Consider the flow system of Fig.6.7. A mechanical energy balance between points 1 and 5 of the system gives
ð6:12ị
Consider the individual terms in this expression.
Fig. 6.6 Friction factor vs. Reynolds number for flow through packed beds
1. Friction term,ΣF.Because the packed bed section has a very large interfacial area, its frictional loss is usually much greater than that of the rest of the piping system. Thus, we often can consider its frictional loss alone, or
ΣFtotalffiΣFpacked section
2. The potential energy term, gΔz. For gases this is usually negligible, but for liquids this can be an important term in the mechanical energy balance.
3. The work flow term in systems with small density changes, Ð
dp/ρ. When the density of the fluid does not vary much as it passes through the packed bed, one can use an average fluid density in the system. Thus,
ðdp ρ ffiΔp
ρ ð6:13ị
This condition is satisfied for all liquids and for gases where the relative pressure variation is less than 10 %, or where
Δp<0:1p
4. The flow work term for gases experiencing large density changes,Ð
dp/ρ. When the frictional pressure drop is large, meaning when Δp > 0:1p; one should properly account for the change of density with pressure. Thus, combining equation (9) with the mechanical energy balance of equation (6.12), dropping the kinetic and potential energy terms, dividing by u20, and introducing the superficial mass velocity
Fig. 6.7 Flow system which includes a packed bed
G0ẳn mw_ð ị A ẳkg
m2ẳu0ρẳuερgive ðρdp
G20 ỵ150 1ð εị2μL ε3d2pG0
ỵ1:75 1ð εịu20L ε3dp
ẳ0 ð6:14ị
If the gas expands reversibly as it flows through the packed bed, it would cool.
Here, however, flow is not reversible, friction causes heating, so it probably is best to assume isothermal flow. With this and the ideal gas assumption, integra- tion from point 3 to point 4 gives
ðmwị
2G20RTp24p23
ỵ150 1ð εị2μL ε3d2pG0
ỵ1:75 1ð εịu0L ε3dp
ẳ0 ð6:15ị
Note how this expression parallels equation (3.21) without the kinetic energy term.
This isothermal packed bed expression is the same one used in gas flow systems experiencing large pressure changes.
5. The kinetic energy term, (Δu2/2). This is usually negligible for both liquids and gases because one rarely gets to very high velocities in packed beds. However, if one has to include this factor, one should use the approach of Chap. 3 [see equations (3.12) and (3.21)] and add the correct term to equation (6.15).
6. The work term, Ws.Taking an energy balance about the whole system from point 1 to point 5 gives the shaft work directly. Alternatively, knowingp1andp2, a mechanical energy balance about the ideal compressor alone (KE, PE, andΣFall ignored) gives the energy actually received by the flowing fluid as
Wsẳ ð2
1
dp ρ
J
kg ð6:16ị
For liquids and for gases experiencing a small fractional change in pressure (orΔ p<0.1p), this expression simplifies to give
WsẳΔp
ρ ð6:17ị
For gases experiencing a high fractional change in pressures orð Δp>0:1pị, equation (6.16) is integrated to give equations (1.9), (1.10), (1.11), (1.12), (1.13) and (1.14).
7. Comments.Although most workers will agree that the form of the Ergun equation reasonably represents the frictional loss in packed beds, some have suggested that we do not trust too strongly the values of the Ergun constants, 150 and 1.75.
For example, in blast furnace work with very large particles, Standish and Williams (1975) find that the Ergun constants should both be doubled or tripled.
Again, Macdonald et al. (1979) recently suggested that the 150 be replaced by 180 and that the 1.75 be replaced by 1.8 for smooth particles and 4.0 for very rough particles.
However, keeping in mind that most of the data points vary by a factor of two from the mean, we prefer, for the time being, to stay with the original Ergun equation with its considerable backing of experimental verification.
Example 6.1 A Laboratory Packed Bed Experiment
In a second floor laboratory of the East China University of Science and Technology, in Shanghai, is a 0.22-m-i.d. glass tube packed to a depth of 1 m with 10-mm spheres as shown below. What will be the superficial velocity of water at 20C through the bed if the water level is kept 3 m above the exit of the bed?
Solution
We can choose to take the mechanical energy balance between various pairs of points in the system, thus, between points 1 and 3, 1 and 4, 2 and 3, or 2 and 4. However, in all cases the packed bed section should be included, because that is where most of the frictional loss occurs. Let us take the balance between points 1 and 4, in which caseΔp/ρ and the kinetic energy terms drop out. Thus we obtain
(continued)
(continued)
ðiị
With no pump or turbine in the systemWsẳ0, so we are left with ΣFỵg zð 4z1ị ẳ0
Now the frictional loss from points 1 to 4 consists of the resistance of the packed bed section 2–3 and the empty pipe sections 1–2 and 3–4. However, it is reasonable to assume that the resistance of the empty pipe sections is negligible compared to that of the packed section. Thus, applying the Ergun equation (6.9), and estimating the bed voidage to be 0.38 from Figs.6.8and 6.9, we obtain
150 1ð 0:38ị2103 u0
0:38
ð ị3ð0:01ị2ð1,000ị ỵ1:75 1ð 0:38ịu20ð ị1 0:38
ð ị3ð0:01ị ỵð ị9:8 ð03ị ẳ0 ðiiị or
10:51u0ỵ1, 977u2029:4ẳ0 ðiiiị Solving the quadratic gives
= 0.119 m/s u0
NOTE: The above solution uses the complete Ergun equation. Had we guessed that the turbulent losses dominated, we would have dropped the linear term in the above quadratic, equation (iii), and this would give
u0ẳ 29:4 1,977 1=2
ẳ0:122 m=s 2ð :5%highị
A Reynolds number check would then show Repẳdpu0ρ
μ ẳð0:01ịð0:122ịð1,000ị
103 ẳ1222
which justifies the assumption that turbulent losses dominate in this situation.
The solution to this problem is quite sensitive to the value chosen for the bed voidage because the voidage, in effect, appears to the fourth and fifth
(continued)
(continued)
powers in equation (ii). So if we had chosenεẳ0.42, instead ofεẳ0.38, we would have found for equation (i)
6:81u0ỵ1,370u2029:4ẳ0 or
= 0.144 m/s (21% high) u0
This large uncertainty in readingεfrom Figs.6.8and6.9is a reflection of the large variation of voidage obtainable in a packed bed. How you pack the bed, whether you shake and tap as you pour the solids in, and so on can give widely different values for voidages.
In the setups above, with any value forxandy, the driving force remains at 3 m, the resistance is unchanged—still at 1 m of packed bed—so the solution is the same as presented above.
Fig. 6.8 Packed bed partially above exit in a U tube
Fig. 6.9 Packed bed below exit in a U tube
Problems on Packed Beds
6.1. I predict that “Happy Bananas” will soon sweep the country. They will be produced by passing readily absorbed nitrous oxide (laughing gas) through a packed bed of green but full-grown Central American bananas. In developing this process we will need to know the pressure drop in these beds of bananas.
To achieve this, estimate the effective banana size dp from the geometric considerations shown here:
6.2. We plan to pack a tower with Raschig rings of dimensions shown below.
Determine the effective particle sizedpof this packing material.
NOTE: The advantage of these specially designed packings is that they give small particle behavior coupled with large voidage—and high voidage gives a low pressure drop.
6.3. For pressure drop purposes a packed bed of spheres of what size would behave like a randomly packed mixture of equal weights of 1 and 2 mm spheres?
Assume the same voidage in both beds.
6.4. Air at about 20C and 1 atm passes upward through a fixed bed (Lẳ0.8 m, dbedẳ0.1 m,εẳ0.4) of spherical particles (ρsẳ3,000 kg/m3,dpẳ10 mm) at a superficial velocity of 1.5 m/s. Find the pressure drop across the bed.
6.5. Water flows downward through a tube inclined 30 from the horizontal and packed 10 m in length with metal spheres (dpẳ1 mm, ρsẳ5,200 kg/m3, εẳ0.34). At a particular flow velocity, the pressure is 3 atm at both ends of the bed.
The tube is now tipped horizontally, the packed length is reduced to 5 m, and the water flows at the same rate through the bed. If the pressure at the entrance of the bed is 3 atm, what is it at the exit of the bed?
6.6. Water flows downward through a vertical tube packed 10 m deep with metal spheres (dpẳ1 mm,ρsẳ5,200 kg/m3,εẳ0.34). At a particular flow velocity, the pressure just above the bed is 3 atm. Just below the bed it is also 3 atm.
The tube is now tipped 45from the vertical, and water at the same flow rate flows upward through the bed. If the pressure just at the entrance is 3 atm, what is it at the exit of the bed?
6.7. For the regenerator of Example 15.1, air is compressed, cooled back to 20C, and then passed at a mass velocityG0ẳ4.8 kg/m2s through a vessel 54.5 m high, 1 m2 in cross section, and filled with close to spherical packing (dpẳ0.05 m,εẳ0.4). Air leaves the packed bed at 20C and 1 atm. What size of ideal compressor will provide this airflow rate?
6.8. Gas flow in packed bed catalytic reactors.Packed beds of catalyst with wall cooling are widely used in the process industry to effect strongly exothermic reactions; for design purposes die engineer must be able to develop a model which reasonably approximates what is happening in these reactors. Such a model should consider a number of phenomena: the flow of generated heat
outward to the cooling walls, the dispersion of flowing material radially and axially, and also the velocity distribution of the flowing gas.
This flow can be pictured in various ways, from the very simple to the more real but complex. For example, the simplest model assumes plug flow of gas, which means that all fluid elements move at exactly the same velocity through the packed bed with no overtaking. The second-stage model assumes plug flow of gas with small random velocity fluctuations superimposed. This is called the axial dispersion model.
Now it is well known that the voidage by the walls of a packed bed is higher than in the main body of the bed, so the third-stage model visualizes two flow regions, a central core of lower voidage surrounded by an annular wall region of higher voidage and one-particle diameter in thickness. Plug flow is assumed to occur in each region but with a higher velocity in the high-voidage region.
More precise models may try to incorporate the true velocity profile in the packed bed, a profile which in fact deviates rather significantly from plug flow; however, this profile is not reliably known today. Schlu¨nder (1978) gives a good accounting of the state of knowledge of the factors involved in the proper modeling of fixed bed reactors.
To go back to the third-stage model, if the voidage in the main body of the packed bed of spherical catalyst particles is 0.36 and if it is 0.5 in the wall region, find the velocity ratio of fluid flowing in these regions for:
(a) Very small particles and slow flow (b) For very large particles and high flow rate
If the packed bed reactor is 8 particles in diameter, what fraction of the fluid flows up the wall region:
(c) For very small particles and slow flow?
(d) For very large particles and high flow rate?
6.9. In one design of a direct-contact crossflow gas–solid heat exchanger, the solids are conveyed horizontally on a screen conveyor while gas percolates upward through the solid. Suppose that hot solidsdpẳ10 mm,εẳ0:4
are conveyed at 0.2 m/s in a layer 0.2 m thick while cold air flows upward through the solid mass from a high-pressure chamber (pressure is 2 kPa above atmospheric) to atmospheric pressure, the whole exchanger being at an average temperature of 100C. Find the direction of airflow through the layer of solid, and give this as the angle from the verticalθ, as shown below.
6.10. The critical step in Motorola’s proposed process for producing ultra pure solar-cell grade silicon is the reaction
Si commercial grade, solid, 98%pure
þSiF4
gas, pure
!2 SiF2
gas, pure
: : :exothermic
The SiF2is then decomposed elsewhere to produce pure silicon. Thermody- namics says that at the chosen operating temperature of 1,350 K, the reaction should only become appreciable when the pressure falls below 100 Pa. The reactor will be a vertical 0.4-m-i.d. tube packed with 6-mm silicon particles (εẳ0.5). Reactant gas (pure SiF4) will enter the reactor at 2,000 Pa and 1,350 K, and 1.2 m downstream we expect the pressure to have dropped to 100 Pa. From here on reaction takes off and conversion to product is soon complete. Estimate the production rate of silicon (kg/h) obtainable from this reactor.
Data:
μSiF4ẳ4105kg=m s estimatedð ị ðmwịSiẳ 0:028
ðmwịSiF4ẳ0:104 Cp=Cv
SiF4 ẳ1:15 estimatedð ị
6.11.Nitrogen fixation by the Haber–Bosch process(1909–1914). This remarkable hydrogenation discovered or invented by Prof. Haber and commercialized by Dr. Bosch makes ammonia (a key ingredient in fertilizers and in explosives) from nitrogen and hydrogen by the very high-pressure reaction
N2þ3H2!hundreds of atmospheres 2 NH3
Its use in agriculture is directly responsible for saving millions of people from starvation these last 70 years. On the other hand, it also allowed the First World War to drag on for three extra years, resulting in untold casualties and misery.
As part of this process, suppose a mixture of 3 parts H2and 1 part N2at 0 C and 2.5 MPa flows at a superficial velocity of u0ẳ1 m/s downward through a packed bed of spherical particles which rest on a holey plate which in turn rests on 8 L-shaped supports, evenly spaced around the tube wall.
Find the downward force (in newtons and in kilograms) exerted on each of the L-shaped supports by the gas flowing through the packed bed of catalyst.
Ignore the resistance of the holey plate.
Data:Particles: spherical,dpẳ2 mm.
Gas: At these conditions the feed mixture behaves as an ideal gas with μẳ2105kg/m ã s.
Bed: diameterẳ1 m, heightẳ300 mm, voidageẳ0.4, total weight of packing and with its support plateẳ700 kg.
6.12.Rotating disk heat exchanger. Our firm is having difficulty in designing a rotating disk heat exchanger. Heat exchanger test books do not seem to help, so I am turning to you for help.
cumbustor
rotating wheel dust deposits
not rotating
dust is blown back in
cold air
not rotating clean air
This unit is to transfer heat from dusty combustion gas leaving a fluidized combustor (60 mol/s, 800C, Cpẳ60 J/mol ã K,) to the incoming air entering the combustor (60 mol/s, 0C, Cpẳ60 J/mol ã K). The big advantage to this device is that dusty gas is returned to the combustor by the rotating wheel which is to be made of metal honeycomb (Cpẳ1,000 J/kg ã K, massẳ1,000 kg):
(a) What should be the rotation rate of the wheel in RPM?
(b) Would you design the wheel to be a large flat pancake unit or a longer smaller diameter unit? Can you explain why?
References
G.G. Brown et al.,Unit Operations,Chapter 16, (Wiley, New York, 1950) S. Ergun, Fluid flow through packed columns. Chem. Eng. Prog.48, 89 (1952)
M. Leva, Articles on fixed, moving and fluidized systems. Chem. Eng. Prog.48, 89 (1952) Flow thru packings and moving beds. Chem. Eng. 204, February, pp. 263–268, March,
pp. 261–265, (1957)
Gas-liquid flow in stacked towers. Chem. Eng.267–270 (1957) Pressure drop in stacked towers. Chem. Eng. 269–272 (1957) Fluidized heat transfer nomograph. Chem. Eng. 254–257 (1957)
Equipment for fixed and moving beds. Chem. Eng. 258–262 (1957). Also see Fig. 6.4 Variables in fixed bed systems. Chem. Eng. 263–266 (1957)
Correlations in fixed bed systems. Chem. Eng. 245–248 (1957) Flow behavior in fluidized systems. Chem. Eng. 289–293 (1957) Liquid flow in pack towers. Chem. Eng. 267–272 (1957)
I.F. Macdonald, M.S. El-Sayed, K. Mow, F.A.L. Dullien, Flow through porous media—the Ergun equation revisited. Ind. Eng. Chem. Fundam.18, 199 (1979)
N. Standish, I.D. Williams,Proceedings of a Blast Furnace Aerodynamics Symposium, (Aus.I.M.
M., Wollongong, 1975)
X. Schlu¨nder, Chemical Reaction Engineering Reviews—Houston, ACS Symposium Series No. 72, p. 110 (1978)
Flow in Fluidized Beds