Yang Fluidization, Solids Handling, and Processing
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796 Fluidization, Solids Handling, and Processing
From Fig. 3 of the text, a velocity of 75 ft/sec and its corresponding NS of 4.5 would satisfy (2), in which case, from Eq. (1):
LW |
=12 |
6328 |
= 9 inches |
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2.5 |
×75x60 |
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therefore, inlet height = 22.5", and from Fig. 19, gas outlet tube I.D. = 9 × 2 = 18"; assuming 1/4" wall thickness, gas outlet tube O.D. = 18.5". Therefore, barrel I.D. = 2(9) + 18.5 = 36.5" and from Fig. 19, Lmin = 54". Minimum dipleg I.D., d, conservatively assuming 100% collection efficiency and 20% safety factor is calculable from the dimensionless gravity flow equation:
6328 × 80 × 4 |
× 144 |
= 1.2 × 43 d ½ |
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60 × 7000 d |
2 |
3.14 |
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from which d = 1.8" or a design minimum of 1.2 × 1.8 = 2.2" which from Fig. 19 calls for a 4" industrial minimum, and thus in summary, a cyclone having the dimensions shown in Fig. 20.
The overall collection efficiency shown in Table 1 is determined from Fig. 2 of the text for the feed distribution given in Fig. 18. From Fig. 4 of the text at 80 grains/cu.ft., EL = 98.7%, therefore:
Loss Rate = 6328 × 60 ×80(1− 0.987 ) = 56.4 lbs/hr
7000
which meets specifications on performance (i.e., less than 60 lbs/hr loss rate).
The anticipated pressure drop is calculable from the relationships in Fig. 21. Assuming the cyclone is located within, or attached externally to, the shell of an 8' diameter fluidized bed reactor operating at a superficial gas velocity of 2 ft/sec, then:
(22.5 × 9)4
Area ratio = ( )2 = 0.028; hence K = 0.5
144 8 3.14
798 Fluidization, Solids Handling, and Processing
Table 1. Prediction of Cyclone Collection Efficiency
Inlet contraction loss:
P(s-i)G = 0.00298(0.1)[1.5(75)2-(2)2]
= 2.513" H2O
Solids acceleration loss:
P(s-i)G = 80(75)2/7000(167)
= 0.385" H2O
Barrel loss:
dhi = 4(22.5 × 9)/2(22.5 + 9) = 12.857"
800 Fluidization, Solids Handling, and Processing
Re = |
12.857 |
× 75× 0.1 |
= 597,888 |
12 × 0.02 |
× 0.000672 |
from any Fanning friction factor chart for smooth drawn tubing, f ~ 0.0033
PB = 0.0033(36.5)(0.1)(4.5)(75)2/26.2(12.857)
= 0.905" H2O
Reversal loss:
Pr = 0.1 (75)2/335
= 1.679 “ H2O Exit contraction loss:
Ve = 6328(4)(144)/3.14(60)(18)2 = 59.68 ft/sec
VB = 6328(4)(144)/3.14(60)36.52 = 14.5 ft/sec
Area ratio = (18/36.5)2 = 0.243; hence K = 0.415
Pe = 0.00298(0.1)[1.415(59.68)2-(14.5)2 ] = 1.439" H2O
Total calculated P across cyclone
= 2.513 + 0.385 + 0.905 + 1.679 + 1.439 = 6.921" H2O
if specific solids inlet loading were >10 lbs/sec × ft2.
Specific loading = |
6328 × 80 ×144 |
= 0.857 lbs/sec x ft2 |
60 × 7000 × 22.5 × 9 |
From Fig. 21 at an abscissa of 0.01 + 0.857 = 0.867, the ordinate equals 1.32; therefore:
Cyclone Design 801
True cyclone P = 1.32 × 6.921
= 9.136" H2O, meeting specifications
Example B. Suppose in the previous example that the loss rate had been specified as not to exceed 40 lbs/hr. To minimize the increase in pressure drop accompanying any increase in inlet velocity necessary to reduce Dth to a value which would bring performance up to the desired level, it might be more expedient to instead increase the gas discharge velocity.
To find EL required to meet 40 lbs/hr loss rate set:
40 = (1 - EL)6328(60)(80)/7000; from which EL = 99%.
From text Fig. 4 at 80 grains/ft3, required EO = 94%. The correspondingly required Dth is estimated by ratio from Example A:
Dth ~ 1− 0.94
9.71− 0.9086
from which Dth = 6.3 microns (estimated) and now checked against the correlation in Fig. 2 of the text for the feed distribution in Fig. 18 (see Table 2)
The 95.41% is slightly better than the required 94% so that the foregoing could be repeated with a new estimate of Dth slightly greater than 6.3 microns. However, since the difference between 94% and 95.41% is probably within the accuracy of any practical means of measurement, the value of 6.3 microns will be accepted as the design criterion, in which case, from Fig. 4, EL ~99.14% leading to a predicted loss rate of:
(1 - 0.9914)6328(60)(80)/7000 = 37.3 lbs/hr
which meets the specified 40 lbs/hr limit.
From the definition of Dth in Fig. 1 of the text:
6.3 |
= |
9(0.02)0.000672(9) |
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12(25,400) |
3.14N V (85 − 0.1)12 |
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s |
802 Fluidization, Solids Handling, and Processing
or
NS Ve = 796.15
which from Fig. 3 suggests Ve = 140 ft/sec and NS = 5.7, so that DO ~11.75" and from Fig. 19, L = 82", which results in a unit dimensioned as shown in Fig. 22.
Table 2. Prediction of Collection Efficiency for Cyclone in Example B
Cyclone Design 803
Figure 22. Cyclone design dimensions for Example B.
The cyclone in Fig. 22 will have a greater contraction loss than that in Fig. 20:
Area ratio = (11.75/36.5)2 = 0.1036; hence K ~ 0.47
Pe = 0.00298(0.1)[1.47(140)2 - (14.5)2]
= 8.52" H2O
P (calc’d) = 2.513 + 0.385 + 0.905 + 1.679 + 8.52 = 14" H2O
804 Fluidization, Solids Handling, and Processing
From Fig. 21, at a solids inlet rate of 0.857 lbs/sec·ft2, the true P/14 = 1.32, so true P = 18.48" H2O which exceeds the specified maximum of 10" H2O.
14.0 ALTERNATE APPROACH TO SOLVING EXAMPLE B
If, as an alternative, the Dth of 6.3 were to be achieved by increasing the inlet velocity, then Eq. (2) of Example A becomes
NS (Vi )1.5 = 6895
which from Fig. 3 suggests Vi = 117.7 ft/sec and NS = 5.4,
then: |
L = 12 |
ACFM |
= 7 3/16" |
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W |
150 V1 |
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therefore: inlet height = 2.5(7 3/16) = 18"; DO = 143/8"; DB = 29.25" and L = 67" which results in a cyclone dimensioned as shown in Fig. 23.
Pressure drop through the cyclone in Fig. 23 is again calculated from the relationships in Fig. 21. As in Example A,
18× 7.1895× 4
Area ratio = ( )2 = 0.0179; K = 0.5
3.14 8 144
Inlet contraction loss:
P(s-i)G = 0.00298(0.1)[1.5(117.4)2 - (2)2]
= 6.160" H2O
Solids acceleration loss:
P(s-i)P = 80(117.4)2/7000 × 167
= 0.943" H2O