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A pilot fluidized bed is to be used to test a chemical reaction. The
bed diameter is 91.4 cm. You wish to process 28.3 x 10 3
cm 3 /s of gaseous
material. The average particle diameter is 100  m. The
reactor height is 10 ft. Allowing for a disengaging height of 7 ft, this means that
we have a maximum bed height of 91.4 cm. The distributor plate is a porous disk.
What is the maximum weight of solids (i.e., holdup) in the bed? Other data:
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Color of pellet: brown
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Solution:
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The amount of solids in the reactor is given by
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(CD12-19)
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The two parameters that need to be found are and .
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A. Calculation of
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(CD12-29)
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1. Gravity term:
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2. Cross-sectional area:
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Superficial velocity:
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Porosity at minimum fluidization [Equation (CD12-9)]:
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B. Calculation of volume fraction of bubbles
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(CD12-46)
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Here we see that we must calculate u
mf
and u b .
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Step 1. First the minimum fluidization velocity is obtained
from Equation (CD12-25):
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(CD12-35)
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Step 2. To calculate u b
we must know the size of the bubble d b
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(CD12-36):
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(CD12-36)
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Step 3. The average size of the bubble, db , is determined by evaluating Equation (CD12-37) at h/2:
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(CD12-37)
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where d bm
and d b0
are given in Equations (CD12-38) and (CD12-39), respectively
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Maximum bubble diameter:
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(CD12-38)
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Minimum bubble diameter:
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(CD12-39)
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Solving for d b
yields
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At the top of the bed (h = 91.4 cm), d b = 8.86 cm.
For purposes of the Kunii-Levenspiel model, we shall take the
bubble diameter to be 5 cm
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Step 4. We can now return to calculate the velocity of bubble
rise and the fraction of bed occupied by bubbles from Equation (CD12-36). We have
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From Figure CD12-6 we see that a 100-m particle corresponds
to a value of of 0.5. Substituting this value into Equation (CD12-46),
the fraction of the bed occupied by the bubble is
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Thus 94% of the bed is in the emulsion phase plus the wakes.
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C. Amount of solids holdup,
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or
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W s = 678 lb of solid particles
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