Unformatted text preview: file, for the
tear variables and the total liquid flow rates are:
Stage j
1 (Total condenser)
2
3
4
5 (Partial reboiler) Vj , kmol/h
0
2600
2600
2600
2600 Tj , oC
64.7
70.6
76.4
82.3
88.1 Lj , kmol/h
2000
2000
3000
3000
400 Compute Kvalues at 1 atm and the above temperatures for each component by Eq. (244),
K = Pis/P. From the 7th edition of Perry's Handbook, on page 250 to 254, in Table 26, vapor
pressure data are fitted to an extended ClausiusClapeyron equation:
P s = exp C1 + C2
+ C 3 ln T + C 4T C 5 , P s is in Pa and T is in K , where the constants are:
T Component
M
E
P C1
81.768
74.475
88.134 C2
6876
7164.3
8498.6 C3
8.7078
7.327
9.0766 C4
7.1926E6
3.1340E6
8.3303E18 C5
2
2
6 Exercise 10.11 (continued)
Analysis: (continued)
Computed Kvalues at the five assumed stage temperatures for P = 101300 Pa are:
Stage
Temperature, K
Kvalues:
M
E
P 1
337.9 2
343.7 3
349.6 4
355.4 5
361.3 1.0086
0.5705
0.2505 1.2639
0.7309
0.3294 1.5702
0.9272
0.4281 1.9350
1.1654
0.5504 2.3665
1.4521
0.7002 A plot of the Kvalues as a function of temperature is shown on the next page.
The component material balance equations, combined with the equilibrium equations to
eliminate vaporphase mole fractions, for each of the 5 stages are as follows: Stage 1: − D + L1 xi ,1 + V2 Ki ,2 xi ,2 = 0
− 2600 xi ,1 + 2600 Ki ,2 xi ,2 = 0 Stage 2: L1 xi ,1 − V2 Ki ,2 + L2 xi ,2 + V3 Ki ,3 xi ,3 = 0
2000 xi ,1 − 2600 Ki ,2 + 2000 xi ,2 + 2600 Ki ,3 xi ,3 = 0 Stage 3: (3) L3 xi ,3 − V4 Ki ,4 + L4 xi ,4 + V5 Ki ,5 xi ,5 = 0
3000 xi ,3 − 2600 Ki ,4 + 3000 xi ,4 + 2600 Ki ,5 xi ,5 = 0 Stage 5: (2) L2 xi ,2 − V3 Ki ,3 + L3 xi ,3 + V4 Ki ,4 xi ,4 = − F3 zi ,3
2000 xi ,2 − 2600 Ki ,3 + 2000 xi ,3 + 2600 Ki ,4 xi ,4 = −1000zi ,3 Stage 4: (1) (4) L4 xi ,4 − V5 Ki ,5 + B xi ,5 = 0
3000 xi ,4 − 2600 Ki ,5 + 400 xi ,5 = 0 (5) For each component, the Kvalues for each stage are substituted from the table above. The
resulting equations form 3 tridiagonal matrix equations, one each for each of the 3 components.
These equations are as follows, as produced from a spreadsheet: Exercise 10.11 (continued)
Analysis: (continued) Analysis: (continued) Exercise 10.11 (continued) ___________________________________________________________________________________________ Exercise 10.11 (continued)
Analysis: (continued)
A new set of stage temperatures is computed by running bubblepoint temperature
calculations to find a temperature that satisfies Eq. (1021), using the normalized mole fractions
with the vapor pressure equations and a pressure of 1 atm. The results are as follows compared
to the initially assumed values: Stage j
1 (Total condenser)
2
3
4
5 (Partial reboiler) Tj , o C
Tj , o C
initially from 1st iteration
assumed
64.7
67.2
70.6
69.6
76.4
73.1
82.3
77.8
88.1
84.1 Exercise 10.12
Subject: Solving a block tridiagonal matrix equation.
Given: 9 linear equations in 9 unknowns. Find: Values of the 9 unknowns using the block form of the Thomas algorithm.
Analysis: To form...
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 Spring '11
 Levicky
 The Land

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