diffusion of liquids through stagnant non diffusing air grp report

5. Data and Results:

Course: CHEP510L1 Experiment No: 5

Group No: Section: CH51FA1

Group Members: Date Performed: 03 August 2015

Caparros, Shamir Date Submitted: 24 August 2015

Escobia, Shiela Mae Instructor: Engr. Robert E. Delfin

Liwanag, Mary Christine

Mendoza, Kinski

Racho, Carlo Angelo

Togonon, Patricia

Villanueva, Kristine Ann

Trial 1: 50 0C

Liquid

Chapman and Engskog Equation

(m2/s)

Other Empirical Equation

(m2/s)

Capillary Tube Method (m2/s)

Ethanol 1.406334968 x 10-5 1.44596281 x 10-5 5.5647848 x 10-4

Ethyl acetate 9.955377601 x 10-6 1.00416571 x 10-5 1.583214884 x 10-4

Methanol 1.787372092 x 10-5 1.752859925 x 10-5 5.500658651 x 10-4

Trial 2: 65 0C

Liquid


(m2/s)


(m2/s)


Ethanol 1.528793491 x 10-5 1.44596281 x 10-5 7.185481761 x 10-4


Methanol 1.945884181 x 10-5 1.898230604 x 10-5 1.60063811 x 10-4

Trial 3: 80 0C

Liquid


(m2/s)


(m2/s)


Ethanol 1.656290956 x 10-5 1.688995378 x 10-5 9.893326824 x 10-5


Methanol 2.110761861 x 10-5 2.048029245 x 10-5 4.55712521 x 10-5

6. Calculations:

TRIAL 1: 50 0C

A. ETHANOL Given: T = 323.15 K zl = 0.03 m

zt = 0.0205 m t = 10 min = 600 s PA1 = 0.2921063 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 42.07 kg/kmol MB = 29 kg/kmol ρA, L = 762.8798 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method Solution: a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M

From Appendix 19 of MSH:

ε/k (K) σ M

Ethanol 362.6 4.530 42.07

Air 78.6 3.711 29

Thus,

AB

4.530 + 3.711σ = = 4.1205

2

AB A B

AB

ε /k = ε /k × ε /k = 362.6 × 78.6

ε /k = 168.8204964

kT 323.15 K =

ε 168.8204964 K

kT = 1.914163309

ε

From Appendix 19 of MSH, σD = 1.091167338 Substituting to the working equation:

3 1/27 2

AB 2

2-5

AB

1.8583 10 × 323.15 K 1 1D = +

42.07 291 atm 4.1205 A 1.091167338

mD = 1.406334968 × 10 s

b) Fuller Method Working Equation:

12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v

From Table 6.2-2 of Geankoplis,

A

A

B

Ethanol:

v = 2(16.5) + 6(1.98) + 1(5.48)

v = 50.36

Air:

v = 20.1

Substituting to the working equation:

12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 323.15 K +

42.07 29D =

1.0 atm 50.36 + 20.1

mD = 1.44596281 × 10 s

c) Capillary Tube Method Working Equation:

2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2

B,LM

A1

A2

B,LM

B,LM

0.292

P - P - P - PP =

P - Pln

P - P

1 - - 1 - 0P =

1 - ln

1 - 0

P = 0.84

1063

0.2

5554243

9

1

21063

atm


3

2 23

AB

2-4

AB

-3 kg m -atm762.8798 0.8455542431 atm 323.15 K 0.03 m - 0.0205 mkmol-KmD =

kg 2600 s 1atm 42.07 - 0 atmkmol

mD = 5.564647848 × 10

82.057 x 10

0.29

s

21063

B. ETHYL ACETATE Given: T = 323.15 K zl = 0.0205 m zt = 0.009 m t = 10 min = 600 s PA1 = 0.3712193 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 88.11 kg/kmol MB = 29 kg/kmol ρA, L = 863.3743 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M

From Table 2.2 of Dutta:

ε/k (K) σ M

Ethyl acetate 521.3 5.205 88.11

Air 78.6 3.711 29

Thus,

AB

5.205 + 3.711σ = = 4.458

2

AB A B

AB

ε /k = ε /k × ε /k = 521.3 × 78.6

ε /k = 202.4207993

kT 323.15 K =

ε 202.4207993 K

kT = 1.596426855

ε


3 1/27 2

AB 2

2-6

AB

1.168071

1.8583 10 × 323.15 K 1 1D = +

88.11 291 atm 4.458 A

mD = 9.955377601 × 10

9

s

44


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Ethanol:

v = 4(16.5) + 8(1.98) + 2(5.48)

v = 92.8

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 323.15 K +

88.11 29D =

1.0 atm 92.8 + 20.1

mD = 1.00416571 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2

B,LM

A1

A2

B,LM

B,LM

0.371

P - P - P - PP =

P - Pln

P - P

1 - - 1 - 0P =

1 - ln

1 - 0

P = 0.80

2193

0.3

0088614

7

3

12193

atm


3

2 23

AB

2-4

AB

-3 kg m -atm0.8000886143 atm 323.15 K 0.0205 m - 0.009 mkmol-KmD =

kg 2600 s 1atm 88.11

863.3743 82.057 x 10

0.3 - 0 atmkmol

mD = 1.583214884 × 10

7121

9

s

3

C. METHANOL Given: T = 323.15 K zl = 0.032 m zt = 0.014 m t = 10 min = 600 s PA1 = 0.5480075 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 32.04 kg/kmol MB = 29 kg/kmol ρA, L = 764.9058 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method Solution: a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Methanol 481.8 3.626 32.04

Air 78.6 3.711 29

Thus,

AB

3.626 + 3.711σ = = 3.6685

2

AB A B

AB

ε /k = ε /k × ε /k = 481.8 × 78.6

ε /k = 194.6008222

kT 323.15 K =

ε 194.6008222 K

kT = 1.660578801

ε


3 1/27 2

AB 2

2-5

AB

1.8583 10 × 323.15 K 1 1D = +

32.04 291 atm 3.6685 A 1.150249512

mD = 1.787372092 × 10 s


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Methanol:

v = 1(16.5) + 4(1.98) + 1(5.48)

v = 29.9

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 323.15 K +

42.07 29D =

1.0 atm 29.9 + 20.1

mD = 1.752859925 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2

B,LM

A1

A2

B,LM

B,LM

0.54

P - P - P - PP =

P - Pln

P - P

1 - - 1 - 0P =

1 - ln

1 - 0

P = 0.

80075

0.5

69010781

4

2

80075

atm


3

2 23

AB

2-4

AB



mD = 5.500659651 × 10

82.057 x 10

0.54

s

80075

TRIAL 2: 65 0C

A. ETHANOL Given: T = 338.15 K zl = 0.047 m zt = 0.0265 m t = 10 min = 600 s PA1 = 0.5778703 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 42.07 kg/kmol

MB = 29 kg/kmol ρA, L = 748.2964 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method Solution: a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Ethanol 362.6 4.530 42.07

Air 78.6 3.711 29

Thus,

AB

4.530 + 3.711σ = = 4.1205

2

AB A B

AB

ε /k = ε /k × ε /k = 362.6 × 78.6

ε /k = 168.8204964

kT 338.15 K =

ε 168.8204964 K

kT = 2.003015079

ε


3 1/27 2

AB 2

2-5

AB

1.8583 10 × 338.15 K 1 1D = +

42.07 291 atm 4.1205 A 1.074457286

mD = 1.528793491 × 10 s


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Ethanol:

v = 2(16.5) + 6(1.98) + 1(5.48)

v = 50.36

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 338.15 K +

42.07 29D =

1.0 atm 50.36 + 20.1

mD = 1.565457487 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2

B,LM

A1

A2

B,LM

B,LM

0.577

P - P - P - PP =

P - Pln

P - P

1 - - 1 - 0P =

1 - ln

1 - 0

P = 0.67

8703

0.5

0039090

7

9

78703

atm


3

2 23

AB

2-4

AB



mD = 7.185481761 × 10

82.057 x 10

0.57

s

78703

B. ETHYL ACETATE Given: T = 338.15 K zl = 0.0265 m zt = 0.018 m t = 10 min = 600 s PA1 = 0.6566834 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 88.11 kg/kmol MB = 29 kg/kmol ρA, L = 844.4323 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Ethyl acetate 521.3 5.205 88.11

Air 78.6 3.711 29

Thus,

AB

5.205 + 3.711σ = = 4.458

2

AB A B

AB

ε /k = ε /k × ε /k = 521.3 × 78.6

ε /k = 202.4207993

kT 338.15 K =

ε 202.4207993 K

kT = 1.670529912

ε


3 1/27 2

AB 2

2-5

AB

1.147662

1.8583 10 × 338.15 K 1 1D = +

88.11 291 atm 4.458 A

mD = 1.084603723 × 10

2

s

23


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Ethanol:

v = 4(16.5) + 8(1.98) + 2(5.48)

v = 92.8

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 338.15 K +

88.11 29D =

1.0 atm 92.8 + 20.1

mD = 1.087150181 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2

B,LM

A1

A2

B,LM

B,LM

0.656

P - P - P - PP =

P - Pln

P - P

1 - - 1 - 0P =

1 - ln

1 - 0

P = 0.61

6834

0.6

4238175

5

1

66834

atm


3

2 23

AB

25

AB

-3 kg m -atm0.6142381751 atm 338.15 K 0.0265 m - 0.018 mkmol-KmD =

kg 2600 s 1atm 88.11

844.4323 8

0.6566834 - 0 atmkmol

mD = 7.840488455 10 s

2.057 x 10


3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Methanol 481.8 3.626 32.04

Air 78.6 3.711 29

Thus,

AB

3.626 + 3.711σ = = 3.6685

2

AB A B

AB

ε /k = ε /k × ε /k = 481.8 × 78.6

ε /k = 194.6008222

kT 338.15 K =

ε 194.6008222 K

kT = 1.737659668

ε


3 1/27 2

AB 2

2-5

AB

1.8583 10 × 338.15 K 1 1D = +

32.04 291 atm 3.6685 A 1.13096168

mD = 1.945884181 × 10 s


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Methanol:

v = 1(16.5) + 4(1.98) + 1(5.48)

v = 29.9

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 338.15 K +

42.07 29D =

1.0 atm 29.9 + 20.1

mD = 1.898230604 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2B, M

B, M

B, M

P + PP =

2

1.018446 atm + 0 atmP =

2

P = 0.509223 atm


3

2 23

AB

2-4

AB

-3 kg m -atm749.2904 0.509223 atm atm 338.15 K 0.041 m - 0.033 mkmol-KmD =


mD = 1.60063811 × 10

82.057 x 10

1.0

s

18446

TRIAL 3: 80 0C

A. ETHANOL Given: T = 353.15 K zl = 0.0225 m zt = 0.006 m t = 10 min = 600 s PA1 = 1.068375 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 42.07 kg/kmol MB = 29 kg/kmol ρA, L = 733.0312 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method

Solution: a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Ethanol 362.6 4.530 42.07

Air 78.6 3.711 29

Thus,

AB

4.530 + 3.711σ = = 4.1205

2

AB A B

AB

ε /k = ε /k × ε /k = 362.6 × 78.6

ε /k = 168.8204964

kT 353.15 K =

ε 168.8204964 K

kT = 2.09186685

ε


3 1/2-7 2

AB 2

2-5

AB

1.8583 × 10 × 353.15 K 1 1D = +

42.07 291 atm 4.1205 A 1.058463967

mD = 1.656290956 × 10 s


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Ethanol:

v = 2(16.5) + 6(1.98) + 1(5.48)

v = 50.36

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 353.15 K +

42.07 29D =

1.0 atm 50.36 + 20.1

mD = 1.688995378 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2B, M

B, M

B, M

P + PP =

2

1.068375 atm + 0P =

2

P = 0.5341875 atm


-3 3

2 23

AB

2-5

AB

kg m -atm0.5341875 atm 353.15 K 0.0225 m - 0.006 mkmol-KmD =

kg 2600 s 1atm 42.07

733.0312 82.057 x 10

1.0683 - 0 atmkmol

mD = 9.893326824 × 10 s

75

B. ETHYL ACETATE Given: T = 353.15 K zl = 0.033 m zt = 0.007 m t = 10 min = 600 s

PA1 = 1.095756 atm PA2 = 0 atm P = 1 atm R = 82.057 x 10-3 m3-atm/kmol-K MA = 88.11 kg/kmol MB = 29 kg/kmol ρA, L = 824.8012 kg/m3 Required: DAB by a) Chapman and Engskog Equation, b) Fuller Method, and c) Capillary Tube Method a) Chapman and Engskog Equation Working Equation:

3-7 2

AB 2

AB D,AB A B

1.8583×10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Ethyl acetate 521.3 5.205 88.11

Air 78.6 3.711 29

Thus,

AB

5.205 + 3.711σ = = 4.458

2

AB A B

AB

ε /k = ε /k × ε /k = 521.3 × 78.6

ε /k = 202.4207993

kT 353.15 K =

ε 202.4207993 K

kT = 1.744632969

ε


3 1/27 2

AB 2

2-5

AB

1.129288

1.8583 10 × 353.15 K 1 1D = +

88.11 291 atm 4.458 A

mD = 1.176400387 × 10

0

s

87


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Ethanol:

v = 4(16.5) + 8(1.98) + 2(5.48)

v = 92.8

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 353.15 K +

88.11 29D =

1.0 atm 92.8 + 20.1

mD = 1.172942507 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2B, M

B, M

B, M

1.095756

P + PP =

2

P = 2

P = 0.

atm + 0atm

547878 atm


-3 3

2 23

AB

2-4

AB


kg 2600 s 1atm 88.11

824.8012 8

1.095756 - 0 atmkmol

mD = 1.17549565 × 10

2.057

s

x 10


3-7 2

AB 2

AB D,AB A B

1.8583 × 10 × T 1 1D = +

Pσ Ω M M


ε/k (K) σ M

Methanol 481.8 3.626 32.04

Air 78.6 3.711 29

Thus,

AB

3.626 + 3.711σ = = 3.6685

2

AB A B

AB

ε /k = ε /k × ε /k = 481.8 × 78.6

ε /k = 194.6008222

kT 353.15 K =

ε 194.6008222 K

kT = 1.814740534

ε


3 1/27 2

AB 2

2-5

AB

1.8583 10 × 353.15 K 1 1D = +

32.04 291 atm 3.6685 A 1.112757083

mD = 2.110761861 × 10 s


12

-7 1.75

A B

AB 21 1

3 3A B

1 11.00 × 10 T +

M MD =

P v + v


A

A

B

Methanol:

v = 1(16.5) + 4(1.98) + 1(5.48)

v = 29.9

Air:

v = 20.1


12

1.75-7

AB 21 1

3 3

2-5

AB

1 11.00 × 10 353.15 K +

42.07 29D =

1.0 atm 29.9 + 20.1

mD = 2.048029245 × 10 s


2 2A,L B,LM l t

AB

A A1 A2

ρ P RT z - zD =

tPM P - P 2

Solving for PB, LM:

A1 A2B, M

B, M

B, M

P + PP =

2

1.783045 atm + 0P =

2

P = 0.8915225 atm


-3 3

2 23

AB

2-5

AB


kg 2600 s 1atm 32.04

732.9796

1.783045 - 0 atmkmol

mD = 4.557712521 × 10 s

82.057 x 10

Technological Institute of the Philippines – Manila

363 P. Casal Street, Quiapo, Manila

College of Engineering and Architecture

Chemical Engineering Department

CHEMICAL ENGINEERING LABORATORY 2

(Diffusion of Liquids through Stagnant Non-Diffusing Air)

Submitted by:

CAPARROS, SHAMIR

ESCOBIA, SHIELA MAE

LIWANAG, MARY CHRISTINE

MENDOZA, KINSKI

RACHO, CARLO ANGELO

TOGONON, PATRICIA

VILLANUEVA, KRISTINE ANN

Submitted to:

ENGR. ROBERT E. DELFIN

Date Submitted:

24 AUGUST 2015

diffusion of liquids through stagnant non diffusing air grp report

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