y1 0.33:= T 100 degC⋅:= Guess: x1 0.33:= P 100 kPa⋅:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

x1

P

⎛⎜⎝

⎞

⎠Find x1 P,( ):= x1 0.169= Ans. P 92.156kPa= Ans.

(c) Given: x1 0.33:= P 120 kPa⋅:= Guess: y1 0.5:= T 100 degC⋅:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

y1

T

⎛⎜⎝

⎞

⎠Find y1 T,( ):= y1 0.542= Ans. T 103.307degC= Ans.

Chapter 10 - Section A - Mathcad Solutions

10.1 Benzene: A1 13.7819:= B1 2726.81:= C1 217.572:=

Toluene: A2 13.9320:= B2 3056.96:= C2 217.625:=

Psat1 T( ) e

A1B1

TdegC

C1+−

kPa⋅:= Psat2 T( ) e

A2B2

TdegC

C2+−

kPa⋅:=

(a) Given: x1 0.33:= T 100 degC⋅:= Guess: y1 0.5:= P 100 kPa⋅:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

y1

P

⎛⎜⎝

⎞

⎠Find y1 P,( ):= y1 0.545= Ans. P 109.303kPa= Ans.

(b) Given:

312

x1

y1

⎛⎜⎝

⎞

⎠Find x1 y1,( ):= x1 0.282= Ans. y1 0.484= Ans.

(f) z1 0.33:= x1 0.282= y1 0.484=

Guess: L 0.5:= V 0.5:=

Given z1 L x1⋅ V y1⋅+=

L V+ 1=

L

V⎛⎜⎝

⎞⎠

Find L V,( ):= Vapor Fraction: V 0.238= Ans.

Liquid Fraction: L 0.762= Ans.

(g) Benzene and toluene are both non-polar and similar in shape andsize. Therefore one would expect little chemical interactionbetween the components. The temperature is high enough andpressure low enough to expect ideal behavior.

(d) Given: y1 0.33:= P 120 kPa⋅:= Guess: x1 0.33:= T 100 degC⋅:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

x1

T

⎛⎜⎝

⎞

⎠Find x1 T,( ):= x1 0.173= Ans. T 109.131degC= Ans.

(e) Given: T 105 degC⋅:= P 120 kPa⋅:= Guess: x1 0.33:= y1 0.5:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

313

0 0.5 160

70

80

90

100

110

120

130

140

T x1( )T x1( )

x1 y'1 x1( ),

0 0.5 10

50

100

150

P x1( )P x1( )

x1 y1 x1( ),

x1 0 0.05, 1.0..:=

y'1 x1( )x1 Psat1 T x1( )( )⋅

x1 Psat1 T x1( )( )⋅ 1 x1−( ) Psat2 T x1( )( )⋅+:=

T x1( ) root x1 Psat1 t( )⋅ 1 x1−( ) Psat2 t( )⋅+ P'− t,⎡⎣ ⎤⎦:=

t 90:=Guess t for root function:

P' 90:=T-x-y diagram:

y1 x1( )x1 Psat1 T( )⋅

P x1( ):=P x1( ) x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+:=

T 90:=P-x-y diagram:

Psat2 T( ) exp A2B2

T C2+−

⎛⎜⎝

⎞

⎠:=

Psat1 T( ) exp A1B1

T C1+−

⎛⎜⎝

⎞

⎠:=

C2 212.300:=B2 3259.93:=A2 13.9726:=

C1 217.572:=B1 2726.81:=A1 13.7819:=

Antoine coefficients: Benzene=1; Ethylbenzene=2(a)

Pressures in kPa; temperatures in degC10.2

314

0 0.5 170

77.5

85

92.5

100

107.5

115

122.5

130

T x1( )T x1( )

x1 y'1 x1( ),

0 0.5 120

66.67

113.33

160

P x1( )P x1( )

x1 y1 x1( ),

x1 0 0.05, 1.0..:=

y'1 x1( )x1 Psat1 T x1( )( )⋅

x1 Psat1 T x1( )( )⋅ 1 x1−( ) Psat2 T x1( )( )⋅+:=

T x1( ) root x1 Psat1 t( )⋅ 1 x1−( ) Psat2 t( )⋅+ P'− t,⎡⎣ ⎤⎦:=

t 90:=Guess t for root function:

P' 90:=T-x-y diagram:

y1 x1( )x1 Psat1 T( )⋅

P x1( ):=P x1( ) x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+:=

T 90:=P-x-y diagram:

Psat2 T( ) exp A2B2

T C2+−

⎛⎜⎝

⎞

⎠:=Psat1 T( ) exp A1

B1

T C1+−

⎛⎜⎝

⎞

⎠:=

C2 211.700:=B2 3174.78:=A2 13.8635:=

C1 218.265:=B1 2723.73:=A1 13.7965:=

Antoine coefficients: 1-Chlorobutane=1; Chlorobenzene=2 (b)

315

0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.850

0.5

1

V z1( )

x1 y1

z1

V is obviously linear in z1:

V z1( )z1 x1−

y1 x1−:=z1 x1 x1 0.01+, y1..:=

z1 x1 1 V−( )⋅ y1 V⋅+=

For a given pressure, z1 ranges from the liquid composition at the bubblepoint to the vapor composition at the dew point. Material balance:

y1 0.89=y1x1 Psat1 T( )⋅

P:=x1 0.5:=

Since for Raoult's law P is linear in x, at the specified P, x1 must be 0.5:

P 104.349=PPsat1 T( ) Psat2 T( )+

2⎛⎜⎝

⎞⎠

:=T 55:=

Psat2 T( ) exp A2B2

T C2+−

⎛⎜⎝

⎞

⎠:=Psat1 T( ) exp A1

B1

T C1+−

⎛⎜⎝

⎞

⎠:=

C2 216.432:=B2 2911.26:=A2 13.8622:=

C1 232.014:=B1 2451.88:=A1 13.7667:=

Antoine coefficinets: n-Pentane=1; n-Heptane=2(a)

Pressures in kPa; temperatures in degC10.3

316

V 0 0.1, 1.0..:=

0 0.5 10

50

100

150

P V( )

V0 0.5 10

0.5

1

x1 V( )

y1 V( )

V

10.4 Each part of this problem is exactly like Problem 10.3, and is worked inexactly the same way. All that is involved is a change of numbers. Infact, the Mathcad solution for Problem 10.3 can be converted into thesolution for any part of this problem simply by changing one number, thetemperature.

10.7 Benzene: A1 13.7819:= B1 2726.81:= C1 217.572:=

Ethylbenzene A2 13.9726:= B2 3259.93:= C2 212.300:=

Psat1 T( ) e

A1B1

TdegC

C1+−

kPa⋅:= Psat2 T( ) e

A2B2

TdegC

C2+−

kPa⋅:=

(b) At fixed T and z1, calculate x1, y1 and P as functions of fraction vapor (V).

z1 0.5:=

Guess: x 0.5:= y 0.5:= pPsat1 T( ) Psat2 T( )+

2⎛⎜⎝

⎞⎠

:=

Given Three equations relate x1, y1, & P for given V:

p x Psat1 T( )⋅ 1 x−( ) Psat2 T( )⋅+=

y p⋅ x Psat1 T( )⋅=

z1 1 V−( ) x⋅ V y⋅+=

f V( ) Find x y, p,( ):=

x1 V( ) f V( )1:= y1 V( ) f V( )2:= P V( ) f V( )3:=

Plot P, x1 and y1 vs. vapor fraction (V)

317

P 66.38 kPa⋅=

(d) T 72.43 deg_C⋅= P 36.02 kPa⋅=

To calculate the relative amounts of liquid and vapor phases, one mustknow the composition of the feed.

10.8 To increase the relative amount of benzene in the vapor phase, thetemperature and pressure of the process must be lowered. For parts (c)and (d), the process must be operated under vacuum conditions. Thetemperatures are well within the bounds of typical steam and cooling watertemperatures.

10.9(1) = benzene(2) = toluene(3) = ethylbenzene

A

13.7819

13.9320

13.9726

⎛⎜⎜⎜⎝

⎞⎟⎠

:= B

2726.81

3056.96

3259.93

⎛⎜⎜⎜⎝

⎞⎟⎠

:= C

217.572

217.625

212.300

⎛⎜⎜⎜⎝

⎞⎟⎠

:=

(a) n rows A( ):= i 1 n..:= T 110 degC⋅:= P 90 kPa⋅:= zi1n

:=

Psat i T,( ) e

AiBi

TdegC

Ci+−

kPa⋅:= kiPsat i T,( )

P:= Guess: V 0.5:=

(a) Given: x1 0.35:= y1 0.70:= Guess: T 116 degC⋅:= P 132 kPa⋅:=

Given x1 Psat1 T( )⋅ 1 x1−( ) Psat2 T( )⋅+ P=

x1 Psat1 T( )⋅ y1 P⋅=

T

P⎛⎜⎝

⎞⎠

Find T P,( ):= T 134.1degC= Ans. P 207.46kPa= Ans.

For parts (b), (c) and (d) use the same structure. Set the defined variablesand change the variables in the Find statement at the end of the solveblock.

(b) T 111.88 deg_C⋅= P 118.72 kPa⋅=

(c) T 91.44 deg_C⋅=

318

y

0.441

0.333

0.226

⎛⎜⎜⎜⎝

⎞⎟⎠

=

P 100 kPa⋅=

(c) T 110 deg_C⋅= V 0.352= x

0.238

0.345

0.417

⎛⎜⎜⎜⎝

⎞⎟⎠

= y

0.508

0.312

0.18

⎛⎜⎜⎜⎝

⎞⎟⎠

=

P 110 kPa⋅=

(d) T 110 deg_C⋅= V 0.146= x

0.293

0.342

0.366

⎛⎜⎜⎜⎝

⎞⎟⎠

= y

0.572

0.284

0.144

⎛⎜⎜⎜⎝

⎞⎟⎠

=

P 120 kPa⋅=

10.10 As the pressure increases, the fraction of vapor phase formed (V)decreases, the mole fraction of benzene in both phases increases and thethe mole fraction of ethylbenzene in both phases decreases.

Given

1

n

i

zi ki⋅

1 V ki 1−( )⋅+∑=

1= Eq. (10.17)

V Find V( ):= V 0.836= Ans.

yizi ki⋅

1 V ki 1−( )⋅+:= Eq. (10.16) y

0.371

0.339

0.29

⎛⎜⎜⎜⎝

⎞⎟⎠

= Ans.

xiyi P⋅

Psat i T,( ):= x

0.142

0.306

0.552

⎛⎜⎜⎜⎝

⎞⎟⎠

= Ans.

(b) T 110 deg_C⋅= V 0.575= x

0.188

0.334

0.478

⎛⎜⎜⎜⎝

⎞⎟⎠

=

319

y1 0.805= Ans.

xiyi P⋅

Psat i T,( ):= x1 0.644= Ans.

ry1 V⋅

z1:= r 0.705= Ans.

(b) x1 0.285= y1 0.678= V 0.547= r 0.741=

(c) x1 0.183= y1 0.320= V 0.487= r 0.624=

(d) x1 0.340= y1 0.682= V 0.469= r 0.639=

10.11 (a) (1) = acetone(2) = acetonitrile A

14.3145

14.8950⎛⎜⎝

⎞⎠

:= B2756.22

3413.10⎛⎜⎝

⎞⎠

:= C228.060

250.523⎛⎜⎝

⎞⎠

:=

n rows A( ):= i 1 n..:=

z1 0.75:= T 340 273.15−( ) degC⋅:= P 115 kPa⋅:=

z2 1 z1−:=

Psat i T,( ) e

AiBi

TdegC

Ci+−

kPa⋅:= kiPsat i T,( )

P:=

Guess: V 0.5:=

Given

1

n

i

zi ki⋅

1 V ki 1−( )⋅+∑=

1= Eq. (10.17)

V Find V( ):= V 0.656= Ans.

Eq. (10.16) yizi ki⋅

1 V ki 1−( )⋅+:=

320

γ1 x1 x2,( ) exp A x22⋅( ):= γ2 x1 x2,( ) exp A x1

2⋅( ):=

P x1 x2,( ) x1 γ1 x1 x2,( )⋅ Psat1⋅ x2 γ2 x1 x2,( )⋅ Psat2⋅+:=

(a) BUBL P calculation: x1 z1:= x2 1 x1−:=

Pbubl P x1 x2,( ):= Pbubl 56.745= Ans.

DEW P calculation: y1 z1:= y2 1 y1−:=

Guess: x1 0.5:= P'Psat1 Psat2+

2:=

Given y1 P'⋅ x1 γ1 x1 1 x1−,( )⋅ Psat1⋅=

P' x1 γ1 x1 1 x1−,( )⋅ Psat1⋅1 x1−( ) γ2 x1 1 x1−,( )⋅ Psat2⋅+

...=

x1

Pdew

⎛⎜⎝

⎞

⎠Find x1 P',( ):= Pdew 43.864= Ans.

10.13 H1 200 bar⋅:= Psat2 0.10 bar⋅:= P 1 bar⋅:=

Assume at 1 bar that the vapor is an ideal gas. The vapor-phase fugacitiesare then equal to the partial presures. Assume the Lewis/Randall ruleapplies to concentrated species 2 and that Henry's law applies to dilutespecies 1. Then:

y1 P⋅ H1 x1⋅= y2 P⋅ x2 Psat2⋅= P y1 P⋅ y2 P⋅+=

x1 x2+ 1= P H1 x1⋅ 1 x1−( ) Psat2⋅+=

Solve for x1 and y1:

x1P Psat2−

H1 Psat2−:= y1

H1 x1⋅

P:=

x1 4.502 10 3−×= y1 0.9= Ans.

10.16 Pressures in kPaPsat1 32.27:= Psat2 73.14:= A 0.67:= z1 0.65:=

321

A 0.95:=

γ1 x1 x2,( ) exp A x22⋅( ):= γ2 x1 x2,( ) exp A x1

2⋅( ):=

P x1 x2,( ) x1 γ1 x1 x2,( )⋅ Psat1⋅ x2 γ2 x1 x2,( )⋅ Psat2⋅+:=

y1 x1( )x1 γ1 x1 1 x1−,( )⋅ Psat1⋅

P x1 1 x1−,( ):=

(a) BUBL P calculation: x1 0.05:= x2 1 x1−:=

Pbubl P x1 x2,( ):= Pbubl 47.971= Ans.y1 x1( ) 0.196=

(b) DEW P calculation: y1 0.05:= y2 1 y1−:=

Guess: x1 0.1:= P'Psat1 Psat2+

2:=

The pressure range for two phases is from the dewpoint to thebubblepoint: From 43.864 to 56.745 kPa

(b) BUBL P calculation: x1 0.75:= x2 1 x1−:=

y1 x1( )x1 γ1 x1 1 x1−,( )⋅ Psat1⋅

P x1 1 x1−,( ):=

The fraction vapor, by material balance is:

Vz1 x1−

y1 x1( ) x1−:= V 0.379= P x1 x2,( ) 51.892= Ans.

(c) See Example 10.3(e).

α12.0γ1 0 1,( ) Psat1⋅

Psat2:= α12.1

Psat1γ2 1 0,( ) Psat2⋅

:=

α12.0 0.862= α12.1 0.226=

Since alpha does not pass through 1.0 for 0<x1<1, there is noazeotrope.

10.17 Psat1 79.8:= Psat2 40.5:=

322

Ans.

10.18 Psat1 75.20 kPa⋅:= Psat2 31.66 kPa⋅:=

At the azeotrope: y1 x1= and γ iP

Psati=

Thereforeγ2

γ1

Psat1Psat2

= x1 0.294:= x2 1 x1−:=

lnγ1 A x22⋅= lnγ2 A x1

2⋅= lnγ2

γ1

⎛⎜⎝

⎞

⎠A x1

2 x22−( )⋅=

Whence A

lnPsat1Psat2

⎛⎜⎝

⎞

⎠x2

2 x12−

:= A 2.0998=

For x1 0.6:= x2 1 x1−:=

Given y1 P'⋅ x1 γ1 x1 1 x1−,( )⋅ Psat1⋅=

P' x1 γ1 x1 1 x1−,( )⋅ Psat1⋅1 x1−( ) γ2 x1 1 x1−,( )⋅ Psat2⋅+

...=

x1

Pdew

⎛⎜⎝

⎞

⎠Find x1 P',( ):= Pdew 42.191=

Ans.x1 0.0104=

(c) Azeotrope Calculation:

Guess: x1 0.8:= y1 x1:= PPsat1 Psat2+

2:=

Giveny1

x1 γ1 x1 1 x1−,( )⋅ Psat1⋅

P= x1 0≥ x1 1≤ x1 y1=

P x1 γ1 x1 1 x1−,( )⋅ Psat1⋅ 1 x1−( ) γ2 x1 1 x1−,( )⋅ Psat2⋅+=

xaz1

yaz1

Paz

⎛⎜⎜⎜⎜⎝

⎞⎟⎟

⎠

Find x1 y1, P,( ):=

xaz1

yaz1

Paz

⎛⎜⎜⎜⎜⎝

⎞⎟⎟

⎠

0.857

0.857

81.366

⎛⎜⎜⎜⎝

⎞⎟⎠

=

323

Vz1 x1−

y1 x1−= For 0 V≤ 1≤ 0.6013 z1≤ 0.65≤ Ans. (a)

(c) Azeotrope calculation:

Guess: x1 0.6:= y1 x1:= PPsat1 Psat2+

2:=

γ1 x1( ) exp A 1 x1−( )2⋅⎡⎣ ⎤⎦:= γ2 x1( ) exp A x12⋅( ):=

Given P x1 γ1 x1( )⋅ Psat1⋅ 1 x1−( ) γ2 x1( )⋅ Psat2⋅+=

y1x1 γ1 x1( )⋅ Psat1⋅

P= x1 0≥ x1 1≤ x1 y1=

x1

y1

P

⎛⎜⎜⎜⎝

⎞

⎟

⎠

Find x1 y1, P,( ):=

x1

y1

P

⎛⎜⎜⎜⎝

⎞

⎟

⎠

0.592

0.592

1.673

⎛⎜⎜⎜⎝

⎞⎟⎠

= Ans.

γ1 exp A x22⋅( ):= γ2 exp A x1

2⋅( ):= P x1 γ1⋅ Psat1⋅ x2 γ2⋅ Psat2⋅+:=

y1x1 γ1⋅ Psat1⋅

P:= P 90.104kPa= y1 0.701= Ans.

10.19 Pressures in bars: Psat1 1.24:= Psat2 0.89:=

A 1.8:= x1 0.65:= x2 1 x1−:=

γ1 exp A x22⋅( ):= γ2 exp A x1

2⋅( ):=

P x1 γ1⋅ Psat1⋅ x2 γ2⋅ Psat2⋅+:= y1x1 γ1⋅ Psat1⋅

P:=

y1 0.6013= P 1.671= Answer to Part (b)

By a material balance,

324

γ2 x1 x2,( ) exp A x12⋅( ):=

P x1 T,( ) x1 γ1 x1 1 x1−,( )⋅ P1sat T( )⋅1 x1−( ) γ2 x1 1 x1−,( )⋅ P2sat T( )⋅+

...:=

y1 x1 T,( )x1 γ1 x1 1 x1−,( )⋅ P1sat T( )⋅

P x1 T,( ):= F 1:=

Guesses: V 0.5:= L 0.5:= T 100:=

Given

F L V+= z1 F⋅ x1 L⋅ y1 x1 T,( ) V⋅+= p P x1 T,( )=

L

V

T

⎛⎜⎜⎜⎝

⎞⎟⎠

Find L V, T,( ):=

L

V

T

⎛⎜⎜⎜⎝

⎞⎟⎠

0.431

0.569

59.531

⎛⎜⎜⎜⎝

⎞⎟⎠

=

T 59.531= (degC) y1 x1 T,( ) 0.307= Ans.

10.20 Antoine coefficients: P in kPa; T in degC

Acetone(1): A1 14.3145:= B1 2756.22:= C1 228.060:=

Methanol(2): A2 16.5785:= B2 3638.27:= C2 239.500:=

P1sat T( ) exp A1B1

T C1+−

⎛⎜⎝

⎞

⎠:= P2sat T( ) exp A2

B2

T C2+−

⎛⎜⎝

⎞

⎠:=

A 0.64:= x1 0.175:= z1 0.25:= p 100:= (kPa)

γ1 x1 x2,( ) exp A x22⋅( ):=

325

Ans.P 0.137bar=Px1 γ1⋅ Psat1 T( )⋅

y1:=

Ans.T 376.453K=T Find T( ):=Psat1 T( )

Psat2 T( )

x2 γ2⋅ y1⋅

x1 γ1⋅ y2⋅=Given

γ2 e0.93 x1

2⋅:=γ1 e

0.93 x22⋅

:=y2 1 y1−:=x2 1 x1−:=

Psat2 T( ) e

A2B2TK⎛⎜⎝

⎞⎠

−

bar⋅:=Psat1 T( ) e

A1B1TK⎛⎜⎝

⎞⎠

−

bar⋅:=

B2 6254.0:=A2 11.63:=B1 2572.0:=A1 10.08:=

T 300 K⋅:=Guess:y1 0.95:=x1 0.002:=10.22

326