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A 50 mm steel plate quenched from 800 °C in oil at 60 °C: centre 625.0 °C and surface 565.4 °C after 60 s by the exact series, centre at 200 °C after 334.6 s

ToolTransient Conduction
SourceCarslaw, H. S., and Jaeger, J. C. (1959). Conduction of Heat in Solids, 2nd ed. Oxford University Press, Oxford (the series solution of the slab with a convecting surface); Abramowitz, M., and Stegun, I. A. (eds.) (1964). Handbook of Mathematical Functions. NBS Applied Mathematics Series 55, Table 4.20 (roots of cot x = lambda x). Illustrative case with the carbon-steel properties of NUREG-1805 Table 2-3.
LevelUndergraduate

Problem

A carbon-steel plate 50 mm thick (k = 54 W/(m K), density 7,850 kg/m3, specific heat 465 J/(kg K)), uniformly at 800 °C, is plunged into oil at 60 °C; the convection coefficient on both faces is taken constant at 500 W/(m2 K). Find the Biot and Fourier numbers after 60 s, the temperature at the mid-plane, at a quarter of the thickness and at the surface, the mean temperature, the heat removed per square metre of plate, and the time for the mid-plane to fall to 200 °C. Say whether the plate may be treated as a lumped mass.

Inputs to Enter

  • Analysis: plane wall, exact series; material carbon steel 0.5 % C (k 54, density 7,850, specific heat 465)
  • Initial temperature 800 °C, fluid 60 °C, h = 500 W/(m2 K), elapsed time 60 s, target 200 °C
  • Half-thickness L = 25 mm, position 12.5 mm from the mid-plane, wall face area 1 m2
Open the Transient Conduction tool (loaded with this example)

Answers

  • Diffusivity 54/(7,850 x 465) = 14.79 mm2/s; Bi = 500 x 0.025/54 = 0.2315; Fo = 14.79e-6 x 60/0.025^2 = 1.420
  • First root of zeta tan zeta = Bi: zeta_1 = 0.46333, C_1 = 1.03559; the second term of the series is below 1e-7 at this time, so one term is enough (Fo above 0.2)
  • Mid-plane theta = 0.76345, T = 624.95 °C; at 12.5 mm 609.86 °C; surface theta = 0.68296, T = 565.39 °C; volume mean 604.96 °C
  • Independent Crank-Nicolson finite-difference solution (200 cells): 624.95, 609.86 and 565.39 °C
  • Q/Q_o = 1 - 0.73643 = 0.2636; Q_o = 7,850 x 465 x 0.05 x 740 = 135.1 MJ per square metre of plate, so 35.60 MJ has left in 60 s
  • Mid-plane reaches 200 °C (theta = 0.1892) at Fo = 7.919, t = 334.6 s; the surface passes 200 °C at 312.6 s
  • Lumped check: Bi on V/A = L is 0.2315, above 0.1, so the lumped model does not hold (it would give 592.7 °C for the whole plate, against 565 to 625 °C)

Notes

The tool's default inputs are this plate; press Submit. The case is illustrative (a real quench has a varying coefficient and temperature-dependent properties), not a published example. The method is checked by test_transient_conduction.php (1,944 checks): the roots against 316 printed values of Abramowitz and Stegun Tables 4.19, 4.20 and 9.7, the series against scipy, a method-of-lines solution and the tool's own finite-difference solver on 20 seeded cases, the semi-infinite, lumped and product solutions against scipy and two-dimensional numerical solutions (References/Transient-Conduction/checks/).