Chap 10 Solns-6E

# Chap 10 Solns-6E - CHAPTER 10 PHASE TRANSFORMATIONS IN...

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CHAPTER 10 PHASE TRANSFORMATIONS IN METALS PROBLEM SOLUTIONS 10.1 The two stages involved in the formation of particles of a new phase are nucleation and growth . The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth. The growth stage is simply the increase in size of the new phase particles. 10.2 This problem calls for us to compute the length of time required for a reaction to go to 99% completion. It first becomes necessary to solve for the parameter k in Equation (10.1). Rearrangement of this equation leads to k = - λν(1 - ψ 29 τ ν = - λν(1 - 0.529 (100 σ29 1.7 = 2.76 ξ10 -4 Now, solving for the time to go to 99% completion t = - λν(1 - ψ 29 κ 1/ν = - λν(1 - 0.9929 2.76 ξ 10 -4 1/1.7 = 305 σ 10.3 This problem asks that we compute the rate of some reaction given the values of n and k in Equation (10.1). Since the reaction rate is defined by Equation (10.2), it is first necessary to determine t 0.5 , or the time necessary for the reaction to reach y = 0.5. Solving for t 0.5 from Equation (10.1) leads to t 0.5 = - λν(1 - ψ 29 κ 1/ν 22

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= - λν(1 - 0.529 5 ξ 10 -4 1/2 = 37.23 σ Now, the rate is just rate = 1 t 0.5 = 1 37.23 s = 2.69 x 10 -2 (s) -1 10.4 This problem gives us the value of y (0.30) at some time t (100 min), and also the value of n (5.0) for the recrystallization of an alloy at some temperature, and then asks that we determine the rate of recrystallization at this same temperature. It is first necessary to calculate the value of k in Equation (10.1) as k = - λν(1 - ψ 29 τ ν = - λν(1 - 0.329 (100 μιν29 5 = 3.57 10 -11 At this point we want to compute t 0.5 , the value of t for y = 0.5, also using Equation (10.1). Thus t 0.5 = - λν(1 - 0.529 κ 1/ν = - λν(1 - 0.529 3.57 ξ 10 -11 1/5 = 114.2 μιν And, therefore, from Equation (10.2), the rate is just rate = 1 t 0.5 = 1 114.2 min = 8.76 x 10 -3 (min) -1 10.5 For this problem, we are given, for the austenite-to-pearlite transformation, two values of y and two values of the corresponding times, and are asked to determine the time required for 95% of the austenite to transform to pearlite. The first thing necessary is to set up two expressions of the form of Equation (10.1), and then to solve simultaneously for the values of n and k . In order to expedite this process, we will 23
rearrange and do some algebraic manipulation of Equation (10.1). First of all, we rearrange as follows: 1 - ψ = εξπ -κτ ν ( 29 Now taking natural logarithms ln (1 - ψ29 = - κτ ν Or - ln (1 - y) = kt n which may also be expressed as ln 1 1 - ψ = κτ ν Now taking natural logarithms again, leads to ln ln 1 1 - ψ = λνκ + ν λντ which is the form of the equation that we will now use. The two equations are thus ln ln 1 1 - 0.2 = λνκ + νλν (280 σ29 ln ln 1 1 - 0.6 = λνκ + νλν (425 σ29 Solving these two expressions simultaneously for n and k yields n = 3.385 and k = 1.162 x 10 -9 .

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## This homework help was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at Arizona.

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Chap 10 Solns-6E - CHAPTER 10 PHASE TRANSFORMATIONS IN...

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