CHE 425
__________________
LAST NAME, FIRST
Problem set #4
1-8 Copy the program UnitOp4.exe from the CHE 425 class distribution folder to your flash
or H drive. You can also download the program from the website:
https://www.csupomona.edu/~tknguyen/che425/homework.htm.
Run the program and choose problem 1-8 in any order. Solve the problems with the data
provided by the program, copy the problem statement to Word. The program will check your
answer and provide an answer code when you click on “Check”. Copy the answer code and
paste them after the problem statement. You need to present all your work with a diagram in
details to get full credit. Your work should look like this:
CHE425
Problem set #1
NGUYEN, THUAN
1. A vapor at the dew point and 200 kPa containing a mole fraction of 0.40 benzene (1) and
0.60 toluene (2) and 100 kmol total is brought into contact with 110 kmol of a liquid at the
boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams
are contacted in a single stage, and the outlet streams leave in equilibrium with each other.
Assume constant molar overflow, calculate the amounts and compositions of the exit
streams.
Data: Vapor pressure, Psat, data: ln Psat = A  B/(T + C), where Psat is in kPa and T is in K.
Compound
Benzene (1)
Toluene (2)
A
14.1603
14.2515
B
2948.78
3242.38
C
 44.5633
 47.1806
Problem 1: Correct, Code =4313336481
Solution
x0 = 0.3, L0 = 110 kmol, x and y are mole fraction of benzene in the liquid and vapor phase,
respectively.
L0
V1
Equilibrium stage
L1
V2
y2 = 0.4, V2 = 100 kmol
For CMO, L1 = L0 = 110 kmol, V2 = V1 = 100 kmol. Making a balance on benzene gives
1/1
L0x0 + V2y2 = L1x1 + V1y1
110(0.30) + 100(0.40) = 110x1 + 100y1
11x1 + 10y1 = 7.3  y1 = 0.73  1.1x1
Since the two streams V1 and L1 are in equilibrium, we have
y1 P1sat
=
 200y1 = x1exp(14.1603  2948.78/(T  44.5633))
x1 200
1  y1 P2sat
=
 200(1  y1) = (1  x1)exp(14.2515  3242.38/(T  47.1806))
1  x1 200
(E-1)
(E-2)
(E-3)
The three equations (E-1,2,3) can be solved for T, x1, and y1 either by graphical or numerical method.
4.2 Construct a Txy and a xy phase equilibrium diagram for n-hexane (1) and n-octane (2) at
14.7 psia. The equilibrium data for these two species are given:
T(R) = 615.67, x =
T(R) = 619.51, x =
T(R) = 623.35, x =
T(R) = 627.19, x =
T(R) = 631.03, x =
T(R) = 634.87, x =
T(R) = 638.71, x =
T(R) = 642.55, x =
T(R) = 646.38, x =
T(R) = 650.22, x =
T(R) = 654.06, x =
T(R) = 657.90, x =
T(R) = 661.74, x =
T(R) = 665.58, x =
T(R) = 669.42, x =
T(R) = 673.25, x =
T(R) = 677.09, x =
T(R) = 680.93, x =
T(R) = 684.77, x =
T(R) = 688.61, x =
T(R) = 692.45, x =
T(R) = 696.29, x =
T(R) = 700.13, x =
T(R) = 703.96, x =
T(R) = 707.80, x =
T(R) = 711.64, x =
1.000, y =
0.934, y =
0.872, y =
0.814, y =
0.759, y =
0.708, y =
0.660, y =
0.615, y =
0.573, y =
0.532, y =
0.494, y =
0.458, y =
0.424, y =
0.391, y =
0.359, y =
0.329, y =
0.301, y =
0.273, y =
0.247, y =
0.221, y =
0.196, y =
0.172, y =
0.149, y =
0.126, y =
0.104, y =
0.083, y =
1.000
0.989
0.978
0.965
0.951
0.937
0.921
0.903
0.885
0.865
0.843
0.820
0.796
0.769
0.741
0.710
0.678
0.643
0.606
0.567
0.524
0.479
0.432
0.381
0.326
0.269
T(R) = 715.48, x =
T(R) = 719.32, x =
T(R) = 723.16, x =
T(R) = 727.00, x =
0.061, y =
0.041, y =
0.020, y =
0.000, y =
0.208
0.143
0.073
0.000
Note: Use Matlab to plot the diagrams with tie lines and label the figures with your name
using the Title command. (See example 4.1-1).
7. Read the problem statement in the program UnitOp4.exe. You can follow the
following procedure assuming 100 kmol feed.
- Assume a temperature
- Calculate equilibrium ratios
- Calculate V/F
- Calculate y1
- Calculate the amount of toluene in the vapor. If the calculate value is
(0.1)(0.6)(100) = 6 kmol, then temperature guess is correct. Otherwise, guess
another T, and repeat the above steps.