Page 246 - Separation process principles 2

6.5 Stage Efficiency 211

lbmolh

performance data, given below, for a bubble-cap tray absorber Lean Rich
Component Gas Oil Total Out Total In
located in a Texas petroleum refinery, were reported by Drickamer
and Bradford [6]. Based on these data, back-calculate the overall c 1 439.9 7.0 446.9 447.5
stage efficiency for n-butane and compare the result with both the
c2 77.1 6.1 83.2 83.2
~rickamer-Bradford and O'Connell correlations. Lean oil and rich c; 40.4 8.7 49.1 49.2
gas enter the tower; rich oil and lean gas leave the tower. c3 170.4 43.0 213.4 213.8
c; 18.4 17.5 35.9 35.9
Performance Data
nC4 35.0 35.0 70.0 70.0
Number of plates 16 nC5 5.7 21.1 26.8 28.4
Plate spacing, in. 24 nC6 - - - -
18.0
0.0
18.0
18.0
Tower diameter, ft 4
786.9 156.4 943.3 946.0
Tower pressure, psig 79
Lean oil temperature, "F 102 Again, we see excellent agreement. The largest difference is 6% for
I .
Rich oil temperature, OF 126 pentanes. Plant data are not always so consistent.
Rich gas temperature, OF 108 For the back-calculation of stage efficiency from the perfor-
Lean gas temperature, "F 108 mance data, the Kremser equation is applied to compute the num-
Lean oil rate, lbmolh 368 ber of equilibrium stages required for the measured absorption of
Rich oil rate, lbmolh 525.4 n-butane.
Rich gas rate, lbmolh 946
35
Lean gas rate, lbmolh 786.9 Fraction of nC4 absorbed = - = 0.50
70
Lean oil molecular weight 250
Lean oil viscosity at 116OF, cP 1.4 AN+' - A
From (6-1 3),
Lean oil gravity, "API 2 1 0.50 = AN+, - 1
I Stream Compositions, MoI% where A = absorption factor = -
L
KV
Component Rich Gas Lean Gas Rich Oil Lean Oil
Because L and V vary greatly through the column, let
CI
c2 368 + 525.4
c, L = average liquid rate = = 446.7 lbmolh
2
c3
c,= and let
nC4 946 + 786.9
V = average vapor rate = = 866.5 lbmollh
nCs 2
nC6
Oil absorbent Assume average tower temperature = the average of inlet and
outlet temperatures = (102 + 126 + 108 + 108)/4 = 111°F. Also
Totals
assume that the viscosity of the lean oil at 116OF equals the
viscosity of the rich oil at 11 1°F. Therefore, p = 1.4 cP.
SOLUTION Assume the ambient pressure is 14.7 psia. Then
Before computing the overall stage efficiency for n-butane, it is Tower pressure = 79 + 14.7 = 93.7 psia
worthwhile to check the consistency of the plant data by examining
From Figure 2.8, at 93.7 psia and 11l0F, KnC4 = 0.7. Thus,
the overall material balance and the material balance for each com-
ponent. From the above stream compositions, it is apparent that the
compositions have been normalized to total 100%.
The overall material balance is
0.736N+' - 0.736
Total flow into tower = 368 + 946 = 1,314 lbmolh Therefore, 0.50 =
0.736N+' - 1
Total flow from tower = 525.4 + 786.9 = 1,3 12.3 lbmolth
Solving, N = N, = 1.45
These two totals agree to within 0.13%. This is excellent
agreement. From the performance data, N, = 16

The component material balance for the oil absorbent is 1.45
=
From (6-2 l), E, = - 0.091 or 9.1%
16
Total oil in = 368 lbmoVh
Total oil out = (0.7024)(525.4) = 369 lbmolh Equation (6-22) is applicable to n-butane, because that compo-
nent is absorbed to the extent of about 50% and thus can be consid-
These two totals agree to within 0.3%. Again, this is excellent
ered one of the key components. Other possible key components
agreement.
are butenes and n-pentane.
Component material balances for other hydrocarbons from
spreadsheet calculations are as follows. From (6-22), E, = 19.2 - 57.8 log(1.4) = 10.8%