Question
1,8,9
Directions: Additional information to complete these problems can be found in your Examination Booklet. -Given Values- Energy for Vacancy Formation (eV/atom): = 0.67 Metal: = Sodium Temperature (C) = 990 Alloy Metal A = Barium Alloy Metal B = Vanadium Metal- X = Aluminum Metal- Y = Lead Magnification = 462 ASTM Gram Size = 9 Gram per Square Inches = 836 1) Calculate the fraction of atom sites that are vacant for given metal and the temperature shown above. Assume an energy for vacancy formation shown above (answer in Engineering Notation) 2) For the metal given, the equilibrium number of vacancies at 600 C is 1 10^25 m^-3. Calculate the fraction of vacancies for this metal at 600 C (answer in Engineering Notation): 3) For an alloy that consists of 75 at% of Alloy Metal and 25 % of Alloy Metal B, what is the concentration of Alloy Metal A in wt %? 4) For an alloy that consists of 75 at% of Alloy Metal and 25% of Alloy Metal B, what is the concentration of Alloy Metal B in wt %? 5) Metal-X forms a substitutional solid solution with Metal-Y. Compute the number of Metal-X atoms per cubic centimetier for a Metal-Y-Metal-X alloy that contains 33wt% Metal-X and 67 6) For the ASTM Gram Size listed above, approximately how many grains would there be per square inch at 100 magnification? 7) For the ASTM Gram Size listed above, approximately how many grains would there be per square inch at the listed magnification above? (write answer to the hundredth place) 8) For the ASTM Gram Size listed above, approximately how many grains would there be per square inch with no magnification? (answer in Engineer mg Notation): 9) A photomicrograph was taken of a specimen at a magnification of 100X, and it was determined that the average number of grams per square inch was the number listed above. What is this spec Your Score = 33.100
Explanation / Answer
Ans1) Here Temperature =9900C = 1263K
Energy for vaccancy formation Ev = 0.67eV
Let f be the fraction of atom sites
k= boltzmann constant=8.62x10-5
f=exp(-Ev/kT)
=exp(-0.67/8.62x10-5x1263)
= 2.13x10-4
Ans 8) Let x be the ASTM grain size
x=9
let A be the number of grains per square inch
M be the magnification
Here there is no magnification
M=1
Therefore
N(M/100)2 = 2x-1
N(1/100)2 =29-1
N= 2560000 grains/in2
Ans 9) Here Number of grains per square inch A=836
Magnification M=100
Let x be the specimen ASTM grain size number
The equation for Number of grains per square inch is given by
A(M/100)2 = 2x-1
836(100/100)2 =2x-1
836=2x-1
taking log on both sides
log(836) =log 2x-1
log(836)=x-1log(2)
2.92=(x-1) (0.3010)
2.92=0.3010x-0.3010
3.22=0.3010x
x=10.96
Therefore the ASTM grain size = 10.69
