Page 126 - Applied Process Design For Chemical And Petrochemical Plants Volume II

Distillation 115

which is estimated by assuming equilibrium at the bottom The other estimates in column 5 of Table 8-10 are
of the tower. Se is estimated from ST and SB. Note that ST calculated in a similar manner. Note that the Cjs are
= S AT and SB = AB. A trial and error solution for the assumed to be completely absorbed for this first iter-
number of theoretical stages is effected by using Equation ation.
8-217 (or 8-216 and Figure 8-58). Values of v1 for the non- 2. Inlet rate of rich oil.
keys can be calculated by using these relationships direct- The maximum mol fraction n-C4 in the leaving liq-
ly with t!he calculated value of n. If necessary, the entire uid is taken as that in equilibrium with the incoming
procedure can be repeated, using the better estimates of gases. Thus, for n-C4,
the component flowrates in the leaving streams that were
estimated in the first iteration.

Example 8-33: Absorption of Hydrocarbons with Lean Oil
A material balance of 1-44 yields
The gas stream shown in Table 8-10 is fed to an isother-
mal absorber operating at 90°F and 75 psia. 90% of the n- 1, = 0.002 Lo + (0.9) 33.6
butane is to be removed by contact with a lean oil stream With the absorption efficiencies assumed above,
consisting of 98.7 mol% non-volatile oil and the light com-
ponents shown in Column 2 of Table 8-10. Estimate the Ln = Lo + 126.34
composition of the product streams and the required Combining the above equations yields the estimate
number of theoretical stages if an inlet rate of 1.8 times of the minimum lean oil rate:
the minimum is used.
Solution: 0.002 (L,, )min + (0.9) 33.6 = 0.02617
(Lo)min + 126.38
1. Initial estimates of extent of absorption of non-keys. or (Lo)min = 1,114.3 mols/hr
As a rough approximation, assume the fraction
absorbed of a given component is inversely propor- Thus, Lo = 1.8 (1,114.3) = 2,005.7 mols/hr
tional LO its K value (Equation 8-202). For example: 3. Effective Absorption Factor for 1144.
n-C4, in off gas = 33.6 - (0.9) (33.6) = 3.36 The total rich oil out is estimated as
L, = 1,975 + 2,005.7 - 1,848.66 = 2,132.04
C1, in off gas = 1,639 - (0.9) (1,639.2) = 1,616.6
The absorption factors are calculated by
AT = 2,005.7/(.65) (1,848.66) = 1.669
C2, in off gas 5 165.8 - (0.9) - 165.8 = 152.5
AB = 2,132.04/(.65) (1,975) = 1.661

Table 8-10
Calculation Summaries for Example 8-33

Initial
Estimate Off-Gas
Lean oil Of Net -.
90°F Feed m Amount Initial AfterFirst Aftersecond AfterThird Rich oil
K Gas In (Mol. Absorbed Estimate Iteration Iteration Iteration out
Component 75 psia (mols/h) Fraction) (mols/hr) (mols/h) (mols/hr) (mols/hr) (mols/hr) (mols/hr)
. ... -. .~
Methane 42.5 1,639.2 - 22.6 1,616.6 1,597.5 1,598.4 1,598.4 40.8
Ethane 7.3 163.8 - 13.3 152.5 141.2 141.8 141.8 24.0
Propane 2.2.5 94.9 - 24.67 70.23 49.84 30.76 50.76 - 44.14
i-Butane 0.88 17.8 0.001 11.83 5.97 3.10 3.13 3.13 1.91 16.58
n-Butane 0.63 33.6 0.002 30.24 3.36 3.36 3.36 3.36 3.83 34.07
i-Pentane 0.28 7.9 0.004 7.9 0 2.08 2.03 2.03 7.66 13.53
n-Pentane 0.225 15.8 0.006 15.8 0 2.51 2.44 2.44 11.49 24.85
-
--
Heavy Oil 0 0.987 - 0 0 0. 0. 1,889.83 1,889.83
0
-
1,975.0 1.000 126.34 1,848.66 1,799.57 1,801.92 1,801.92 1,914.72 2,087.78
.- ~ .- _- ~ .~ . .. . .