FLUID FOW FOR CHEMICAL ENGINEERING EKC 212 FLUID FOW FOR CHEMICAL ENGINEERING (ALIRAN BENDALIR KEJURUTERAAN KIMIA) CHAPTER 7 (PART 3) Dr Mohd Azmier Ahmad Tel: 04-5996459 Email: chazmier@eng.usm.my

Goals Describe forces that act on a bed of particles. Describe how pressure drop and bed height vary with fluid velocity. Apply basic equations to compute pressure drop across the bed, the bed height and the diameter of the bed. List advantages and disadvantages of fluidized beds.

SYSTEM OF FLUIDIZATION Operation in which a fluid transforms fine solids into a fluid-like state. Minimum fluidization velocity (Vmf) : When a fluids flow upward through a packed bed particles at low velocity, the particle remain stationary (fixed bed). As the fluid velocity is increased, the drag force just equals to the gravitational force of the particle. Upon further increase in velocity, the particle just begin to move & this is the onset or minimum fluidization (MF). Fluid velocity at this stage is called the minimum fluidization velocity (Vmf).

Application of fluidized bed Heat exchange Drying Coating (example, metals with polymer) Solidification Granulation Oil cracking Zinc roasting Coal combustion Gas desulfurization Adsorption/desorption

TYPE OF FLUIDIZATION Fluidization regimes depend on; (1) fluid properties; (2) fluid velocity; (3) pipe diameter; (4) solid properties. The following figures show the kind of contact between solids and fluid, from a packed bed to lean phase fluidization as the fluid density is increases. Fixed bed Minimum Smooth Bubbling Slugging Lean phase fluidization (MF) fluidization fluidization fluidization fluidization

Undesirable phenomena in beds of large particles

Fixed Bed : Fluid at low velocity does not impart enough drag to overcome gravity and particles do not move. Fluid flow through the void spaces (pores) without disturbing the bed. Fluidized Bed : At high enough velocities, fluid drag plus buoyancy overcomes the gravity force and the bed expands. p increases with constant bed height until onset of fluidization. Beyond that p becomes constant.

Fluidized bed designs

FLUIDIZATION EXPERIMENT Vertical tube containing particles Control valve for gas flow Manometer to measure pressure drop (ΔP)

EQUATION IN FLUIDIZATION MF occurs when; Drag force by upward moving fluid = Net gravity force of particle Fd = Fgrav – Fbuoy Drag force, Fd by the upward fluid is : Fd = ΔP x A ΔP = pressure drop in fluid flow across the bed A = bed cross section area From Bernoulli’s equation, ΔP = ρf g hL hL = bed height …..(14) …..(15) …..(16)

At MF, the net downward force, Fgrav - Fbuoy = where; Lmf = bed height at MF; εmf = voidage (porosity) at MF Combine Eqs. (14-17) and simplify :- or …..(17) …..(18) …..(19)

Voidage (porosity) Voidage at minimum fluidization, εmf for various particles type:- The εmf can be experimentally determined by subjecting the bed to a rising fluid stream. Fluid velocity is increased until the MF occurs & measure the bed height, Lmf. Then as the velocity is further increased, ΔP decreases very slightly & then remains unchanged as the bed continuously expanding.

Since the volume hLA (1- ε) is equal to the total volume of solids :- hL1A (1- ε1 ) = hL2A (1- ε2) hL1 = height of bed with voidage ε1 hL2 = height of bed with voidage ε2 …..(20)

IRREGULAR-SHAPED OF PARTICLE (ISP) For ISP, use particle size & shaped factor, s. From Ergun Equation for turbulent flow:- …..(21) DP = particle size of a sphere having the same volume as the irregular-shaped particle. s= shape factor @ sphericity L = hL

Vmf for V; εmf for ε; and Lmf for L Eq. (21) can be used to calculate the minimum fluid velocity, Vmf, by substituting; Vmf for V; εmf for ε; and Lmf for L Combine the result with Eq. (18) :- …..(22) Defining Equation (22) becomes :- …..(23)

Example 7.19 Solid particles having a size of 0.12mm, a shape factor S of 0.88, and a density of 1000 kg/m3 are to be fluidized using air at 2.0 atm and 25oC. The voidage at minimum fluidizing (εmf) condition is 0.42. If the cross-section of the empty bed is 0.3 m2 and the bed contains 300 kg of solid, calculate the (i) Lmf, (ii) ΔPmf & (iii) Vmf. Given: μf = 1.845 x 10-5 Pa.s; f = 2.374 kg/m3; P = 2.02 x 105 Pa Solution i. Volume of solids = 300 kg / 1000kg/m3 = 0.30 m3 The height of the solids occupy in the bed if ε1 = 0 :-

ΔPmf = Lmf (1-εmf) (ρp-ρf )g = 1.724 (1-0.42) (1000-2.374) (9.81) From Eq. 20 :- hL1 = L1 hL2 = Lmf ε1 = 0 ε2 = εmf hL1 = height of bed with voidage ε1 hL2 = height of bed with voidage ε2 ii) Given data : - DP = 0.00012 m p = 1000 kg/m3 From Eq. 18 : - ΔPmf = Lmf (1-εmf) (ρp-ρf )g = 1.724 (1-0.42) (1000-2.374) (9.81) = 0.0978 x 105 Pa S = 0.88 εmf = 0.42

Vmf = 0.005029 m/s iii) To calculate Vmf :- Substitute all known values in Eq. 23 to get value of NRe,mf Hence NRe,mf = 0.07764 Vmf = 0.005029 m/s

Example 7.20 Solid particles having a size of 0.14mm, a shape factor S of 0.88, & a density of 900 kg/m3 are to be fluidized using air at 2.0 atm and 25oC. The voidage at minimum fluidizing (εmf) condition is 0.42. If the cross-section of the empty bed is 0.42 m2 and the bed contains 380 kg of solid, calculate the (i) Lmf, (ii) ΔPmf & (iii) Vmf. Given: μf = 1.845 x 10-5 Pa.s; f = 2.374 kg/m3; P = 2.02 x 105 Pa Solution i. Volume of solids = 380 kg / 900kg/m3 = 0.422 m3 The height of the solids occupy in the bed if ε1 = 0 :-

ii) Given data : - DP = 0.00014 m p = 900 kg/m3 From Eq. 18 : - ΔPmf = Lmf (1-εmf) (ρp-ρf )g = 1.613 (1-0.42) (900-2.374) (9.81) = 8806 Pa S = 0.88 εmf = 0.38

Vmf = 4.26x10-3 m/s iii) To calculate Vmf :- Substitute all known values in Eq. 23 to get value of NRe,mf Hence NRe,mf = 0.0769 Vmf = 4.26x10-3 m/s

WORK IN PAIR 5: Solid particles having a size of 0.15 mm, a shape factor s of 0.72, & a density of 875 kg/m3 are to be fluidized using air at 2.0 atm & 25oC. The bed diameter is 0.74 m & the bed contains 410 kg of solids. The minimum fluidized bed height , Lmf is 1.724 m.   Calculate the voidage at minimum fludizing condition. Calculate the pressure drop at minimum fludizing condition, ΔPmf. Calculate the minimum velocity for fludization, Vmf. Given: μf = 1.845 x 10-5 Pa.s; f = 2.374 kg/m3; P = 2.02 x 105 Pa HINT ΔPmf = Lmf (1-εmf) (ρp-ρf )g