Rhodes Solutions Chaptr7

SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION EXERCISE 7.1: A packed bed of solid particles of density 2500 kg/m3, occupies a depth of 1 m in a vessel of cross-sectional area 0.04 m2. The mass of solids in the bed is 50 kg and the surface-volume mean diameter of the particles is 1 mm. A liquid of density 800 kg/m3 and viscosity 0.002 Pas flows upwards through the bed. (a) Calculate the voidage (volume fraction occupied by voids) of the bed. (b) Calculate the pressure drop across the bed when the volume flow rate of liquid is

1.44 m3/h. (c) Calculate the pressure drop across the bed when it becomes fluidized. SOLUTION TO EXERCISE 7.1: (a) Bed voidage (volume fraction occupied by the voids) is calculated from Text-Equation 7.24: mass of solids in the bed, M = 1− ε( )ρpAH

Hence, voidage, ε =1 −50

2500 × 0.04 ×1= 0.5

(b) Pressure drop across the bed when the flow rate is 1.44 m3/h: Assume firstly that the bed is not fluidized at this flow rate. Estimate the pressure drop from the Ergun Equation (Text-Equation 7.3): (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2 +1.75(1 − ε)

ε3ρfU

2

xsv

Superficial liquid velocity, U =1.44

0.04 × 3600= 0.01 m / s

μ = 0.002 Pa.s; ε = 0.5; ρf = 800 kg/m3; H = 1.0 m; xsv = 10-3 m. Hence, −Δp( ) = 6560 Pa

(c) Check if the bed is fluidized: When fluidized, the apparent weight of the bed will be supported by the pressure difference. Hence (Text-Equation 7.2), Δp = H(1− ε)(ρp − ρf )g

SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION Page 7.1

−Δp( ) = 1.0 × (1 − 0.5) × 2500 − 800( )× 9.81 = 8338.5 Pa.

So the assumption in part (b) is correct and the answer to part (c) is 8338.5 Pa. EXERCISE 7.2: 130 kg of uniform spherical particles with a diameter of 50 μm and particle density 1500 kg/m3 are fluidized by water (density 1000 kg/m3, viscosity 0.001 Pas.) in a circular bed of cross-sectional area 0.2 m2. The single particle terminal velocity of the particles is 0.68 mm/s and the voidage at incipient fluidization is known to be 0.47. (a) Calculate the bed height at incipient fluidization. (b) Calculate the mean bed voidage when the liquid flow rate is 2 x 10-5 m3/s. SOLUTION TO EXERCISE 7.2: (a) Bed height at incipient fluidization. From Text-Equation 7.24: mass of solids in the bed, M = 1− εmf( )ρpAHmf

Therefore, with M = 130 kg, εmf = 0.47, ρp = 1500 kg/m3 and A = 0.2 m2, Hmf =

1300.2 × (1− 0.47) ×1500

= 0.818 m

Bed height at incipient fluidization, Hmf = 0.818 m. (b) Bed height when liquid flow rate is 2 x 10-5 m3/s: Use Richardson-Zaki equation (Equation 7.21), U = UTεn

To determine exponent n, calculate single particle Reynolds number, Rep at U=UT:

Rep =UTρfx

μ=

(0.68 ×10−3) ×1000 × (50 ×10−6 )0.001

= 0.034, which is less than 0.3. Hence, n = 4.65 (Text-Equation 7.22) Hence, applying the Richardson-Zaki equation, 1 ×10−4 = (0.68 ×10−3)ε4.65


which gives, ε = 0.6622 hence, bed voidage at a liquid flow rate of 2 x 10-5 m3/s is ε = 0.6622 EXERCISE 7.3: 130 kg of uniform spherical particles with a diameter of 60 μm and particle density 1500 kg/m3 are fluidized by water (density 1000 kg/m3, viscosity 0.001 Pas.) in a circular bed of cross-sectional area 0.2 m2. The single particle terminal velocity of the particles is 0.98 mm/s and the voidage at incipient fluidization is known to be 0.47. (a) Calculate the bed height at incipient fluidization. (b) Calculate the mean fluidized bed voidage when the liquid flow rate is 2 x 10-5 m3/s. SOLUTION TO EXERCISE 7.3: (a) Bed height at incipient fluidization. From Text-Equation 7.24: mass of solids in the bed, M = 1− εmf( )ρpAHmf

Therefore, with M = 130 kg, εmf = 0.47, ρp = 1500 kg/m3 and A = 0.2 m2, Hmf =

1300.2 × (1− 0.47) ×1500

= 0.818 m

Bed height at incipient fluidization , Hmf = 0.818 m. (b) Bed height when liquid flow rate is 2 x 10-5 m3/s: Use Richardson-Zaki equation (Equation 7.21), U = UTεn

To determine exponent n, calculate single particle Reynolds number Rep at U=UT:

Rep =UTρfx

μ=

(0.98 ×10−3) ×1000 × (60 ×10−6 )0.001

= 0.0588, which is less than 0.3. Hence, n = 4.65 (Text-Equation 7.22) Hence, applying the Richardson-Zaki equation, 1 ×10−4 = (0.98 × 10−3)ε4.65


which gives, ε = 0.6121 hence, bed voidage at a liquid flow rate of 2 x 10-5 m3/s is ε = 0.6121 EXERCISE 7.4: A packed bed of solid particles of density 2500 kg/m3, occupies a depth of 1 m in a vessel of cross-sectional area 0.04 m2. The mass of solids in the bed is 59 kg and the surface-volume mean diameter of the particles is 1 mm. A liquid of density 800 kg/m3 and viscosity 0.002 Pas flows upwards through the bed. (a) Calculate the voidage (volume fraction occupied by voids) of the bed. (b) Calculate the pressure drop across the bed when the volume flow rate of liquid is 0.72 m3/h. (c) Calculate the pressure drop across the bed when it becomes fluidized. SOLUTION TO EXERCISE 7.4: (a) Bed voidage (volume fraction occupied by the voids) is calculated from text-Equation 7.24: mass of solids in the bed, M = 1− ε( )ρpAH

Hence, voidage, 41.0104.02500

591 =××

−=ε

(b) Pressure drop across the bed when the flow rate is 0.72 m3/h: Assume firstly that the bed is not fluidized at this flow rate. Estimate the pressure drop from the Ergun Equation (Text-Equation 7.3): (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2 +1.75(1 − ε)

ε3ρfU

2

xsv

Superficial liquid velocity, U = m/s 005.0360004.0

72.0=

×

μ = 0.002 Pa.s; ε = 0.41; ρf = 800 kg/m3; H = 1.0 m; xsv = 10-3 m. Hence, ( ) Pa7876p =Δ−


(c) Check if the bed is fluidized: When fluidized, the apparent weight of the bed will be supported by the pressure difference. Hence (Text-Equation 7.1), Δp = H(1− ε)(ρp − ρf )g

( ) ( ) Pa. 983981.98002500)41.01(0.1p =×−×−×=Δ−

So the assumption in part (b) is correct and the answer to part (c) is 9839 Pa. EXERCISE 7.5: 12 kg of spherical resin particles of density 1200 kg/m3 and uniform diameter 70 μm are fluidized by water (density 1000 kg/m3 and viscosity 0.001 Pas.) in a vessel of diameter 0.3 m and form an expanded bed of height 0.25 m. (a) Calculate the difference in pressure between the base and the top of the bed. (b) If the flow rate of water is increased to 7 cm3/s, what will be the resultant bed height and bed voidage (liquid volume fraction)? State and justify the major assumptions. SOLUTION TO EXERCISE 7.5: (a) The frictional pressure loss is given by the force balance over the fluidized bed

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρρ

−=ρρ

−=−=Δ−p

f

p

f 1MggMMgupthrustweightAp

Hence, ( ) Pa. 5.277

4)3.0(

81.912001000112

p 2 =π

×⎥⎦⎤

⎢⎣⎡ −×

=Δ−

Frictional pressure drop (-Δp) = 277.5 Pa. However, the measured pressure drop across the bed will include the hydrostatic head of the liquid in the bed. Applying the mechanical energy equation between the bottom (1) and the top (2) of the fluidized bed: p1 − p2

ρfg+

U12 − U2

2

2g+ (z1 − z2 ) = friction head loss =

277.5ρfg

U1 = U2; z1 - z2 = - H = - 0.25 m.


Hence, p1 - p2 = 2730 Pa. Difference in pressure between the base and the top of the bed = 2730 Pa. (b) Calculate bed height and mean bed voidage at a flow rate of 7 cm3/s. Apply Richardson-Zaki equation (Text-Equation 7.21), U = UTεn

Superficial liquid velocity, U = m/s 109.9

43.0

107area tionalsec cross

rate flow volume 52

6−

−

×=π×

=

To determine the single particle terminal velocity, UT, assume Stokes Law (Text-Equation 2.13)

UT =x2g ρp − ρf( )

18μ

with x = 70 μm, ρp = 1200 kg/m3, pf = 1000 kg/m3 and μ = 0.001 Pa.s, UT = 5.34 x 10-4 m/s. To determine exponent n, calculate single particle Reynolds number Rep at U=UT.

Rep =UTρfx

μ=

(5.34 × 10−4) ×1000 × (70 × 10−6)0.001

= 0.037, which is less than 0.3. Hence, n = 4.65 (Text-Equation 7.22) Hence, applying the Richardson-Zaki equation, 9.9 ×10−5 = (5.34 ×10−4 )ε4.65 gives, ε = 0.696 From Equation 7.24, mass of solids in the bed, M = 1− ε( )ρpAH

Hence, bed height, m. 465.0

43.0)696.01(1200

12H2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ π×−×

=

EXERCISE 7.6: A packed bed of solids of density 2000 kg/m3 occupies a depth of 0.6m in a cylindrical vessel of inside diameter 0.1m. The mass of solids in the bed is 5kg and


the surface-volume mean diameter of the particles is 300 μm. Water (density 1000 kg/m3 and viscosity 0.001 Pas) flows upwards through the bed. a) What is the voidage of the packed bed? b) Use a force balance over the bed to determine the bed pressure drop when fluidized. c) Hence, assuming laminar flow and that the voidage at incipient fluidization is the same as the packed bed voidage, determine the minimum fluidization velocity. Verify the assumption of laminar flow. SOLUTION TO EXERCISE 7.6:

(a) Cross-sectional area of bed, A =π0.12

4= 7.85 ×10−3 m2

From Equation 7.24, calculate bed voidage: mass of solids in the bed, M = 1− ε( )ρpAH

Hence, voidage, ε =1 −

52000 × 7.85 ×10−3 × 0.6

= 0.4692

(b) Force balance on bed. Apply Text-Equation 7.2: Δp = H(1− ε)(ρp − ρf )g

−Δp( ) = 0.6 × (1− 0. 4692) × 2000 −1000( )× 9.81 = 3124 Pa.

Pressure drop across the bed when fluidized = 3124 Pa. (c) Assuming laminar flow through the bed, we apply only the laminar component of the Ergun equation.

Hence, (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2

With −Δp( ) = 3124 Pa ; μ = 0.001 Pa.s; ρf = 1000 kg/m3; H = 0.6 m; xsv = 300 x 10-6 m, and assuming the voidage of the bed at minimum fluidization is equal to the packed bed voidage, ε = 0.4692 then: U = Umf = 1.145 x 10-3 m/s


Check Reynolds number for use of laminar flow in packed bed.

R ′ e =Umfρf xsv

μ 1 − ε( )= 0.647, which is less than 10, the nominal upper limit for laminar

flow. Hence the assumption of laminar flow is justified and Umf = 1.145 mm/s. EXERCISE 7.7: A packed bed of solids of density 2000 kg/m3 occupies a depth of 0.5m in a cylindrical vessel of inside diameter 0.1m. The mass of solids in the bed is 4kg and the surface-volume mean diameter of the particles is 400 μm. Water (density 1000 kg/m3 and viscosity 0.001 Pas) flows upwards through the bed. a) What is the voidage of the packed bed? b) Use a force balance over the bed to determine the bed pressure drop when fluidized. c) Hence, assuming laminar flow and that the voidage at incipient fluidization is the same as the packed bed voidage, determine the minimum fluidization velocity. Verify the assumption of laminar flow. SOLUTION TO EXERCISE 7.7:

(a) Cross-sectional area of bed, A =π0.12

4= 7.85 ×10−3 m2

From Text-Equation 7.24, mass of solids in the bed, M = 1− ε( )ρpAH

Hence, voidage, ε =1 −

42000 × 7.85 ×10−3 × 0.5

= 0.4907

(b) Force balance on bed. Apply Text-Equation 7.2: Δp = H(1− ε)(ρp − ρf )g

−Δp( ) = 0.5 × (1− 0.4907) × 2000 −1000( ) × 9.81 = 2498 Pa.

Pressure drop across the bed when fluidized = 2498 Pa. (c) Assuming laminar flow through the bed, we apply only the laminar component of the Ergun equation.


Hence, (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2

With −Δp( ) = 2498 Pa ; μ = 0.001 Pa.s; ρf = 1000 kg/m3; H = 0.5 m; xsv = 400 x 10-6 m, and assuming the voidage of the bed at minimum fluidization is equal to the packed bed voidage, ε = 0.4907, then: U = Umf = 2.43 x 10-3 m/s Check Reynolds number for use of laminar flow in packed bed:

R ′ e =Umfρf xsv

μ 1 − ε( )=1.907, which is less than 10, the nominal upper limit for laminar

flow. Hence the assumption of laminar flow is justified and Umf = 2.43 mm/s. EXERCISE 7.8: By applying a force balance, calculate the incipient fluidizing velocity for a system with particles of particle density 5000 kg/m3 and mean volume diameter 100μm and a fluid of density 1.2 kg/m3 and viscosity 1.8 x 10-5 Pas. Assume that the voidage at incipient fluidization is 0.5. If in the above example the particle size is changed to 2mm, what is Umf? SOLUTION TO EXERCISE 7.8: Force balance on bed. Apply Text-Equation 7.2: Δp = H(1− ε)(ρp − ρf )g

−Δp( )H

= (1− 0.5) × 5000 −1.2( )× 9.81 = 24519 Pa.

With such small particles in liquid we can assume laminar flow through the bed and so apply only the laminar component of the Ergun equation.

Hence, (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2

With ; μ = 1.8 x 10-5 Pa.s; ρf = 1.2 kg/m3; −Δp( ) = 24519 Paxsv = 100 x 10-6 m, and given that the voidage of the bed at minimum fluidization is voidage, ε = 0.5, then


U = Umf = 0.0454 m/s Check Reynolds number for use of laminar flow in packed bed.

R ′ e =Umfρf xsv

μ 1 − ε( )= 0.6 , which is less than 10, the nominal upper limit for laminar

flow. Hence the assumption of laminar flow is justified and Umf = 4.54 cm/s. For a particle size of 2 mm: Flow is unlikely to be fully laminar, so we will use the full Ergun equation: (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2 +1.75(1 − ε)

ε3ρfU

2

xsv

From the force balance, −Δp( )H

= (1− 0.5) × 5000 −1.2( )× 9.81 = 24519 Pa.

Hence, with −Δp( ) H = 24519 Pa / m ; μ = 1.8 x 10-5 Pa.s; ρf = 1.2 kg/m3; xsv = 2 x 10-3 m, and given that the voidage of the bed at minimum fluidization is voidage, ε = 0.5, then 24519 = 1350Umf + 4200Umf

2

Solving, gives Umf = 2.26 m/s EXERCISE 7.9: A powder of mean sieve size 60 μm and particle density 1800 kg/m3 is fluidized by air of density 1.2 kg/m3 and viscosity 1.84 x 10-5 Pas in a circular vessel of diameter 0.5 m. The mass of powder charged to the bed is 240 kg and the volume flowrate of air to the bed is 140 m3/hr. It is known that the average bed voidage at incipient fluidization is 0.45 and correlation reveals that the average bubble rise velocity under the conditions in question is 0.8 m/s. Estimate: (a) the minimum fluidization velocity, Umf (b) the bed height at incipient fluidization (c) the visible bubble flow rate (d) the bubble fraction (e) the particulate phase voidage (f) the mean bed height


(g) the mean bed voidage SOLUTION TO EXERCISE 7.9: (a) Minimum fluidization velocity, Umf:

Using the Ergun equation with a voidage of 0.45 at incipient fluidization (given in question): (−Δp)

H= 150

(1 − ε)2

ε3μUxsv

2 +1.75(1 − ε)

ε3ρfU

2

xsv

From the force balance, −Δp( )H

= (1− 0.45) × 1800 −1.2( )× 9.81 = 9705.4 Pa / m.

Hence, with −Δp( ) = 9705.4 Pa ; μ = 1.84 x 10-5 Pa.s; ρf = 1.2 kg/m3; xsv = 60 x 10-6 m, and given that the voidage, ε = 0.45, then: 9705.4 = 2.545 ×106 Umf + 0.2112 ×106 Umf

2

Solving, gives Umf = 3.8 x 10-3 m/s Using the Wen and Yu correlation gives Umf = 2.13 x 10-3 m/s. However, for gas fluidization the Wen and Yu correlation is often taken as being most suitable for particles larger than 100 μm, whereas the correlation of Baeyens, shown in Text-Equation 7.11, is best for particles less than 100 μm.

U mf =(ρp − ρf )0.934 g0.934xp

1.8

1110μ0.87ρg0.066 (Text-Equation 7.11)

Umf =(1800 −1.2)0.9349.810.934(60 × 10−6)1.8

1110(1.84 ×10−5)0.87(1.2)0.066

Umf = 2.73 x 10-3 m/s (b) Bed height at incipient fluidization: Applying Text-Equation 7.24, with voidage at Umf = 0.45, mass of solids in the bed, M = 1− εmf( )ρpAHmf

Bed cross-sectional area, A =πD2

4=

π(0.5)2

4= 0.1963 m2

hence, 240 = 1− 0.45( ) ×1800 × 0.1963 × Hmf


and so Hmf = 1.235 m (c) Visible bubble flow rate: From the two-phase theory, QB = U − Umf( )A

Superficial gas velocity, U =QA

=1403600

⎡ ⎣

⎤ ⎦

10.1963

= 0.198 m / s

hence, Q B = 0.198 − 0.0038( )× 0.1963 = 0.0381 m3 / s (d) Bubble fraction: From the Two-Phase theory, bubble fraction, εB =

QBAUB

, where UB is the mean

bubble rise velocity, given in the question as 0.8 m/s. This gives, bubble fraction = 0.245. (e) Particluate phase voidage: The Two-Phase theory assumes that the gas flow through the particulate phase is that equivalent to the flow at incipient fluidization. We may assume therefore that the vodage of the particulate phase is the same as the bed voidage at incipient fluidization, εmf. Hence particulate phase voidage = 0.45. (f) Mean bed height: From the Two-Phase theory, one expression for bubble fraction (Text-Equation 7.28) is:

εB =H − Hmf

H, where H is the mean bed height.

With εB = 0.245 and Hmf = 1.235 m, mean bed height, H = 1.636 m. (g) Mean bed voidage: From Text-Equation 7.24, mass of solids in the bed, M = 1− ε( )ρpAH

Hence, voidage, ε =1 −240

1800 × 0.1963 ×1.636= 0.5848

Mean bed voidage = 0.5848. EXERCISE 7.10: A batch fluidized bed process has an initial charge of 2000 kg of solids of particle density 1800 kg/m3 and with the size distribution shown below:


size range number (i) size range (micron) mass fraction in feed 1 15 - 30 0.10 2 30 - 50 0.20 3 50 - 70 0.30 4 70 - 100 0.40

The bed is fluidized by a gas of density 1.2 kg/m3 and viscosity 18.4 x 10-5 Pas at a superficial gas velocity of 0.4 m/s. The fluid bed vessel has a cross-sectional area of 1 m2. Using a discrete time interval calculation with a time increment of 5 minutes, calculate: (a) the size distribution of the bed after 50 minutes (b) the total mass of solids lost from the bed in that time (c) the maximum solids loading at the process exit (d) the entrainment flux above the transport disengagement height of solids in size range 1 (15 - 30 μm) after 50 minutes. Assume that the process exit is positioned above TDH and that none of the entrained solids are returned to the bed. SOLUTION TO EXERCISE 7.10: (a) The size distribution of the bed after 50 minutes: First calculate the elutriation rate constants for the four size ranges under these conditions from the Zenz and Weil correlation (Text-Equation 7.46). The value of particle size x used in the correlation is the arithmetic mean of each size range: x1 = 22.5 x 10-6 m; x2 = 40 x 10-6 m; x3 = 60 x 10-6 m; x4 = 85 x 10-6 m With U = 0.40 m/s, ρp = 1800 kg/m3 and ρf = 1.2 kg/m3 K1∞

* = 0.83 kg/m2s K2∞

* = 0.281 kg/m2s K3∞

* = 0.131 kg/m2s K4∞

* = 0.068 kg/m2s From Text-Equation 7.38, the entrainment rate for particles in size range i is

Ri = −ddt

(MBmBi ) = Ki∞* AmBi


where K elutriation rate constant i∞* =

MB = total mass of solids in the bed

A = area of bed surface mBi = fraction of the bed mass with size xi at time t.

and total rate of entrainment, RT = Ri∑ = Ki∞

* AmBi∑ (7.40)

For batch operation it can best be solved by writing Equation 7.38 in finite increment form: −Δ(mBiMB) = Ki∞

* AmBiΔt (7.41)

where Δ(mBiMB) is the mass of solids in size range i entrained in time increment Δt.

Then total mass entrained in time Δt = Δ(mBiMB){ }i=1

4

∑ (7.42)

and mass of solids remaining in the bed

at time t+Δt = MB( )t − Δ mBiMB( t{ ) }i=1

4

∑ (7.43)

(where subscript t refers to the value at time t.)

Bed composition at time t+Δt = mBi( )t+Δ t =mBiMB( )t − Δ mBiMB( )t[ ]MB( )t − Δ mBiMB( )t{ }

i=1

4

∑ (7.44)

Solution to a batch entrainment problem is by sequential application of Equations 7.41 to 7.44 for the required time period. Using a time increment of 300 seconds: Mass of solids in size range 1 entrained in time increment Δt = Δ(mB1MB) = 0.83 x 1 x 0.1 x 300 = 24.9 kg Mass of solids in size range 2 entrained in time increment Δt = Δ(mB2MB) = 0.281 x 1 x 0.2 x 300 = 16.86 kg Mass of solids in size range 3 entrained in time increment Δt = Δ(mB3MB)


= 0.131 x 1 x 0.3 x 300 = 11.79 kg Mass of solids in size range 4 entrained in time increment Δt = Δ(mB4MB) = 0.068 x 1 x 0.4 x 300 = 8.16 kg Total mass entrained in first 300 seconds = 61.71 kg. Mass of solids remaining in the bed, MB = 2000 - 61.71 = 1938.29 kg. Bed composition after the first 300 seconds (Equation 7.44):

mB1( )0+300 =0.1 × 2000 − 24.9

2000 − 61.71= 0.09034

mB2( )0+300 =

0.2 × 2000 −16.862000 − 61.71

= 0.1977

mB3( )0+300 =0.3 × 2000 −11.79

2000 − 61.71= 0.3035

mB4( )0+300 =0.4 × 2000 − 8.16

2000 − 61.71= 0.4085

The calculations for the remaining time steps are summarised below: Time (sec)

Bed loss kg

Bed mass kg

mB1 mB2 mB3 mB4

300 61.71 1938.3 0.0903 0.1977 0.3035 0.4085 600 59.42 1878.9 0.0812 0.1951 0.3067 0.4170 900 57.23 1821.6 0.0727 0.1922 0.3097 0.4254 1200 55.15 1766.5 0.0647 0.1890 0.3125 0.4338 1500 53.17 1713.3 0.0573 0.1855 0.3150 0.4421 1800 51.31 1662.0 0.0505 0.1819 0.3173 0.4503 2100 49.56 1612.5 0.0442 0.1779 0.3193 0.4585 2400 47.92 1564.5 0.0386 0.1738 0.3211 0.4665 2700 46.39 1518.1 0.0334 0.1695 0.3226 0.4745 3000 44.96 1473.2 0.0288 0.1649 0.3238 0.4824 Therefore size distribution of bed after 50 minutes is:

size range number (i) size range (micron) mass fraction in bed


1 15 - 30 0.0288 2 30 - 50 0.1649 3 50 - 70 0.3238 4 70 - 100 0.4824

(b) Total mass of solids entrained from the bed in 50 minutes = 2000 - 1473.2 = 527 kg. (c) Maximum solids loading at the process exit. This will occur at the start of the 50 minutes period when the concentration of fines in the bed is a maximum.

initial rate of carryover = 61.71300

= 0.2057 kg / s

gas flow rate at exit, Q = UA = 0.4 x 1 = 0.4 m3/s

solids loading at exit = 0.2057

0.4= 0.514 kg / m3

(d) the entrainment flux above the transport disengagement height of solids in size range 1 (15 - 30 μm) after 50 minutes. From Text-Equation 7.38, entrainment rate of solids in size range 1 (above TDH) after 50 minutes (noting that mB1 at this time is 0.0288): = K1∞

* AmB1 = 0.83 ×1× 0.0288 kg/s = 0.0239 kg/s

hence solids flux = 0.0239

A=

0.02391.0

= 0.0239 kg/m2.s


EXERCISE 7.11: A powder having a particle density of 1800 kg/m3 and the following size distribution

size range number (i) size range (micron) mass fraction in feed 1 20 - 40 0.10 2 40 - 60 0.35 3 60 - 80 0.40 4 80 - 100 0.15

is fed into a fluidized bed 2m in diameter at a rate of 0.2 kg/s. The cyclone inlet is 4m above the distributor and the mass of solids in the bed is held constant at 4000 kg by withdrawing solids continuously from the bed. The bed is fluidized using dry air at 700 K (density 0.504 kg/m3 and viscosity 3.33 x 10-5 Pas) giving a superficial gas velocity of 0.3 m/s. Under these conditions the mean bed voidage is 0.55 and the mean bubble size at the bed surface is 5 cm. For this powder, under these conditions, Umf = 0.155 cm/s and Umb = 0.447 m/s. Assuming that none of the entrained solids are returned to the bed, estimate (a) the flow rate and size distribution of the entrained solids entering the cyclone (b) the equilibrium size distribution of solids in the bed (c) the solids loading of the gas entering the cyclone (d) the rate at which solids are withdrawn from the bed. SOLUTION TO EXERCISE 7.11: (a) The flow rate and size distribution of the entrained solids entering the cyclone First estimate the transport disengagement height, TDH: From the Horio correlation (Text-Equation 7.37), TDH = = 4.47dBvs

0.5 = 4.47 × 0.050.5 =1.0 m.

The graphical method of Zenz (Text-Figure 7.12) gives TDH = 0.25 m. U - Umb = 0.145 m/s = 0.476 ft/s db = 0.05 m = 1.97 inches (for safety, take db = 3 inches) TDH = 10 inches = 0.254 m. From Text-Equation 7.24, mass of solids in the bed, MB = 1 − ε( )ρpAH


Hence, given ε = 0.55, ρp = 1800 kg/m3, MB = 4000 kg and vessel cross-sectional

area A=π22

4= 3.142 m2 :

Applying Text-Equation 7.24, bed height, H =4000

1800 × 1 − 0.55( )× 3.142= 1.57 m.

Since the cyclone entrance is 4 metres above the distributor and the "worst case" estimate of TDH is 1m, then the cyclone entrance may be considered to be above TDH. Hence we may use K values to estimate carryover. i∞

*

Now calculate the elutriation rate constants for the four size ranges under these conditions from the Zenz and Weil correlation (Text-Equation 7.46). The value of particle size x used in the correlation is the arithmetic mean of each size range: x1 = 30 x 10-6 m; x2 = 50 x 10-6 m; x3 = 70 x 10-6 m; x4 = 90 x 10-6 m With U = 0.30 m/s, ρp = 1800 kg/m3 and ρf = 0.504 kg/m3 K1∞

* = 5.16 ×10−2 kg/m2s; K2∞

* = 1.975 ×10−2 kg/m2s; K3∞

* =1.049 ×10−2 kg/m2s; K4∞

* = 6.54 ×10−3 kg/m2s The overall and component material balances over the fluidized bed system are: Overall balance: F = Q + R (Solution Manual Equation: 7.11.1) Component balance: FmFi = QmQi + RmRi (Solution Equation: 7.11.2)

where F, Q and R are the mass flow rates of solids in the feed, withdrawal and filter discharge respectively and mFi, mQi and mRi are the mass fractions of solids in size range i in the feed, withdrawal and filter discharge respectively. From Text-Equation 7.39 the entrainment rate of size range i at the gas exit from the freeboard is given by: Ri = Rm (Solution Manual Equation: 7.11.3) Ri = Ki∞

* AmBi

and R (Solution Manual Equation: 7.11.4) = Ri∑ = RmRi∑


Combining these equations with the assumption that the bed is well mixed (mQi = mBi),

mBi =FmFi

F − R + Ki∞* A

(Solution Manual Equation: 7.11.5)

Now both mBi and R are unknown. However, noting that mBi∑ =1, we have:

0.2 × 0.1

0.2 − R + (5.16 ×10−2 × 3.142)+

0.2 × 0.350.2 − R + (1.975 ×10−2 × 3.142)

+0.2 × 0.40

0.2 − R + (1.049 ×10−2 × 3.142)+

0.2 × 0.150.2 − R + (6.54 ×10−3 × 3.142)

=1.0

Solving for R by trial and error, R = 0.0485 kg/s Substituting R = 0.0485 kg/s in Solution Manual-Equation 7.11.5, mB1 = 0.0638; mB2 = 0.328; mB3 = 0.433 and mB4 = 0.174 Therefore size distribution of bed (answer to question b) is:

size range number (i) size range (micron) mass fraction in bed 1 20 - 40 0.0638 2 40 - 60 0.328 3 60 - 80 0.433 4 80 - 100 0.174

From Solution Manual-Equation 7.11.3, knowing R and mBi, we can calculate mRi:

mR1 =K1∞

* AmB1R

=5.16 ×10−2 × 3.142 × 0.0638

0.0485= 0.213

similarly, mR2 = 0.420; mR3 = 0.294; mR4 = 0.074 Therefore size distribution of solids entering the cyclone is: size range number (i) size range (micron) mass fraction entering filter

1 20 - 40 0.213 2 40 - 60 0.420 3 60 - 80 0.294 4 80 - 100 0.074


(c) Solids loading for gas entering the filter, mass flow of solidsvolume flow of gas

=R

UA=

0.04850.3 ×3.142

= 0.0515 kg / m3

(d) From Solution Manual-Equation 7.11.1, the rate of withdrawal of solids from the bed, Q = 0.152 kg/s EXERCISE 7.12: A gas phase catalytic reaction is performed in a fluidized bed operating at a superficial gas velocity equivalent to 10xUmf. For this reaction under these conditions it is known that the reaction is first order in reactant A. Given the following information,

kHmf(1-εp)/U = 100; χ =KCHUB

= 1.0, use the reactor model of Orcutt et al. to

determine: (a) the conversion of reactant A, (b) the effect on the conversion found in (a) of doubling the inventory of catalyst (c) the effect on the conversion found in part (a) of halving the bubble size by using suitable baffles (assuming the interphase mass transfer coefficient is inversely proportional to the bubble diameter) If the reaction rate were two orders of magnitude smaller, comment on the wisdom of installing baffles in the bed with a view to improving conversion. SOLUTION TO EXERCISE 7.12: (a) From section 7.9 the model of Orcutt et al. gives for a first order reaction:

Conversion, 1 −CHC0

= 1− βe−χ( )−1 −βe−χ( )2

kHmf 1 − εp( )U

+ 1 − βe−χ( ) (Text-EQ. 7.65)

where, χ =KCHUB

and β = (U - Umf)/U

From the information given in the question,



00; = 1.0 and U

Umf=10 χ =

KCHUB

kHmf(1-εp)/U = 1

Hence, β = 0.9 So, from Text-Equation 7.65, conversion = 0.6645

Hmf are doubled. Thus, assuming all else , β = 0.9 and kHmf(1-εp)/U = 200

nd so the new conversion = 0.8744

the base case in (a) and KC is versely proportional to bubble diameter, then KC increases by a factor of 2, causing

iving conversion = 0.8706 [i.e. 87.06% conversion of reactant A, compared with

ler:

a):

we introduce baffles causing the bubble size to halve, then KC will double, = 2.0.

ence, and so conversion = 0.468 [i.e. 46.8% conversion of A

n rates it is the reaction rate which is controlling the conversion, whereas at high reaction rates it is the interphase mass transfer which controls the conversion.

[i.e. 66.45% conversion of reactant A]. (b) If the inventory of catalyst in the bed is doubled, both the operating bed height H and the height at incipient fluidizationremains constant, under the new conditions χ = 2.0a[i.e. 87.44% conversion; increasing from 66.45%] (c) If the bubble size is halved (compared withinχ to increase by a factor of 2. Hence χ = 2.0. G66.45% in case (a)] (d) If the reaction rate were two orders of magnitude smal then kHmf(1-εp)/U = 1. So, for the conditions in part ( conversion = 0.4 [i.e. 40% conversion of reactant A] Ifgiving χ

1 −βe−χ = 0.8782hcompared with 40% without baffles] So at low reaction rate, the introduction of baffles to reduce bubble size and improve interphase mass transfer, has a much smaller effect than at high reaction rates. This is because at low reactio

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