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Level 5 — MasteryMaterials Chemistry (Aerospace)

Materials Chemistry (Aerospace)

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Level 5 — Mastery (cross-domain: chemistry + physics + mathematics + computation) Time limit: 90 minutes Total marks: 60

Instructions: Answer all three questions. Show all working. Physical constants: Stefan–Boltzmann σ=5.67×10−8 W m−2 K−4\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}; universal gas constant R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}.

Question 1 — Precipitation Hardening & Diffusion Kinetics of 7075 Al (20 marks)

The 7075 aluminium alloy is strengthened by precipitation of η′\eta' (MgZn2_2) phase during artificial ageing. The volume fraction of precipitate transformed, X(t)X(t), follows Johnson–Mehl–Avrami–Kolmogorov (JMAK) kinetics:

X(t)=1−exp⁡ ⁣(−k t n),k=k0exp⁡ ⁣(−QRT)X(t) = 1 - \exp\!\left(-k\,t^{\,n}\right), \qquad k = k_0 \exp\!\left(-\frac{Q}{RT}\right)

(a) Ageing at T1=393 KT_1 = 393\ \mathrm{K} gives X=0.50X=0.50 after t=6.0 ht=6.0\ \mathrm{h}; at T2=433 KT_2 = 433\ \mathrm{K} the same X=0.50X=0.50 is reached after t=0.90 ht=0.90\ \mathrm{h}. Assuming the Avrami exponent nn and k0k_0 are temperature-independent, derive an expression for the activation energy QQ in terms of the two times and temperatures, and evaluate it (kJ/mol). (7)

(b) Given n=1.8n = 1.8, compute the rate constant kk at T1T_1 (with tt in hours). Then determine the time to reach X=0.90X = 0.90 at T1T_1. (6)

(c) Explain, at the microstructural level, why over-ageing (holding too long at temperature) reduces yield strength, and relate this to the Orowan mechanism (qualitative: how inter-precipitate spacing controls strength). (4)

(d) State one reason why 7075 is more susceptible to stress-corrosion cracking than 2024, and name a temper (heat-treatment) designation used to mitigate it. (3)

Question 2 — Thermal Protection: Radiative Equilibrium of a UHTC Leading Edge (22 marks)

A sharp leading edge on a hypersonic vehicle is coated with a ZrB2_2–SiC ultra-high-temperature ceramic (UHTC). At steady state, the convective heat flux from the boundary layer is balanced by re-radiation ("radiative cooling"):

qconv=ε σ Tw4q_{\text{conv}} = \varepsilon\,\sigma\,T_w^{4}

where TwT_w is the wall temperature and ε\varepsilon the emissivity.

(a) For an incoming convective flux qconv=1.20 MW m−2q_{\text{conv}} = 1.20\ \mathrm{MW\,m^{-2}} and ε=0.85\varepsilon = 0.85, compute the equilibrium wall temperature TwT_w. Comment on whether ZrB2_2–SiC (melting point ≈3245 K\approx 3245\ \mathrm{K}) survives. (6)

(b) The convective flux at a stagnation point scales (Fay–Riddell) as qconv∝1/Rnq_{\text{conv}} \propto \sqrt{1/R_n}, where RnR_n is the nose radius. If sharpening the edge from Rn=50 mmR_n = 50\ \mathrm{mm} to Rn=5 mmR_n = 5\ \mathrm{mm}, by what factor does qconvq_{\text{conv}} increase, and what new TwT_w results (same ε\varepsilon, base flux from part (a))? Explain the design trade-off between aerodynamic sharpness and thermal survivability. (6)

(c) Write a short, self-contained Python function wall_temperature(q, eps) that returns TwT_w from the radiative-equilibrium balance, and show the two lines of code that would reproduce your answers to (a) and (b). (4)

(d) Contrast the radiative/reusable TPS strategy (silica tiles, UHTCs) with the ablative strategy (PICA, AVCOAT). Give one governing physical mechanism for each, and one mission scenario where the ablative choice is superior. (6)

Question 3 — Laminate Theory of a CFRP Panel & Corrosion Coupling (18 marks)

A unidirectional carbon/epoxy ply has longitudinal modulus E1=138 GPaE_1 = 138\ \mathrm{GPa}, transverse modulus E2=9.0 GPaE_2 = 9.0\ \mathrm{GPa}, in-plane shear modulus G12=6.9 GPaG_{12} = 6.9\ \mathrm{GPa}, and major Poisson ratio ν12=0.30\nu_{12} = 0.30.

(a) Using the rule of mixtures, the ply has fibre volume fraction VfV_f. Given carbon fibre modulus Ef=230 GPaE_f = 230\ \mathrm{GPa} and matrix modulus Em=3.5 GPaE_m = 3.5\ \mathrm{GPa}, determine VfV_f from the measured E1=138 GPaE_1 = 138\ \mathrm{GPa}. (4)

(b) For the transverse direction, the inverse rule of mixtures gives 1E2=VfEf+1−VfEm.\frac{1}{E_2} = \frac{V_f}{E_f} + \frac{1-V_f}{E_m}. Compute the predicted E2E_2 using your VfV_f from (a), and compare with the measured 9.0 GPa9.0\ \mathrm{GPa}. Comment on the discrepancy. (5)

(c) Derive the minor Poisson ratio ν21\nu_{21} using the reciprocity relation ν12E1=ν21E2\dfrac{\nu_{12}}{E_1} = \dfrac{\nu_{21}}{E_2} and evaluate it. (3)

(d) A CFRP wing skin is fastened to an aluminium spar. Explain the galvanic corrosion risk this creates, why carbon fibre is the more noble member, and state one surface treatment or design measure used to prevent it. (6)

End of paper

Answer keyMark scheme & solutions

Question 1

(a) [7 marks] At a fixed transformed fraction X=0.5X=0.5, ln⁡(1/(1−X))=ktn\ln(1/(1-X)) = k t^n is constant. Therefore at both temperatures: k1t1n=k2t2n=−ln⁡(1−0.5)=ln⁡2.k_1 t_1^n = k_2 t_2^n = -\ln(1-0.5) = \ln 2. Since k=k0e−Q/RTk = k_0 e^{-Q/RT}: k0e−Q/RT1t1n=k0e−Q/RT2t2nk_0 e^{-Q/RT_1} t_1^n = k_0 e^{-Q/RT_2} t_2^n ⇒(t1t2)n=e−Q/RT2+Q/RT1=exp⁡ ⁣[QR(1T1−1T2)].\Rightarrow \left(\frac{t_1}{t_2}\right)^n = e^{-Q/RT_2 + Q/RT_1} = \exp\!\left[\frac{Q}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]. (2 marks setup)

Taking logs: Q=nRln⁡(t1/t2)(1T1−1T2).Q = \frac{n R \ln(t_1/t_2)}{\left(\tfrac{1}{T_1}-\tfrac{1}{T_2}\right)}. (2 marks derivation)

Note QQ here is independent of nn-cancellation? — check: the nn multiplies ln⁡(t1/t2)\ln(t_1/t_2), but this expression as written requires nn. However, the ratio at equal X actually eliminates the exponent because both sides carry tnt^n: we can keep nn explicit. Using n=1.8n=1.8 (from part b), R=8.314R=8.314: 1T1−1T2=1393−1433=2.5445×10−3−2.3095×10−3=2.350×10−4 K−1.\frac{1}{T_1}-\frac{1}{T_2} = \frac{1}{393}-\frac{1}{433} = 2.5445\times10^{-3}-2.3095\times10^{-3} = 2.350\times10^{-4}\ \mathrm{K^{-1}}. ln⁡(t1/t2)=ln⁡(6.0/0.90)=ln⁡(6.667)=1.8971.\ln(t_1/t_2) = \ln(6.0/0.90) = \ln(6.667) = 1.8971. Q=1.8×8.314×1.89712.350×10−4=28.3942.350×10−4=1.208×105 J/mol≈121 kJ/mol.Q = \frac{1.8 \times 8.314 \times 1.8971}{2.350\times10^{-4}} = \frac{28.394}{2.350\times10^{-4}} = 1.208\times10^{5}\ \mathrm{J/mol} \approx 121\ \mathrm{kJ/mol}. (3 marks: correct number ≈ 121 kJ/mol.) [Accept 118–124 kJ/mol depending on rounding.]

(b) [6 marks] k1t1n=ln⁡2k_1 t_1^n = \ln 2 with t1=6.0t_1 = 6.0 h, n=1.8n=1.8: k1=ln⁡26.01.8=0.693125.107=0.02761 h−1.8.k_1 = \frac{\ln 2}{6.0^{1.8}} = \frac{0.6931}{25.107} = 0.02761\ \mathrm{h^{-1.8}}. (3 marks)

For X=0.90X=0.90: k1t1.8=ln⁡(1/0.10)=2.3026k_1 t^{1.8} = \ln(1/0.10) = 2.3026. t1.8=2.30260.02761=83.40⇒t=83.401/1.8=83.400.5556.t^{1.8} = \frac{2.3026}{0.02761} = 83.40 \Rightarrow t = 83.40^{1/1.8} = 83.40^{0.5556}. ln⁡t=0.5556×ln⁡(83.40)=0.5556×4.4237=2.4576⇒t=e2.4576=11.68 h.\ln t = 0.5556\times\ln(83.40) = 0.5556\times4.4237 = 2.4576 \Rightarrow t = e^{2.4576} = 11.68\ \mathrm{h}. (3 marks; ≈ 11.7 h.)

(c) [4 marks]

  • Over-ageing coarsens precipitates: fewer, larger, more widely spaced particles (Ostwald ripening). (1)
  • Orowan strengthening: dislocations bow between and loop around non-shearable precipitates; the stress increment Δτ∝Gb/λ\Delta\tau \propto Gb/\lambda, inversely proportional to inter-precipitate spacing λ\lambda. (2)
  • Over-ageing increases λ\lambda, so Δτ\Delta\tau falls → yield strength drops. (1)

(d) [3 marks]

  • 7075 (Al-Zn-Mg-Cu) is more SCC-susceptible because Zn-rich grain-boundary precipitates and anodic PFZ (precipitate-free zones) provide active corrosion paths under tensile stress. (2)
  • Mitigating temper: T73 (or T7351) over-aged temper trades some strength for greatly improved SCC resistance. (1)

Question 2

(a) [6 marks] Tw=(qconvεσ)1/4=(1.20×1060.85×5.67×10−8)1/4.T_w = \left(\frac{q_{\text{conv}}}{\varepsilon\sigma}\right)^{1/4} = \left(\frac{1.20\times10^{6}}{0.85\times5.67\times10^{-8}}\right)^{1/4}. Denominator =4.8195×10−8= 4.8195\times10^{-8}. Ratio =2.4899×1013= 2.4899\times10^{13}. Tw=(2.4899×1013)0.25=2233 K.T_w = (2.4899\times10^{13})^{0.25} = 2233\ \mathrm{K}. (4 marks) Since 2233 K<3245 K2233\ \mathrm{K} < 3245\ \mathrm{K} melting point, the UHTC survives with comfortable margin. (2 marks)

(b) [6 marks] q∝Rn−1/2q \propto R_n^{-1/2}, so factor =50/5=10=3.162= \sqrt{50/5} = \sqrt{10} = 3.162. (2) New flux q′=3.162×1.20=3.795 MW m−2q' = 3.162\times1.20 = 3.795\ \mathrm{MW\,m^{-2}}. Tw′=(3.795×1064.8195×10−8)1/4=(7.874×1013)1/4=2977 K.T_w' = \left(\frac{3.795\times10^{6}}{4.8195\times10^{-8}}\right)^{1/4} = (7.874\times10^{13})^{1/4} = 2977\ \mathrm{K}. Alternatively Tw′=Tw×(3.162)1/4=2233×1.334=2978 K.T_w' = T_w\times(3.162)^{1/4} = 2233\times1.334 = 2978\ \mathrm{K}. (2) Trade-off: sharper edges give better aerodynamics/lower drag/less shock stand-off, but drive the wall temperature up steeply (T∝Rn−1/8T\propto R_n^{-1/8}); at 2977 K the margin to melting shrinks to ~270 K — hence sharp leading edges demand UHTCs. (2)

(c) [4 marks]

def wall_temperature(q, eps):
    sigma = 5.67e-8
    return (q / (eps * sigma)) ** 0.25
 
# part (a)
print(wall_temperature(1.20e6, 0.85))          # ~2233 K
# part (b)
print(wall_temperature(1.20e6 * 10**0.5, 0.85))# ~2977 K

(2 marks function, 2 marks correct reproduction lines.)

(d) [6 marks]

  • Radiative/reusable (tiles, UHTCs): rejects heat by re-radiation at high surface temperature; material stays chemically intact → reusable. Mechanism: Stefan–Boltzmann radiation + low thermal conductivity to protect substructure. (2)
  • Ablative (PICA, AVCOAT): absorbs heat by endothermic pyrolysis/charring and mass removal; ablated gases block convective heating (transpiration/blowing) and carry heat away. Consumed → single-use. (2)
  • Ablative superior for: very high heat-flux, short-duration events such as planetary entry / capsule re-entry (e.g. Apollo, Stardust, Mars) where fluxes far exceed radiative-equilibrium capability. (2)

Question 3

(a) [4 marks] Rule of mixtures (longitudinal): E1=EfVf+Em(1−Vf)E_1 = E_f V_f + E_m(1-V_f). 138=230Vf+3.5(1−Vf)=3.5+226.5Vf.138 = 230 V_f + 3.5(1-V_f) = 3.5 + 226.5 V_f. Vf=138−3.5226.5=134.5226.5=0.5938≈0.59.V_f = \frac{138 - 3.5}{226.5} = \frac{134.5}{226.5} = 0.5938 \approx 0.59. (4 marks)

(b) [5 marks] 1E2=0.5938230+0.40623.5=2.582×10−3+0.11606=0.11864 GPa−1.\frac{1}{E_2} = \frac{0.5938}{230} + \frac{0.4062}{3.5} = 2.582\times10^{-3} + 0.11606 = 0.11864\ \mathrm{GPa^{-1}}. E2=8.43 GPa.E_2 = 8.43\ \mathrm{GPa}. (3 marks) Comparison: predicted 8.438.43 GPa is close to measured 9.09.0 GPa but slightly low. The inverse rule of mixtures assumes uniform stress and ignores matrix/fibre Poisson constraint and fibre packing; real transverse stiffness is dominated by the compliant matrix, and semi-empirical models (Halpin–Tsai) give better agreement. Discrepancy ~6%. (2 marks)

(c) [3 marks] ν21=ν12E2E1=0.30×9.0138=0.30×0.06522=0.01957≈0.020.\nu_{21} = \nu_{12}\frac{E_2}{E_1} = 0.30\times\frac{9.0}{138} = 0.30\times0.06522 = 0.01957 \approx 0.020. (3 marks)

(d) [6 marks]

  • Carbon fibre is electrically conductive and electrochemically noble (high position in galvanic series); aluminium is highly active (anodic). (2)
  • In presence of an electrolyte (moisture/salt), a galvanic cell forms: Al becomes the anode and corrodes preferentially, with the large-cathode(carbon)/small-anode area ratio accelerating attack. (2)
  • Prevention (any one): apply an insulating barrier (e.g. fibreglass/GFRP isolation ply or sealant/primer between the joint), anodise the aluminium to form a protective oxide, use titanium fasteners, or wet-install with corrosion-inhibiting sealant. (2)
[
  {"claim":"Q1(a) activation energy ~121 kJ/mol", "code":"n=1.8;Rg=8.314;import math;Q=n*Rg*math.log(6.0/0.90)/(1/393-1/433);result = abs(Q/1000-121)<3"},
  {"claim":"Q1(b) k1 ~0.02761 and t(0.90) ~11.7 h", "code":"import math;k1=math.log(2)/6.0**1.8;t=(math.log(1/0.10)/k1)**(1/1.8);result = abs(k1-0.02761)<0.001 and abs(t-11.7)<0.3"},
  {"claim":"Q2(a) Tw ~2233 K", "code":"Tw=(1.20e6/(0.85*5.67e-8))**0.25;result = abs(Tw-2233)<15"},
  {"claim":"Q2(b) sharpened Tw ~2977 K", "code":"Tw=(1.20e6*10**0.5/(0.85*5