The empirical equation for average particle size (d) in μm is ______
Posted: Thu Jul 14, 2022 1:33 pm
a) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{0.45}(1000\frac{f_1}{f_a})^{1.5}\)
b) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{1.5}(1000\frac{f_1}{f_a})^{0.45}\)
c) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{0.45}(1000\frac{f_a}{f_1})^{1.5}\)
d) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{0.45}(10000\frac{f_1}{f_a})^{1.5}\)
b) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{1.5}(1000\frac{f_1}{f_a})^{0.45}\)
c) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{0.45}(1000\frac{f_a}{f_1})^{1.5}\)
d) d=\(A\frac{\sqrt{\sigma}}{v\sqrt{\rho}}+B\Big(\frac{\sqrt{\mu}}{\sigma \rho}\Big)^{0.45}(10000\frac{f_1}{f_a})^{1.5}\)