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 TextEquation 7.24: mass of solids in the bed, M = (1− ε )ρ pAH Hence, voidage, ε = 1 −
50 = 0.5 2500 × 0.04 × 1
(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): (1 − ε)2 μU (1 − ε) ρ f U 2 (− Δp) = 150 + 1.75 2 H ε3 ε 3 xsv xsv Superficial liquid velocity, U =
1.44 = 0.01 m / s 0.04 × 3600
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)(ρ p − ρ f )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− ε mf )ρ p AH mf Therefore, with M = 130 kg, εmf = 0.47, ρp = 1500 kg/m3 and A = 0.2 m2, H mf =
130 = 0.818 m 0.2 × (1− 0.47) × 1500
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 = UT ε n To determine exponent n, calculate single particle Reynolds number, Rep at U=UT:
Re p =
U Tρ f x (0.68 × 10−3 ) × 1000 × (50 × 10 −6 ) = μ 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)ε 4.65
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.2
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− ε mf )ρp AH mf
Therefore, with M = 130 kg, εmf = 0.47, ρp = 1500 kg/m3 and A = 0.2 m2, H mf =
130 = 0.818 m 0.2 × (1− 0.47) × 1500
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 = UT ε n To determine exponent n, calculate single particle Reynolds number Rep at U=UT: Re p =
U Tρ f x (0.98 × 10−3 ) × 1000 × (60 × 10 −6 ) = μ 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 )ε 4.65
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.3
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) (b) (c)
Calculate the voidage (volume fraction occupied by voids) of the bed. Calculate the pressure drop across the bed when the volume flow rate of liquid is 0.72 m3/h. 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 )ρ pAH Hence, voidage, ε = 1 −
59 = 0.41 2500 × 0.04 × 1
(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): (1 − ε)2 μU (1 − ε) ρf U 2 (− Δp) = 150 + 1.75 2 ε3 ε 3 xsv xsv H Superficial liquid velocity, U =
0.72 = 0.005 m/s 0.04 × 3600
μ = 0.002 Pa.s; ε = 0.41; ρf = 800 kg/m3; H = 1.0 m; xsv = 10-3 m. Hence, (− Δp ) = 7876Pa
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.4
(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)(ρ p − ρ f )g
(− Δp) = 1.0 × (1 − 0.41) × (2500 − 800 ) × 9.81 = 9839 Pa. 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 ⎡ ρ ⎤ g = Mg⎢1 − f ⎥ ρp ⎣⎢ ρ p ⎦⎥
(− Δp)A = weight − upthrust = Mg − M ρ f
⎡ 1000 ⎤ × 9.81 12 × ⎢1 − 1200 ⎥⎦ ⎣ = 277.5 Pa. Hence, (− Δp ) = π(0.3) 2 4
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 − p 2 U12 − U22 277.5 + + (z1 − z 2 ) = friction head loss = ρf g 2g ρf g
U1 = U2; z1 - z2 = - H = - 0.25 m.
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.5
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 = UT ε n Superficial liquid velocity, U =
volume flow rate 7 × 10 −6 = = 9.9 × 10 −5 m/s cross sec tional area π0.3 2 4
To determine the single particle terminal velocity, UT, assume Stokes Law (TextEquation 2.13) x2 g ρp − ρf UT = 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. Re p =
U Tρ f x (5.34 × 10− 4 ) × 1000 × (70 × 10− 6 ) = μ 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 )ε 4.65 gives, ε = 0.696 From Equation 7.24, mass of solids in the bed, M = (1− ε )ρ pAH Hence, bed height, H =
12 ⎛ π0.3 2 1200 × (1 − 0.696 ) × ⎜⎜ ⎝ 4
⎞ ⎟ ⎟ ⎠
= 0.465 m.
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 SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.6
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 −3 2 = 7.85 × 10 m 4
From Equation 7.24, calculate bed voidage: mass of solids in the bed, M = (1− ε )ρ pAH Hence, voidage, ε = 1 −
5 = 0.4692 −3 2000 × 7.85 × 10 × 0.6
(b) Force balance on bed. Apply Text-Equation 7.2: Δp = H(1− ε)(ρ p − ρ f )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. (1 − ε)2 (− Δp) = 150 Hence, H ε3
μU x 2sv
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 SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.7
Check Reynolds number for use of laminar flow in packed bed. U ρ x Re ′ = mf f sv = 0.647, which is less than 10, the nominal upper limit for laminar μ(1 − ε ) 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 −3 2 = 7.85 × 10 m 4
From Text-Equation 7.24, mass of solids in the bed, M = (1− ε )ρ pAH Hence, voidage, ε = 1 −
4 2000 × 7.85 × 10
−3
× 0.5
= 0.4907
(b) Force balance on bed. Apply Text-Equation 7.2: Δp = H(1− ε)(ρ p − ρ f )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.
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.8
Hence,
(1 − ε)2 μU (− Δp) = 150 H ε 3 x 2sv
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: U ρ x Re ′ = mf f sv = 1.907, which is less than 10, the nominal upper limit for laminar μ(1 − ε ) 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)(ρ p − ρ f )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,
(1 − ε)2 (− Δp) = 150 H ε3
μU x 2sv
With (− Δp) = 24519 Pa ; μ = 1.8 x 10-5 Pa.s; ρf = 1.2 kg/m3; xsv = 100 x 10-6 m, and given that the voidage of the bed at minimum fluidization is voidage, ε = 0.5, then SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.9
U = Umf = 0.0454 m/s Check Reynolds number for use of laminar flow in packed bed. U ρ x Re ′ = mf f sv = 0.6 , which is less than 10, the nominal upper limit for laminar μ(1 − ε ) 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: (1 − ε)2 μU (1 − ε) ρ f U 2 (− Δp) = 150 2 + 1.75 ε3 H ε 3 xsv xsv From the force balance, (− Δp) = (1− 0.5) × (5000 − 1.2) × 9.81 = 24519 Pa. H 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 + 4200U2mf 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 SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.10
(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): (1 − ε)2 μU (1 − ε) ρ f U 2 (− Δp) = 150 + 1.75 2 H ε3 ε 3 xsv xsv From the force balance, (− Δp) = (1− 0.45) × (1800 − 1.2) × 9.81 = 9705.4 Pa / m. H 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 U2mf 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 =
(ρ p − ρf ) 0.934 g0.934 x1.8 p 1110μ 0.87ρ 0.066 g
(Text-Equation 7.11)
(1800 − 1.2)0.934 9.810.934 (60 × 10−6 )1.8 1110(1.84 × 10−5 )0.87 (1.2)0.066 Umf = 2.73 x 10-3 m/s U mf =
(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− ε mf )ρ p AH mf πD2 π(0.5)2 2 = = 0.1963 m 4 4 hence, 240 = (1− 0.45) × 1800 × 0.1963 × Hmf
Bed cross-sectional area, A =
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.11
and so Hmf = 1.235 m (c) Visible bubble flow rate: From the two-phase theory, QB = (U − Umf )A
Superficial gas velocity, U =
Q ⎡ 140 ⎤ 1 = = 0.198 m / s A ⎣ 3600 ⎦ 0.1963
3 hence, QB = (0.198 − 0.0038) × 0.1963 = 0.0381 m / s
(d) Bubble fraction:
QB , where UB is the mean From the Two-Phase theory, bubble fraction, ε B = AU B 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: H − Hmf εB = , where H is the mean bed height. H 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 )ρ pAH 240 = 0.5848 Hence, voidage, ε = 1 − 1800 × 0.1963 × 1.636
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:
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.12
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 K*1∞ = 0.83 kg/m2s K*2∞ = 0.281 kg/m2s K*3∞ = 0.131 kg/m2s K*4∞ = 0.068 kg/m2s From Text-Equation 7.38, the entrainment rate for particles in size range i is d * R i = − (MBm Bi ) = K i∞Am Bi dt
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.13
where K*i∞ = elutriation rate constant 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 =
∑ R i = ∑ K*i∞ AmBi
(7.40)
For batch operation it can best be solved by writing Equation 7.38 in finite increment form: − Δ(m Bi MB ) = K*i∞ Am Bi Δt
(7.41)
where Δ(mBiMB) is the mass of solids in size range i entrained in time increment Δt. 4
Then total mass entrained in time Δt =
∑ {Δ(m BiM B)}
(7.42)
i=1
and mass of solids remaining in the bed at time t+Δt = (M B)t −
4
∑ {Δ (m BiMB )t }
(7.43)
i=1
(where subscript t refers to the value at time t.) Bed composition at time t+Δt = (m Bi )t +Δt =
(m Bi M B)t − [Δ (m Bi MB )t ] 4
(MB )t − ∑ {Δ(m Bi MB )t }
(7.44)
i=1
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) SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.14
= 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): =
0.1 × 2000 − 24.9 = 0.09034 2000 − 61.71
(m B2 )0+300 =
0.2 × 2000 − 16.86 = 0.1977 2000 − 61.71
(m B3 )0 +300
=
0.3 × 2000 − 11.79 = 0.3035 2000 − 61.71
(m B4 )0+300 =
0.4 × 2000 − 8.16 = 0.4085 2000 − 61.71
(m B1 )0 +300
The calculations for the remaining time steps are summarised below: mB2 mB3 Time Bed loss Bed mass mB1
mB4
(sec)
kg
kg
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)
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
mass fraction in bed Page 7.15
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. 61.71 = 0.2057 kg / s 300 gas flow rate at exit, Q = UA = 0.4 x 1 = 0.4 m3/s 0.2057 = 0.514 kg / m 3 solids loading at exit = 0.4 initial rate of carryover =
(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∞ Am B1 = 0.83 × 1× 0.0288 kg/s = 0.0239 kg/s hence solids flux =
0.0239 0.0239 = = 0.0239 kg/m2.s 1.0 A
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.16
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), 0.5 0.5 TDH = = 4.47d Bvs = 4.47 × 0.05 = 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, M B = (1 − ε )ρ pAH
SOLUTION TO CHAPTER 7 EXERCISES: FLUIDIZATION
Page 7.17
Hence, given ε = 0.55, ρp = 1800 kg/m3, MB = 4000 kg and vessel cross-sectional π2 2 2 area A = = 3.142 m : 4 4000 Applying Text-Equation 7.24, bed height, H = = 1.57 m. 1800 × (1 − 0.55) × 3.142 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 i∞ values to estimate carryover. 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; * −2 K 2∞ = 1.975 × 10 kg/m2s; *
K*3∞ = 1.049 × 10 −2 kg/m2s; K*4∞ = 6.54 × 10 −3 kg/m2s
The overall and component material balances over the fluidized bed system are: Overall balance:
F=Q+R
Component balance: Fm Fi = Qm Qi + Rm Ri
(Solution Manual Equation: 7.11.1) (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 Ri = Ki∞ AmBi
and R =
∑ R i = ∑ Rm Ri
(Solution Manual Equation: 7.11.3) (Solution Manual Equation: 7.11.4)
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Combining these equations with the assumption that the bed is well mixed (mQi = mBi), Fm Fi m Bi = (Solution Manual Equation: 7.11.5) F − R + K*i∞ A Now both mBi and R are unknown. However, noting that 0.2 × 0.1
−2
∑ m Bi = 1 , we have:
0.2 × 0.35
+
−2 0.2 − R + (5.16 × 10 × 3.142) 0.2 − R + (1.975 × 10 × 3.142) 0.2 × 0.40 0.2 × 0.15 + + = 1.0 −2 0.2 − R + (1.049 × 10 × 3.142) 0.2 − R + (6.54 × 10 −3 × 3.142)
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: m R1 =
* Am −2 K1∞ B1 = 5.16 × 10 × 3.142 × 0.0638 = 0.213 R 0.0485
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
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(c) Solids loading for gas entering the filter, 0.0485 R mass flow of solids = = = 0.0515 kg / m3 volume flow of gas UA 0.3 × 3.142 (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, χ=
kHmf(1-εp)/U = 100;
KCH = 1.0, use the reactor model of Orcutt et al. to UB
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:
(
) ) (
2
1 − βe − χ CH −χ = 1− βe − Conversion, 1 − C0 kHmf 1 − ε p + 1 − βe − χ U KCH and β = (U - Umf)/U where, χ = UB
(
)
(
)
(Text-EQ. 7.65)
From the information given in the question,
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kHmf(1-εp)/U = 100; χ =
U KCH = 1.0 and = 10 U mf UB
Hence, β = 0.9 So, from Text-Equation 7.65, conversion = 0.6645 [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 fluidization Hmf are doubled. Thus, assuming all else remains constant, under the new conditions χ = 2.0, β = 0.9 and kHmf(1-εp)/U = 200 and so the new conversion = 0.8744 [i.e. 87.44% conversion; increasing from 66.45%] (c) If the bubble size is halved (compared with the base case in (a) and KC is inversely proportional to bubble diameter, then KC increases by a factor of 2, causing χ to increase by a factor of 2. Hence χ = 2.0. Giving conversion = 0.8706 [i.e. 87.06% conversion of reactant A, compared with 66.45% in case (a)] (d) If the reaction rate were two orders of magnitude smaller: then kHmf(1-εp)/U = 1. So, for the conditions in part (a): conversion = 0.4 [i.e. 40% conversion of reactant A] If we introduce baffles causing the bubble size to halve, then KC will double, giving χ = 2.0. hence, 1 − βe −χ = 0.8782 and so conversion = 0.468 [i.e. 46.8% conversion of A compared 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 reaction 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.
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