Hand Calculations logo Hand Calculations All help pages ▾

Thermal Protection Thickness⚠ unverified

Aerospace / Reentry · Compute the required thermal-protection-system (TPS) thickness

Parameters

InputSymbolUnitDefaultDescription
qqW/m**21.0Incident heat flux
tts1.0Heating duration
kk1.0Thermal conductivity of the TPS material, in watts per metre kelvin (W/(m.K))
cpcp1.0Specific heat capacity of the TPS material, in joules per kilogram kelvin (J/(kg.K))
rhoρkg/m**31.0Density of the TPS material
T_maxTmaxK1.0Maximum allowable temperature
OutputSymbolUnitDescription
resultthicknessmRequired TPS thickness, in metres (m)

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Model the heat shield as a lumped heat sink: a slab of area $A$, thickness $\delta$, that absorbs the incident heat by warming uniformly, with negligible loss out the back (conservative) and negligible internal temperature gradient (the lumped assumption). The total heat absorbed over the entry, per unit area, is the heat load $q\,t$.

The energy stored in raising the slab from $T_0$ to at most $T_{max}$ is its mass times specific heat times the temperature rise, per unit area:

$$E_{stored} = (\rho\,\delta)\,c_p\,(T_{max} - T_0).$$

Setting the absorbed heat equal to the stored energy at the limiting condition (back face just reaching $T_{max}$):

$$q\,t = \rho\,\delta\,c_p\,(T_{max} - T_0) \quad\Longrightarrow\quad \delta = \frac{q\,t}{\rho\,c_p\,(T_{max} - T_0)}. \qquad\blacksquare$$

This is a conservative lower bound on thickness for a heat sink: it ignores re-radiation from the hot surface (which helps) and conduction gradients (which hurt), and it says nothing about ablative shields (which carry heat away with departing mass, Ablation Rate) or insulative shields (which limit back-face temperature by low conductivity). It is the right first estimate for a solid heat-sink layer.

Dimensional check. $$\left[\frac{q\,t}{\rho\,c_p\,\Delta T}\right] = \frac{(\text{W}/\text{m}^2)(\text{s})}{(\text{kg}/\text{m}^3)(\text{J}/(\text{kg}\cdot\text{K}))(\text{K})} = \frac{\text{J}/\text{m}^2}{\text{J}/\text{m}^3} = \text{m}.\ \checkmark$$

History and Development

Related Concepts: Reentry Heat Load, Stagnation Heat Flux, Ablation Rate, Peak Heating Altitude, Reentry Time, Thermal Stress, Reentry Heat Flux

Notes: Registry calculator thermal-protection-thickness (unverified). Heat-sink TPS model $\delta = q t/[\rho c_p(T_{max}-300)]$. Unused input $k$ (conductivity — not part of a heat-sink model); $c_p$ units-label bug (is J/(kg·K)); reference $T_0 = 300\,\text{K}$ hard-coded (defaults give negative thickness). Conservative lower bound; does not model ablative or insulative TPS. Flagged in Known Issues.

← Back to the workspace  ·  All help pages  ·  Getting started