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0.5237
0.8866
0.9943
0.9999873
1.0000 x2
5.000
4.3911
4.0569
4.0032
4.0000
4.0000 (b) Starting guess is x1 = 4, x2 = 5
The NewtonRaphson method fails. After only one iteration, x1 = 6.2135, x2 = 10.7719
(c) Starting guess is x1 = 1, x2 = 1
The NewtonRaphson method fails. After only one iteration, x1 = 15.4910, x2 = 24.9880
(d) Starting guess is x1 = 8, x2 = 1
The NewtonRaphson method fails. After only one iteration, x1 = 5.1320, x2 = 0.4449
Note that by a homotopy continuation method, two roots in the positive domain are obtained:
x1 = 1, x2 = 4
x1 = 4.0715, x2 = 0.65027
If the NewtonRaphson method is damped by using a line search, the method fails less often. Exercise 10.10
Subject: Solving a set of two nonlinear equations by the NewtonRaphson method.
Given: Two nonlinear equations in x1 and x2:
x2
− x1 = 0
2
1
1
1−
+ exp 1
− 1 − 2 x1 + x2 = 0
4π
4π sin πx1 x2 −
exp 2 x1 Find: Solutions for three different starting guesses
Analysis:
(a) Starting guess is x1 = 0.4, x2 = 0.9
The NewtonRaphson method converges in 4 iterations:
Iteration
x1
x2
Start
0.4000 0.9000
1
0.1415 0.8394
2
0.2806 0.9073
3
0.2984 0.9026
4
0.2994 0.9030
(b) Starting guess is x1 = 0.6, x2 = 0.9
The NewtonRaphson method converges to a different solution in just 3 iterations.
Iteration
x1
x2
Start
0.6000 0.9000
1
0.4901 1.0235
2
0.4999 1.0002
3
0.5000 1.0000
(c) Starting guess is x1 = 1, x2 = 1
The NewtonRaphson method converges with some oscillations in 14 iterations.
Iteration
x1
x2
Iteration
x1
x2
Start
1.0000
1.0000
8
1.0042 0.4451
1
0.9091
0.6914
9
0.7160 1.2837
2
1.0720
0.2987
10
0.6514 0.9886
3
0.9727
0.5500
11
0.5135 1.0452
4
1.1633 0.0252
12
0.5045 1.0008
5
1.0511
0.3321
13
0.5001 1.0001
6
0.9512
0.6053
14
0.5000 1.0000
7
1.0876
0.2072 Exercise 10.11
Subject: Calculations for the first iteration of the BubblePoint method
Given: Feed of 1000 kmol/h of a bubblepoint mixture of 60 mol% methanol (M), 20 mol%
ethanol (E), and 20 mol% npropanol (P) is sent to the middle stage of a column with a total
condenser, 3 equilibrium stages in the column, and a partial reboiler, and operating at 1 atm. The
distillate rate = 600 kmol/h and the reflux rate = 2,000 kmol/h.
Assumptions: Ideal solutions with Raoult's law Kvalues. Constant molar overflow. Find: Liquidphase component mole fractions and stage temperatures from the first iteration of
the bubblepoint method.
Analysis: By material balance, the bottoms flow rate = 1000  600 = 400 kmol/h. The liquid
rate above the feed = 2000 kmol/h. The liquid rate below the feed = 3000 kmol/h. The boilup
rate over the entire column = 3000  400 = 2600 kmol/h. Based on the distillate rate, methanol
will be mainly in the distillate, with ethanol and npropanol mainly in the bottoms. The boiling
points of these three components are, respectively, 64.7oC, 78.4oC, and 97.8oC. As a first
approximation, assume a distillate temperature of 64.7oC and a bottoms temperature of (78.4 +
97.8)/2 = 88.1oC. Therefore, the initial guesses, assuming a linear temperature pro...
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 Spring '11
 Levicky
 The Land

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