Question

Target temperatures for Heat Treatment Surface T = T(x = 0,) 1500 K Inner part 7(x = 0,004 m, 1) <1200 K Initial temperature 9.", W/m? - 787 = 0) = 300 K IIIIIIII TEX0 Properties of metallic materials are largely dependent on composition and phase and the phase changes with temperature. To achieve desirable properties by transforming a material phase, heat treatment is conducted. Here, a thick slab of stainless steel is treated by supplying a constant heat flux (93", W/m2) on a surface. The initial temperature of the slab is T = 300 K. and we want to treat its surface (reaching over 1500 K), while maintaining an inner temperature below 1200 K. The slab can be considered as a semi-infinite solid, and other heat transfer modes such as convection or radiation are ignored. Properties of Steel (a) For 7 0,/ 100s) 1500 K, 4.'" -8000 kg/ m W ith the resulting 4.". Co 500 J/kg-K Tex=0.004 m, 1 = 100 s)? k-16 W/m-K (b) For T - 7(x=0,1-25 s) - 1500 K.4,"? a-4.0 x 10-6 m/s with the resulting 4", Tex=0.004 m, 1 = 25 s)? (a) To reach the surface temperature of 1500 K (T, at x = 0) within ta = 100 sec, what should the minimum supplied heat flux (9," W/m2) be? With this 9.2", find the temperature at 4 mm below the surface after 100 sec of heating, T (x = 0.004 m, t = 100 sec). (b) If we want to reach the surface temperature of 1500 K (T, at x = 0) four times faster, i.e., within 25 sec (t = 25 sec = t1/4), what should the minimum supplied heat flux (95.2", W/m2) be? Find the temperature at 4 mm below the surface after 25 sec of heating with 9.2",

Answer 1

Initial temp. To = 3ook Surface temp. Ts = 1500K: 4N properties are, 2=4mm = 0,004 m of steel slab P= 8000 kg/m3 cp = 500 Ilegok K = 16 W/mak a = 4x100 ml Given time ,t, = 100 sec. minimum supplied heat flux, q = -2koo 8 = Ti-Ts = 300 - 1500 = -1200. 7," = -2016) (-1200) June = 1.0832 x l.8 wan? 11 = 108.32 MW/m2 for semin infinite solid, temp. distribution is given by of = exffu) where, u= 2 2V at us 0.004 2V 4x16x loo from tables, exf(u) = 0.1125

= 0.1125 T- Is Ti-Ts = 0.1125 T-1500 300 - 1500 T= 1365k Temp. of and Surface below 4mm, loo see of heating b) Here, = 255€ = toly qo'=2K 8; v =-20/10) qd' = 5.4162 x 10 W/ 3 = 54.16 mW/mz u= = a1 zrat 0.004 - zvux156x25 = 0.2 from Tables, erf (w = 0.2227 T-1500 = 0.2227 300-1500 T= 1233 R Temp. 4 mm below the surface and 25 sec of heating