Unformatted text preview: 5
20
3.600
nC 5
35
0.000
Total:
100
46.725 Bottoms
0.000
0.000
1.875
16.400
35.000
53.275 The results with Kvalues and relative volatilities are as follows, using the SRK equation of state
for Kvalues and the bubblepoint equation, (412): Component
C3
iC 4
nC 4
iC 5
nC 5 K,
distillate,
140oF
2.100
1.056
0.818
0.401
0.329 αi,iC5 , 140oF K, bottoms,
230oF αi,iC5 , 230oF 5.24
2.63
2.04
1.00
0.82 3.880
2.260
1.863
1.068
0.922 3.63
2.12
1.74
1.00
0.86 αi,iC5 ,
geometric
mean
4.36
2.36
1.88
1.00
0.84 Exercise 9.17 (continued) Analysis: (continued)
(a) Using the Fenske equation (912), with nC4 as the LK and iC5 as the HK,
log
N min = d nC4 biC5
d iC5 bnC4 log log ( α nC4,iC5 )avg = ( 23.125 ) (16.4 )
( 3.6 ) (1.875 )
log1.88 = 6.38 (b) To compute the distribution of nonkey components at total reflux, use for the lighter than
light key components, LLK, Eq. (915), and use Eq. (916) for the heavier than heavy key, HHK,
with iC5 as the reference component, r, and the above value of Nmin.
fi
fi
=
(1)
Thus, for the LLK, i, bi =
d iC5 N min 1 + 3.6 α 6.38
i,iC5
1+
α i,iC5
16.4
b
iC5 fi
di = For the HHK, i, 1+ d iC 5
biC5
d iC5
biC5 α iN min
,iC 5 3.6 6.38
α i ,iC5
16.4
=
3.6 6.38
1+
α i ,iC5
16.4
fi α iN min
,iC5 (2) Using the above geometric mean values of αi, iC5 , followed by use of Eqs. (1) and (2) with the
material balance, fi = di + bi , the following results are obtained:
lbmol/h:
Component
Feed Distillate Bottoms
C3
5
4.998
0.002
iC 4
15
14.720
0.280
nC 4
25
23.125
1.875
iC 5
20
3.600
16.400
nC 5
35
2.356
32.644
Total:
100
48.799
51.201
(c) Because the split of the LK and HK components is not sharp and because iC4 boils close to
nC4, and iC5 boils close to nC5, it is possible that both iC4 and nC5 will distribute at minimum
reflux. It is questionable that C3 will distribute. Nevertheless, first assume a Class 1
Underwood minimum reflux and use the following rearrangement of Eq. (921) for a feed at the
bubble point: F
Lmin = d nC 4
f nC 4 − α nC 4 ,iC5
α nC4 ,iC5 − 1 d iC5
f iC5 100
= 23125
.
3.6
− 188
.
25
20
188 − 1
. = 66.7 l Exercise 9.17 (continued)
Analyze: (c) (continued)
Now, use the Class 2 Underwood equation, Eq. (928),which for a bubblepoint feed is:
α i ,iC5 zi , F
.
.
4.36(0.05) 2.36(015) 188(0.25) 1.00(0.20) 0.84(0.35)
= 1− q = 0 =
+
+
+
+
.
4.36 − θ
2.36 − θ
188 − θ
1.00 − θ
0.84 − θ
i α i ,iC5 − θ (3) There are four roots to Eq. (3), which from a spreadsheet calculation are:
θ = 0.9248, 1.317, 2.173, 3.996
Assume that the C3 does not distribute. Then, the 3.996 root is not needed and dC3 = 5 lbmol/h.
Eq. (929) is applied in the following form for each of the three remaining values of θ, in terms
of the three unknowns, diC4 , dnC5 , and Lmin. The form used is obtained by multiplying Eq. (929)
by the total distillate rate, D, which converts distillate mole fractions to component distillate
rates, and the reflux ratio, R to L.
2.36 (d iC 4 ) 1.88 (23125)
0.84 (d nC5 )
4.36 (5)
.
1.00 (3.6)
D + Lmin =
+
+
+
+
4.36 − 0.9248 2.36 − 0.9248 188 − 0.9248 1.00 − 0.9248 0.84 − 0.9248
.
= 6.346 + 1644(d iC4 ) + 45514 +...
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 Spring '11
 Levicky
 The Land

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