Question 5 of 9: Samurai-Sword Clay Quench — Critical Heat Flux, and Cooling Rates for F+P and Martensite

Question 5: Samurai-Sword Clay Quench — Critical Heat Flux, and Cooling Rates for F+P and Martensite (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\rho=7860\ \text{kg/m}^3$, $C_p=450\ \text{J/kg K}$, $k=80\ \text{W/m K}$, sword (half-)thickness quenched symmetrically from $L=5$ mm, $T_i=800\,{}^{\circ}\text{C}$, bainite-start nose at $T_{nose}\approx550\,{}^{\circ}\text{C}$, $t_{nose}\approx40$ s (diagram reading), $Bi<0.1$ (given).

Find. (a) The maximum surface heat flux $q''$ the clay-coated body can be quenched at while still reaching the bainite region (rather than missing the nose and cooling straight to $M_s=325\,{}^{\circ}\text{C}$, forming martensite). (b) The cooling rates that deliver (1) an F+P structure of 242 HB and (2) a fully martensitic structure of 554 HB, read from the same diagram.

Approach. Because the cooling curve for the body must intersect the bainite-start curve rather than pass to its left (which would bypass all diffusive transformation and produce martensite at $M_s$), the fastest tolerable cooling is the one whose average rate from $T_i$ to the nose exactly matches the nose’s own time budget. With $Bi\ll0.1$ the sword cools essentially uniformly through its cross-section, so a lumped surface-flux balance converts this critical cooling rate directly into a critical heat flux.

1. Critical cooling rate to just reach the bainite nose. Cooling faster than this rate reaches 550°C before 40 s — i.e. passes to the left of (misses) the bainite-start curve and continues straight down to $M_s$, forming martensite. $$\left|\dfrac{dT}{dt}\right|_{crit}=\dfrac{T_i-T_{nose}}{t_{nose}}=\dfrac{800-550}{40}=\boxed{6.25\ ^{\circ}\text{C/s}}$$
2. Lumped-capacitance surface-flux balance. With low Biot number the sword is isothermal through its cross-section at any instant; for a slab of thickness $L$ quenched symmetrically from both faces, each face effectively drains half the thickness’s thermal mass: $$q''=\rho\left(\dfrac{L}{2}\right)C_p\left|\dfrac{dT}{dt}\right|_{crit}$$
3. Substitute. $$q''=(7860)(0.0025)(450)(6.25)$$ $$\boxed{q''=55{,}300\ \text{W/m}^2\approx0.055\ \text{MW/m}^2}$$
4. Sanity-check the Biot assumption. $h\approx q''/(T_i-T_{nose})=55{,}300/250\approx221\ \text{W/m}^2\text{K}$, giving $Bi=h(L/2)/k=221\times0.0025/80\approx0.0069\ll0.1$ — the lumped assumption is self-consistent.

Part (b) — cooling rates for 242 HB (F+P) and 554 HB (fully martensitic). Each solid S-shaped curve drawn on the diagram is one continuous-cooling path, and the circled number where it leaves the bottom of the chart is the hardness that path produces. So the two rates asked for are the slopes of the two paths whose end-circles read 242 and 554. Reading a path means picking off its $(t,T)$ coordinates, which was done by digitizing the reproduced page against the chart’s own printed decade and 100°C gridlines rather than by eye.

1. Rate for the fully martensitic structure (554 HB). Martensite is secured once the path has reached $M_s$ without ever touching the bainite/pearlite C-curves, so the rate that matters is the average from the austenitizing temperature down to $M_s$: $$\left|\dfrac{dT}{dt}\right|_{554}=\dfrac{800-325}{76}\Rightarrow\boxed{\approx 6.3\ ^{\circ}\text{C/s}}$$ Checking that this path really does miss the nose: at the nose temperature ($\approx550\,{}^{\circ}\text{C}$) the digitized 554 HB path is at $\approx21$ s, comfortably to the left of the nose’s $\approx40$ s, so no diffusive transformation can start — consistent with the diagram’s own 554 HB label.
2. Rate for the ferrite + pearlite structure (242 HB). On this path austenite decomposes to F+P between $Ac_1=670\,{}^{\circ}\text{C}$ and roughly $590\,{}^{\circ}\text{C}$, i.e. the structure is secured by the time the path reaches 600°C at $\approx1.2\times10^4$ s: $$\left|\dfrac{dT}{dt}\right|_{242}=\dfrac{800-600}{1.2\times10^4}\Rightarrow\boxed{\approx 0.017\ ^{\circ}\text{C/s}\ (\approx1\ ^{\circ}\text{C/min})}$$ Taken right down to 300°C the same path averages $(800-300)/(1.8\times10^4)\approx0.028\ ^{\circ}\text{C/s}$, so anywhere in the range 0.02–0.03°C/s — a few degrees per minute — describes this anneal.
3. Interpretation. The two rates differ by a factor of roughly 375: a 6°C/s quench (minutes from red heat to hand-warm) gives 554 HB martensite, while about 1°C per minute (a furnace or still-air anneal taking some six hours to fall to 100°C) gives soft 242 HB ferrite + pearlite. The intermediate circled hardnesses on the diagram (299, 322, 327, 294, 270) are mixed bainite/martensite paths that fall between these two bounds.

This low value (≈0.055 MW/m², far below the several MW/m² typical of direct water quenching) is exactly why the traditional technique insulates the body with a thick clay coating while leaving the edge bare: the clay throttles the heat flux at the body down toward this critical ceiling so the thick section transforms to tough upper bainite, while the exposed, uninsulated edge sees a much higher flux, misses the nose, and hardens to martensite.