Rhodes Solutions Ch6

SOLUTIONS TO CHAPTER 6: FLOW THROUGH A PACKED BED OF PARTICLES EXERCISE 6.1: 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, which is restrained at its upper surface. (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. SOLUTION TO EXERCISE 6.1: (a) Knowing the mass of particles in the bed, the density of the particles and the volume of the bed, the bed voidage may be calculated: mass of bed, M = M = AH 1− ε( )ρp

giving voidage, ε =1 −50

2500 × 0.04 ×1= 0.5

(b) With a liquid flow rate of 1.44 m3/h, the superficial liquid velocity through the bed, U is given by:

U =1.44

3600 × A= 0.01 m / s

Use the Ergun equation (Text-Equation 6.15) to estimate the pressure drop across the bed at this flow rate:

−Δp( )H

= 150μUxsv

2(1− ε)2

ε3 +1.75ρf U2

xsv

(1− ε)ε3

With μ = 0.002 Pa.s, ρf = 800 kg/m3, xsv = 1 mm and H = 1 m, −Δp( ) = 600 ×103U + 5.6 × 106U2 = 6560 Pa

Checking the Reynolds number, R ′ e =Uρf xsvμ 1 − ε( )

= 8 (Text-Equation 6.12). Since the

Reynolds number is less than 10, we might estimate the pressure drop using the Carman Kozeny equation (Text-Equation 6.16): −Δp( )H

= 180μUxsv

2(1− ε)2

ε3 = 7200 Pa.

Alternatively we could use the laminar part of the Ergun equation, which gives, (-Δp) = 6000 Pa.

SOLUTIONS TO CHAPTER 6: FLOW THROUGH A PACKED BED OF PARTICLES p. 6.1

EXERCISE 6.2: 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) Calculate the superficial liquid velocity at which the frictional pressure drop across the bed is 4130 Pa. SOLUTION TO EXERCISE 6.2: (a) Knowing the mass of particles in the bed, the density of the particles and the volume of the bed, the bed voidage may be calculated: mass of bed, M = AH 1− ε( )ρp giving voidage, ε =1 −

5

2000 × π4

(0.1)2 × 0.6= 0.4695

Use the Ergun equation (Text-Equation 6.15) to estimate the relationship between pressure drop across the bed and superficial liquid velocity:

−Δp( )H

= 150μUxsv

2(1− ε)2

ε3 +1.75ρf U2

xsv

(1− ε)ε3

With μ = 0.001 Pa.s, ρf = 1000 kg/m3, xsv = 300 μm and H = 0.6 m, (−Δp) = 2.72 ×106 × U2 +17.9 ×106 U With (-Δp) = 4130 Pa, we solve the quadratic for U: U = 1.5 x 10-3 m/s, i.e. 1.5 mm/s (positive root) EXERCISE 6.3: A gas absorption tower of diameter 2 m contains ceramic Raschig rings randomly packed to a height of 5 m. Air containing a small proportion of sulphur dioxide passes upwards through the absorption tower at a flow rate of 6 m3/s. The viscosity and density of the gas may be taken as 1.80 x 10-5 Pas and 1.2 kg/m3 respectively. Details of the packing is given below: Ceramic Raschig rings

Surface area per unit volume of packed bed, SB = 190 m2/m3. Voidage of randomly packed bed = 0.71

(a) Calculate the diameter, xsv, of a sphere with the same surface-volume ratio as the Raschig rings.


(b) Calculate the frictional pressure drop across the packing in the tower. (c) Discuss how this pressure drop will vary with flow rate of the gas within ±10% of the quoted flow rate. (d) Discuss how the pressure drop across the packing would vary with gas pressure and temperature. SOLUTION TO EXERCISE 6.3: (a) From Text-Equation 6.6: SB = S 1 − ε( ) , where S surface area per unit volume of rings.

Therefore, S =SB

1− ε( )=

1901− 0.71( )

= 655.2 m2 / m3

If xsv is the diameter of a sphere with the same surface-volume ratio as the rings, πxsv

2

π6

xsv3

= 655.2 m2 / m3

Hence, xsv = 9.16 mm (b) Superficial gas velocity, U =

Q

π D2

4

=6

π 22

4

=1.91 m / s

Using the Ergun equation (Text-Equation 6.15) to describe the relationship between gas velocity and pressure drop across the packed bed,

−Δp( )H

= 150μUxsv

2(1− ε)2

ε3 +1.75ρf U2

xsv

(1− ε)ε3

With μ = 1.8 x 10-5 Pa.s, ρf = 1.2 kg/m3, xsv = 9.16 x 10-3 m and H = 5 m,

−Δp( )5

= 1501.8 ×10−5 ×1.91(9.16 ×10−3)2 ×

(1 − 0.71)2

0.713 +1.751.2 ×1.912

9.16 ×10−3 ×(1 − 0.71)

0.713

which gives −Δp( ) = 72.0 + 3388.4 = 3460.4 Pa . (c) We note that the turbulent component makes up 98% of the total. Hence, within

of the quoted flow rate the pressure drop across the bed will increase with the square of the superficial velocity and hence with the square of the flow rate: ±10%

−Δp( ) ∝ Q2


(d) Variation with gas pressure: Pressure increase affects only the gas density (except at very high pressure which are not relevant here). And gas density is directly proportional to absolute gas pressure (if we assume ideal gas behaviour). Since the flow is predominantly turbulent, then: −Δp( ) ∝ absolute gas pressure

Variation with gas temperature: Since the flow is predominantly turbulent, gas viscosity has almost no affect (see Ergun equation). Therefore, variation in gas temperature will influence only the gas density. Assuming ideal gas behaviour, ρf ∝

1absolute temperature, T

Hence, −Δp( ) ∝1T

EXERCISE 6.4: A solution of density 1100 kg/m3 and viscosity 2 x 10-3 Pas is flowing under gravity at a rate of 0.24 kg/s through a bed of catalyst particles. The bed diameter is 0.2 m and the depth is 0.5 m. The particles are cylindrical, with a diameter of 1 mm and length of 2 mm. They are packed to give a voidage of 0.3. Calculate the depth of liquid above the top of the bed. [Hint: apply the mechanical energy equation between the bottom of the bed and the surface of the liquid] SOLUTION TO EXERCISE 6.4: Calculate the frictional pressure loss through the bed. Superficial liquid velocity, U =

0.24

1100 × π4

0.2( )2= 6.94 ×10−3 m / s

Surface-volume diameter of particles, xsv:

Volume of one cylindrical particle = π2

mm3

Surface area of one cylindrical particle = 2.5π mm2 Surface-volume ratio of particles =

2.5ππ 2

= 5 mm2/mm3

For a sphere of diameter xsv, surface-volume ratio = 6

xsv

Hence, diameter of sphere which has the same surface-volume ratio as the particles, xsv = 1.2 mm


Checking the Reynolds number (Text-Equation 6.12),

R ′ e =Uρf xsvμ 1 − ε( )

=6.94 ×10−3 ×1100 ×1.2 ×10−3

2 ×10−3 × (1 − 0.3)= 6.5

The Reynolds number is less than 10 and so we can assume that laminar flow dominates . The Ergun equation (Text-Equation 6.15) reduces to: −Δp( )H

= 150μUxsv

2(1− ε)2

ε3

With μ = 0.002 Pa.s, ρf = 1100 kg/m3, xsv = 1.2 mm, 3.0=ε and H = 0.5 m,

−Δp( )0.5

= 1502 × 10−3 × 6.94 ×10−3

(1.2 ×10−3)2 ×(1 − 0.3)2

0.33 = 26240 Pa / m

which gives −Δp( ) = 13120 Pa . Expressed in terms of head of liquid, friction head loss through the bed, hloss =

131201100 × 9.81

= 1.216 m

Applying the mechanical energy balance between the liquid surface (position 1) and the bottom of the packed bed (position 2): (Solution Manual-Figure 6.4.1)

z1 +U1

2

2g+

p1ρfg

= z2 +U2

2

2g+

p2ρfg

+ hloss

Assuming that p1 = p2 = atmospheric, and that U1 = U2, z1 − z2 = hloss = 1.216 m The height of the packed bed is 0.5 m and so the depth of liquid above the bed is 0.716 m (1.216 - 0.5 m). EXERCISE 6.5: In the regeneration of an ion exchange resin, hydrochloric acid of density 1200 kg/m3 and viscosity 2 x 10-3 Pas flows upwards through a bed of resin particles of density 2500 kg/m3 resting on a porous support in a tube 4 cm in diameter. The particles are spherical, have a diameter 0.2 mm and form a bed of void fraction 0.5. The bed is 60 cm deep and is unrestrained at its upper surface. Plot the frictional pressure drop across the bed as function of acid flow rate up to a value of 0.1 l/min. SOLUTION TO EXERCISE 6.5: Assuming laminar flow, the Ergun equation (Text-Equation 6.15) reduces to: −Δp( )H

= 150μUxsv

2(1− ε)2

ε3


With μ = 0.002 Pa.s, ε = 0.5, xsv = 0.2 mm and H = 0.6 m,

−Δp( )0.6

= 1502 × 10−3 × U(0.2 ×10−3)2 ×

(1 − 0.5)2

0.53

which gives . −Δp( ) = 9 ×106 × U Pa Referring to Chapter 7, the packed bed will fluidize when the buoyant weight of the particles is supported. Using Text-Equation 7.2, the pressure drop at which this occurs is: ( ) ( )( ) Pa. 3826g1Hp fp =ρ−ρε−=Δ− Using the reduced Ergun equation, the superficial liquid velocity at which the packed bed pressure drop is 3826 Pa is U = 4.25 x 10-4 m/s. Checking the Reynolds number (Text-Equation 6.12),

R ′ e =Uρf xsvμ 1 − ε( )

=4.25 ×10−4 ×1200 × 0.2 × 10−3

2 ×10−3 × (1− 0.5)= 0.102

The Reynolds number is less than 10 and so our assumption that laminar flow dominates is valid. In summary then, the pressure drop increases linearly with liquid flow rate up to a pressure drop of 3826 Pa at a superficial liquid velocity of 4.25 x 10-4 m/s. Beyond this velocity the pressure drop will remain essentially constant with increasing liquid flow rate, since the bed is fluidized (see Chapter 7). EXERCISE 6.6: The reactor of a catalytic reformer contains spherical catalyst particles of diameter 1.46 mm. The packed volume of the reactor is to be 3.4 m3 and the void fraction is 0.25. The reactor feed is a gas of density 30 kg/m3 and viscosity 2 x 10-5 Pas flowing at a rate of 11,320 m3/h. The gas properties may be assumed constant. The pressure loss through the reactor is restricted to 68.95 kPa. Calculate the cross-sectional area for flow and the bed depth required. SOLUTION TO EXERCISE 6.6: Using the Ergun equation (Text-Equation 6.15) to describe the relationship between gas velocity and pressure drop across the packed bed,

−Δp( )H

= 150μUxsv

2(1− ε)2

ε3 +1.75ρf U2

xsv

(1− ε)ε3

With μ = 2.0 x 10-5 Pa.s, ρf = 30 kg/m3, xsv = 1.46 x 10-3 m, −Δp( )=68.75 kPa and ε = 0.25,


−Δp( )

H= 150

2.0 ×10−5 × U(1.46 ×10−3)2 ×

(1 − 0.25)2

0.253 +1.7530 × U 2

1.46 ×10−3 ×(1 − 0.25)

0.253

which gives 68.75 ×103

H= 50666U +1.726 ×106U2

Reactor volume, V = AH = 3.4 m3.

Gas volumetric flowrate, Q = UA = 113203600

= 3.144 m3 / s

Substituting gives: 0.681H2 + 21.467H3 = 1.0 Solving, bed depth, H = 0.35 m, and so cross-sectional area, A = 9.71 m2. EXERCISE 6.7: A leaf filter has an area of 2 m2 and operates at a constant pressure drop of 250 kPa. The following results were obtained during a test with an incompressible cake: Volume of filtrate collected (litre)

280 430 540 680 800

Time (min) 10 20 30 45 60 Calculate: (a) the time required to collect 1200 litre of filtrate at a constant pressure drop of 400 kPa with the same feed slurry. (b) the time required to wash the resulting filter cake with 500 litre of water (same properties as the filtrate) at a pressure drop of 200 kPa. SOLUTION TO EXERCISE 6.7: For filtration at constant pressure drop we use Text-Equation 6.27, which indicates that if we plot t/V versus V a straight line will have a gradient

rcφμ2A2 −Δp( )

and an intercept rcφμ

A2 −Δp( )Veq on the t/V axis.

Using the data given in the question: V(m3) 0.28 0.43 0.54 0.68 0.80 t /V (sec/m3) 2142 2790 3333 3971 4500 This is plotted in Solution Manual-Figure 6.7.1.


From the plot : gradient = 4625 s/m6 intercept = 800 s/m3 hence,

rcφμ2A2 −Δp( )

= 4625

and

rcφμA2 −Δp( )

Veq = 800

which, with A = 2 m2 and (-Δp) = 250 x 103 Pa, gives Pas/m2 rcφμ = 9.25 ×109

and Veq = 0.0865 m3 Substituting in Text-Equation 6.27, t

V=

9.25 ×109

4 −Δp( ) 0.5V + 0.0865( )

which applies to the filtration of the same slurry in the same filter at any pressure drop. (a) To calculate the time required to pass 1200 litre (1.2 m3) of filtrate at a pressure drop of 400 kPa, we substitute V = 1.2 m3 and (-Δp) = 400 x 103 Pa in the above equation, giving: t = 4763 sec (or 79.4 minutes) (b) During the filtration the cake thickness is continuously increasing and, since the pressure drop is constant, the volume flow rate of filtrate will continuously decrease. The filtration rate is given by Text-Equation 6.26.

( )( )φ+μ

Δ−=

eqc VVrAp

dtdV

A1

Substituting the volume of filtrate passed at the end of the filtration period (V = 1.2 m3), r Pas/m2, Veq = 0.0865 m3 and (-Δp) = 400 x 103 Pa, we find

the filtration rate at the end of the filtration period is cφμ = 9.25 ×109

dVdt

= 1.34 ×10−4 m3 / s


If we assume that the wash water has the same physical properties as the filtrate, then during a wash period at a pressure drop of 400 kPa the wash rate would also be 1.34 x 10-4 m3/s. However, the applied pressure drop during the wash cycle is 200 kPa. According to Text-Equation 6.26 the liquid flow rate is directly proportional to the applied pressure drop, and so:

flowrate of wash water (at 200 kPa) = 1.34 ×10−4 ×200 ×103

400 ×103⎛

⎝ ⎜ ⎞

⎠ ⎟ = 6.7 ×10−5 m3 / s

Hence, the time needed to pass 0.5 m3 of wash water at this rate is 7462 sec (or 124.3 minutes) EXERCISE 6.8: A laboratory leaf filter has an area of 0.1 m2, operates at a constant pressure drop of 400 kPa and produces the following results during a test on filtration of a slurry: Volume of filtrate collected (litre)

19 31 41 49 56 63

Time (sec) 300 600 900 1200 1500 1800 (a) Calculatethe time required to collect 1.5 m3 of filtrate during filtration of the same slurry at a constant pressure drop of 300 kPa on a similar full-scale filter with and area of 2 m2. (b) Calculate the rate of passage of filtrate at the end of the filtration in part (a). (c) Calculate the time required to wash the resulting filter cake with 0.5 m3 of water at a constant pressure drop of 200 kPa. Assume the cake is incompressible and that the flow properties of the filtrate are the same as those of the wash solution.) SOLUTION TO EXERCISE 6.8: For filtration at constant pressure drop we use Text-Equation 6.27, which indicates that if we plot t/V versus V a straight line will have a gradient

rcφμ2A2 −Δp( )



Using the data given in the question: V(m3) 0.019 0.031 0.041 0.049 0.056 0.063 t /V (sec/m3) 15789 19355 21951 24490 26786 28571 This is plotted in Solution Manual-Figure 6.8.1.


From the plot : gradient = 2.904 x 105 s/m6 intercept = 10300 s/m3 hence,

rcφμ2A2 −Δp( )

= 2.904 x 105

and

rcφμA2 −Δp( )

Veq = 10300

which, with A = 0.1 m2 and (-Δp) = 400 x 103 Pa, gives Pas/m2 rcφμ = 2.323 ×109

and Veq = 0.0177 m3 Substituting in Text-Equation 6.27, t

V=

2.323 ×109

4 −Δp( ) 0.5V + 0.0177( )

which applies to the full-scale filter (area 2 m2) using the same slurry at any pressure drop. (a) To calculate the time required to pass 1.5 m3 of filtrate through the full-scale filter at a pressure drop of 300 kPa, we substitute V = 1.5 m3 and (-Δp) = 300 x 103 Pa in the above equation, giving: t = 2229 sec (or 37.1 minutes) (b) During the filtration the cake thickness is continuously increasing and, since the pressure drop is constant, the volume flow rate of filtrate will continuously decrease. The filtration rate is given by Text-Equation 6.26.

( )( )φ+μ

Δ−=

eqc VVrAp

dtdV

A1

Substituting the volume of filtrate passed at the end of the filtration period (V = 1.5 m3), r Pas/m2, Veq = 0.0177 m3 and (-Δp) = 300 x 103 Pa, we find

the filtration rate at the end of the filtration period is cφμ = 2.323 ×109

dVdt

= 3.40 ×10−4 m3 / s


If we assume that the wash water has the same physical properties as the filtrate, then during a wash period at a pressure drop of 300 kPa the wash rate would also be 3.40 x 10-4 m3/s. However, the applied pressure drop during the wash cycle is 200 kPa. According to Text-Equation 6.26 the liquid flow rate is directly proportional to the applied pressure drop, and so: flowrate of wash water (at 200 kPa)

=3.40 ×10−4 ×200 ×103

300 ×103⎛

⎝ ⎜ ⎞

⎠ ⎟ = 2.27 ×10−4 m3 / s

Hence, the time needed to pass 0.5 m3 of wash water at this rate is 2202 sec (or 36.7 minutes) EXERCISE 6.9: A leaf filter has an area of 1.73 m2, operates at a constant pressure drop of 300 kPa and produces the following results during a test on filtration of a slurry: Volume of filtrate collected (litre)

0.19 0.31 0.41 0.49 0.56 0.63

Time (sec) 300 600 900 1200 1500 1800 Assuming that the cake is incompressible and that the flow properties of the filtrate are the same as those of the wash solution, calculate: (a) the time required to collect 1 m3 of filtrate during filtration of the same slurry at a constant pressure drop of 400 kPa. (c) the time required to wash the resulting filter cake with 0.8 m3 of water at a constant pressure drop of 250 kPa. SOLUTION TO EXERCISE 6.9: For filtration at constant pressure drop we use Text-Equation 6.27, which indicates that if we plot t/V versus V a straight line will have a gradient

rcφμ2A2 −Δp( )



Using the data given in the question: V(m3) 0.19 0.31 0.41 0.49 0.56 0.63 t /V (sec/m3) 1579 1935 2195 2449 2679 2857 This is plotted in Solution Manual-Figure 6.9.1.


From the plot : gradient = 2960 s/m6 intercept = 1000 s/m3 hence,

rcφμ2A2 −Δp( )

= 2960

and

rcφμA2 −Δp( )

Veq = 1000

which, with A = 1.73 m2 and (-Δp) = 300 x 103 Pa, gives Pas/m2 rcφμ = 5.32 ×109

and Veq = 0.169 m3 Substituting in Text-Equation 6.27, tV

=5.32 ×109

1.732 −Δp( )0.5V + 0.169( )

which applies to the filtration of the same slurry in the same filter at any pressure drop. (a) To calculate the time required to pass 1.0 m3 of filtrate at a pressure drop of 400 kPa, we substitute V = 1.0 m3 and (-Δp) = 400 x 103 Pa in the above equation, giving: t = 2973 sec (or 49.5 minutes) (b) During the filtration the cake thickness is continuously increasing and, since the pressure drop is constant, the volume flow rate of filtrate will continuously decrease. The filtration rate is given by Text-Equation 6.26.

( )( )φ+μ

Δ−=

eqc VVrAp

dtdV

A1

Substituting the volume of filtrate passed at the end of the filtration period (V = 1.0 m3), Pas/m2, Veq = 0.169 m3 and (-Δp) = 400 x 103 Pa, we find

the filtration rate at the end of the filtration period is

rcφμ = 5.32 ×109

dVdt

= 1.923 ×10−4 m3 / s


If we assume that the wash water has the same physical properties as the filtrate, then during a wash period at a pressure drop of 400 kPa the wash rate would also be 1.923 x 10-4 m3/s. However, the applied pressure drop during the wash cycle is 250 kPa. According to Text-Equation 6.26 the liquid flow rate is directly proportional to the applied pressure drop, and so: flowrate of wash water (at 250 kPa) =

1.923 ×10−4 ×250 ×103

400 ×103⎛

⎝ ⎜ ⎞

⎠ ⎟ =1.202 ×10−4 m3 / s

Hence, the time needed to pass 0.8 m3 of wash water at this rate is 6656 sec (or 110.9 minutes)


Figure 6.4.1: Application of the mechanical energy equation-Exercise 6.4.

Figure 6.7.1: Plot of t/V versus V for Exercise 6.7.


Figure 6.8.1: Plot of t/V versus V for Exercise 6.8.

Figures 6.9.1: Plot of t/V versus V for Exercise 6.9.


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Rhodes Solutions Ch6