Ch11 - CH 11 PHASE TRANSFORMATIONS 1 Rewrite the expression...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CH 11 PHASE TRANSFORMATIONS 1. Rewrite the expression for the total free energy change for nucleation (Equation 11.1) for the case of a cubic nucleus of edge length a (instead of a sphere of radius r). Now differentiate this expression with respect to a (per Equation 11.2) and solve for both the critical cube edge length, a*, and also ?G*. dG = a^3*dG_v + 6a^2*gam Find a* using dG/da = 0, a Then dG* = (a*)^3*dG_v + 6(a*)^2*gam =============== 2. (a) For the solidification of nickel, calculate the critical radius r* and the activation free energy ?G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are 2.53 109 J/m3 and 0.255 J/m2, respectively. (b) calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.360 nm for solid nickel at its melting temperature. A) r* = (-2*gam*T_m/dH_f)(1/(T_m-T)) = -2(.355)(1455+273)/(-2.53E9)*1/319 dG* = (16*pi*gam^3*T_m^3/3dH_f^2)(1/(T_m-T)^2) = 16pi(.255)^3)(1445+273)^2/(3*-2.53E9)^2 * 1/(319) B) # unit cells/part = crit nuc vol / unit cell vol = (4/3pi*r^3)/a^3 =116 4 atoms / FCC = 116(4) = 464 atoms/crit nuc =============== 3. (a) Assume for the solidification of nickel that nucleation is homogeneous, and the number of stable nuclei is 10^6 nuclei per cubic meter. Calculate the critical radius and the number of stable nuclei that exist at 200 degrees of supercooling (b) What is significant about the magnitudes of these critical radii and the numbers of stable nuclei? A) For 200K Supercooling, r* = (-2*gam*T_m/dH_f)(1/dT) = (-(2)(.255)(1455+273)/(-2.53E9))(1/200) Then, K_1 = n*/exp(-dG*/kT) [use n* for homogeneus nucleation, 10^6] and dG*=(16*pi*gam^3*T_m^2/[3dH_f^2](1/(dT)^2) so # stable nuclei= n_200* = K_1*exp(-dG*/kT) B) Small changes in radaii and inc in temp result in much higher number of stable nuclei. =============== 5. The kinetics of the austenite-to-pearlite transformation obey the Avrami relationship. Using the fraction transformedtime data given here (.2, 280; .6, 425), determine the total time required for 95% of the austenite to transform to pearlite. y = y-1 = -e^(-kt^n), t. t = (ln(1-y)/-k)^(1/n) [[1]] Need n and k. ln(1-y)=-kt^n lnln(1/1-y)=ln(kt^n) lnln(1/1-y)=lnk + n*ln(t). Plug in y = .2, .6 and t = 280, 425 and solve system to get n and k; plug back into [[1]] =============== 7. (a) Briefly describe the phenomena of superheating and supercooling. (b) Why do these phenomena occur? (a) Superheating and supercooling correspond to heating or cooling above or below a phase transition temperature without the occurrence of the transformation. (b) These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling. =============== 8. Suppose that a steel of eutectoid composition is cooled to 675C (1250F) from 760C (1400F) in less than 0.5 s and held at this temperature. (a) How long will it take for the austenite-to-pearlite reaction to go to 50% completion? To 100% completion? (b) Estimate the hardness of the alloy that has completely transformed to pearlite. (a) From Figure 11.23, a horizontal line at 675?C intersects the 50% and reaction completion curves at about 80 and 300 seconds, respectively; these are the times asked for in the problem statement. (b) The pearlite formed will be coarse pearlite. From Figure 11.31(a), the hardness of an alloy of composition 0.76 wt% C that consists of coarse pearlite is about 205 HB (93 HRB ============== 9. What is the driving force for the formation of spheroidite? The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area. SPHERODIDE IS A COMBO OF PERALITE AND BAINITIC PHASES. ================ 18. Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite. The hardness and strength of iron-carbon alloys that have microstructures consisting of a-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online