Question

a glass of bead wih a diameter of 5.0 mm and a specific gravity of 2.5 was dropped into a dilute guar gum/water solution. It took the bead an average of 43 seconds to fall 12 inch through the solution.Assuming that the solution density was the same as density of pure water.Estimate the apparent viscosity of the mixture. Express your answer in pascal seconds

Answer 1

Here the diameter of the sphere is very low, so this indicates that it's a Stokes law region(Re < 1). So, stokes law applicable in this region. The stokes is given by

= glo:-P.)d? V 18μ where V = terminal settling velocity of the solid particle g = gravitational acceleration P. = density of settling particle P = density of water d = diameter of particle u = dynamic viscosity

F1 w E R T Y Distance (x) = 12 inch = 12x 2.54cm. Given that: Diameter (do -5mm = 5x10m. -3 Density of bed = 2.5.5.9= 2500 kg/h Time (t) = 43 seconds = 30.48cm -0.30m of solution (Pw) = 1000 kg 1m² we that Distance velocity Time 0.30 43 -3 V = 6.97X10 m/s. we Know that Vs = g (Ps-Pw) d² Density Know 184 Mi g(Ps-Pw)d? 18 Vs (M = 2.93 Pa.s.