Thermal aging rule of thumb for solid dielectrics As the operating temperature of a solid dielectric increases by 10 °C, how does the rate of thermal deterioration (aging) typically change?

Difficulty: Easy

Correct Answer: It approximately doubles

Explanation:


Introduction / Context:
Designers of insulation systems use empirical thermal-life rules to estimate aging. A common engineering heuristic, derived from Arrhenius-type behavior, states that every 10 °C rise roughly doubles the rate of deterioration for many insulating materials (the “10-degree rule”).


Given Data / Assumptions:

  • Solid organic dielectrics under steady thermal stress below decomposition temperatures.
  • Chemical degradation follows Arrhenius kinetics with an effective activation energy.
  • Comparison is approximate; exact factors vary by material and temperature range.


Concept / Approach:

The Arrhenius relation for reaction rate k is k = k0 exp(−Ea/(R T)). A small temperature increase ΔT changes k by a factor exp(Ea/R * (1/T − 1/(T+ΔT))). For typical activation energies of polymer aging, ΔT ≈ 10 °C leads to a factor near 2 in the practical operating range. Hence lifetime L, inversely related to rate, roughly halves per 10 °C rise.


Step-by-Step Solution:

Assume Arrhenius: k ∝ exp(−Ea/(R T)).Compute ratio k(T+10)/k(T) ≈ exp(Ea/R * (1/T − 1/(T+10))).For typical Ea → ratio ≈ 2 → aging rate doubles.


Verification / Alternative check:

IEC thermal index testing and Montsinger’s rule in transformer insulation practice reflect the same trend: 10 °C rise ≈ 2× aging rate (though factors from ~1.8 to ~2.5 are observed).


Why Other Options Are Wrong:

Values 1.5×, 3×, or 4× can occur for specific materials/temperatures, but the widely used general rule is “approximately doubles”.


Common Pitfalls:

Treating the rule as exact for all ranges; always consult material-specific thermal life data for critical designs.


Final Answer:

It approximately doubles

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