Answer
$d = 0.109 mm$
Work Step by Step
Given:
refer to the given table
Required:
grain diameter after 200 min at 650°C
Solution:
Using modified Equation 7.9:
$d = \sqrt d_{0}^{2} + Kt = \sqrt {(0.031 mm)^{2} }+ (5.44 \times 10^{-5} mm^{2}/min)(200 min) = 0.109 mm$
