chemical and phase

7/27/2019 Chemical and Phase

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CHEMICAL AND PHASE

EQUILIBRACOURSEWORK

Name: Yap Wei Khang

Student ID:UNIMKL-007852

Lecturer: Dr.Sergey SportarModule Code:H82CPE

Due Date:13th December 2011


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Part#1: Example Sheet #1, Part B

Question 8

------

{ [ ()]}

() () *

+

(

)

[ ( ) ()]

() ()

() (

)


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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)


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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


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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)


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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)


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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)


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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


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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


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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


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To find PA and PW:

To find PTOTAL:

To find Y1 and Y2:


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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


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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


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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


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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


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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


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Microsoft Excel Screenshots

Part 1

Question 9


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Question 10


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Part 2


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Part 3

chemical and phase

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