plasma-expert/spark-lessons/lessons/03-spark-physics/05-thermal-memory.md

id	title	section	difficulty	estimated_time	prerequisites	objectives	tags
phys-05	Thermal Memory and Channel Persistence	Spark Growth Physics	intermediate	40	[phys-03 phys-04]	[Understand thermal diffusion time constants for plasma channels Calculate channel persistence times for different diameters Recognize the role of convection in extending channel lifetime Apply thermal memory concepts to QCW vs burst mode operation]	[thermal-diffusion convection channel-persistence time-constants ionization-memory]

Thermal Memory and Channel Persistence

Once formed, a plasma channel doesn't instantly disappear. It has thermal memory - the channel stays hot and partially ionized for some time after power is removed. Understanding these timescales is crucial for optimizing operating modes.

Temperature Regimes

Plasma channels exist in different temperature regimes depending on current and power density:

Streamers (Cold Plasma)

Temperature: T ≈ 1000-3000 K
- Weakly ionized (few % ionization)
- Mostly neutral gas with some ions/electrons
- Purple/blue color (N₂ molecular emission)
- Low conductivity

Leaders (Hot Plasma)

Temperature: T ≈ 5000-20,000 K
- Fully ionized plasma
- White/orange color (blackbody + line emission)
- High conductivity
- Approaching temperatures of stellar photospheres!

Temperature comparison:

• Room temperature: 300 K
• Candle flame: 1500 K
• Thin streamers: 1000-3000 K
• Thick leaders: 5000-20,000 K
• Sun's photosphere: 5800 K

Leaders are literally as hot as the surface of the Sun!

Thermal Diffusion Time

Heat diffuses radially outward from the hot channel core according to:

τ_thermal = d² / (4α_thermal)

where:
  d = channel diameter [m]
  α_thermal ≈ 2×10⁻⁵ m²/s (thermal diffusivity of air)

Physical meaning: Time for heat to diffuse a distance d through air by conduction.

Examples for Different Channel Sizes

Thin streamer (d = 100 μm):

τ = (100×10⁻⁶)² / (4 × 2×10⁻⁵)
  = 10⁻⁸ m² / (8×10⁻⁵ m²/s)
  = 1.25×10⁻⁴ s
  = 0.125 ms
  ≈ 0.1-0.2 ms

Medium channel (d = 2 mm):

τ = (2×10⁻³)² / (4 × 2×10⁻⁵)
  = 4×10⁻⁶ m² / (8×10⁻⁵ m²/s)
  = 0.05 s
  = 50 ms

Thick leader (d = 5 mm):

τ = (5×10⁻³)² / (4 × 2×10⁻⁵)
  = 25×10⁻⁶ m² / (8×10⁻⁵ m²/s)
  = 0.3125 s
  = 312 ms
  ≈ 0.3-0.6 s

Key insight: Thermal diffusion time scales as d² - thicker channels persist much longer!

Why Observed Persistence is Longer

Pure thermal diffusion predicts cooling in 0.1-300 ms, but channels persist longer due to additional effects:

1. Convection (Buoyancy)

Hot gas is less dense and rises:

Buoyancy velocity: v ≈ √(g × d × ΔT/T_amb)

where:
  g = 9.8 m/s² (gravity)
  d = channel diameter
  ΔT = temperature excess above ambient
  T_amb = ambient temperature (≈300 K)

Example: 2 mm channel at ΔT = 10,000 K

v ≈ √(9.8 × 0.002 × 10000/300)
  ≈ √(9.8 × 0.002 × 33.3)
  ≈ √(0.653)
  ≈ 0.81 m/s

The hot channel rises at ~0.8 m/s, creating a continuously renewing hot column!

Effect on persistence:

• Rising column remains coherent (doesn't diffuse sideways as fast)
• Maintains hot gas path for seconds
• Why Tesla coil sparks leave visible "smoke trails"
• Enhances thermal memory significantly

2. Ionization Memory

Even after thermal cooling begins, ions and electrons persist:

Recombination time: τ_recomb = 1/(α_recomb × n_e)

where:
  α_recomb ≈ 10⁻¹³ m³/s (recombination coefficient)
  n_e = electron density [m⁻³]

Typical: τ_recomb ≈ 10 μs to 10 ms

Effect on persistence:

• Channel remains partially ionized after cooling
• Lower resistance than cold air
• Easier to re-ionize than virgin air
• "Memory" of previous discharge path

3. Broadened Effective Diameter

Turbulence and mixing increase effective channel size:

d_effective > d_initial (due to turbulence)

Larger diameter → longer τ_thermal

Effective Persistence Times

Combining all effects:

Thin streamers:

Pure thermal: ~0.1-0.2 ms
With convection: ~1-5 ms
Ionization memory: ~0.1-1 ms
Effective persistence: ~1-5 ms

Thick leaders:

Pure thermal: ~50-300 ms
With convection: seconds (buoyant column maintained)
Ionization memory: ~1-10 ms
Effective persistence: seconds

Visual evidence: High-speed photography shows spark channels glowing and rising for seconds after power is removed.

{image:spark-channel-persistence-sequence}

QCW Advantage

QCW ramp times (5-20 ms) are designed to exploit channel persistence:

Timeline of QCW Growth

t = 0 ms:
  - Initial streamers form from topload
  - Thin, fast, purple channels
  - Temperature: ~2000 K

t = 0.5-1 ms:
  - Current begins flowing through streamers
  - Joule heating: P = I²R
  - Temperature rising

t = 1-2 ms:
  - Channel heats to 5000+ K
  - Thermal ionization becomes dominant
  - Leader formation begins at base

t = 2-5 ms:
  - Leader established and growing
  - Hot channel maintained by continuous power
  - New growth builds on existing ionization
  - Temperature: 10,000-20,000 K

t = 5-20 ms:
  - Leader continues extending
  - Persistence time >> growth time
  - Channel stays hot entire duration
  - Efficient energy use: no re-ionization needed

t > 20 ms (after ramp ends):
  - Power removed
  - Channel begins cooling
  - Buoyancy carries hot gas upward
  - Visible glow for seconds

Key advantage: The ramp duration (5-20 ms) is shorter than thermal diffusion time (50+ ms for leaders), so the channel NEVER cools during growth!

Energy Efficiency Mechanism

QCW flow:

Energy → Initial ionization (startup cost)
       → Heating to leader temperature
       → Maintaining hot channel (low cost)
       → Extending length (efficient)

Result: Most energy after startup goes into extension
ε_QCW ≈ 5-15 J/m (low, efficient)

Burst Mode Problem

Burst mode pulses are short (50-500 μs) with long gaps (ms):

Timeline of Burst Mode

t = 0 μs:
  - High voltage, cold air
  - Streamer inception

t = 0-100 μs:
  - First pulse (high peak power)
  - Bright streamers form
  - Some heating but limited
  - Temperature reaches ~3000-5000 K

t = 100 μs (pulse ends):
  - Power removed
  - Channel begins cooling immediately
  - Thermal diffusion time ~0.1-0.5 ms for thin channels

t = 0.1-1 ms:
  - Channel cools significantly
  - Temperature drops to ~1000 K
  - Ionization recombines
  - Channel approaching cold air

t = 1-10 ms (between pulses):
  - Next pulse arrives
  - Must re-ionize mostly cold gas
  - Energy wasted on re-heating
  - Little thermal memory remains

Result: Each pulse restarts from nearly cold conditions!

Energy inefficiency mechanism:

Energy → Initial ionization (EVERY pulse)
       → Heating (REPEATED)
       → Brief brightening
       → Cooling (wasted)
       → Re-ionization overhead (high)

Result: Energy into repeated startup, not cumulative growth
ε_burst ≈ 30-100+ J/m (high, inefficient)

Analogy: Boiling Water

QCW (efficient):

Turn stove on and keep it on
Water heats up once
Maintain boiling continuously
Minimal energy to sustain

Burst (inefficient):

Pulse stove on/off rapidly
Water heats briefly
Water cools between pulses
Must reheat repeatedly
High energy for little sustained boiling

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WORKED EXAMPLE: Thermal Time Constants

Given:

• Channel diameter: d = 2 mm (typical leader)
• Air thermal diffusivity: α = 2×10⁻⁵ m²/s
• Temperature excess: ΔT = 8000 K
• Ambient temperature: T_amb = 300 K

Find: (a) Pure thermal diffusion time (b) Convection velocity (c) QCW ramp time recommendation

Solution

Part (a): Thermal diffusion time

τ_thermal = d² / (4α)
          = (2×10⁻³)² / (4 × 2×10⁻⁵)
          = 4×10⁻⁶ m² / (8×10⁻⁵ m²/s)
          = 0.05 s
          = 50 ms

Part (b): Convection velocity

v ≈ √(g × d × ΔT/T_amb)
  ≈ √(9.8 × 0.002 × 8000/300)
  ≈ √(9.8 × 0.002 × 26.67)
  ≈ √(0.523)
  ≈ 0.72 m/s

Upward velocity of ~0.7 m/s helps maintain hot column.

Part (c): QCW ramp recommendation

τ_thermal = 50 ms

For efficient QCW operation:
T_ramp << τ_thermal (finish before significant cooling)

Recommended: T_ramp = 0.1 × τ to 0.4 × τ
           = 5-20 ms

Sweet spot: ~10 ms (20% of τ_thermal)

Reasoning:

• If T_ramp >> τ_thermal (e.g., 200 ms):

  • Channel cools during ramp
  • Must reheat repeatedly
  • Loses QCW efficiency advantage

• If T_ramp << τ_thermal (e.g., 1 ms):

  • May not form thick leaders
  • Closer to burst behavior
  • Doesn't exploit full persistence

• Optimal: T_ramp ≈ 10-20 ms

  • Channel stays hot throughout
  • Leaders form and persist
  • Maximum efficiency

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WORKED EXAMPLE: Burst vs QCW Timing

Given:

• Burst pulse: 200 μs every 5 ms (5 ms period)
• QCW ramp: 15 ms continuous
• Both use same average power

Find: (a) Why burst is inefficient for thin channels (d = 100 μm) (b) Why QCW is efficient for thick channels (d = 3 mm)

Solution

Part (a): Burst with thin streamers

Channel diameter: d = 100 μm
Thermal time: τ = (100×10⁻⁶)² / (8×10⁻⁵) = 0.125 ms

Timeline:
t = 0: Pulse starts, channel forms
t = 200 μs: Pulse ends (0.2 ms)
           Channel cooling for: 0.125 ms ≈ τ/1
t = 5 ms: Next pulse
         Channel has cooled for: 5 ms = 40 × τ
         COMPLETELY COLD

Result: Each pulse re-ionizes from scratch
        High ε (inefficient)

Part (b): QCW with thick leaders

Channel diameter: d = 3 mm
Thermal time: τ = (3×10⁻³)² / (8×10⁻⁵) = 112 ms

Timeline:
t = 0: Ramp starts, initial streamers
t = 2 ms: Heating → leader formation begins
t = 5 ms: Leader well-established (hot)
t = 15 ms: Ramp ends
          Total time elapsed: 15 ms = 0.13 × τ

Cooling fraction: exp(-15/112) ≈ exp(-0.13) ≈ 0.88

Result: Channel stays at 88% of peak temperature!
        Leader persists throughout ramp
        Low ε (efficient)

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Key Takeaways

• Thermal diffusion time: τ = d²/(4α), scales quadratically with diameter
• Thin streamers: τ ≈ 0.1-0.2 ms (fast cooling)
• Thick leaders: τ ≈ 50-600 ms (slow cooling)
• Convection: Hot gas rises at ~0.5-1 m/s, maintains hot column for seconds
• Ionization memory: Partial ionization persists 0.1-10 ms after thermal cooling
• Effective persistence: 1-5 ms for streamers, seconds for leaders
• QCW advantage: Ramp time (5-20 ms) << leader thermal time (~50+ ms)
• Burst problem: Gap between pulses (ms) >> streamer thermal time (~0.1 ms)

Practice

{exercise:phys-ex-05}

Problem 1: A streamer has d = 150 μm. Calculate τ_thermal. If burst pulse width is 500 μs with 10 ms between pulses, does the channel cool significantly?

Problem 2: Why do thick leaders persist longer than thin streamers? Give two physical reasons with approximate timescales.

Problem 3: A QCW coil uses 25 ms ramps. For a 3 mm diameter leader (τ ≈ 100 ms), estimate the fraction of peak temperature remaining at end of ramp (use exponential cooling approximation).

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Next Lesson: Streamers vs Leaders