1 00:00:18,720 --> 00:00:24,970 Welcome to the seventh lecture of module one which is on diffusion mass transfer. So, before 2 00:00:24,970 --> 00:00:31,470 going to this lecture, let us have recap on the previous lecture what we have discussed. 3 00:00:31,470 --> 00:00:37,440 In our previous lecture, we have considered the diffusion coefficient measurements and 4 00:00:37,440 --> 00:00:44,440 predictions. First, we have considered the gas phase diffusion coefficient measurements. 5 00:01:01,880 --> 00:01:08,880 As we have discussed, that there are several methods available to measure the gas-phase 6 00:01:15,800 --> 00:01:22,800 diffusivity, but we have considered only two simple methods to determine experimentally 7 00:01:24,430 --> 00:01:31,430 the gas-phase diffusivity; one is twin bulb method, and the second one we have considered 8 00:01:44,100 --> 00:01:51,100 is Stefan tube method. So, for both the cases we have obtained the governing equations to 9 00:02:02,100 --> 00:02:08,310 calculate the diffusivity from the experiments, and we have explained both the methods with 10 00:02:08,310 --> 00:02:15,260 relevant examples. So, in this lecture we will continue with our discussion of gas-phase 11 00:02:15,260 --> 00:02:21,920 diffusivity prediction by different correlations available in the literature. 12 00:02:21,920 --> 00:02:28,920 In this lecture we will consider gas-phase diffusivity prediction, and then we will consider 13 00:02:45,620 --> 00:02:52,620 liquid-phase diffusivity measurements, and liquid-phase diffusivity prediction. Let us 14 00:03:13,950 --> 00:03:20,950 consider the first case gas-phase diffusivity predictions. 15 00:03:29,270 --> 00:03:36,270 Although there are several experimental data available for different gaseous components, 16 00:03:36,349 --> 00:03:43,349 there are many components, which are required experimental data or shortage of experimental 17 00:03:43,379 --> 00:03:49,910 data. In those cases, instead of experimental data we can use the empirical correlations 18 00:03:49,910 --> 00:03:55,300 by which we can determine the diffusivity of the gas-phase components. 19 00:03:55,300 --> 00:04:02,300 A very simple, and reasonably accurate empirical equations which was suggested by Fuller, Schettler 20 00:04:04,709 --> 00:04:11,709 and Giddings in 1966 for binary, binary gas-phase diffusivity, gas-phase diffusivity up to moderate 21 00:04:38,240 --> 00:04:45,240 pressure, say up about 20 atmosphere. The correlations they proposed is diffusivity 22 00:04:57,560 --> 00:05:04,560 of component A to diffusivity in B is equal to 10 to the power minus 7 T to the power 23 00:05:10,440 --> 00:05:17,440 1.75 divided by P t into summation over V A to the power one-third plus summation over 24 00:05:32,540 --> 00:05:39,540 V B to the power one-third whole square multiplied by square root of 1 by M A plus 1 by M B and 25 00:05:50,970 --> 00:05:57,970 the unit in meter square per second, where T is equal to temperature in Kelvin; M A, M B, these are the molecular 26 00:06:15,320 --> 00:06:22,320 weights of A and B; P t, total pressure in atmosphere; V A and V B are the atomic diffusion 27 00:06:46,180 --> 00:06:53,180 volume in meter cube. So, the diffusivity obtained from this relation is in meter square 28 00:07:08,300 --> 00:07:09,180 per second. 29 00:07:09,180 --> 00:07:16,180 Now, let us take an example to calculate the diffusivity from this correlation. For example, 30 00:07:17,330 --> 00:07:24,330 for mixtures of carbon monoxide and hydrogen, predict the diffusivity using Fuller et al. 31 00:07:25,280 --> 00:07:32,280 1966 formula for the following conditions. The first one is at 1 atmosphere pressure 32 00:07:32,460 --> 00:07:38,410 and 100 degree centigrade temperature; the second problem is at 2 atmosphere pressure 33 00:07:38,410 --> 00:07:44,340 and 100 degree centigrade temperature and the third one, at 1 atmosphere pressure, 200 34 00:07:44,340 --> 00:07:45,669 degree centigrade temperature. 35 00:07:45,669 --> 00:07:52,669 So, let us consider the first one at 1 atmosphere presser and 100 degree centigrade temperature. 36 00:07:53,639 --> 00:08:00,639 We know that diffusivity D AB will be equal to 10 to the power minus 7 T to the power 37 00:08:04,960 --> 00:08:11,960 1.75 divided by P t summation over V A to the power one-third plus summation over V 38 00:08:26,319 --> 00:08:33,319 B to the power one-third whole square and into root over bar 1 by M A plus 1 by M B. 39 00:08:45,970 --> 00:08:52,970 So, the atomic values, values, can be obtained from text book. 40 00:09:03,140 --> 00:09:10,140 Let us consider A is equal to carbon monoxide and B is equal to hydrogen. Now, summation 41 00:09:18,920 --> 00:09:25,920 of V A would be 1 into carbon plus 1 into oxygen, so this will be equal to 1. The molecular 42 00:09:42,579 --> 00:09:49,579 volume of carbon is 16.5 plus for oxygen it is 5.48, so it will be equal to 21.98. And 43 00:10:05,520 --> 00:10:12,520 molecular weight of carbon monoxide M A would be 1 into 12 plus 1 into 16, would be 28 kg 44 00:10:20,860 --> 00:10:27,860 per kmol. Similarly, for molecular volume for V B would be 2 into hydrogen, 2 into molecular 45 00:10:36,420 --> 00:10:43,420 volume 2.31, it is 4.62 and M B is 2 into 1 is equal to 2 kg per kmol. 46 00:10:57,189 --> 00:11:04,189 The temperature is given as 100 degree centigrade, is equal to 100 plus 273, this much Kelvin, 47 00:11:14,480 --> 00:11:21,480 so 373 Kelvin. And P t is given 1 atmosphere, so if we substitute, D AB would be equal to 10 to the power minus 48 00:11:36,110 --> 00:11:43,110 7, 373 temperature to the power 1.75 divided by 1 pressure into molecular volume of A is 49 00:11:54,679 --> 00:12:01,679 21.98 to the power one-third plus, for hydrogen 4.62 to the power one-third whole square into 50 00:12:12,589 --> 00:12:19,589 root over 1 by 28 plus 1 by 2 meter square per second. So, this will give you 1.159 into 51 00:12:28,920 --> 00:12:35,920 10 to the power minus 4 meter square per second. Now, let us consider the second problem, that 52 00:12:50,130 --> 00:12:57,130 is, at 2 atmosphere and 100 degree centigrade temperature. So, for this case, since we have 53 00:13:04,120 --> 00:13:10,360 already determined for the same system the diffusivity at a particular temperature and 54 00:13:10,360 --> 00:13:17,360 pressure, there is no need to calculate the diffusion coefficient for the same system 55 00:13:18,760 --> 00:13:25,760 at different conditions from scratch. So, we can use the diffusion co-efficient determined 56 00:13:28,290 --> 00:13:35,290 in the earlier problem and calculate with respect to the temperature and pressure corrections. 57 00:13:36,760 --> 00:13:43,760 So, keeping all the properties same for the components in problem two, at constant temperature, 58 00:13:46,150 --> 00:13:53,150 which is temperature is 100 degree centigrade remains same, the diffusivity D AB from the 59 00:13:53,860 --> 00:14:00,860 Fullers equation is inversely proportional with the total pressure. So, we can write, 60 00:14:03,110 --> 00:14:10,110 at that temperature it is inversely proportional with pressure, so we can write D AB1 or D 61 00:14:14,339 --> 00:14:21,339 AB2 by D AB1 will be P t 1 by P t 2. 62 00:14:26,079 --> 00:14:33,079 So, in this problem we can write D AB2 will be equal to P t 1 by P t 2 into D AB1. So, 63 00:14:45,059 --> 00:14:52,059 P t 1 is 1 atmosphere, P t 2 is 2 atmosphere, and D AB is we have obtained 1.156 into 10 64 00:15:07,240 --> 00:15:14,240 to the power minus 4 meter square per second. So, we can calculate D AB2 is equal to 1 by 65 00:15:23,500 --> 00:15:30,500 2 into 1.156 into 10 to the power minus 4 meter square per second, which is equal to 66 00:15:35,900 --> 00:15:42,900 0.58 into 10 to the power minus 4 meter square per second. 67 00:15:47,890 --> 00:15:54,890 So, for the third problem, similarly, since the temperature T 2 is given as 200 degree 68 00:16:02,910 --> 00:16:09,910 centigrade temperature and P t 2 is 1 atmosphere, which is remain same for the first problem, 69 00:16:12,160 --> 00:16:19,160 so we can write D AB is directly proportional to the T to the power of 1.75. So, we can 70 00:16:26,740 --> 00:16:33,740 write D AB2 by D AB1 will be equal to T 1, T 2 by T 1 to the power 1.75. 71 00:16:41,919 --> 00:16:48,919 T 2 is given as 200 plus 273 is equal to 473 K and D AB1 is 1.159 into 10 to the power 72 00:17:10,890 --> 00:17:17,890 minus 4 meter square per second and T 1 is 100 degree centigrade, which is 373 Kelvin. 73 00:17:26,069 --> 00:17:33,069 So, D AB2 would be 473 by 373 to the power 1.75 into 1.159 into 10 to the power minus 74 00:17:48,700 --> 00:17:55,700 4 meter square per second, which will be about 1.761 into 10 to the power minus 4 meter square 75 00:17:57,669 --> 00:18:03,809 per second. So, this is different conditions for the same systems. 76 00:18:03,809 --> 00:18:10,809 There is another important useful correlation, which is Chapman-Enskog equation. In this 77 00:18:12,679 --> 00:18:19,679 case, the equation is derived based on the kinetic theory of gasses and the diffusion 78 00:18:22,769 --> 00:18:29,580 coefficient calculated from this equation is strongly dependent on binary interaction 79 00:18:29,580 --> 00:18:36,580 parameters of the mixture pair. So the equation, which is given D AB is equal 80 00:18:41,730 --> 00:18:48,730 to 1.858 into 10 to the power minus 7 T to the power 1.5 divided by P t sigma AB square 81 00:19:02,850 --> 00:19:09,850 omega D into square root of 1 by M A plus 1 by M B, this is in meter square per second, 82 00:19:17,649 --> 00:19:24,649 where T is the temperature in Kelvin; M A, M B are the molecular weights of A and B; 83 00:19:40,029 --> 00:19:47,029 P t is the total pressure, total pressure, in atmosphere; sigma AB the characteristic 84 00:19:56,240 --> 00:20:03,240 length, characteristic length, of binary mixture in angstrom, this we can calculate using sigma 85 00:20:21,759 --> 00:20:28,759 A plus sigma B divided by 2, and omega D is the collision integral, integral, which is 86 00:20:39,919 --> 00:20:46,919 a function of K T by epsilon AB. This epsilon AB can be calculated from root over epsilon 87 00:20:59,940 --> 00:21:04,159 A into epsilon B. 88 00:21:04,159 --> 00:21:11,159 Now, let us take an example, very simple example of mixture of ammonia and hydrogen for this. 89 00:21:13,779 --> 00:21:20,779 Predict the diffusivity using this formula at 1 atmosphere pressure and 100 degree centigrade 90 00:21:21,269 --> 00:21:27,730 temperature. Now, let us consider A is ammonia, so M A 91 00:21:27,730 --> 00:21:34,730 is equal to NH 3 is equal to 17 and M B is hydrogen is 2, T is given, 100 degree centigrade, 92 00:21:48,940 --> 00:21:55,940 so which is equal to 100 plus 273 So, 373 Kelvin and P t is the total pressure, which 93 00:22:04,529 --> 00:22:11,529 is 1 atmosphere. Now, the Lennard-Jones potential parameters 94 00:22:30,850 --> 00:22:37,850 we can calculate for ammonia from the literature, we can, from the text book we can get 2.900 95 00:22:43,539 --> 00:22:50,539 angstrom, so that we can calculate the epsilon A by K is 558.3 and for hydrogen sigma B is 96 00:23:04,090 --> 00:23:11,090 2.827 angstroms and epsilon B by K is 59.7. 97 00:23:17,750 --> 00:23:24,750 So, using this value for this ammonia-hydrogen pair, for NH 3, which is A and H 2, which 98 00:23:37,470 --> 00:23:44,470 is B, this pair, we can calculate sigma AB, which is sigma A plus sigma B by 2 is equal 99 00:23:53,639 --> 00:24:00,639 to 2.900 plus 2.827 by 2 is equal to 2.8635 angstrom. So, epsilon AB by K we can obtain, 100 00:24:03,299 --> 00:24:10,299 epsilon A by K, epsilon B by K to the power half. So, it will be 558.3 into 59.7 to the 101 00:24:34,720 --> 00:24:41,720 power half is equal to 182.6. So, with this value K T by epsilon AB will be 373 by 182.6 102 00:24:54,230 --> 00:25:01,230 will give you 2.04. With these values from the collision integral 103 00:25:16,529 --> 00:25:23,529 we can calculate omega D, we can obtain 1.075. 104 00:25:24,419 --> 00:25:31,419 Now, if we substitute all these values in the equation D AB, will be 1.858 into 10 to 105 00:25:35,799 --> 00:25:42,799 the power minus 7, T is 100 degree centigrade, is 373 to the power 1.5 divided by 1 into 106 00:25:53,519 --> 00:26:00,519 2.8635 square into 1.075 multiplied by root over bar 1 by 17 plus 1 by 2. So, this will 107 00:26:14,190 --> 00:26:21,190 give 1.134 into 10 to the power minus 4 meter square per second. 108 00:26:25,220 --> 00:26:31,779 Let us consider another technique for liquid phase diffusion coefficient measurements. 109 00:26:31,779 --> 00:26:38,779 There are many methods available to calculate the liquid phase diffusivity; one of the simplest 110 00:26:38,889 --> 00:26:45,889 methods is known as the diaphragm cell method. In this case, a simple cell is divided into 111 00:26:48,620 --> 00:26:55,620 two parts with a diaphragm, this diaphragm porous in nature and in two compartments, 112 00:26:59,350 --> 00:27:06,350 compartment 1 and comportment 2. We can take two different concentrations of the liquids 113 00:27:11,279 --> 00:27:18,279 and allow them to diffuse between the compartments. After certain period of time take the sample 114 00:27:19,139 --> 00:27:26,139 from different compartments and analyze, and then from the mole balance we can calculate 115 00:27:27,169 --> 00:27:34,169 the diffusivity. The assumption is, that one very dilute solution 116 00:27:34,169 --> 00:27:41,169 is placed in compartment 1 and relatively little lower concentrations are placed in 117 00:27:43,179 --> 00:27:50,179 compartment 2, and molecular diffusions take place through the diaphragm. Let the volume 118 00:27:51,480 --> 00:27:58,480 of the compartment 1 is D A1 and the concentration in comportment 1 is C A1 and the volume of 119 00:27:59,129 --> 00:28:06,129 compartment 2 is D 2 and the concentration is C A2. Both the comportments are well mixed. 120 00:28:06,549 --> 00:28:13,549 Let the area of diaphragm is A, and porosity is epsilon, so effective area for diffusion for diffusion is A epsilon. The diffusion 121 00:28:50,309 --> 00:28:57,309 path is not uniform, so to account the non-uniformity of the diffusion path we can use a term tortuosity 122 00:29:04,090 --> 00:29:11,090 factor tau, tortuosity, tortuosity factor tau. Since the solute concentration is very 123 00:29:20,690 --> 00:29:27,690 low, solute concentration is very low, so the bulk flow term can be neglected and taken 124 00:29:53,710 --> 00:30:00,710 as 0. So, in that case we can write flux equation N A will be minus D AB dC A divided by dh, 125 00:30:18,369 --> 00:30:25,369 which is equal to C A1 minus C A2 divided by h 2 minus h 1 into tau, which is equal 126 00:30:39,879 --> 00:30:46,879 to D AB C A1 minus C A2 by x tau, where x is equal to h 2 minus h 1, the thickness of 127 00:31:00,220 --> 00:31:02,679 the diaphragm. 128 00:31:02,679 --> 00:31:09,679 So, we can write flux equations, flux equations for both sides of the diaphragm. For compartment 129 00:31:29,049 --> 00:31:36,049 one it is v 1 d CA 1 dt would be equal to epsilon N A and for compartment 2 will be 130 00:31:43,700 --> 00:31:50,700 v 2 d C A2 dt is equal to A epsilon N A. So, combining these two equations we can get, 131 00:31:55,830 --> 00:32:02,830 d C A1 minus C A2 dt would be equal to A epsilon D AB by x d tau C A1 minus C A2 into 1 by 132 00:32:25,509 --> 00:32:32,509 v 1 plus 1 by v 2. So, if the initial concentrations are between the compartments C A1, 0 and C A2, 133 00:32:51,320 --> 00:32:58,320 0, then with the following conditions at t is equal to 0, C A1 will be C A1, 0; and C 134 00:33:06,580 --> 00:33:13,580 A2 will be C A2, 0; t is equal t F, C A1 will be C A1, F and C A2 will be C A2, F. 135 00:33:24,999 --> 00:33:31,999 Using this condition we can integrate the flux equations and it will give D AB will 136 00:33:36,470 --> 00:33:43,470 be x d tau by A epsilon t F 1 by v 1 plus 1 by v 2 inverse ln C A1, 0 minus C A2, 0 divided 137 00:34:04,799 --> 00:34:11,799 by C A1, F minus C A2, F, which we can write 1 by alpha t F ln C A1, 0 minus C A2, 0 divided 138 00:34:28,460 --> 00:34:35,460 by C A1, F minus C A2, F, where alpha is the cell constant, constant, which is equal to 139 00:34:47,210 --> 00:34:54,210 A epsilon by x d tau 1 by v 1 plus 1 by v 2. So, this can be determined, alpha can be 140 00:35:14,690 --> 00:35:21,690 obtained using a solute of known diffusion coefficient. 141 00:35:34,400 --> 00:35:41,400 Now, let us consider a simple example. To measure the diffusivity of acetone in water 142 00:35:45,500 --> 00:35:52,500 at 20 degree centigrade a diaphragm cell is used. Initially, compartment one of volume 143 00:35:52,569 --> 00:35:59,569 50 centimetre cube is filled with 0.5 molar acetones in water and the compartment 2 of 144 00:36:03,660 --> 00:36:10,660 volume 55 centimetre cube is filled with water. The molar concentration of acetone dropped 145 00:36:11,569 --> 00:36:18,569 to 0.4 molar in compartment 1 after 40 hours. The cell constant is given as 0.3 per centimetre 146 00:36:21,619 --> 00:36:26,790 square. Calculate the diffusivity of acetone. 147 00:36:26,790 --> 00:36:33,790 So, the data, which are given is, volume of compartment 1 is 50 centimetre cube and for 148 00:36:39,440 --> 00:36:46,440 compartment 2 is 55 centimetre cube. t F, total time, which is 40 hours is 40 into 60 149 00:36:57,900 --> 00:37:04,900 into 60 seconds, so it will be 14400 seconds. Alpha cell constant is given, 0.3 centimetre 150 00:37:12,349 --> 00:37:19,349 minus 2; at t is equal to 0, C A1, 0 is equal to 0.5 and C A2, 0, because of pure water 151 00:37:31,480 --> 00:37:38,480 it is 0. Concentration of acetone is 0 at t is equal to t F, C A1, F is 0.4, dropped 152 00:37:46,059 --> 00:37:53,059 in compartment 1 and C A2, F we have to calculate from the material balance. 153 00:37:53,660 --> 00:38:00,660 So, if you do the material balance, v 1 C A1, 0 plus v 2, plus v 2 into C A2, 0 will 154 00:38:07,700 --> 00:38:14,700 be v 1 C A1, F plus v 2 C A2, F. So, 50 into 0.5 plus 55 into 0 will be equal to 50 into 155 00:38:29,609 --> 00:38:36,609 104 plus 55 into C A2, F, so C A2, F is equal to 0.091. 156 00:38:43,359 --> 00:38:50,359 So, then diffusivity D AB we can use, 1 by alpha t F ln concentration of C A1, 0 minus 157 00:39:01,589 --> 00:39:08,589 C A2, 0 by C A1, F minus C A2, F. Now, putting the value 1 by 0.3 into 144000 seconds ln 158 00:39:22,150 --> 00:39:29,150 0.5 minus 0 divided by 0.4 minus 0.091, so this will give 1.11 into 10 to the power minus 159 00:39:39,970 --> 00:39:46,970 5 centimetre square per second, so which is equal to 1.11 into 10 to the power of minus 160 00:39:48,910 --> 00:39:54,390 9 meter square per second. 161 00:39:54,390 --> 00:40:01,390 There is some empirical correlations available to determine the liquid phase diffusivity 162 00:40:03,089 --> 00:40:10,089 and one of the simplest correlations, which is known as Wilke-Chang equation and it gives 163 00:40:12,700 --> 00:40:19,700 D AB is equal to 1.173 into 10 to the power minus 16 phi M B to the power 0.5 into T divided 164 00:40:33,200 --> 00:40:40,200 by mu B v A to the power 0.6 in meter square per second. 165 00:40:44,980 --> 00:40:51,980 Here, D AB is the diffusivity of solute A in B, solvent B; phi is the association factor 166 00:40:55,920 --> 00:41:02,920 M B is the molecular weight of solvent B, solvent B, and T is the temperature in Kelvin; 167 00:41:30,099 --> 00:41:37,099 mu B is the solution viscosity in kg per meter second; v A is the molar volume at normal 168 00:41:52,670 --> 00:41:59,670 boiling point in meter cube per kmol. 169 00:42:04,400 --> 00:42:11,400 So, let us consider simple example to predict the diffusivity of acetone in water at 20 170 00:42:13,940 --> 00:42:20,940 degree centigrade using Wilke-Chang equation. The association factor for water is given 171 00:42:21,619 --> 00:42:28,619 2.26, the molar volume of acetone is given 0.074 meter cube per kmol and viscosity of 172 00:42:29,789 --> 00:42:36,789 water at 20 degree centigrade is 1.002 into 10 to the power minus 3 kg per meter second. 173 00:42:39,990 --> 00:42:46,990 So, using this values we can calculate, D AB will be 1.173 into 10 to the power of minus 174 00:42:53,510 --> 00:43:00,510 16 into phi M B to the power 0.5 into T divided by mu B v A to the power 0.6. The temperature is 20 degree 175 00:43:26,980 --> 00:43:33,980 centigrade, which is 20 plus 273 Kelvin, so 293 Kelvin. So, if you substitute, D AB would 176 00:43:39,109 --> 00:43:46,109 be 1.173 into 10 to the power of minus 16 into association factor phi is given 2.26 177 00:43:48,039 --> 00:43:55,039 and M B is 18 for water. So, substituting 2.26 into 18 to the power 0.5 into 293 divided 178 00:44:01,220 --> 00:44:08,220 by viscosity is given, 1.002 into 10 to the power minus 3 multiplied by the volume, molar 179 00:44:12,010 --> 00:44:19,010 volume is given 0.074 to the power 0.6. This will give 1.04 into 10 to the power minus 180 00:44:26,450 --> 00:44:30,700 9 meter square per second. 181 00:44:30,700 --> 00:44:37,700 There is another correlation, which is Stokes-Einstein equation, which also says that the liquid 182 00:44:40,059 --> 00:44:46,700 phase diffusivity for different components varies linearly with absolute temperature, 183 00:44:46,700 --> 00:44:53,700 and inversely proportional to the viscosity of the medium. So, we can write D AB mu by 184 00:44:56,750 --> 00:45:12,300 T as constant. So, this is end of lecture 7 and in the next lecture we will continue 185 00:45:12,329 --> 00:45:16,480 diffusivity for the multi-component mixtures. Thank you.