chemical and phase
TRANSCRIPT
-
7/27/2019 Chemical and Phase
1/21
1
CHEMICAL AND PHASE
EQUILIBRACOURSEWORK
Name: Yap Wei Khang
Student ID:UNIMKL-007852
Lecturer: Dr.Sergey SportarModule Code:H82CPE
Due Date:13th December 2011
-
7/27/2019 Chemical and Phase
2/21
2
Part#1: Example Sheet #1, Part B
Question 8
------
{ [ ()]}
() () *
+
(
)
[ ( ) ()]
() ()
() (
)
-
7/27/2019 Chemical and Phase
3/21
3
Question 9
To determine value of , the following equation must be applied ()
can be found by plotting graph of specific volume of gas (
) versus temperature (
). Value
is determined by calculating the gradient of the graph.From steam table, at 100 and 1 atm, value of is 1.673 Table 1: Values specific volume of superheated steam, , at different temperatures (K) at a constantpressure of 1 atm
Temperature (K) Specific volume,
(
)
373.15 1.673398.15 1.793
423.15 1.912
448.15 2.039
473.15 2.145
498.15 2.260
523.15 2.375
548.15 2.490
Figure 1: Graph of specific volume, vg, against temperature, K, is shown as above.
Comment: Specific volume of superheated steam increases with temperature (K) and behaves in a
linearity pattern.
y = 0.0047x - 0.0592
0
0.5
1
1.5
2
2.5
3
370 420 470 520 570
Specificvolume(m3/kg)
Temperature (K)
Specific Volume(Vg) vs. Temperature (T)
Data Points
Linear (Data Points)
-
7/27/2019 Chemical and Phase
4/21
4
From Figure 1, the gradient of the graph, Then substitute the value back into equation (2) giving,
()
To determine the value of, using the equation of
P
(PV)
VP
1-
P
1=
T
P
(PV)
T
can be found by plotting graph of PV (m2/s
2) versus pressure (KPa). Value
P
(PV)
T
is
determined by calculating the gradient of the graph.
Table 2: Values of PV and specific volumes at different pressure (KPa) under constant temperature of
100Pressure (KPa) Specific volume, () PV (m2/s2)
1 172.1800 172.180
10 17.1900 171.900
20 8.5847 171.694
30 5.7144 171.432
40 4.2792 171.168
50 3.4181 170.905
75 2.2698 170.235
100 1.6955 169.550
101.325 1.6730 169.517
Example calculations when determining PV
If pressure =
and specific volume, , therefore the units of pressure x volume (PV) will be:
For first data, P= 1KPa and Vg =172.1800
-
7/27/2019 Chemical and Phase
5/21
5
Figure 2: Graph of PV (m2/s
2) against pressure (KPa) is shown as above.
Comment: Values of PV decreases in terms of pressure, KPa. The graph shows a linear negative slope
with a gradient of -0.0265.
From Figure 2, the gradient of the graph
P
(PV)
T
Next, substitute the value of
P
(PV)
T
back into equation of
P
(PV)
VP
1-
P
1=
T
y = -0.0265x + 172.21
169
169.5
170
170.5
171
171.5
172
172.5
0 20 40 60 80 100 120
PV,m2/s2
Pressure, KPa
PV (m2/s2) vs. Pressure (KPa)
Data Points
Linear (Data Points)
-
7/27/2019 Chemical and Phase
6/21
6
Question 10
To show that , a graph consists of specific volume against temperature must beplotted. From the graph, its gradient can be determined and will be the value for Table 3: Values of specific volume at different temperature (K) at a constant pressure of 1 bar
Temperature, T (K) Specific volume, V (m3/kg)
423.15 1.937
448.15 2.055
473.15 2.173
498.15 2.290
523.15 2.406
548.15 2.523
573.15 2.639
623.15 2.871
673.15 3.103
723.15 3.334
773.15 3.565
Figure 3: Graph of specific volume (Vg) against temperature (K) is shown as above
Comment: Specific volume increases directly with temperature, K. The graph shows a linear positive
slope with a gradient of 0.0046, approximately.
y = 0.0046x - 0.0269
0
1
2
3
4
0 100 200 300 400 500 600 700 800 900S
pecificVolume(m3/kg)
Temperarture (K)
Specific Volume(Vg) vs. Temperature (K)
Data Points
Linear (Data
Points)
-
7/27/2019 Chemical and Phase
7/21
7
From Figure 3, the gradient of the graph, 3/kg.KSince at 1 bar, a graph of entropy, S, against Ln P is plotted and gradient of thegraph, can be determined.Table 4: Values of entropy at different pressures at a constant temperature of 250
Pressure, KPa Entropy, Sg (KJ/KgK) Ln P
0.6112 10.390 -0.4923
1 10.163 0.0000
5 9.420 1.6094
10 9.100 2.3026
50 8.355 3.9120
75 8.167 4.3175
100 8.033 4.6052
101.325 8.027 4.6183
150 7.843 5.0106
Figure 4: Graph of entropy, Sg, against ln P is shown as above
Comment: Values of entropy of gas decreases as the values of Ln P increases. The graph shows a
linear negative slope with a gradient of -0.04626.
From Figure 4, the gradient of the graph,
Having the value of , then substitute it into equation m3/kg KThus, is proven
y = -0.4626x + 10.164
0
2
4
6
8
10
12
-1 0 1 2 3 4 5 6
Entropy(KJ/KgK)
ln P
Entropy (Sg) vs. ln P
Data Points
Linear (Data
Points)
-
7/27/2019 Chemical and Phase
8/21
8
Part#2: Example Sheet#2, Part B, Question A
Coefficient of fugacity, , must be determined first before calculating the fugacity, . Therefore, to calculatethe coefficient:
However, the compressibility factor, must be determined from P=0 bar to the saturationpressure, which in this case, 146.11 bar before calculating for .
,
Table 5: Values of pressure and specific volume for gas at 340P( bar) Vg (m
3/kg) Vg (m
3/kmol) z (z-1)/P P (KPa) ((z-1)/P) P
0 0 0 0 0 0 0
0.006112 463 8340.945 0.988 8.23E-05 0.6112 5.03134E-05
0.01 282.98 5097.885 1 3.09E-05 0.3888 1.19953E-05
0.05 56.592 1019.505 0.9999 -8.01E-06 4 -0.00003203
0.1 28.294 509.7164 0.9998 -1.11E-05 5 -0.000055425
0.5 56.592 1019.505 0.9999 -4.00E-06 40 -0.000160148
0.75 3.7678 67.87692 0.9986 -1.82E-05 25 -0.0004551 2.8246 50.88517 0.9982 -1.81E-05 25 -0.0004525
1.01325 2.7872 50.21141 0.998 -1.95E-05 1.325 -2.58375E-05
1.5 1.881 33.88622 0.9971 -1.94E-05 48.675 -0.000941861
2 1.4092 25.38674 0.996 -2.00E-05 50 -0.0009995
3 0.93764 16.89158 0.9941 -2.97E-05 100 -0.0029675
4 0.70194 12.64545 0.9922 -1.94E-05 100 -0.0019395
5 0.5416 9.756924 0.9569 -8.60E-05 100 -0.0086025
6 0.45036 8.113235 0.9549 -7.51E-05 100 -0.0075128
7 0.38516 6.938657 0.9528 -6.74E-05 100 -0.0067444
8 0.33628 6.058084 0.9507 -6.16E-05 100 -0.0061611
9 0.2982 5.372073 0.9484 -5.73E-05 100 -0.0057294
10 0.2678 4.824417 0.9464 -5.36E-05 100 -0.0053614
15 0.17642 3.178206 0.9352 -4.32E-05 500 -0.021605520 0.13074 2.355281 0.9241 -3.80E-05 500 -0.018987
30 0.08992 1.619909 0.9533 -1.56E-05 1000 -0.015563
40 0.06184 1.114048 0.8742 -3.15E-05 1000 -0.031462
50 0.04794 0.863639 0.8471 -3.06E-05 1000 -0.030584
60 0.0386 0.695379 0.8185 -3.03E-05 1000 -0.030257
70 0.03178 0.572517 0.7862 -3.05E-05 1000 -0.030549
80 0.02899 0.522255 0.8196 -2.26E-05 1000 -0.022551
90 0.02486 0.447853 0.7907 -2.33E-05 1000 -0.023258
100 0.02149 0.387142 0.7594 -2.40E-05 1000 -0.024046
-
7/27/2019 Chemical and Phase
9/21
9
110 0.01866 0.33616 0.7254 -2.50E-05 1000 -0.024966
120 0.01621 0.292023 0.6874 -2.60E-05 1000 -0.026048
130 0.01403 0.25275 0.6446 -2.73E-05 1000 -0.027342
140 0.01238 0.22300 0.6124 -2.77E-05 1000 -0.027683
146.11 0.01137 0.204831 0.5871 -2.83E-05 611 -0.017267471
Total -0.400264564
After calculating all the necessaries, to find fugacity,
:
Ln -0.400264564 = exp(-0.400264564)= 0.6701
Part#2: Example Sheet#2, Part B, Question B
Given values: Temperature (T) = 613.15K and pressure (P) is 146.11 bar
Value of critical temperature of steam, TC = 374.15 Value of critical pressure of steam, PC = 222.12 bar
Source of TC and PC values :http://www.engineeringtoolbox.com/critical-point-water-steam-d_834.html
To find the coefficient of fugacity,
, reduced temperature ( Tr) and reduced pressure (Pr) is determined
Tr = Pr=
Tr = 0.947 Pr= 0.658
http://www.engineeringtoolbox.com/critical-point-water-steam-d_834.htmlhttp://www.engineeringtoolbox.com/critical-point-water-steam-d_834.htmlhttp://www.engineeringtoolbox.com/critical-point-water-steam-d_834.htmlhttp://www.engineeringtoolbox.com/critical-point-water-steam-d_834.html -
7/27/2019 Chemical and Phase
10/21
10
Using the generalized fugacity chart, coefficient of fugacity can be found
Figure 5: Fugacity coefficients of gases and vapors (from O.A Hougen and K.M. Watson. Chemical Process
Principle Charts)
From Figure 5, at Pr = 0.658 and Tr = 0.947, the value of is Since
Percentage error calculated from:http://www.marshu.com/articles/calculate-percent-error-formula.php
0.71
http://www.marshu.com/articles/calculate-percent-error-formula.phphttp://www.marshu.com/articles/calculate-percent-error-formula.phphttp://www.marshu.com/articles/calculate-percent-error-formula.phphttp://www.marshu.com/articles/calculate-percent-error-formula.php -
7/27/2019 Chemical and Phase
11/21
-
7/27/2019 Chemical and Phase
12/21
12
To find PA and PW:
To find PTOTAL:
To find Y1 and Y2:
-
7/27/2019 Chemical and Phase
13/21
13
Table 6: Values for ,,PTOTAL, Pa, Pw, Pa*, Pw*, Y1, Y2X1 X2 Pa Pw Pa* Pw* PTOTAL Y1 Y20.01 0.99 7.750373 1.000283 189.5396 627.4815 2445.555 633.6405 817.0211 0.231989 0.76
0.04 0.96 6.589635 1.004441 644.6125 610.9961 2445.555 633.6405 1255.609 0.513386 0.48
0.07 0.93 5.658159 1.013379 968.6135 597.1696 2445.555 633.6405 1565.783 0.618613 0.38
0.1 0.9 4.903962 1.026927 1199.291 585.6321 2445.555 633.6405 1784.923 0.671901 0.32
0.13 0.87 4.288175 1.044981 1363.306 576.064 2445.555 633.6405 1939.37 0.702963 0.29
0.16 0.84 3.78145 1.0675 1479.639 568.1855 2445.555 633.6405 2047.824 0.722542 0.270.19 0.81 3.361414 1.094493 1561.899 561.7471 2445.555 633.6405 2123.646 0.73548 0.2
0.22 0.78 3.010863 1.126017 1619.911 556.5223 2445.555 633.6405 2176.433 0.744296 0.25
0.25 0.75 2.716454 1.162176 1660.809 552.3011 2445.555 633.6405 2213.11 0.750441 0.24
0.28 0.72 2.467754 1.203111 1689.808 548.8849 2445.555 633.6405 2238.693 0.754819 0.24
0.31 0.69 2.256543 1.249008 1710.735 546.0813 2445.555 633.6405 2256.816 0.75803 0.2
0.34 0.66 2.076297 1.300088 1726.417 543.7004 2445.555 633.6405 2270.117 0.760497 0.23
0.37 0.63 1.921801 1.356612 1738.952 541.5506 2445.555 633.6405 2280.502 0.76253 0.2
0.4 0.6 1.788862 1.418878 1749.904 539.4351 2445.555 633.6405 2289.339 0.764371 0.23
0.43 0.57 1.674083 1.487224 1760.446 537.1484 2445.555 633.6405 2297.594 0.766213 0.23
0.46 0.54 1.574698 1.562028 1771.465 534.4728 2445.555 633.6405 2305.938 0.768219 0.23
0.49 0.51 1.488444 1.643709 1783.635 531.1754 2445.555 633.6405 2314.81 0.770532 0.220.52 0.48 1.413455 1.732726 1797.475 527.0043 2445.555 633.6405 2324.479 0.773281 0.22
0.55 0.45 1.348189 1.829587 1813.389 521.6853 2445.555 633.6405 2335.074 0.776587 0.22
0.58 0.42 1.291364 1.934844 1831.698 514.9182 2445.555 633.6405 2346.616 0.78057 0.2
0.61 0.39 1.241907 2.049099 1852.663 506.3728 2445.555 633.6405 2359.035 0.785348 0.21
0.64 0.36 1.198923 2.173005 1876.5 495.6853 2445.555 633.6405 2372.185 0.791043 0.20
0.67 0.33 1.161654 2.307271 1903.396 482.4536 2445.555 633.6405 2385.849 0.797785 0.20
0.7 0.3 1.129465 2.452666 1933.518 466.2326 2445.555 633.6405 2399.751 0.805716 0.19
0.73 0.27 1.101815 2.610019 1967.02 446.5297 2445.555 633.6405 2413.55 0.81499 0.1
0.76 0.24 1.078245 2.780226 2004.05 422.7993 2445.555 633.6405 2426.849 0.825783 0.17
0.79 0.21 1.058367 2.964252 2044.752 394.4367 2445.555 633.6405 2439.188 0.838292 0.16
0.82 0.18 1.041847 3.163137 2089.272 360.7726 2445.555 633.6405 2450.045 0.852749 0.140.85 0.15 1.028403 3.378003 2137.762 321.0659 2445.555 633.6405 2458.828 0.869423 0.13
0.88 0.12 1.017792 3.610053 2190.378 274.4971 2445.555 633.6405 2464.875 0.888637 0.11
0.91 0.09 1.009809 3.860581 2247.285 220.1598 2445.555 633.6405 2467.445 0.910774 0.08
0.94 0.06 1.004279 4.130978 2308.658 157.0533 2445.555 633.6405 2465.711 0.936305 0.06
0.97 0.03 1.001051 4.422736 2374.682 84.07274 2445.555 633.6405 2458.754 0.965807 0.03
0.99 0.01 1.000116 4.629894 2421.379 29.33688 2445.555 633.6405 2450.716 0.988029 0.01
-
7/27/2019 Chemical and Phase
14/21
14
Figure 6: Graph of against x at a constant temperature of 95Comment: The intersection point of the graph is approximately around 0.45. For
, the graph increases with
the x- values but shows a decrease in correspond to the increase of x-values.
Figure 7: Diagram of x-y at a constant temperature of 95Comment: The graph behaved in an increasing manner values of x and y increases.
0
1
2
3
4
5
6
7
8
9
0 0.2 0.4 0.6 0.8 1 1.2
x
Graph of vs x
Y2
Y1
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
y
x
x-y Diagram
acetonewater
-
7/27/2019 Chemical and Phase
15/21
15
Figure 8: Diagram of P-x-y for acetone at a constant temperature of 95Comment: Graph increases as x-y values increases. However, it never crosses through the point (1,2500).
Figure 9: P-x-y diagram for water at a constant temperature of 95Comment: The graph is exactly vice versa from Figure 7. It is decreasing as x-y values increases, however, values
do pass through 2500mmHg benchmark.
Part B
At a given pressure of 1200mm Hg, consider the function T= T(x)
Using the Antoine equation, and by rearrange it in terms of T, we get
0
500
1000
1500
2000
2500
3000
0 0.2 0.4 0.6 0.8 1 1.2
Pto
tal,bar
x-y
P-x-y Diagram for Acetone
acetoneacetone
0
500
1000
1500
2000
2500
3000
0 0.2 0.4 0.6 0.8 1 1.2
Ptotal,bar
x-y
P-x-y Diagram for Water
water
water
-
7/27/2019 Chemical and Phase
16/21
16
The T value above, we can assume its value is near since T(x=0.01) should be approximately close to thevalue 0. Then using the Goal Seek tool in Microsoft Excel, the value of T can be calculated by targeting P at the
initial value of. The equation used for targeting P is shown as below. ( ) ( )
Table 7: Values for X1, X2, ,, PTotal ,T, PA and Y1X1 X2 PTotal T PA Y1
0.01 0.99 7.750373 1.000283 1200.000006 106.4946902 3280.043006 0.211846
0.04 0.96 6.589635 1.004441 1200.000404 93.56980589 2354.392676 0.517153
0.07 0.93 5.658159 1.013379 1200.000013 86.45901815 1939.250818 0.640068
0.1 0.9 4.903962 1.026927 1200.000158 82.17031923 1717.774997 0.701992
0.13 0.87 4.288175 1.044981 1200.000615 79.44109203 1587.408962 0.737434
0.16 0.84 3.78145 1.0675 1200 77.64863286 1506.070785 0.7593510.19 0.81 3.361414 1.094493 1200 76.44946285 1453.495116 0.773585
0.22 0.78 3.010863 1.126017 1200.000001 75.6381911 1418.746434 0.783136
0.25 0.75 2.716454 1.162176 1200.000001 75.08465455 1395.411975 0.789703
0.28 0.72 2.467754 1.203111 1200.000001 74.70276412 1379.488896 0.794323
0.31 0.69 2.256543 1.249008 1200.000002 74.43380074 1368.359858 0.797672
0.34 0.66 2.076297 1.300088 1200.000002 74.23685261 1360.255245 0.800217
0.37 0.63 1.921801 1.356612 1200.000002 74.08304113 1353.951894 0.802292
0.4 0.6 1.788862 1.418878 1200.000003 73.95187622 1348.594688 0.80415
0.43 0.57 1.674083 1.487224 1200.000003 73.82886798 1343.585706 0.80599
0.46 0.54 1.574698 1.562028 1200.000003 73.70391136 1338.512297 0.807972
0.49 0.51 1.488444 1.643709 1200.000004 73.57016386 1333.098586 0.8102320.52 0.48 1.413455 1.732726 1200.000004 73.42324835 1327.171634 0.812889
0.55 0.45 1.348189 1.829587 1200.000005 73.26067648 1320.637118 0.816048
0.58 0.42 1.291364 1.934844 1200.000007 73.08142589 1313.461425 0.819809
0.61 0.39 1.241907 2.049099 1200.000009 72.88562765 1305.658217 0.824266
0.64 0.36 1.198923 2.173005 1200.000012 72.67433469 1297.278259 0.829513
0.67 0.33 1.161654 2.307271 1200.000015 72.44935126 1288.401692 0.835645
0.7 0.3 1.129465 2.452666 1200.000016 72.21311015 1279.132237 0.842762
0.73 0.27 1.101815 2.610019 1200.000018 71.9685884 1269.592978 0.850971
0.76 0.24 1.078245 2.780226 1200.00002 71.71925533 1259.923512 0.860387
0.79 0.21 1.058367 2.964252 1200.000022 71.46904978 1250.278329 0.871141
0.82 0.18 1.041847 3.163137 1200.000023 71.22238508 1240.826399 0.883380.85 0.15 1.028403 3.378003 1200.000024 70.98418255 1231.751999 0.897272
0.88 0.12 1.017792 3.610053 1200.000025 70.75993663 1223.256921 0.913016
0.91 0.09 1.009809 3.860581 1200.000024 70.55581781 1215.564337 0.930845
0.94 0.06 1.004279 4.130978 1200.000023 70.37882335 1208.924754 0.951043
0.97 0.03 1.001051 4.422736 1200.000021 70.2369918 1203.624804 0.973953
0.99 0.01 1.000116 4.629894 1200.000003 70.1665385 1200.998898 0.990939
-
7/27/2019 Chemical and Phase
17/21
17
Figure 10: T-x-y diagram for acetone at a constant pressure 1200mm Hg
Comment: Both graphs are behaving in a decreasing manner. Through the graph, acetone will only reach anequilibrium around 70K at the highest mole fraction.
Figure 11: x-y diagram for acetone at a constant pressure of 1200mm Hg
Comment: Overall the graph behaves in an increasing pattern as the x-values increases.
0
20
40
60
80
100
120
0 0.2 0.4 0.6 0.8 1 1.2
Temperature,K
x-y
T-x-y Diagram
acetone
acetone
0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
y(x)
x
x-y Diagram
Data Points
-
7/27/2019 Chemical and Phase
18/21
18
Microsoft Excel Screenshots
Part 1
Question 9
-
7/27/2019 Chemical and Phase
19/21
19
Question 10
-
7/27/2019 Chemical and Phase
20/21
20
Part 2
-
7/27/2019 Chemical and Phase
21/21
21
Part 3