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

Before any symbol appears in a calculation, here is the full toolkit in plain words:

The melting-point rule from the parent, re-stated so we can plug into it:

The scenario matrix

Every cell below is a distinct kind of question this topic can throw. The examples that follow are tagged with the cell they cover.

#	Case class	What is "extreme" about it	Example
A	Normal input — plug a mid-range $E_{co h}$ into $T_{m}$	nothing tricky, calibrate intuition	Ex 1
B	High end — highest- $T_{m}$ metal (W)	check the estimator near the top	Ex 2
C	Low end / comparison — lower- $T_{m}$ metal (Mo) vs W	does the trend come out right?	Ex 3
D	Degenerate PBR < 1 — oxide too thin	limiting failure mode	Ex 4
E	PBR in protective band 1–2	the "good" case for contrast	Ex 5
F	Volatility overrides PBR — decent PBR but oxide boils away	the special catch for W/Mo	Ex 6
G	Real-world word problem — pick a metal for a live nozzle	multi-criterion selection	Ex 7
H	Exam-style twist — "just use pure W" trap + unit trap	catches the naive answer	Ex 8

Worked examples

Worked example Ex 1 — Case A: a mid-range metal (Ta), normal input

Tantalum has $E_{co h} \approx 8.1 eV/atom$ . Estimate its melting point and compare to the true $3290 K$ .

Forecast: with $c = 0.032$ , do you expect roughly 2000 K, 3000 K, or 5000 K? Guess before reading.

Multiply bond energy by $c$ : $0.032 \times 8.1 = 0.2592 eV$ . Why this step? $c$ is the fraction of the bond well-depth that vibrations must reach before an atom escapes its cage. This is the "escape energy."
Convert energy to temperature: divide by $k_{B}$ : $T_{m} \approx \frac{0.2592 eV}{8.617 \times 1 0 ^{- 5} eV/K} \approx 3008 K .$ Why this step? Dividing eV by (eV/K) leaves K — the temperature whose thermal jiggle equals the escape energy.

Verify: $3008 K$ vs true $3290 K$ — same order of magnitude, within $\sim 9%$ . The estimator is honest for a mid-range case. Units: $eV / (eV/K) = K$ ✓.

Worked example Ex 3 — Case C: comparison / low end (Mo vs W), is the TREND right?

Molybdenum, $E_{co h} \approx 6.8 eV/atom$ . Estimate $T_{m}$ and confirm Mo comes out below W.

Forecast: Mo is lighter and cheaper but melts lower. Predict: above or below the 3305 K we got for W?

Escape energy: $0.032 \times 6.8 = 0.2176 eV$ .
To temperature: $\frac{0.2176}{8.617 \times 1 0 ^{- 5}} \approx 2525 K$ . Why this step? Same conversion — we want a number we can rank against W's.
Compare: $2525 K (Mo) < 3305 K (W)$ . Lower $E_{co h}$ ⇒ lower $T_{m}$ . Why this step? The whole point of the estimator is that it reproduces the d-band cohesion trend: half-filled band ⇒ big $E_{co h}$ ⇒ high $T_{m}$ .

Verify: predicted $2525 K$ vs true Mo $2896 K$ — right size, and crucially Mo < W in both prediction and reality. The ordering survives, which is what a selection engineer actually needs.

Now we switch tools — from melting to the oxidation skin. The next three examples all use the Pilling–Bedworth ratio, so here is exactly what each symbol in it means, anchored to a picture.

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

Worked example Ex 4 — Case D: PBR < 1 (a magnesium-like non-protector), the degenerate limit

A metal forms oxide $M O$ ( $n = 1$ ) with $M_{M} = 24.3$ , $M_{o x} = 40.3$ , $ρ_{M} = 1.74$ , $ρ_{o x} = 3.58 g/cm^{3}$ . Compute PBR and classify.

Forecast: the oxide is denser than the light metal — do you expect a big or small PBR?

Numerator $M_{o x} ρ_{M} = 40.3 \times 1.74 = 70.12$ . Why this step? This is "how much oxide volume one mole makes" (mass over its own density comes next).
Denominator $n M_{M} ρ_{o x} = 1 \times 24.3 \times 3.58 = 86.99$ . Why this step? This is "how much metal volume got eaten."
Divide: $PBR = 70.12/86.99 \approx 0.806$ . Why this step? Ratio of the two volumes = the PBR by definition.

Verify: $0.806 < 1$ → oxide too thin, cracks, non-protective. This is the classic Mg case and matches the parent's rule "PBR < 1 cracks." Dimensionless (g·(g/cm³) / g·(g/cm³) cancels) ✓.

Worked example Ex 6 — Case F: decent PBR but VOLATILE — the W/Mo catch

Tungsten oxide $W O_{3}$ : $n = 1$ , $M_{M} = 183.8$ , $M_{o x} = 231.8$ , $ρ_{M} = 19.3$ , $ρ_{o x} = 7.16 g/cm^{3}$ . Compute PBR, then explain why W still needs a coating.

Forecast: the number will look "protective." Trap: does a good PBR guarantee protection?

Numerator: $M_{o x} ρ_{M} = 231.8 \times 19.3 = 4473.74$ .
Denominator: $n M_{M} ρ_{o x} = 1 \times 183.8 \times 7.16 = 1315.9$ .
Divide: $PBR = 4473.74/1315.9 \approx 3.399$ . Why this step? Standard PBR — but note the dense metal (19.3) over a light oxide (7.16) inflates it.

Verify: $3.399 > 2$ → by the rule the oxide should spall — already non-protective. And worse: $W O_{3}$ sublimes/boils away above ~500 °C, so even a perfect PBR would fail. This is the parent's key point — for W/Mo, volatility overrides geometry. Coatings (Thermal barrier coatings and ablatives, iridium liners) are mandatory. Contrast with the protective $1.29$ of Ex 5.

Worked example Ex 7 — Case G: real-world word problem, full selection

Design a throat insert for a short-burn (25 s) apogee motor: $T \approx 3000 K$ , fuel-rich (reducing) exhaust with little free O₂, and a hard thermal-shock hit at ignition. Which metal/alloy, and why?

Forecast: name your metal before reading. Highest $T_{m}$ ? Lightest? Toughest?

Temperature margin: need $T_{m} ≫ 3000 K$ . From Ex 2/3, only W ( $3695 K$ ) gives a comfortable $\sim 700 K$ margin; Mo ( $2896 K$ ) would be below the gas temperature — instant fail. Why this step? You never run a nozzle at or above the metal's $T_{m}$ ; you leave hundreds of K of headroom for hot-spots.
Thermal shock: pure W is brittle below its ductile-brittle transition and cracks on the ignition thermal spike. Alloy in Re → W–25Re (the rhenium effect lowers that transition). Why this step? Toughness, not just $T_{m}$ , decides survival of the first millisecond.
Oxidation: reducing exhaust + only 25 s ⇒ little $W O_{3}$ forms and little time to lose it, so the Ex-6 volatility problem is tolerable with a thin/short-life coat. Why this step? Match the coating effort to the environment — a reducing, short burn is the one case bare-ish W is defensible.

Verify: answer = W–25Re throat, radiation-cooled — exactly real apogee-motor practice. Every criterion (margin, toughness, oxidation) is satisfied, and the choice is consistent with Ex 2 (W margin) and Ex 6 (why short/reducing dodges the oxide problem).

Worked example Ex 8 — Case H: exam twist — the "pure W" trap + a unit trap

Part (a): A student writes: "Nozzle at 3000 K in air, long 10-minute burn. Highest $T_{m}$ wins, so use pure tungsten." Mark this. Part (b): They also convert $E_{co h} = 8.9 eV$ using $k_{B} = 8.617 \times 1 0^{- 5}$ but forget the factor $c$ , getting $T_{m} = 8.9/8.617 \times 1 0^{- 5}$ . What "answer" do they get and why is it absurd?

Forecast: where does each argument break?

(a) The chemistry trap: in air (oxidising) for 10 minutes, W grows volatile $W O_{3}$ (Ex 6) that boils off continuously → the throat erodes fast. High $T_{m}$ does not buy chemical inertness. Why this step? Selection is multi-criterion: $T_{m}$ , toughness and oxidation. Pick the failing criterion. Fix: coat the W (iridium/silicide) or switch to a ceramic like ZrB₂/HfC for sustained oxidising service.
(b) The unit trap: $\frac{8.9}{8.617 \times 1 0 ^{- 5}} \approx 1.03 \times 1 0^{5} K$ . Why this step? Dropping $c = 0.032$ multiplies the temperature by $1/0.032 \approx 31$ , giving $\sim 103, 000 K$ — hotter than the surface of many stars and $\sim 28 \times$ the real value. The physical role of $c$ (only a small fraction of the bond depth is needed to escape) is what keeps $T_{m}$ realistic.

Verify: (a) correct mark = wrong, because volatile-oxide erosion beats melting margin here; (b) the absurd $\approx 1.03 \times 1 0^{5} K$ flags the missing $c$ . Both traps are resolved by remembering all the criteria and every factor in the formula.

Active recall

Recall Which matrix cells did we cover? (hide)

A/B/C — melting-point estimator at mid/high/low $E_{co h}$ (Ta, W, Mo) ::: Ex 1, 2, 3
D — PBR < 1, non-protective cracking oxide ::: Ex 4
E — PBR 1–2, protective skin (alumina-like) ::: Ex 5
F — good-looking PBR but volatile oxide (W/Mo catch) ::: Ex 6
G — full multi-criterion nozzle selection ::: Ex 7
H — exam trap: pure-W fallacy + dropped- $c$ unit blunder ::: Ex 8

Recall Numbers worth remembering (hide)

$T_{m}^{es t}$ (W) from $E_{co h} = 8.9$ eV ::: $\approx 3305 K$ (true 3695)
$T_{m}^{es t}$ (Mo) from $E_{co h} = 6.8$ eV ::: $\approx 2525 K$ (true 2896)
PBR of $W O_{3}$ ::: $\approx 3.4$ — yet volatility still kills it
Protective PBR band ::: $1$ to $2$

Connections

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