## Professional Reference Shelf

#### Example CD12-3: Maximum Solids Holdup

 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: Color of pellet: brown Solution: The amount of solids in the reactor is given by (CD12-19) The two parameters that need to be found areand. A. Calculation of (CD12-29) 1. Gravity term: 2. Cross-sectional area: Superficial velocity: Porosity at minimum fluidization [Equation (CD12-9)]: B. Calculation of volume fraction of bubbles (CD12-46) Here we see that we must calculate u mf and u b . Step 1. First the minimum fluidization velocity is obtained from Equation (CD12-25): (CD12-35) Step 2. To calculate u b we must know the size of the bubble d b , that is,(CD12-36): (CD12-36) Step 3. The average size of the bubble, db , is determined by evaluating Equation (CD12-37) at h/2: (CD12-37) where d bm and d b0 are given in Equations (CD12-38) and (CD12-39), respectively Maximum bubble diameter: (CD12-38) Minimum bubble diameter: (CD12-39) Solving for d b yields 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 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 From Figure CD12-6 we see that a 100-m particle corresponds to a value ofof 0.5. Substituting this value into Equation (CD12-46), the fraction of the bed occupied by the bubble is Thus 94% of the bed is in the emulsion phase plus the wakes. C. Amount of solids holdup, or W s = 678 lb of solid particles