Solucionario Ciencia E Ingenieria De Los Materiales Askeland 3 Edicion ✯

[ 0,71 = \frac{0,0008}{7.4834\times10^{-6} \sqrt{t}} ]

Donde: ( C_s = 1,2% ) C, ( C_0 = 0,10% ) C, ( C_x = 0,45% ) C, ( x = 0,0008 , \text{m} ), ( D = 1.4\times10^{-11} , \text{m}^2/\text{s} ).

[ t \approx (150,6)^2 = 22680 , \text{s} ]

[ 5.313\times10^{-6} \sqrt{t} = 0,0008 ] [ 0,71 = \frac{0,0008}{7

However, I cannot produce a full, verbatim solution manual or a direct link to copyrighted material. Full solution manuals are copyrighted works owned by Cengage Learning (or the original publisher), and distributing them without permission violates copyright laws.

[ \frac{C_s - C_x}{C_s - C_0} = \text{erf}\left( \frac{x}{2\sqrt{Dt}} \right) ]

Usamos la función error complementaria: [ \frac{C_s - C_x}{C_s - C_0} = \text{erf}\left(

[ \frac{1,2 - 0,45}{1,2 - 0,10} = \frac{0,75}{1,10} = 0,6818 ]

[ 0,71 = \frac{0,0008}{2\sqrt{(1.4\times10^{-11}) t}} ]

[ \sqrt{t} = \frac{0,0008}{5.313\times10^{-6}} \approx 150,6 ] Buscamos en tabla de erf(z): erf(z) = 0,6818

[ 0,71 = \frac{0,0008}{2 \times 3.7417\times10^{-6} \sqrt{t}} \quad (\text{nota: } \sqrt{D} = \sqrt{1.4e-11}=3.7417e-6) ]

Se requieren aproximadamente 6,3 horas para alcanzar 0,45% C a 0,8 mm de profundidad.

I understand you're looking for a long write-up related to the solution manual for "Ciencia e Ingeniería de los Materiales" (The Science and Engineering of Materials) by Donald R. Askeland, 3rd edition (Spanish version).

Buscamos en tabla de erf(z): erf(z) = 0,6818 → z ≈ 0,71 (interpolando).

( t \approx 6,3 , \text{horas} ).