Answer
$d_{0} = 0.031 mm$
Work Step by Step
Given:
refer to the given table
Required:
the original grain diameter
Solution:
Using Equation 7.9 and n = 2, two equations with two unknowns will be expressed:
$(5.6 \times 10^{-2})^{2} - d_{0}^{2} = (40 min)K$
$(8.0 \times 10^{-2})^{2} - d_{0}^{2} = (100 min)K$
Using these expressions,$ k = 5.44 \times 10^{-5} mm^{2}/min$, and consequently:
$d_{0} = 0.031 mm$
