Refractory metals — W, Mo, Ta, Re for rocket nozzles: Exercises

Constants you will reuse (memorise these two):

Level 1 — Recognition

Q1. From the parent table, list W, Re, Ta, Mo in order of decreasing melting point and give each $T_{m}$ in K.

Recall Solution Q1

Read straight from the table: $W (3695) > R e (3459) > T a (3290) > M o (2896) K$ Mnemonic check: "We Realise Tantalising Molten metals" = W, Re, Ta, Mo. ✓

Q2. Which of these four has (a) the highest density, (b) the lowest density? Why does density matter for a nozzle?

Recall Solution Q2

Highest density: Re, $21.0 g/cm^{3}$ .
Lowest density: Mo, $10.2 g/cm^{3}$ . Density matters because a rocket must lift its own structure: a heavier nozzle steals payload mass. That is exactly why Mo (a third of W's mass) is chosen when peak temperature allows.

Level 2 — Application

Q3. Estimate the melting point of molybdenum, given $E_{co h} (Mo) \approx 6.8 eV/atom$ , using $k_{B} T_{m} \approx 0.032 E_{co h}$ .

Recall Solution Q3

Rearrange for $T_{m}$ — why? we want a temperature out, so divide the calibrated bond fraction by the exchange rate $k_{B}$ : $T_{m} \approx \frac{0.032 \times 6.8 eV}{8.617 \times 1 0 ^{- 5} eV/K} = \frac{0.2176}{8.617 \times 1 0 ^{- 5}} \approx 2525 K$ True value 2896 K → same order of magnitude, and correctly lower than W's ~3300 K estimate. The cohesive-energy logic holds. ✓

Q4. Tantalum has $E_{co h} \approx 8.1 eV/atom$ . Estimate $T_{m}$ and compare to its true value (3290 K). Is the estimate high or low?

Recall Solution Q4

$T_{m} \approx \frac{0.032 \times 8.1}{8.617 \times 1 0 ^{- 5}} = \frac{0.2592}{8.617 \times 1 0 ^{- 5}} \approx 3008 K$ Estimate 3008 K vs true 3290 K → the model reads low by about 280 K (≈9%). The single fixed constant $c = 0.032$ was calibrated near W, so it slightly under-predicts elsewhere — that is expected for a one-parameter model.

Q5. Compute the Pilling–Bedworth ratio for the oxidation of molybdenum to $M o O_{3}$ . Data: $M_{Mo} = 95.95 g/mol$ , $M_{M o O_{3}} = 143.95 g/mol$ , $ρ_{Mo} = 10.2 g/cm^{3}$ , $ρ_{M o O_{3}} = 4.69 g/cm^{3}$ , and $n = 1$ Mo atom per $M o O_{3}$ .

Recall Solution Q5

Recall the definition (parent note): $PBR = \frac{M _{o x} ρ _{M}}{n M _{M} ρ _{o x}}$ Why this form? it is (volume of oxide made) ÷ (volume of metal eaten); >1 means the oxide is bulkier than the metal it replaced. Substitute: $PBR = \frac{143.95 \times 10.2}{1 \times 95.95 \times 4.69} = \frac{1468.29}{450.01} \approx 3.26$ $PBR \approx 3.26 > 2$ → the oxide is too bulky, spalls off, giving no protection. Combined with $M o O_{3}$ being volatile, uncoated Mo oxidises catastrophically. ✓

Level 3 — Analysis

Q6. Compute PBR for tantalum → $T a_{2} O_{5}$ and compare with the Mo result. Which metal's oxide geometry is better, and does that alone make it "protective"? Data: $M_{Ta} = 180.95$ , $M_{T a_{2} O_{5}} = 441.89 g/mol$ , $ρ_{Ta} = 16.7$ , $ρ_{T a_{2} O_{5}} = 8.2 g/cm^{3}$ , $n = 2$ .

Recall Solution Q6

$PBR = \frac{441.89 \times 16.7}{2 \times 180.95 \times 8.2} = \frac{7379.6}{2967.6} \approx 2.49$ So $T a_{2} O_{5}$ gives PBR ≈ 2.49 — just above the "2" spall threshold, versus Mo's 3.26. Geometrically Ta's oxide is closer to protective, and crucially $T a_{2} O_{5}$ is not volatile (unlike $M o O_{3}$ / $W O_{3}$ ), so it can actually stick and slow further attack. But PBR alone doesn't guarantee protection — adhesion, cracking, and volatility all matter. This is exactly why Ta beats Mo/W in oxidising service. ✓

Q7. Using the half-filled d-band idea, explain in your own words why $T_{m}$ peaks around W/Re and falls toward gold — and predict whether iron (Group 8) should melt higher or lower than molybdenum (Group 6, same period-neighbour logic).

Recall Solution Q7

Chain of reasoning: melting point tracks cohesive energy; cohesive energy tracks how many bonding d-orbitals are filled while anti-bonding ones stay empty. A half-filled d-band (Groups 5–7) maximises bonding-orbital occupation and minimises anti-bonding occupation → strongest cohesion → highest $T_{m}$ . Past half-filling (toward Cu, Au, Hg) electrons must enter anti-bonding states, which cancel bonding → cohesion collapses → $T_{m}$ falls (Hg is liquid!). Prediction: iron sits at Group 8, past the half-filled peak, so its d-band is filling anti-bonding levels → weaker cohesion than the half-filled Group-6 Mo. Predict $T_{m} (Fe) < T_{m} (Mo)$ . ✓ (Reality: Fe 1811 K < Mo 2896 K.) See the schematic below.

Figure — Refractory metals — W, Mo, Ta, Re for rocket nozzles

Level 4 — Synthesis

Q8. A short-burn (25 s), fuel-rich (reducing) apogee motor runs at $T \approx 3000$ K. Design the throat material: pick the base metal, decide any alloying, and justify the coating decision. Give a one-line reason per choice.

Recall Solution Q8

Base metal — W. Its $T_{m} = 3695$ K clears 3000 K with ~700 K margin, so it stays solid. Why W not Mo? Mo's 2896 K is below the operating temperature — it would melt.
Alloy — W–25Re. Ignition thermal shock cracks brittle W; the "rhenium effect" lowers the ductile-brittle transition and raises recrystallisation temperature, so the throat tolerates the shock.
Coating — minimal / short-life only. The exhaust is reducing (little free $O_{2}$ ) and the burn is short, so volatile $W O_{3}$ loss is limited; a thin or no coating is acceptable. Design: radiation-cooled W–25Re throat insert — essentially real apogee-motor practice. ✓

Q9. For a regeneratively cooled thrust-chamber liner you must machine intricate coolant channels and survive a mildly oxidising, corrosive exhaust at ~2500 K. Justify choosing Ta–10W over pure W.

Recall Solution Q9

Ductility / fabricability: Ta is ductile and machinable at room temperature (W is brittle below ~250–450 °C), so complex cooling channels can actually be formed.
Corrosion resistance: Ta forms a non-volatile, adherent $T a_{2} O_{5}$ (PBR ≈ 2.49, non-volatile) — far better in oxidising service than W's volatile $W O_{3}$ .
The 10% W: raises strength and $T_{m}$ back up, recovering some peak-temperature capacity lost by using ductile Ta. Trade-off logic: you sacrifice a little peak temperature to gain manufacturability + corrosion resistance — the correct 80/20 when geometry is complex and the environment is oxidising. ✓

Level 5 — Mastery

Q10. Full chain. You are handed a candidate metal "X" with $E_{co h} = 7.4 eV/atom$ for a nozzle running at $T = 2700$ K in an oxidising exhaust. Its oxide $X O_{2}$ is volatile with PBR data $M_{X} = 92.9$ , $M_{X O_{2}} = 124.9 g/mol$ , $ρ_{X} = 8.57$ , $ρ_{X O_{2}} = 4.47 g/cm^{3}$ , $n = 1$ . (a) Estimate $T_{m}$ . (b) Does X survive thermally with margin? (c) Compute PBR. (d) Give the final verdict and one mitigation. (This is niobium, Nb.)

Recall Solution Q10

(a) Melting estimate. $T_{m} \approx \frac{0.032 \times 7.4}{8.617 \times 1 0 ^{- 5}} = \frac{0.2368}{8.617 \times 1 0 ^{- 5}} \approx 2748 K$ (b) Thermal margin. $2748 K - 2700 K \approx 48 K$ of margin — dangerously thin. Real Nb melts at 2750 K, so the estimate is spot-on, but ~50 K margin is not safe for a hot throat: X barely survives. (c) PBR. $PBR = \frac{124.9 \times 8.57}{1 \times 92.9 \times 4.47} = \frac{1070.39}{415.26} \approx 2.58$ $> 2$ → oxide too bulky, spalls; and it is volatile, so no protection at all. (d) Verdict. X (Nb) has insufficient thermal margin AND a non-protective volatile oxide in an oxidising exhaust → reject bare X. Mitigation: either drop to a lower- $T$ / reducing environment, or apply a silicide/ceramic oxidation barrier and back it with a higher- $T_{m}$ substrate (e.g. W–Re) — pushing you toward ceramic-matrix composites or coatings. ✓

Connections

5.4.02 Refractory metals — W, Mo, Ta, Re for rocket nozzles (Hinglish)
Metallic bonding and the electron sea model
d-block trends — melting points and cohesive energy
Oxidation kinetics and the Pilling–Bedworth ratio
Creep and recrystallisation in metals
Ceramic-matrix composites — alternatives to refractory metals (ZrB2, HfC)
Thermal barrier coatings and ablatives