tutorial sheet 04 answers 2014

8/12/2019 Tutorial Sheet 04 Answers 2014

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Tutorial Sheet No. 4: A nswers

Process Equipment Design for Phase-Equilibria Systems.

1) a) Calculate the bubble point pressure and composition of the vapour in equilibrium with aliquid at 40 C containing:

5 mole % methane (CH 4)20 mole % ethane (C 2H6)25 mole % propane (C 3H8)20 mole % isobutane (C 4H10)30 mole % n-butane (C 4H10)

Use an initial guess of 1000 kPa.[12.5 marks]

b) Calculate the dew point pressure and composition of the liquid in equilibrium with avapour at 40 C containing:

5 mole % methane (CH 4)20 mole % ethane (C 2H6)25 mole % propane (C 3H8)20 mole % isobutane (C 4H10)30 mole % n-butane (C 4H10)

Use an initial guess of 1000 kPa.

[12.5 marks]

Answer:a) Use high temperature nomograph with: K i = yi / x i

Therefore yi = xi K i

Guess P until yi = 1.0

P = 1000 kPa P = 2500 kPa P = 2700 kPa Component x i K i y i K i y i K i y i methane 0.05 18 0.90 7.8 0.39 7.2 0.36ethane 0.20 3.9 0.78 1.9 0.38 1.8 0.36

propane 0.25 1.35 0.34 0.68 0.17 0.64 0.16isobutane 0.20 0.60 0.12 0.32 0.064 0.31 0.062nbutane 0.30 0.43 0.13 0.23 0.069 0.22 0.066

1.00 2.27 1.07 1.01

The bubble point pressure is 2700 kPa and the composition of the vapour is:

36 mole % methane, 36 mole % ethane, 16 mole % propane, 6.2 mole % isobutane

and 6.6 mole % n-butane[12.5 marks]


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b) Use high temperature nomograph with: K i = yi / x i

Therefore xi = yi / K i

Guess P until xi = 1.0


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P = 1000 kPa P = 820 kPa P = 760 kPa y i K i x i K i x i K i x i

methane 0.05 18 0.00278 21.5 0.00233 23 0.00217ethane 0.20 3.9 0.05128 4.6 0.04348 5 0.04000

propane 0.25 1.35 0.18519 1.6 0.15625 1.7 0.14706isobutane 0.20 0.60 0.33333 0.72 0.27778 0.76 0.26316nbutane 0.30 0.43 0.69767 0.52 0.57692 0.56 0.53571

1.00 1.27 1.06 0.99

The dew point pressure is 760 kPa and the composition of the liquid is:

0.2 mole % methane, 4 mole % ethane, 14.7 mole % propane, 26.3 mole % isobutaneand 53.6 mole % n-butane

[12.5 marks]


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2) Many thermodynamic process engineering design problems consider a system in which

either a vapour phase is just being formed from a boiling liquid phase, or a liquid phase is just

being formed from a saturated vapour phase. In order to take into account molecular

interactions in real systems, the liquid-vapour equilibrium constant, K i, can be used in some

hydrocarbon systems.

a) For a non-ideal system, show that the Design Performance Equation to predict thecomponent mole fraction xi in the liquid stream from a flash vessel processing a liquid

feed stream of component mole fraction x Fi is:

L)-(1 K L x

xi

Fii

where L is the product liquor molar flow from the flash vessel operating at a system

pressure P and a system temperature T . Identify the assumptions used in the

derivation.

[5 marks]

b) A liquid stream containing 15 mol% ethane (C 2H6), 35 mol% propane (C 3H8) and 50mol% n-butane (C 4H10) enters a flash vessel at 40 C. If 40% of the stream remains as

a liquid (based on the molar flow), calculate the pressure of the vessel and thecomposition of the exit streams.

[10 marks]

c) The vapour stream from a gas well is a mixture containing 50 mol% methane, 10mol% ethane, 20 mol% propane and 20 mol% n-butane. The stream is fed to a partial

condenser at a pressure of 17.25 bar and a temperature of 27 C. Determine the mole

fraction of the gas which condenses and the composition of the liquid and gas phases

leaving the condenser.

[10 marks]

Data supplied:

Data Sheet No. (1): De Priester Nomogram for Phase Equilibrium Constants for

Hydrocarbons (High Temperature Range).


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Answer: DIAGRAM

Assumptions: No chemical reaction only physical division between binary phases. Steadystate system at constant temperature, T , and pressure, P . System is at phase equilibrium.

Basis of calculation: F = 1.0 kmol feedstock.

Mass Balance:

0)( 0)(

ONACCUMULATIREACTIONOUT-IN

IN = OUT

Overall Balance:

F = V + L 1.0 = V + L then V = 1 L (1)

Component Balance:

L xV y F x ii Fi

Substitute for basis, F = 1.0 and for V from Eq. (1) to give:

L x L y x ii Fi )-(1 (2)

At phase equilibrium,

iii x K y

Substitute for y i in Eq. (2) to give:

L x L K x x iii Fi )-(1

)-(1

L K L

x x

i

Fii (3)

[5 marks]

Feedstock F kmol

Liquid Feed x Fi

EQUILBRIUMUNIT

OR STAGE

V kmol yi

L kmol xi


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b) Now we know L = 0.4, so guess P and use;

)1( L K L

x x

i

Fii

To calculate xi and then check for xi = 1.

P = 600 kPa P = 820 kPaComponent x F i K i x i K i x i y i ethane 0.15 6 0.0375 4.7 0.0466 0.2190

propane 0.35 2.1 0.2108 1.6 0.2574 0.4118n-butane 0.5 0.67 0.6234 0.54 0.6906 0.3729

0.8717 0.9946 1.0037

Iteration can be done from first guess on component contributing largest amount to xi using:

K i new = K i old xi

Using data P = 600 kPa K n-butane = 0.6234 0.8717 = 0.54

Drawing a line through 40 C and K n-butane = 0.54, gives the second set of data at P = 820 kPaand the required result.


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c) We know that: T = 27 C and P = 17.25 bar (= 17.25 10 5 Pa = 1725 kPa).

Therefore we can draw a line on the nomograph to get the K i values.

We know y Fi values coming in therefore we need to guess L and iterate until xi = 1.

L = 0.5 L = 0.2 L = 0.1 L = 0.142Component y Fi K i x i x i x i x i y i

methane 0.5 10.3 0.0885 0.0592 0.0534 0.0557 0.5737ethane 0.1 2.1 0.0645 0.0532 0.0503 0.0514 0.1079

propane 0.2 0.69 0.2367 0.2660 0.2774 0.2725 0.1880n-butane 0.2 0.21 0.3306 0.5435 0.6920 0.6208 0.1304

1.0 0.7203 0.9219 1.0731 1.0004 1.0000

We know L is between 0 and 1 so iteration is straight forward (final result L = 0.142 and V =0.858 with compositions given in the table).


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3) For a non-ideal system, show that the Design Performance Equation to predict thecomponent mole fraction xi in the liquid stream from a flash vessel processing a liquid feedstream of component mole fraction x Fi is:

L)-(1 K L

x

x i Fi

i

where L is the product liquor molar flow from the flash vessel operating at a system pressure P and a system temperature T . Identify the assumptions used in the derivation.

[5 marks]

A liquid of composition 25 mole % ethane, 15 mole % n-butane and a third unknowncomponent enters a separation vessel which operates at a pressure of 750 kPa andtemperature of 6 C. If 87% of the solution leaves the vessel as the liquid stream, calculate:

a)

The equilibrium constant, K i, for the unknown component. [10 marks] b) The composition of the vapour and liquid leaving the separation unit.

[7 marks]c) Determine the chemical name of the third remaining component in the liquid mixture.

[3 marks]

Answer: DIAGRAM

Assumptions: No chemical reaction only physical division between binary phases. Steadystate system at constant temperature, T , and pressure, P . System is at phase equilibrium.

Basis of calculation: F = 1.0 kmol feedstock.

Mass Balance:

0)( 0)(

ONACCUMULATIREACTIONOUT-IN

IN = OUT

Overall Balance:

Feedstock

F kmol

Liquid Feed x Fi

EQUILBRIUMUNIT

OR STAGEV kmol

yi

L kmol xi


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F = V + L 1.0 = V + L

then V = 1 L (1)

Component Balance:

L xV y F x ii Fi

Substitute for basis, F = 1.0 and for V from Eq. (1) to give:

L x L y x ii Fi )-(1 (2)

At phase equilibrium,

iii x K y

Substitute for y i in Eq. (2) to give:

L x L K x x iii Fi )-(1

)-(1

L K L x

xi

Fii (3)

[5 marks]

a) The mole fraction of the unknown component in the feed stream is:

x Fi of unknown component = 1 0.25 0.15 = 0.6

Next, draw a tie line on the De Priester Chart between P = 750 kPa and T = 6 C. Read off the K i values for ethane and n- butane. Knowing the K i values and that L = 0.87 we can calculatethe mole fractions of ethane and n-butane in the output liquid stream using:

i

Fi

i

Fii K .

x.- K .

x x

0.13870)8701(870

Component x F i K i x i

ethane 0.25 2.6 0.207n-butane 0.15 0.18 0.168unknown 0.6 - -

We know that: xi = 1

Therefore xethane + xn-butane + xunknown = 1

0.207 + 0.168 + xunknown = 1

xunknown = 0.625


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We can now calculate K i as:

i K ..

0.138706.0

6250

0.87 + 0.13 K i = 0.6/0.625


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K i = [(0.6/0.625) 0.87]/0.13 = 0.69[10 marks]

b) Using the calculated data we get:

Component x F i K i x i y i ethane 0.25 2.60 0.207 0.538n-butane 0.15 0.18 0.168 0.030unknown 0.6 0.69 0.625 0.431Total 1.00 - 1.000 0.999

Remember yi = K i xi

Therefore, the stream compositions are:

Liquid: 20.7 mole % ethane, 16.8 mole % n- butane, 62.5 mole % of the unknown compoundVapour: 53.8 mole % ethane, 3.0 mole % n- butane, 43.1 mole % of the unknown compound

[7 marks]

c) Looking on the De Priester chart, the compound which has a K i of 0.69 when P = 750 kPaand T = 6 C is propane.

Therefore, the unknown compound is most likely propane.[3 marks]


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4) Acetone (1) and methanol (2) form an azeotrope boiling at 55.7 C and 760 mmHg pressure, with a mole fraction of 80% acetone. Given the following Antoine equations, where

oi p is in mmHg and T is in C:

Acetone: 229.6641210.595

-7.11714log 110 T po

Methanol:239.726

1582.271 -8.08097log 210 T

p o

a) Determine the Van Laar coefficients from the azeotrope data and the followingequations;

1

2

11

2212 lnln

ln1

x x

A

2

2

22

1121 lnln

ln1

x x

A

[5 marks]

b) Calculate the azeotropic boiling point and composition at P = 1520 mmHg. Use A12 and A21 as calculated above and:

2

2

1

21

12

121

1

ln

x x

A A

A 2

1

2

12

21

212

1

ln

x x

A A

A

[15 marks]

c) Calculate the lowest pressure at which an azeotrope exists.[5 marks]

Answer:a) When T = 55.7 C and P = 1 atm, we are at the azeotrope and oii P/p , therefore:

Acetone:229.66455.7

1210.595 -7.11714 plog o110

mmHg749.65 p o1

1.0138mmHg749.65

mmHg760

pP

o1

1

Methanol:239.72655.7

1582.271 -8.08097 plog o210

mmHg530.97 p o2

1.4313mmHg530.97mmHg760

pP

o1

1


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We know x1 = 0.8, x2 = 0.2, 1 = 1.0138 and 2 = 1.4313, therefore:

2

11

22112 lnx

lnx 1lnA

0.77901.0138ln0.81.4313ln0.2

11.0138lnA2

12

2

22

11221 lnx

lnx 1lnA

0.47671.4313ln0.21.0138ln0.8

11.4313lnA2

21

[5 marks]

b) At the azeotrope:

o1

o2

o2

o1

2

1

p p

P/pP/p

Therefore:

2

1

2

12

21

212

2

1

21

12

12212

1o1

o

2

11lnlnln p

pln

x x

A A

A

x x

A A

A

Let:2

1

21

12

x x

A A

Therefore:

22

21122

212

12o1

o2

1111 p p

ln

A A A A

Need to guess T , calculate o1 p ,o2 p and solve for . Then calculate x1 and 1 and P = 1

o1 p .

Iterate until P = 1520 mmHg

1 st guess: When T = 55.7 C, we know that mmHg749.65 p o1 and P = 760 mmHg, therefore

oldo1

old

newnew

o1 )(pP

P )(p


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mmHg1499.3749.65760

1520 )(p new

o1

C77.507.11714-1499.3log

1210.595 -229.664-T

10

new

mmHg1239.00 p o2

22

14767.07790.0

1499.31239.00

ln

096966.038138.0286042.0 2

Giving: = 2.624803 which results in 1 = 1.0611

mmHg87.15900611.13.1499 pP 1o1

This is too high, therefore we need to iterate.

1 st iteration:

mmHg1432.5129.14991590.87

1520 )(p new

o1

C75.967.11714-1432.51log

1210.595 -229.664-T

10new

mmHg1171.72 p o2

22

14767.07790.0

1432.511171.72

ln

097992.040191.0275779.0 2


mmHg11.15140570.151.1432 pP 1o1

This is slightly too low so can iterate again:

2nd iteration:

mmHg1438.0851.14321514.11

1520 )(p new

o1

C76.097.11714-1438.08log

1210.595 -229.664-T10

new


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mmHg1177.31 p o2

2

2

1

4767.07790.0

1438.08

1177.31ln

097905.040016.0276652.0 2


mmHg50.15200573.108.1438 pP 1o1

Close enough, so T = 76.09 C.

===============================================================We could also have interpolated this value using the first 2 points we had already i.e.:

C08.76)11.151487.1590(

)96.755.77()11.15141520(96.75 oT

===============================================================

Composition:

)1( 1

1

21

12

2

1

21

12

x

x

A

A

x

x

A

A

Rearranging, gives:

6263.0)1(

1)1(

1

4767.0738652.27790.01

2112

A A

x

Therefore, the azeotrope at 1520 mmHg is at 62.63% Acetone.[15 marks]

c) Since x1 decreases as P rises, it will increase as P falls and eventually reach a value of 1.0for P < 760 mmHg. When x 1 = 1.0 then:

2

1

21

12

x x

A A

Therefore:

21221

212

o1

o2

111 p p

ln A A A

This equation can be solved either by trial and error or analytically:


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oioi plog30258.2 pln

Therefore,

21o1o2o1o2o1

o2 plog plog30258.2 pln pln

p p

ln A

30258.2229.6641210.595

-7.11714239.726

1582.271 -8.08097 21

AT T

229.664

1210.595

239.7261582.271

30258.2 7.117148.08097 21

T T A

A21 = 0.4767, so:

229.6641210.595

239.726

1582.271170873.1

T T

239.726)1210.595()229.664(271.1582)229.664)(239.726(170873.1 T T T T

1.2902111210.5957.363390271.1582)55056.43469.39(170873.1 2 T T T T

59.73179676.37164464.07549.60170873.1 2 T T T

08715.52-92.177170873.1 2 T T

Therefore: T = 38.98 C (or -190.94 C which is invalid)

When T = 38.98 C;

Acetone:229.66438.98

1210.595 -7.11714 plog o110

mmHg408.15 po1

So mmHg15.408115.408 pP 1o1

[5 marks]


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5) Acetone (1) and Hexane (2) form an azeotrope containing 41 wt% of Hexane, boiling at

49.8 C at a pressure of 760 mmHg. Pure component vapour pressures may be calculated

using the following equation and Antoine coefficients:

)(log *10 C T

B A p i

Antoine Coefficients RMM

kg / kmolA B C

Acetone (1) 7.1327 1219.97 230.653 58.08

Hexane (2) 7.01051 1246.33 232.988 86.18

with*i p in mmHg and T in C.

a) Calculate the normal boiling points for Acetone and Hexane at 1 atm.[4 marks]

b) Use the azeotropic data and the following equations to estimate the van Laar constantsA12 and A 21.

2

11

22112 ln

ln1ln

x x A

2

22

11221 ln

ln1ln

x x A

[7 marks]

c) Using the Antoine equations and Van Laar coefficients calculated in part (b), estimatethe boiling point and vapour composition at 760 mmHg of a liquid containing 20 %

by mole of Acetone.

2112221

2

2

2

21121 ][

ln x A x A

x A A 2221112

2

1

2

12212 ][

ln x A x A

x A A

[8 marks]

d) Find the dew point of a vapour containing 50 % by mole of Acetone.[6 marks]

Answer:a) To find the boiling point substitute a pressure of 1 atm (=760 mmHg) into the Antoine

equations:


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8950.105.401

7602

Need to work out mole fractions at azeotrope, therefore taking a basis of 100 kg of themixture:

Mass of Hexane = (41/100) 100 = 41 kgMass of Acetone = 100 41 = 59 kg

Moles of Hexane = 41 / 86.18 = 0.4757 kmolsMoles of Acetone = 59 / 58.08 = 1.0158 kmols

Mole fraction of Acetone ( x1) = 1.0158 / (0.4757 + 1.0158) = 0.681Mole fraction of Hexane ( x2) = 1 x1 = 1 0.681 = 0.319

2

11

2112 ln

ln1ln x x A

2216.12535.1ln681.08950.1ln319.0

12535.1ln2

12 A

2

22

1221 ln

ln1ln

x x

A

9678.18950.1ln319.02535.1ln681.0

18950.1ln2

21 A

[7 marks]

c) x1 = 0.2, x2 = 0.8 and P = 760 mmHg.

2112221

22

22112

1 ][ln

x A x A x A A

9154.0)]2.02216.1()8.09678.1[(

8.09678.12216.1ln 222

1

4978.21

2221112

21

21221

2 ][ln

x A x A x A A

0355.0

)]2.02216.1()8.09678.1[(

2.02216.19678.1ln 2

22

2


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0361.12

Now we need to guess T , so as a first guess use the average boiling point, T = 62.6 C.

Acetone: 9726.2)653.2306.62(97.1219

1327.7log*1 p

mmHg86.938*1 p

Hexane: 7941.2)988.2326.62(

33.124601051.7log *2 p

mmHg44.622*2 p

iiii p x p * and i p P

So

mmHg02.4694978.286.9382.01 p

mmHg93.5150361.144.6228.02 p

mmHg95.98493.51502.469 i p P

This value is too high as we need P = 760 mmHg, so iterate temperature value as follows:

iold new p

p p760

)()( *1*1

mmHg44.72495.984

76086.938)( *1 new p

From Antoine equation for Acetone: C87.54 newT .

Hexane: 6808.2)988.23287.54(

33.124601051.7log *2 p

mmHg51.479*2 p


So

mmHg90.3614978.244.7242.01 p


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mmHg46.3970361.151.4798.02 p

mmHg36.75946.39790.361 i p P

This value is slightly too low as we need P = 760 mmHg, so iterate again:

iold new p

p p760

)()( *1*1

mmHg725.0536.759

76044.724)( *1 new p

From Antoine equation for Acetone: C90.54 newT .

Hexane: 6813.2)988.23290.54(

33.124601051.7log *2 p

mmHg06.480*2 p


So

mmHg21.3624978.205.7252.01 p

mmHg91.3970361.106.4808.02 p

mmHg12.76091.39721.362 i p P

Therefore: T = 54.90 C and the vapour composition is found from:

P

p x y iiii

*

477.0760

05.7254978.22.01 y

523.0760

06.4800361.18.02 y

[8 marks]

d) y1 = 0.5, y2 = 0.5, P = 760 mmHg, 2216.112 A and 9678.121 A .

Again guess T = 62.6 C.


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Therefore mmHg86.938*1 p and mmHg44.622*2 p

Also assume 1 = 2 = 1 then

*ii

ii p

P y x

4047.086.938

7605.01 x

6105.044.622

7605.02 x

Recalculatei values using Van Laar equations:

8462.1)]4047.02216.1()6105.09678.1[(

6105.09678.12216.1exp 2

22

1

1821.1)]6105.09678.1()4047.02216.1[(

4047.02216.19678.1exp 2

22

2

Recalculate xi values:

2192.09388462176050

1 .

x

5165.044.622

7605.02 x

7357.0 i x

Therefore: mmHg72.690

0.1

7357.086.938)( *1 new p

From Antoine equation: C5.53 newT

mmHg23.457)( *2 new p

Recalculate i values using Van Laar equations:

1495.2)]2192.02216.1()5165.09678.1[(

5165.09678.12216.1exp 2

22

1


23/23

0893.1)]5165.09678.1()2192.02216.1[(

2192.02216.19678.1exp 2

22

2


2559.072.690

7605.01 x

7630.023.457

7605.02 x

0189.1 i x

Therefore: mmHg77.7030.1

0189.172.690)( *

1

new p

From Antoine equation: C04.54 newT

mmHg93.465)( *2 new p

Again recalculate i values using Van Laar equations:

3091.2

)]2559.02216.1()7630.09678.1[(

7630.09678.12216.1exp 2

22

1

0602.1)]7630.09678.1()2559.02216.1[(

2559.02216.19678.1exp 2

22

2


2338.077.703

7605.01 x

7693.093.465

7605.02 x

0031.1 i x

Could recheck final values with another iteration.[6 marks]

tutorial sheet 04 answers 2014

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