Yang Fluidization, Solids Handling, and Processing

Cyclone Design 805

Figure 23. Alternative cyclone design for Example B.

Barrel loss:

dhi = 4(18 × 7.1875)/2(18 + 7.1875) = 10.27"

PB =0.00314(29.25)(0.1)(5.4)(117.4)2/26.2(10.27)

= 2.54" H2O

806 Fluidization, Solids Handling, and Processing

Reversal loss:

Pr = 0.1(117.4)2/335

= 4.114“ H2O

Exit contraction loss:

Ve = 6328(4)144/60(3.14)(14.375)2 = 93.58 ft/sec

VB = 6328(4)144/60(3.14)(29.25)2 = 22.60 ft/sec

Area ratio = (14.375/29.25)2 = 0.2415; K = 0.415

Pe = 0.00298(0.1)[1.415(93.58)2-(22.6)2] = 3.54" H2O

P (calc’d) = 6.16 + 0.943 + 2.54 + 4.114 + 3.54

= 17.3" H2O

Inlet ldg. = 6328(80)144/60(7000)18(7.1875)

=1.342 lbs/sec·ft2

Therefore from Fig. 21

True P = 1.26 × 17.3 = 21.8" H2O

This again exceeds the specified 10" H2O maximum to a greater degree than the cyclone in Fig. 22, although it represents also a substantially smaller unit.

Since neither design in Figs. 22 or 23 will satisfy the specified pressure drop limitation, the only alternative lies in reducing inlet width without increasing inlet velocity, which requires multiple cyclones operating in parallel as illustrated in the design of Example C.

Example C. Suppose, in the previous Example B, performance must be achieved at a pressure loss less than 10" H2O requiring therefore multiple cyclones in parallel.

808 Fluidization, Solids Handling, and Processing

The pressure drop through 2 parallel units of Fig. 24 at Vi = 97.3 ft/ sec, NS = 5.08, Ve = 76.4 ft/sec, VB = 18.28 ft/sec, with inlet area ratio of 0.0108 so that K = 0.5, from Fig. 21:

Inlet contraction loss:

P(s-i)G = 0.00298(0.1)[1.5(97.3)2-(2)2] = 4.23" H2O

Solids acceleration loss:

P(s-i)P = 80(97.3)2/7000(167) = 0.648" H2O

Barrel loss:

dhi = 4(5.625)(13.875)/2(5.625+13.875) = 8"

Re = 8(97.3)0.1/12(0.02)0.000672 = 482,639; f = 0.00347

PB = 0.00347(23)0.1(5.08)(97.3)2/8(26.2) = 1.831" H2O

Reversal loss:

Pr = 0.1(97.3)2/335 = 2.826" H2O

Exit contraction loss:

Area ratio = (11.25/23)2 = 0.239; K = 0.415

Pe = 0.00298(0.1)[1.415(76.4)2-(18.28)2] = 2.362" H2O

P(calc’d) + 4.23 + 0.648 + 1.831 + 2.826 + 2.362

= 11.897" H2O

Inlet ldg.= 6328(80)144/2(60)7000(5.625)13.875

= 1.11 lbs/sec·ft2

Cyclone Design 809

From graph in Fig. 21:

True P = 1.29 × 11.875 = 15.35" H2O

This still exceeds the design specification of 10" H2O and could have been anticipated, since to meet this pressure drop criterion, Vi from Example A must not exceed about 75 ft/sec which would require 6 cyclones in parallel as opposed to the 2 in this Example C.

15.0 ALTERNATE APPROACH TO SOLVING EXAMPLE C

Assume the designer does not desire to use 6 cyclones in parallel, but must still meet all the specifications in Example B (i.e., cannot accept 15.35" H2O pressure drop) but is willing to accept 4 cyclones in parallel. Could 4 cyclones suffice (by a reduction in inlet velocity compensated in performance by an increase in exit gas velocity)?

Since true P must be ≤ 10" H2O, then P(calc’d) must equal or be < ~10/1.3 = 7.7" H2O. In Example C, P(calc’d) exclusive of exit loss equaled 9.535" H2O so that now the following 3 equations would need to be satisfied with n = 4:

Equations (3), (4) and (5) reduce to:

810 Fluidization, Solids Handling, and Processing

Since NS is a function of Ve, solving Eq. (8) for Vi and substituting in Eq.

(6) leads to

Vi = (3447.3/NSVe)2

which is satisfied with Ve = 87 ft/sec at NS = 4.81 so that

Vi = 67.86 ft/sec

LW = 4.75"

DO = 7.5"

DB = 19.5"

L = 32"

inlet height = 11.75" as summarized in Fig. 25.

Check on performance:

Dth = 12(25400)[9(0.02)0.000672(4.75)/12(3.14)87(85-0.1)4.81]0.5

= 6.3 microns (will meet specification) Check on pressure drop:

Inlet contraction loss:

P(s-i)G = 0.00298(0.1)[1.5(67.86)2-(2)2] = 2.057" H2O

Solids acceleration loss:

P(s-i)P = 80(67.86)2/7000(167) = 0.315" H2O

Cyclone Design 811

Barrel loss:

dhi = 4(4.75 × 11.75)/2(4.75 + 11.75) = 6.77"

Re = 6.77(67.86)0.1/12(0.02)0.000672 = 284,854; f = 0.0039

PB = 0.0039(19.5)0.1(67.86)24.81/26.2(6.77) = 0.95" H2O

Reversal loss:

Pr = 0.1(67.86)2/335 = 1.375" H2O

Exit contraction loss:

Area ratio = (7.5/19.5)2 = 0.148; K = 1.45

VB = 6328(4)144/4(3.14)60(19.5)2 = 12.7 ft/sec

Pe = 0.00298(0.1)[1.45(87)2-(12.7)2] = 3.223" H2O

P(calc’d) = 2.057 + 0.315 + 0.95 + 1.375 + 3.223

= 7.92" H2O

Inlet ldg. = 6328(80)144/4(60)7000(4.75)11.75

= 0.778 lbs/sec·ft2

From the graph in Fig. 21:

True P = 1.33(7.92) = 10.53" H2O

This barely exceeds the specified 10" H2O and can be brought to specification by a moderate increase in inlet height to 12" which reduces Vi to 66.6 ft/sec and results in the overall dimensions shown in Fig. 25.

The foregoing examples illustrate the relationships among the variables as they affect performance (collection efficiency) and pressure drop.

Cyclone Design 813

where: W = lbs of solids collected per second

d = dipleg I.D. inches (minimum)

α = collected solids’ angle of internal friction, degrees

ρB = collected solids’ bulk density, lbs/cu.ft.

ρG = density of gas entering cyclone, lbs/cu.ft.

g = gravitational field, ft/sec2

If the fluid stream is a gas, the last term in the above equation is essentially unity. Unless the cyclone itself is rotating or, for example, located on another planet, g can be taken as 32.2. If the bulk solids’ angle of internal friction is unknown, then taking an average value of 62 degrees, the equation reduces to:

d = (152.8 W/ρB)0.4

W is generally based conservatively on the assumption of 100% collection efficiency.

Pressure Balance. The extent to which the dipleg of a cyclone is filled with exiting solids depends on the pressure balance around the cyclone and its dipleg. Figure 26 shows a cyclone with pressure PB (psi) inside its barrel and with its dipleg immersed Hbed feet into a fluidized bed of density ρbed (lbs/ft3). A simple pressure balance treating the dipleg as one leg of a manometer gives the feet of solids in the dipleg (HDL) from:

PB + (HDL ρDL)/144 = PV = ( ρBed HBed)/144

where ρDL is the lbs/ft3 bulk density of solids in the dipleg and PV is the pressure in the vessel in the dilute phase above the bed, psi. (PV -PB) is determined from the cyclone pressure drop correlations in Fig. 21.

814 Fluidization, Solids Handling, and Processing

Figure 26. Cyclone pressure balance (no restrictive “valve” on dipleg end).

NOTATIONS