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 0 C Liquid Chapman and Engskog Equation (m 2 /s) Other Empirical Equation (m 2 /s) Capillary Tube Method (m 2 /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 0 C Liquid Chapman and Engskog Equation (m 2 /s) Other Empirical Equation (m 2 /s) Capillary Tube Method (m 2 /s) Ethanol 1.528793491 x 10 -5 1.44596281 x 10 -5 7.185481761 x 10 -4 Ethyl acetate 1.084603723 x 10 -5 1.087150181 x 10 -5 7.840488455 x 10 -5 Methanol 1.945884181 x 10 -5 1.898230604 x 10 -5 1.60063811 x 10 -4 Trial 3: 80 0 C Liquid Chapman and Engskog Equation (m 2 /s) Other Empirical Equation (m 2 /s) Capillary Tube Method (m 2 /s) Ethanol 1.656290956 x 10 -5 1.688995378 x 10 -5 9.893326824 x 10 -5 Ethyl acetate 1.176400387 x 10 -5 1.172942507 x 10 -5 1.17549565 x 10 -4 Methanol 2.110761861 x 10 -5 2.048029245 x 10 -5 4.55712521 x 10 -5 6. Calculations: TRIAL 1: 50 0 C A. ETHANOL Given: T = 323.15 K zl = 0.03 m
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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
Chapman and Engskog Equation
(m2/s)
Other Empirical Equation
(m2/s)
Capillary Tube Method (m2/s)
Ethanol 1.528793491 x 10-5 1.44596281 x 10-5 7.185481761 x 10-4
Ethyl acetate 1.084603723 x 10-5 1.087150181 x 10-5 7.840488455 x 10-5
Methanol 1.945884181 x 10-5 1.898230604 x 10-5 1.60063811 x 10-4
Trial 3: 80 0C
Liquid
Chapman and Engskog Equation
(m2/s)
Other Empirical Equation
(m2/s)
Capillary Tube Method (m2/s)
Ethanol 1.656290956 x 10-5 1.688995378 x 10-5 9.893326824 x 10-5
Ethyl acetate 1.176400387 x 10-5 1.172942507 x 10-5 1.17549565 x 10-4
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
Substituting to the working equation:
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
From Appendix 19 of MSH, σD = 1.168071944 Substituting to the working equation:
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
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 = 4(16.5) + 8(1.98) + 2(5.48)
v = 92.8
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 +
88.11 29D =
1.0 atm 92.8 + 20.1
mD = 1.00416571 × 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
0.371
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
Substituting to the working equation:
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
From Appendix 19 of MSH:
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
From Appendix 19 of MSH, σD = 1.150249512 Substituting to the working equation:
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
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
Methanol:
v = 1(16.5) + 4(1.98) + 1(5.48)
v = 29.9
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 29.9 + 20.1
mD = 1.752859925 × 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
0.54
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
Substituting to the working equation:
3
2 23
AB
2-4
AB
-3 kg m -atm764.9058 0.6901078120 atm 323.15 K 0.032 m - 0.014 mkmol-KmD =
kg 2600 s 1atm 32.04 - 0 atmkmol
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
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 338.15 K =
ε 168.8204964 K
kT = 2.003015079
ε
From Appendix 19 of MSH, σD = 1.074457286 Substituting to the working equation:
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
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 × 10 338.15 K +
42.07 29D =
1.0 atm 50.36 + 20.1
mD = 1.565457487 × 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
0.577
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
Substituting to the working equation:
3
2 23
AB
2-4
AB
-3 kg m -atm748.2964 0.6700390909 atm 338.15 K 0.047 m - 0.0265 mkmol-KmD =
kg 2600 s 1atm 42.07 - 0 atmkmol
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
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 338.15 K =
ε 202.4207993 K
kT = 1.670529912
ε
From Appendix 19 of MSH, σD = 1.147662223 Substituting to the working equation:
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
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 = 4(16.5) + 8(1.98) + 2(5.48)
v = 92.8
Air:
v = 20.1
Substituting to the working equation:
12
1.75-7
AB 21 1
3 3
2-5
AB
1 11.00 × 10 338.15 K +
88.11 29D =
1.0 atm 92.8 + 20.1
mD = 1.087150181 × 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
0.656
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
Substituting to the working equation:
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
C. METHANOL Given: T = 338.15 K zl = 0.041 m zt = 0.033 m t = 10 min = 600 s PA1 = 1.018446 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 = 749.2904 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
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
From Appendix 19 of MSH, σD = 1.13096168 Substituting to the working equation:
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
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
Methanol:
v = 1(16.5) + 4(1.98) + 1(5.48)
v = 29.9
Air:
v = 20.1
Substituting to the working equation:
12
1.75-7
AB 21 1
3 3
2-5
AB
1 11.00 × 10 338.15 K +
42.07 29D =
1.0 atm 29.9 + 20.1
mD = 1.898230604 × 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 A2B, M
B, M
B, M
P + PP =
2
1.018446 atm + 0 atmP =
2
P = 0.509223 atm
Substituting to the working equation:
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 =
kg 2600 s 1atm 32.04 - 0 atmkmol
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
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 353.15 K =
ε 168.8204964 K
kT = 2.09186685
ε
From Appendix 19 of MSH, σD = 1.058463967 Substituting to the working equation:
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
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 × 10 353.15 K +
42.07 29D =
1.0 atm 50.36 + 20.1
mD = 1.688995378 × 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 A2B, M
B, M
B, M
P + PP =
2
1.068375 atm + 0P =
2
P = 0.5341875 atm
Substituting to the working equation:
-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
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 353.15 K =
ε 202.4207993 K
kT = 1.744632969
ε
From Appendix 19 of MSH, σD = 1.129288087 Substituting to the working equation:
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
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 = 4(16.5) + 8(1.98) + 2(5.48)
v = 92.8
Air:
v = 20.1
Substituting to the working equation:
12
1.75-7
AB 21 1
3 3
2-5
AB
1 11.00 × 10 353.15 K +
88.11 29D =
1.0 atm 92.8 + 20.1
mD = 1.172942507 × 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 A2B, M
B, M
B, M
1.095756
P + PP =
2
P = 2
P = 0.
atm + 0atm
547878 atm
Substituting to the working equation:
-3 3
2 23
AB
2-4
AB
kg m -atm0.547878 atm 353.15 K 0.033 m - 0.007 mkmol-KmD =
kg 2600 s 1atm 88.11
824.8012 8
1.095756 - 0 atmkmol
mD = 1.17549565 × 10
2.057
s
x 10
C. METHANOL Given: T = 353.15 K zl = 0.029 m zt = 0.026 m t = 10 min = 600 s PA1 = 1.783045 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 = 732.9796 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
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
From Appendix 19 of MSH, σD = 1.112757083 Substituting to the working equation:
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
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
Methanol:
v = 1(16.5) + 4(1.98) + 1(5.48)
v = 29.9
Air:
v = 20.1
Substituting to the working equation:
12
1.75-7
AB 21 1
3 3
2-5
AB
1 11.00 × 10 353.15 K +
42.07 29D =
1.0 atm 29.9 + 20.1
mD = 2.048029245 × 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 A2B, M
B, M
B, M
P + PP =
2
1.783045 atm + 0P =
2
P = 0.8915225 atm
Substituting to the working equation:
-3 3
2 23
AB
2-5
AB
kg m -atm0.8915225 atm 353.15 K 0.029 m - 0.026 mkmol-KmD =
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
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