3: Comparing Operating Modes

Given:

• Two coils both deliver P = 50 kW average
• Coil A: QCW mode, ε_A = 8 J/m
• Coil B: Burst mode, ε_B = 50 J/m
• Both operate for T = 10 ms

Find: Which produces longer sparks?

Solution

Coil A (QCW):

L_A = P × T / ε_A
    = 50,000 W × 0.010 s / 8 J/m
    = 500 J / 8 J/m
    = 62.5 m (voltage-limited in practice)

Coil B (Burst):

L_B = P × T / ε_B
    = 50,000 W × 0.010 s / 50 J/m
    = 500 J / 50 J/m
    = 10 m (still voltage-limited in practice)

Comparison:

Ratio: L_A / L_B = ε_B / ε_A = 50/8 = 6.25×

QCW coil produces 6.25× longer sparks for same power!

Practical reality:

• Both limited by voltage before reaching these lengths
• But ratio still applies: QCW gives much better length efficiency
• Coil A might reach 2.5 m while Coil B reaches 0.4 m
• Burst mode wastes energy on brightness and branching

Why choose burst mode then?

• Spectacular brightness and branches (visual appeal)
• Higher peak current (electromagnetic effects)
• Simpler drive electronics
• Better for musical/modulated output
• Different aesthetic goals than pure length

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Power-Limited vs Voltage-Limited

Understanding the interplay between power and voltage limits:

Voltage-Limited Condition

E_tip < E_propagation
- Field too weak at tip
- Spark cannot extend
- More power → brighter/hotter, not longer
- Common for Tesla coils

Power-Limited Condition

E_tip > E_propagation, but P_stream insufficient
- Field adequate but not enough energy
- Spark grows slowly or stalls before reaching potential
- More voltage doesn't help without more power
- Less common for Tesla coils (usually have excess power)

Practical Implications

For most Tesla coils:

1. Design for adequate voltage (large topload, high primary voltage)
2. Ensure sufficient power (but don't need enormous amounts)
3. Optimize ε by choosing appropriate operating mode
4. Accept that voltage limit dominates final length

Rule of thumb:

• If P × T / ε >> L_actual, you're voltage-limited
• If P × T / ε ≈ L_actual, you might be power-limited
• Most coils fall in first category (voltage-limited)

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Key Takeaways

• ε definition: Energy per meter [J/m], approximately constant for a given mode
• Growth rate: dL/dt = P/ε when field threshold is met
• Total energy: E_total ≈ ε × L (linear scaling)
• Theoretical minimum: ε_theory ≈ 0.3-0.5 J/m (ionization only)
• Practical values: 10-300× higher than theoretical (includes heating, radiation, losses)
• Operating mode matters: QCW has low ε (efficient), burst has high ε (inefficient)
• Voltage limit dominates: Most Tesla coils have more than enough power, limited by voltage

Practice

{exercise:phys-ex-03}

Problem 1: A burst-mode coil has ε = 60 J/m. To reach L = 1.5 m in a 200 μs pulse, what power is required? Is this realistic?

Problem 2: A QCW coil delivers 30 kW average power for 15 ms with ε = 12 J/m. Calculate: (a) Total energy delivered (b) Maximum length if power-limited (c) If actual length is only 1.8 m, what does this tell you?

Problem 3: Explain why practical ε is 50-100× larger than the theoretical ionization minimum. List at least three major energy sinks.

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Next Lesson: Empirical ε Values