# GATE MT 2013 Q48. Common Data for Questions 48 and 49: A steel specimen containing 0.2 wt.% C is carburized in an atmosphere that maintains a carbon content of 1.2 wt.% C at the surface of the specimen. Given: For carbon diffusion in austenite: D0=2.0×10-5 m2 /s Activation energy for diffusion, Q=142 kJ/mol . What is the depth (in m) from the surface of the specimen at which a composition of 0.4 wt.% C is obtained after carburizing at 870°C for 10 h?

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Correct answer is option D. C(x,t)= A- B exp(x/2√Dt) C(0,0)= A-B(0)= Cs A= Cs C(∞,0)= C0= A-B B= Cs- C0 ( Cs- C)/(Cs- C)= exp(x/2√Dt) D=D0- exp(-Q/RT) = 6.48×10-12 m2/s and ( Cs- C)/(Cs- C) = (1.2-0.4)/(1.2-0.2)= 0.8 AE/AD = AB/AC ⇒AE= 0.00628 ⇒E(y)= 0.90628 (x/2√Dt)= 0.90628 x= 8.75 ×10-4 mRead more

Correct answer is option D.

C

_{(x,t)}= A- B exp(x/2√Dt)C

_{(0,0)}= A-B(0)= C_{s}A= C

_{s}C

_{(∞,0)}= C_{0}= A-BB= C

_{s}– C_{0}( C

_{s}– C)/(C_{s}– C)= exp(x/2√Dt)D=D

_{0}– exp(-Q/RT) = 6.48×10^{-12}m^{2}/sand ( C

_{s}– C)/(C_{s}– C) = (1.2-0.4)/(1.2-0.2)= 0.8AE/AD = AB/AC

⇒AE= 0.00628

⇒E(y)= 0.90628

(x/2√Dt)= 0.90628

x= 8.75 ×10

^{-4}mx= 875 μm

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