1)

(a) Given x1 = 0.33 and

BUBL P Calculation

Step 1: Find P1sat

& P

Thus,

Thus,

Step 2: Find P.

Step 3: Calculate y

AkPaP sat

i −=/ln

7819.13/ln 1 =kPaP sat

932.13/ln 2 =kPaP sat

195.5=

P sat .1801 =

3076.4=

P sat 2597.742 =

( 45.18033.0=

sat PxPxP 211 +=

(satPx 11 1−+=

kPa3025.109=

sat

iii PxPy =

PPxysat

111 = =

UNIVERSITI MALAYSIA PERLIS

Pusat Pengajian Kejuruteraan Bioproses

ERT 206: Thermodynamics

Exercise 3 Solutions

= 0.33 and T= 100 °C, find y1 and P.

Calculation:

& P2sat

using Antoine equation.

Calculate y1 using Raoult’s law.

CCT

B

+°/

572.217100

81.27267819

+°−

C

625.217100

96.3056932

+°−

C

195

kPa45.

3076

kPa2597

) ( )( )kPakPa 2597.7433.0145 −+

satP2

) satPx 21−

kPa

( ) 5448.03025.10945.18033.0 == kPakPa

5448

(b) Given y1 = 0.33 and T= 100 °C, find x1 and P.

DEW P Calculation:

Step 1: Find P from Raoult’s Law assuming

Step 2: Find x

(c) Given x1 = 0.33 and P= 120 kPa, find y1 and T.

BUBL T Calculation:

Step 1: Start with α=1, find P2sat

Step 2: Find T using Antoine eq & substitute P2sat

obtained in step 1

∑ =i ix 1

satsat PyPyP

2211 //

1

+=

( ) 2597.74/33.0145.180/33.0

1

−+=

kPa156.92=

sat

iii PxPy =

satPPyx 111 =

( )( ) ( )45.180156.9233.0=

1685.0=

( )( ) ( )kPa

xx

PP sat 120

33.01133.0

120

21

2 =−+

=+

=α

CPA

BT

sat−

−=

2ln

625.217120ln9320.13

96.3056−

−=

C°= 67.116

Step 3: Find new α by substituting T

* find the expression for α first

-

* substituting T=116.67̊C in the equation;

Step 4: Repeat step 1 by using new α until similar value of α is obtained

2nd

iteration: 3rd

iteration:

4th

iteration: 5th

iteration:

Thus, T = 103.307°C

Step 5: Find P1sat

& find y1 using Raoult’s law

572.217/

81.27267819.13/ln 1

+°−=

CTkPaP sat

625.217/

96.3056932.13/ln 2

+°−=

CTkPaP sat

( )625.217/

96.3056

572.217/

81.27261501.0/ln 21

+°+

+°−−=

CTCTPP satsat

625.217/

96.3056

572.217/

81.27261501.0ln

+°+

+°−−=

CTCTα

625.21767.116

96.3056

572.21767.116

81.27261501.0ln

+°+

+°−−=

CCα

8362.0=

3076.2=α

3076.2=α

kPaP sat 8277.832 =

CT °= 0507.104

3985.2=newα

3985.2=α

CT °= 3502.103

40387.2=newα

kPaP sat 10714.822 =

CT °= 3094.103

4042.2=newα

40387.2=α

kPaP sat 00765.822 =4042.2=α

4042.2=newα

kPaP sat 00.822 =

CT °= 307.103

satsat PP 21 α= PPxy sat

111 =

( )kPa00.824042.2=

kPa144.197=

( ) kPakPa 120144.19733.0=

542.0=

(d) Given y1 = 0.33 and P= 120 kPa, find x1 and T.

DEW T Calculation:

Step 1: Start with α=1, find P1sat

Step 2: Find T using Antoine eq & substitute P1sat

obtained in step 1

Step 3: Find new α by substituting T

Step 4: Repeat step 1 by using new α until similar value of α is obtained

2nd

iteration: 3rd

iteration:

4th

iteration: 5th

iteration:

( )α211 yyPP sat +=

( )[ ]167.033.0120 +=

kPa120=

CPA

BT

sat−

−=

1ln

572.217120ln7819.13

81.2726−

−=

C°= 595.85

625.217/

96.3056

572.217/

81.27261501.0ln

+°+

+°−−=

CTCTα

625.217/595.85

96.3056

572.217/595.85

81.27261501.0ln

+°+

+°−−=

CCα

5527.2=α

5527.2=α

kPaP sat 8347.2441 =

CT °= 7005.111

34215.2=newα

34215.2=α

CT °= 8766.108

36250.2=newα

kPaP sat 9091.2271 =

CT °= 1563.109

3604.2=newα

36250.2=α

PaP sat 5446.2291 =3604.2=α

3606.2=newα

kPaP sat 3807.2291 =

CT °= 1284.109

Step 5: Find x1 using Raoult’s law

2) Assuming the validity of the De Priester charts, make the following VLE calculations

for methane(1)/ethylene(2)/ethane(3) system:

(a) BUBL P, given x1 = 0.10, x2 = 0.50, and T = -60 °F

Component xi P = 200 psia P = 250 psia P = 215 psia

Ki yi=Kixi Ki yi=Kixi Ki yi=Kixi

Methane 0.10 5.600 0.560 4.600 0.460 5.150 0.515

Ethylene 0.50 0.700 0.350 0.575 0.288 0.650 0.325

Ethane 0.40 0.445 0.178 0.380 0.152 0.420 0.168

SUM = 1.088 SUM = 0.900

SUM = 1.008

(close enough)

Thus, BUBL P = 215 psia

(b) DEW P, given y1 = 0.50, y2 = 0.25, and T = -60 °F

Thus, DEW P = 210 psia

Component yi P = 190 psia P = 200 psia

Ki xi=yi/Ki Ki xi=yi/Ki

Methane 0.50 5.900 0.085 5.600 0.089

Ethylene 0.25 0.730 0.342 0.700 0.357

Ethane 0.25 0.460 0.543 0.445 0.562

SUM = 0.971 SUM = 1.008

(close enough)

PPxy sat

111 =

satPPyx 111 =

1726.0=

( ) kPakPa 3807.22912033.0=

(c) BUBL T, given x1 = 0.12, x2 = 0.40, and P = 250 psia

Component xi T = -50 F T = -60 F T = -57 F

Ki yi=Kixi Ki yi=Kixi Ki yi=Kixi

Methane 0.12 4.900 0.588 4.600 0.552 4.700 0.564

Ethylene 0.40 0.680 0.272 0.570 0.228 0.615 0.246

Ethane 0.48 0.450 0.216 0.380 0.182 0.405 0.194

SUM = 1.076 SUM = 0.962

SUM = 1.004

(close enough)

Thus, BUBL T = -57 F

(d) DEW T, given y1 = 0.43, y2 = 0.36, and P = 250 psia

Thus, DEW T = -45 F

Component yi T = -40 F T = -50 F T = -45 F

Ki xi= yi /Ki Ki xi= yi /Ki Ki xi= yi /Ki

Methane 0.43 5.200 0.083 4.900 0.088 5.050 0.085

Ethylene 0.36 0.800 0.450 0.680 0.529 0.740 0.486

Ethane 0.21 0.520 0.404 0.450 0.467 0.485 0.433

SUM = 0.937 SUM = 1.084

SUM = 1.005

(close enough)