Question 8.10: Power Demand in Fluidized Beds In designing a fluidized bed......

In practice, the volume of the bed of particles is usually fixed. A typical case is a fluidized bed consisting of catalyst particles for which the volume of bed is dependent on the required chemistry of a particular reaction. The power demand for fluidization is therefore the product of the flow rate and pressure drop over the depth of the bed. In practice, the flow rate is likely to have already been fixed according to the mass balance of the process. The bed voidage is found from

e=1-{\frac{\mathsf{\rho}}{\rho_{p}}}= 1 – \frac{\frac{m}{AL}}{\rho_b} = 1 – \frac{\frac{1800}{\frac{\pi \times 1^2}{4}}}{2100} = 0.68               (8.38)

By combining Equation 8.23 with Equation 8.28, the incipient velocity of the fluidizing gas is

{\frac{\Delta p}{L}}=(1-e)(\rho_{P}-\rho)g            (8.23)

\frac{\dot{Q}}{A}=\frac{1}{K s^{2}}\frac{e^{3}}{\left(1-e\right)^{2}}\frac{\Delta p}{\mu L}               (8.28)

U=\frac{\dot{Q}}{A}=\frac{1}{K s^{2}}\frac{e^{3}}{\left(1-e\right)}\frac{\left(\rho_{p}-\rho\right)g}{\mu}= \frac{0.231}{\frac{\pi \times 1^2}{4}} = 0.294 ms ^{-1}                   (8.39)

Assuming that K has a value of 5 and noting that for the spherical particles, the specific surface area of a particle is:

s={\frac{6}{d_{p}}}={\frac{6}{0.002}}=3000\ \mathrm{m^{2}m^{-3}}         (8.40)

At the incipient gas velocity, the pressure drop over the bed is given by Equation 8.28, which gives an incipient velocity of

U=\frac{1}{5\mu s^{2}}\frac{e^{3}}{(1-e)^{2}}\frac{\Delta p}{L} ={\frac{1}{5\times1.42\times10^{-5}\times3000^{2}}}\times{\frac{0.68^{3}}{\left(1-0.68\right)^{2}}}\times{\frac{\Delta p}{1.6}}=0.294~\mathrm{ms}^{-1}~~,                    (8.41)

Solving, the pressure drop is found to be 98 \ {\mathrm{Nm}}^{-2}. The power required for the flow of gas through the bed is therefore

P_{o}=\dot{Q}\Delta p=0.231\times98=22.6~W          (8.42)