ppk a tugas 1 hendrik panthron 38980
DESCRIPTION
ASPEN SIMULATIONTRANSCRIPT
Homework #1 no.1
Ethanol 97% Dilution Simulation Hendrik Panthron Pangarso Mursid , Chemical Plant Design Class A , 11/320155/TK/38980
ASPEN Simulation Laboratory, Chemical Engineering Universitas Gadjah Mada Jl. Grafika no.2 Bulaksumur, Sleman, Yogyakarta.
Email : [email protected] Problem Identification.
In Pharmatical Industry, there is a ethanol 97% that we want dilute until have ethanol 70%. We need to find how much water added to ethanol feed to make this ethanol solution at concentration we want. Mixing condition at 25oC and atmospheric pressure. Analysis and Hand Calculation. Known:
F = 200 L/day XF = 0,97 XL = 0,7 We assumed that process going on steady state condition and well mixed.
Component Mass Balance of Ethanol in Mixer
Rate of mass in – Rate of mass out = Rate of mass accumulation XF . 𝜌mixF . F − XL. 𝜌mixL . L = 0 0,7. 𝜌mixL . L = 0,97. 𝜌mixF . F
L =𝜌mixF .0,97 .F
𝜌mixL .0,7
L =𝜌mixF .0,97 .200 L/day
𝜌mixL .0,7
L = 277,1429𝜌mixF
𝜌mixL (1)
Mixed Density Calculation. As we know, ethanol-water system is non-ideal mixture. The reason is hydrogen bonds form between the compounds and pull each other closer. So if you combine volumes of both subtances the resulting volume is slighlty less than you would expect because the water being able to fit in the small voids that the larger
ethanol molecules create. For simplicity, we can neglect this phenomenom. We assumed that 𝜌mixF
𝜌mixL is
one or almost one. So the Equation becomes : L = 277,1429 L/day W = L – W W = 277,1429 – 200 W = 77, 1429 L/day If we use dilution formula, VF. MF = VL. ML
VL = VF.MF
ML= 200
L
day.
0,97
0,7= 277,1429 L/day same with W
In other words, ethanol dilution ratio in 2,6 : 1 . .
*From property analysis with ASPEN, we found the value of ethanol density at 25oC is 785,8957 kg/m3
Simulation with ASPEN Plus v7.2
1. The first step is choose the mixer type for ethanol dilution. Choose ”Triangle” type. Then, move the cursor to the desired location on the main flowsheet window and click mouse button. Then, press Ctrl+ M to rename it ; “MIXER”
2. Next you have to add stream to the block, click on the small arrow to the right of the STREAM button (at the lower lef corner of your stream), and choose Material stream by clicking on it. The mixer now show arroes where the stream can be connected; red arrows requred streams and blue just optional streams.
3. To set up the feed stream to the column, move the crosshair on the top of red feed position and
left click once. Now, move the mouse to the left and click again. You should now have a defined feed stream (EtOH 97%), then do it again for solvent stream (Water) and product stream (EtOH 70%).
4. Now that you have defined the unit oper unit operations to be simulated and set up the streams into and out of the process, you must enter the rest of the information required to complete the simulation. Within Aspen Plus, the easiest way to find the next step is to use one of the following equivalent commands: (1) click the Next icon (blue N ->); (2) find 'Next' in the Tools menu; or (3) use keyboard shortcut F4. Any option will open the Data Browser. In the Data Browser, you are required to enter information at locations where there are red semicircles. When you have finished a section, a blue checkmark will appear. However, providing some 'Setup' settings is often desirable. We can also named our simulation then pick the unit we used as shown below. For example, i choose metric cm bar (METCBAR) unit. Then input our components.
5. Choose Property Method, NTRL Method.
*We choose NTRL method caused by this is the one of activity coefficient models. The activity represent the deviation of the mixture form ideality . In other words, non-ideal solution. 6. Now, We press F4 at Stream Folder. We input the parameter condition that we known as well for EtOH 97 % and water. Next, thon forget we choose “Liquid-only”
6. Set Bubble Prop-sets and Mass-fraction
7. Click Run or F5. Wait until results available, no warning or error.
The NTRL Results.
The results no proper to objective variable, ethanol 70% that we want. Some of error, because of the assumption at first time simulation. So, there is trial and error method and get the result as shown below.
So, the water needed at 62,5 L/day to make ethanol 70% from ethanol 97%. The dilution ratio is 3,2 : 1. Then, we look bubble point each stream. Along concentration ethanol decrease from 97% to 70%, the bubble point of ethanol just arise from the shown results, 77,832oC to 79,578oC. In this case, the hydroxyl group makes ethanol polar at that carbon, which allows the ethanol form polar bonds with water molecules. So, if we add more water, there is increase the polar bonds with ethanol and water molecules. The polar bonds is type of strong bonds. The implication is the increasing boiling point that means molecules are breaking away more hardly then the do in high purity (97%).
From figure upper, we know that at 0,7 mole fraction of ethanol, the gibbs free energy of mixing for ethanol-water is -0,2325052 kcal/mol.
Gibbs free energy of mixing for ETHANOL/WATER
Massfrac ETHANOL
DG
MIX
kc
al/m
ol
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
-0,2
-0,1
5-0
,1-0
,05
0,0
25,0 C
Homework #1 no.2
Multicomponent Mixture Simulation Hendrik Panthron Pangarso Mursid , Chemical Plant Design Class A , 11/320155/TK/38980
ASPEN Simulation Laboratory, Chemical Engineering Universitas Gadjah Mada Jl. Grafika no.2 Bulaksumur, Sleman, Yogyakarta.
Email : [email protected] Problem Identification. There is multicomponent system, consist of water, ethanol, and chloroform. We analyze the equilibrium behaviour or non-ideal characteristic (azeotrope) of this system with various property method. (NTRL, PENG-ROB. And UNIQUAC) The NTRL Results
T-xy for ETHANOL/WATER
Liquid/Vapor Molefrac ETHANOL
Tem
per
ature
C
0,0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 0,4 0,45 0,5 0,55 0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 1,0
70,0
80,0
90,0
100,0
110,0
120,0
130,0
140,0
150,0
160,0
170,0
180,0
190,0
200,0
210,0
220,0
T-x 1,0 atm
T-x 10,0 atm
T-x 20,0 atm
T-y 1,0 atm
T-y 10,0 atm
T-y 20,0 atm
P-xy for ETHANOL/WATER
Liquid/Vapor Molefrac ETHANOL
Pre
ssu
re atm
0,0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0
0,2
0,4
0,6
0,8
1,0
P-x 25,0 C
P-x 50,0 C
P-x 75,0 C
P-y 25,0 C
P-y 50,0 C
P-y 75,0 C
Physical Property Model: NRTL Valid Phase: VAP-LIQ
Mixture Investigated For Azeotropes At A Pressure Of 101325 N/SQM
Comp ID Component Name Classification Temperature
CHCL3 CHLOROFORM Stable Node 61,10 C
ETHANOL ETHANOL Stable Node 78,31 C
WATER WATER Stable Node 100,02 C
3 Azeotropes Sorted by Temperature
01
Number Of Components: 2 Temperature 59,29 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
CHCL3 0,8441 0,9335
Residue curve for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,90.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1 CURVE 1,0 (PRES = 1.01325 bar)
CURVE 2,0 (PRES = 1.01325 bar)
CURVE 3,0 (PRES = 1.01325 bar)
CURVE 4,0 (PRES = 1.01325 bar)
CURVE 5,0 (PRES = 1.01325 bar)
CURVE 6,0 (PRES = 1.01325 bar)
CURVE 7,0 (PRES = 1.01325 bar)
CURVE 8,0 (PRES = 1.01325 bar)
CURVE 9,0 (PRES = 1.01325 bar)
CURVE 10,0 (PRES = 1.01325 bar)
Ternary map for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Phase envelope 1 (PRES = 1,01325 bar)
Tie line 1 (PRES = 1,01325 bar)
Tie line 2 (PRES = 1,01325 bar)
Tie line 3 (PRES = 1,01325 bar)
Tie line 4 (PRES = 1,01325 bar)
Tie line 5 (PRES = 1,01325 bar)
Azeotrope (PRES = 1,01325 bar)
ETHANOL 0,1559 0,0665
02
Number Of Components: 2 Temperature 48,61 C
Homogeneous Classification: Unstable Node
MOLE BASIS MASS BASIS
CHCL3 0,6774 0,9330
WATER 0,3226 0,0670
03
Number Of Components: 2 Temperature 78,15 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
ETHANOL 0,8952 0,9562
WATER 0,1048 0,0438
The PENGROB Results
Residue curve for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
CURVE 1,0 (PRES = 1.01325 bar)
CURVE 2,0 (PRES = 1.01325 bar)
CURVE 3,0 (PRES = 1.01325 bar)
CURVE 4,0 (PRES = 1.01325 bar)
CURVE 5,0 (PRES = 1.01325 bar)
CURVE 6,0 (PRES = 1.01325 bar)
CURVE 7,0 (PRES = 1.01325 bar)
CURVE 8,0 (PRES = 1.01325 bar)
CURVE 9,0 (PRES = 1.01325 bar)
CURVE 10,0 (PRES = 1.01325 bar)
CURVE 11,0 (PRES = 1.01325 bar)
CURVE 12,0 (PRES = 1.01325 bar)
CURVE 13,0 (PRES = 1.01325 bar)
CURVE 14,0 (PRES = 1.01325 bar)
Ternary map for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Phase envelope 1 (PRES = 1,01325 bar)
Tie line 1 (PRES = 1,01325 bar)
Tie line 2 (PRES = 1,01325 bar)
Tie line 3 (PRES = 1,01325 bar)
Tie line 4 (PRES = 1,01325 bar)
Tie line 5 (PRES = 1,01325 bar)
Azeotrope (PRES = 1,01325 bar)
Physical Property Model: PENG-ROB Valid Phase: VAP-LIQ
Mixture Investigated For Azeotropes At A Pressure Of 101325 N/SQM
Comp ID Component Name Classification Temperature
CHCL3 CHLOROFORM Saddle 61,47 C
ETHANOL ETHANOL Stable Node 76,98 C
WATER WATER Stable Node 101,90 C
2 Azeotropes Sorted by Temperature
01
Number Of Components: 2 Temperature 57,55 C
Homogeneous Classification: Unstable Node
MOLE BASIS MASS BASIS
CHCL3 0,6596 0,9277
WATER 0,3404 0,0723
02
Number Of Components: 2 Temperature 72,78 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
ETHANOL 0,5959 0,7904
WATER 0,4041 0,2096
The UNIQUAC Results
Residue curve for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
CURVE 1,0 (PRES = 1.01325 bar)
CURVE 2,0 (PRES = 1.01325 bar)
CURVE 3,0 (PRES = 1.01325 bar)
CURVE 4,0 (PRES = 1.01325 bar)
CURVE 5,0 (PRES = 1.01325 bar)
CURVE 6,0 (PRES = 1.01325 bar)
CURVE 7,0 (PRES = 1.01325 bar)
CURVE 8,0 (PRES = 1.01325 bar)
CURVE 9,0 (PRES = 1.01325 bar)
CURVE 10,0 (PRES = 1.01325 bar)
CURVE 11,0 (PRES = 1.01325 bar)
Physical Property Model: UNIQUAC Valid Phase: VAP-LIQ
Mixture Investigated For Azeotropes At A Pressure Of 101325 N/SQM
Comp ID Component Name Classification Temperature
CHCL3 CHLOROFORM Stable Node 61,10 C
ETHANOL ETHANOL Stable Node 78,31 C
WATER WATER Stable Node 100,02 C
3 Azeotropes Sorted by Temperature
01
Number Of Components: 2 Temperature 59,67 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
CHCL3 0,8737 0,9472
ETHANOL 0,1263 0,0528
02
Number Of Components: 2 Temperature 48,78 C
Homogeneous Classification: Unstable Node
MOLE BASIS MASS BASIS
CHCL3 0,6916 0,9370
WATER 0,3084 0,0630
03
Number Of Components: 2 Temperature 78,16 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
ETHANOL 0,8999 0,9583
Ternary map for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Phase envelope 1 (PRES = 1,01325 bar)
Tie line 1 (PRES = 1,01325 bar)
Tie line 2 (PRES = 1,01325 bar)
Tie line 3 (PRES = 1,01325 bar)
Tie line 4 (PRES = 1,01325 bar)
Tie line 5 (PRES = 1,01325 bar)
Azeotrope (PRES = 1,01325 bar)
WATER 0,1001 0,0417
The WILSON Results
The ternary diagram error.
Physical Property Model: WILSON Valid Phase: VAP-LIQ
Mixture Investigated For Azeotropes At A Pressure Of 101325 N/SQM
Comp ID Component Name Classification Temperature
CHCL3 CHLOROFORM Saddle 61,10 C
ETHANOL ETHANOL Stable Node 78,31 C
WATER WATER Stable Node 100,02 C
2 Azeotropes Sorted by Temperature
01
Number Of Components: 2 Temperature 59,33 C
Homogeneous Classification: Unstable Node
MOLE BASIS MASS BASIS
CHCL3 0,8543 0,9383
ETHANOL 0,1457 0,0617
02
Number Of Components: 2 Temperature 77,98 C
Homogeneous Classification: Saddle
MOLE BASIS MASS BASIS
ETHANOL 0,8722 0,9458
WATER 0,1278 0,0542
Residue curve for CHCL3/ETHANOL/WATER
Molefrac CHCL3
Mole
frac
WA
TE
R
Molefra
c ET
HA
NO
L
0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
CURVE 1,0 (PRES = 1.01325 bar)
CURVE 2,0 (PRES = 1.01325 bar)
CURVE 3,0 (PRES = 1.01325 bar)
CURVE 4,0 (PRES = 1.01325 bar)
CURVE 5,0 (PRES = 1.01325 bar)
CURVE 6,0 (PRES = 1.01325 bar)
CURVE 7,0 (PRES = 1.01325 bar)
CURVE 8,0 (PRES = 1.01325 bar)
CURVE 9,0 (PRES = 1.01325 bar)
CURVE 10,0 (PRES = 1.01325 bar)
CURVE 11,0 (PRES = 1.01325 bar)
CURVE 12,0 (PRES = 1.01325 bar)
The difference between Ternary diagram and Residue diagram.
Residual curve describes the change in the composition of the liquid phase to a mixture of chemicals with a particular condition. A residue represents the liquid residue composition with time as the result of a simple, one stage batch distillation. Occasionally, in total reflux condition. Ternary maps are maps with equilateral triangular scheme that describes the composition of a mixture of the three components forming. Residual curve maps and maps ternary function to determine the feasibility of the process of separating a mixture, so that the residual curve maps and maps ternany an important basis to be considered in designing a distillation process. Ternary residue curve maps and the maps are usually used to analyze a ternary mixture that can not be easily separated by distillation because azetrop formed.