Question 8.3: Particle Acceleration A spherical particle of calcium carbo......

The acceleration of a particle is derived from a force balance on a particle and is related to the buoyancy effect and drag resistance expressed as

{\frac{d U}{d t}}=g\left(1-{\frac{\mathrm{\rho}}{\mathrm{\rho}_{s}}}\right)-{\frac{18\mathrm{\mu}U}{\mathrm{\rho}_{s}d_{p}^{2}}}                (8.12)

The terminal velocity is given by Stokes’ law (Equation 8.1), which occurs at an infinite period of time:

U_{t}={\frac{g\left(\rho_{p}-\rho\right)d_{p}^{2}}{18\mu}}             (8.1)

U_{t}={\frac{g\left(\rho_{s}-\rho\right)d^{2}}{18 \mu}}={\frac{9.81\times\left(2800-1000\right)\times\left(100\times10^{-6}\right)^{2}}{18\times0.001}}=9.81\times10^{-3}\ \mathrm{ms}^{-1}                (8.13)

Since the time to attain this velocity takes an infinite period of time, 90% of this value therefore corresponds to a velocity of 0.00883 ms ^{–1} . Integrating Equation 8.12 therefore gives a time to reach this velocity of

t=\frac{1}{\frac{18{\mathrm{\mu}}}{{\rho}_{s}d_{p}^{2}}} \ln \left\lgroup\frac{g\biggl(1-\frac{\rho}{\rho_s}{}\biggr)-\frac{18\mathrm{\mu}U}{{\rho}_{s}d_{p}^{2}}}{g\biggl(1-\frac{\rho}{\rho_s}\biggr)}\right\rgroup             (8.14)

\frac{1}{\frac{18 \times 0.001}{2800 \times (100 \times 10^{-6})^2} } \ln\left\lgroup\frac{9.81 \times \left(1- \frac{1000}{2800} \right) – \frac{18 \times 0.001 \times 0.00883}{2800 \times (100 \times 10^{-6})^2} }{9.81 \times \left\lgroup 1-\frac{1000}{2800} \right\rgroup } \right\rgroup = 3.68 \times 10^{-3} \ s

That is less than four-thousandths of a second.