Carbon-carbon composites (RCC for nose cone - leading edges): Exercises

Two constants recur, so define them once here in plain words:

Definition The five symbols used all over this page

$σ$ = stress — the internal force a material feels spread over the area carrying it, i.e. force per unit area. Think of it as "how hard the material is being squeezed or pulled from inside". Units: pascals ( $Pa$ ), where $1 Pa = 1 N/m^{2}$ and $1 MPa = 1 0^{6} Pa$ .
$ε$ = strain — how much the material actually stretches or shrinks, as a fraction of its original length ( $Δ L / L$ ). It is a pure number (no units) because it's a length divided by a length.
$E$ = Young's modulus — how stiff a material is: the stress you get per unit of strain, $E = σ / ε$ . Big $E$ = hard to deform. Units: pascals ( $Pa$ ), and $1 GPa = 1 0^{9} Pa$ .
$α$ = coefficient of thermal expansion — how much a material grows per degree of heating, as a fraction of its length. Units: per kelvin ( $K^{- 1}$ ). Built fully in Thermal stress and α (coefficient of expansion).
$Δ T$ = temperature change — how many degrees hotter (or colder) the part became, i.e. $Δ T = T_{final} - T_{initial}$ . The capital Greek delta ( $Δ$ ) is standard shorthand for "the change in". Units: kelvin ( $K$ ); a change of $1 K$ equals a change of $1^{\circ} C$ .

Intuition When is this formula allowed? (validity limits)

$σ = E α Δ T$ is a small-strain, linear-elastic model. It assumes three things that you must check before trusting a number:

Linear elasticity — stress stays proportional to strain (Hooke's law), which only holds below the material's yield/fracture point. Once the steel bar in L2.2 exceeds its yield stress, the formula over-predicts — the real material yields and the excess "stress" becomes permanent deformation.
Constant $E$ and $α$ — both actually drift with temperature. Over a huge $Δ T$ you should use averaged values; our single numbers are first-order estimates.
Full constraint — the formula is the maximum case (zero expansion allowed). A partly-free part develops less stress. So treat every answer here as a conservative upper bound.

Level 1 — Recognition

L1.1

Which single property of graphite makes it survive re-entry heat without ever becoming soft, and why is that different from a metal?

Recall Solution

Answer: Graphite has no liquid phase at ordinary pressure — it goes straight from solid to gas (sublimes) near $3600^{\circ} C$ . Why it matters: A metal passes through a molten stage as it heats, and long before melting it goes soft and loses shape. Carbon skips the liquid entirely, so there is never a "soft" temperature window — it holds its geometry right up until it turns to vapour. See Graphite structure and sublimation.

L1.2

Name the coating painted on RCC and the chemistry that lets it self-heal a small crack.

Recall Solution

Answer: The coating is silicon carbide (SiC). When hot oxygen reaches it, the surface oxidises to glassy silica: $SiC (s) + \frac{3}{2} O_{2} (g) \to SiO_{2} (s) + CO (g)$ Self-heal mechanism: the $SiO_{2}$ formed is a glass that softens and flows at re-entry temperatures, so it runs into any hairline crack and plugs it — like a scab sealing a cut. See Silicon Carbide and oxidation-resistant ceramics.

L1.3

RCC "survives 1500 °C" yet we still call it oxidation-prone. State the two different things being confused, in one sentence.

Recall Solution

Answer: Temperature tolerance is not the same as chemical inertness. Carbon can withstand enormous heat, but it also reacts (burns) with oxygen above about $400^{\circ} C$ : $C (s) + O_{2} (g) \to CO_{2} (g)$ . Surviving heat ≠ resisting oxygen.

Level 2 — Application

L2.1

A fully constrained RCC panel is heated by $Δ T = 1200 K$ . Using $E = 30 GPa$ and $α = 2 \times 1 0^{- 6} K^{- 1}$ , compute the thermal stress. Is $72 MPa$ small or large for carbon (compare to typical strengths near $300 MPa$ )?

Recall Solution

What we do: plug straight into $σ = E α Δ T$ . Why this formula: the panel wants to expand by $α Δ T$ but is clamped, so an equal-and-opposite squeeze appears; stiffness $E$ turns that forced strain into stress. $σ = (30 \times 1 0^{9}) (2 \times 1 0^{- 6}) (1200) = 7.2 \times 1 0^{7} Pa = 72 MPa$ Read it: $72 MPa$ is roughly a quarter of the material's strength — comfortably survivable. That headroom is why carbon is chosen.

L2.2

Same $Δ T = 1200 K$ , but for steel: $E = 200 GPa$ , $α = 12 \times 1 0^{- 6} K^{- 1}$ . Compute the stress and the ratio steel/carbon.

Recall Solution

Why the same formula applies: the physics is identical regardless of material — a clamped steel bar also wants to expand by $α Δ T$ , is also prevented, and also develops an equal-and-opposite mechanical strain that its stiffness $E$ converts into stress. Only the two numbers $E$ and $α$ change; the reasoning $σ = E α Δ T$ is universal. So we reuse it and simply substitute steel's values. $σ_{steel} = (200 \times 1 0^{9}) (12 \times 1 0^{- 6}) (1200) = 2.88 \times 1 0^{9} Pa = 2880 MPa$ Ratio: $\frac{2880}{72} = 40$ . Why so big: steel is both stiffer ( $E$ up $\approx 6.7 \times$ ) and expands more ( $α$ up $6 \times$ ); the two factors multiply — $6.7 \times 6 = 40$ . Steel's induced stress is far past its yield point, so it buckles or cracks. Look at the bar chart below: the red bar (steel) towers over black (carbon).

Figure — Carbon-carbon composites (RCC for nose cone - leading edges)

Figure s01 — bar chart of constrained thermal stress for the same $Δ T = 1200 K$ . Horizontal axis: material (carbon vs steel). Vertical axis: induced stress in MPa (linear scale, 0–3200). Black bar = carbon (72 MPa); the single red bar = steel (2880 MPa); the red arrow highlights the 40× jump.

L2.3

Deposit exactly $12 g$ of carbon by CVI using $CH_{4} (g) \sim 1100^{\circ} C C (s) + 2 H_{2} (g)$ . How much methane is consumed and how much hydrogen off-gas is produced? (Molar masses: C = 12, CH₄ = 16, H₂ = 2 g/mol.)

Recall Solution

Step 1 — moles of carbon: $12 g \div 12 g/mol = 1 mol C$ . Step 2 — stoichiometry: the equation is $1 : 1 : 2$ , so $1 mol C$ needs $1 mol CH_{4}$ and makes $2 mol H_{2}$ . Step 3 — masses: $m_{CH_{4}} = 1 \times 16 = 16 g, m_{H_{2}} = 2 \times 2 = 4 g$ Mass-balance check: in $16$ g = out $12 + 4 = 16$ g. ✓ Why it matters: the $4$ g of hydrogen is flammable — CVI furnaces must vent it safely. See Chemical Vapour Infiltration / Deposition (CVI/CVD).

Level 3 — Analysis

L3.1

A phenolic resin has a char yield of $55 %$ (fraction of starting mass left as carbon after pyrolysis). You impregnate a char with $100 g$ of fresh resin. If the resin fully infiltrates the pores, how much carbon does this one cycle add? Then explain qualitatively why cycle 2 adds less.

Recall Solution

This cycle's carbon: $100 g \times 0.55 = 55 g$ of new carbon deposited — if all resin fit in the pores. Why cycle 2 adds less: after cycle 1 the pores are partly filled, so less resin can enter on the next pass. Less resin in $\Rightarrow$ less char out. This is diminishing returns: each cycle can only work on the shrinking pore volume that remains. See Pyrolysis and char yield of polymers. Consequence: density approaches full value asymptotically — you never quite reach $100 %$ , which is why $\approx 4$ – $6$ cycles are used and RCC is slow and costly.

L3.2

Model densification as filling porosity where each cycle removes a fixed fraction $f = 0.55$ of the remaining pore volume. Starting from $30 %$ porosity, compute the porosity left after cycles 1, 2, 3, 4. How many cycles to get below $3 %$ porosity?

Recall Solution

Why a geometric model: "removes a fixed fraction of what's left" means each cycle multiplies the remaining porosity by $(1 - f) = 0.45$ . That is a geometric decay $P_{n} = P_{0} (1 - f)^{n}$ . $P_{0} = 30%$ $P_{1} = 30 \times 0.45 = 13.5%$ $P_{2} = 13.5 \times 0.45 = 6.075%$ $P_{3} = 6.075 \times 0.45 = 2.734%$ $P_{4} = 2.734 \times 0.45 = 1.230%$ First cycle below 3 %: cycle 3 ( $2.73% < 3%$ ). Read the curve: the black porosity curve below drops steeply then flattens — the red dashed $3%$ line is crossed at cycle 3, visually confirming diminishing returns.

Figure s02 — remaining porosity versus densification cycle number. Horizontal axis: cycle $n$ (0–6). Vertical axis: remaining porosity in per cent (linear scale, 0–33), each point labelled with its value. Black curve = model $P_{n} = 30 \times 0.4 5^{n}$ ; the red dashed line marks the 3 % target and the red arrow marks where the curve first drops below it (cycle 3).

L3.3

The SiC coating oxidises to $SiO_{2}$ . The trouble: SiC and the carbon underneath have different expansion coefficients, so cooling from re-entry opens cracks. If the coating $α_{SiC} = 4.5 \times 1 0^{- 6} K^{- 1}$ and carbon $α_{C} = 2 \times 1 0^{- 6} K^{- 1}$ , and both cool by $Δ T = 1000 K$ , estimate the mismatch strain $Δ ε = (α_{SiC} - α_{C}) Δ T$ . Why does this favour the glass sealant?

Recall Solution

Why subtract the $α$ 's: if two bonded layers wanted to shrink by different amounts, the difference is the strain that must be absorbed at the interface — that's what pries a crack open. Watch the units: each $α$ carries $K^{- 1}$ and $Δ T$ carries $K$ , so their product is $K^{- 1} \times K = 1$ — a pure number (dimensionless strain), exactly what a fractional length change should be. Carrying the cancellation through in a single line: $Δ ε = (4.5 - 2) \times 1 0^{- 6} K^{- 1} \times 1000 K = 2.5 \times 1 0^{- 6} \times 1000 = 2.5 \times 1 0^{- 3} = 0.25%$ Interpretation: a $0.25%$ mismatch strain on every thermal cycle repeatedly cracks the brittle SiC. The glassy $SiO_{2}$ sealant is essential precisely because it re-flows and re-plugs these thermally-driven cracks each mission — the coating is not permanent, it is self-repairing.

Level 4 — Synthesis

L4.1

An engineer proposes replacing RCC with a tungsten nose cap (metal, melts at $3422^{\circ} C$ , so "it won't melt at 1500 °C"). Give three first-principles reasons this is a bad idea for a re-entry leading edge, each tied to a property.

Recall Solution

Reason 1 — thermal stress. Tungsten's $α \approx 4.5 \times 1 0^{- 6} K^{- 1}$ and $E \approx 400 GPa$ . For $Δ T = 1200 K$ : $σ = 400 \times 1 0^{9} \times 4.5 \times 1 0^{- 6} \times 1200 = 2.16 \times 1 0^{9} Pa = 2160 MPa$ — enormous versus carbon's $72 MPa$ . It cracks from thermal shock. Reason 2 — weight. Tungsten's density ( $\approx 19.3 g/cm^{3}$ ) is $\sim 10 \times$ carbon's ( $\approx 1.8 g/cm^{3}$ ). A metal nose cap is dead mass on a vehicle where every kilogram costs fuel. Reason 3 — strength-vs-temperature. Metals soften as they approach melting; carbon gains strength up to $\sim 2000^{\circ} C$ . At leading-edge temperatures tungsten is well into its weakening regime while C/C is still hardening. Verdict: "high melting point" is only one requirement — RCC wins on stress, mass, and hot-strength together.

L4.2

Explain, as a connected chain of cause and effect, why RCC needs the SiC coating and why that same interface is its most dangerous weakness. Reference the Space Shuttle Columbia disaster — materials case study.

Recall Solution

Chain of cause → effect:

RCC's matrix and fibres are carbon, chosen because carbon sublimes (no melt) and strengthens when hot — ideal for the temperature it must endure.
But carbon oxidises above ~400 °C — and re-entry air is full of hot oxygen and plasma, so bare carbon would burn away.
So we add an SiC coating that oxidises to a glassy $SiO_{2}$ film, sealing oxygen out and even self-healing hairline cracks (L1.2).
But SiC and carbon have mismatched $α$ (L3.3): every heating/cooling cycle strains the interface by $\approx 0.25%$ and cracks the brittle coating. The protection is therefore only as good as the intact coating — it relies on the glass re-flowing into those cracks each mission.
Columbia (2003), the case study: during launch a chunk of foam insulation broke off and struck the wing's leading edge, punching a hole clean through the RCC panel and its coating. On re-entry, superheated plasma poured through that breach onto the now-unprotected carbon inside the wing; the carbon oxidised and eroded from within, the wing structure failed, and the orbiter broke apart — killing all seven crew. See Space Shuttle Columbia disaster — materials case study. The synthesis: the SiC/carbon interface is simultaneously the enabler (it is the only reason carbon can be used in hot air at all) and the single point of failure (breach it and the very carbon it protects begins to burn). Columbia is the historical proof that this one interface decides whether the vehicle lives or dies.

Level 5 — Mastery

L5.1

Design brief. You must qualify a C/C leading edge to survive a worst-case constrained heating of $Δ T = 1500 K$ with a safety factor of at least 3 against a hot strength of $300 MPa$ . Using $E = 30 GPa$ , find the maximum allowable $α$ . Does real carbon ( $α = 2 \times 1 0^{- 6} K^{- 1}$ ) pass?

Recall Solution

What "safety factor 3" means: the actual stress must stay below strength/3, i.e. $σ_{m a x} = 300/3 = 100 MPa$ . Why invert the formula: we know the stress ceiling and want the property limit, so solve $σ = E α Δ T$ for $α$ : $α_{m a x} = \frac{σ _{m a x}}{E Δ T} = \frac{100 \times 1 0 ^{6}}{( 30 \times 1 0 ^{9} ) ( 1500 )}$ $α_{m a x} = \frac{1 0 ^{8}}{4.5 \times 1 0 ^{13}} = 2.22 \times 1 0^{- 6} K^{- 1}$ Verdict: real carbon's $α = 2 \times 1 0^{- 6} K^{- 1}$ is below the $2.22 \times 1 0^{- 6}$ ceiling → it passes, but only barely (margin ≈ 10 %). This razor-thin margin explains why real designs also allow some expansion (not fully constrained) rather than trusting $α$ alone.

L5.2

Full mass-and-cycle synthesis. A char part has $30 %$ porosity and total volume $500 cm^{3}$ ; pore volume must be filled with deposited carbon (density $2.0 g/cm^{3}$ ). Each densification cycle fills $55 %$ of the remaining pore volume via resin of $55 %$ char yield. (a) What mass of carbon totally fills the pores? (b) After how many cycles is porosity below $3 %$ (reuse L3.2's model)? (c) For cycle 1 only, how much fresh resin must be impregnated to deposit that cycle's carbon?

Recall Solution

(a) Carbon to fill all pores. Pore volume $= 30% \times 500 = 150 cm^{3}$ . Filling with carbon of density $2.0 g/cm^{3}$ : $m_{fill} = 150 \times 2.0 = 300 g$ (b) Cycles below 3 % porosity. Same geometric decay $P_{n} = 30% \times 0.4 5^{n}$ as L3.2 → crosses $3%$ at cycle 3 ( $2.73%$ ). (c) Resin needed for cycle 1. Cycle 1 fills $55%$ of the $150 cm^{3}$ pores $= 0.55 \times 150 = 82.5 cm^{3}$ of carbon. Carbon mass this cycle $= 82.5 \times 2.0 = 165 g$ . That carbon is $55%$ of the impregnated resin mass, so: $m_{resin} = \frac{165}{0.55} = 300 g$ Read it: to lay down $165$ g of carbon you must push in $300$ g of resin and drive off $300 - 165 = 135$ g of volatiles — that lost mass is the porosity that the next cycle must chase.

Wrap-up recall

Recall One-line answers (cover them)

Formula for constrained thermal stress ::: $σ = E α Δ T$ What $Δ T$ means and its units ::: temperature change $T_{final} - T_{initial}$ , in kelvin Steel/carbon stress ratio at same $Δ T$ (Ex L2.2) ::: $40 \times$ CH₄ mass to deposit 12 g C ::: $16$ g, releasing $4$ g H₂ Porosity model per cycle ::: $P_{n} = P_{0} (1 - f)^{n}$ , geometric decay Cycles to reach <3 % porosity from 30 % at f=0.55 ::: 3 cycles Why the SiC coating is also the weak point ::: breach it → carbon oxidises (Columbia) Resin needed for 165 g carbon at 55 % char yield ::: 300 g