3 . The average grain diameter for a brass material was measured as a function o
ID: 943190 • Letter: 3
Question
3 . The average grain diameter for a brass material was measured as a function of time at 650°C, which is tabulated below at two different times: Time (min) Grain Diameter (mm) 30 3.9 × 10–2 90 6.6 × 10–2 (a) What was the original grain diameter? (b) What grain diameter would you predict after 150 min at 650°C?
3. The average grain diameter for a brass material was measured as a function of time at 650°C, which is tabulated below at two different times: Time (min) Grain Diameter (mm) 30 90 (a) What was the original grain diameter? (b) What grain diameter would you predict afer 150 min at 650°C? .9 x 102 6.6 x 102Explanation / Answer
Ans- (a) we can solve the problem, using the data given and Equation(d02 = d2 - Kt), where d is grain diameter at time (t) and d0 is original grain diameter .
we use two equations with d0 and K as unknowns; thus
For 30 min
(0.039 mm)2 - d02 = (30 min)K (i)
For 60 min
(0.066 mm)2 - d02 = (60 min)K (ii)
We can solve these equation as
(0.039 mm)2 - (30 min)K = (0.066 mm)2 - (60 min)K
K = (0.002835 mm2/60 min)
K = 4.73×10-5 mm2/min
Put the value of K in equation(i) we get the value of d0 is 0.01 mm.
(b) At 150 min, the diameter d is computed using equation(d=) as
d2 = d02 + Kt
d = 0.085 mm
Hence the original grain diameter is 0.01 mm and at after 150 min grain diameter would be 0.085 mm.
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