DIFFUSION OF LIQUIDS THROUGH STAGNANT NON-DIFFUSING AIR

TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES 363 P. CASAL ST., QUIAPO, MANILA COLLEGE OF ENGINEERING AND ARCHITECTURE CHEMICAL ENGINEERING DEPARTMENT

EXPERIMENT NO. 1: DIFFUSION OF LIQUIDS THROUGH STAGNANT NON-DIFFUSING AIR

SUBMITTED BY:

SUBMITTED TO:

NOVEMBER 20, 2014

EXPERIMENT NO.1 DIFFUSION OF LIQUIDS THROUGH STAGNANT NON-DIFFUSING AIR Discussion: Diffusion is the movement, under the influence of a physical stimulus, of an individual component thorugh a mixture. Although the usual cause of diffusion is a cncetration gradient, diffusion can also be caused by an activity gradient, as in reverse osmosis, by a pressure gradient, by a temperature gradient, or by the application of an external force field, as in centrifuge. Diffusion is not restricted to molecular transfer through stagnat layers of solid or fluid. Sometime diffusion process is accompanied by bulk flow of the mixture in a direction parrallel to the direction of diffusion, and it is often associated with heat flow. In all mass-transfer operations, diffusion occurs in at least one phase and often in both phases. Diffusivities are best established by experimental measurements, and where such information is available for the system of interes, it should be used directly. One method for determining molecular diffusivity is through capillary tube method. This method evaporates pure liquids in a narrow tube with gas passed through the top of the binary mixture. The difference between the initial and final height was measured with respect to time. The diffusivity is computed from: ( ) ( ) Another is using the Chapman and Enskog equation, which is: ( Where: DAB = diffusivity (m2/s) T = temperature (K) MA and MB = molecular weight of A and B 𝜴DAB = collision integral σAB = average collision Materials:           

Water Bath 5 Capillary Tubes Vernier Caliper Timer 3 Iron Stand 3 Iron clamp Fan Cork Ethanol Ethyl Acetate Tert Butyl Alcohol

)0.5

Procedure: 1. Prepare the water bath and fill it with tap water and set it at 50°C 2. Fill the capillary tube with pure volatile organic liquids and measure the initial height of the liquid. 3. Provide a gentle stream of air by fan. 4. Measure the height of the remaining liquid in the capillary tubes after 10 and 15 minutes. 5. Repeat procedure 2 -4 for trial 2 (water bath temperature = 65 ℃) and trial 3 (water bath temperature = 80 ℃.) 6. Compare the results with those obtained using Chapman and Engskog equation and other empirical equation. Experminental Set-Up:

Data and Results: Trial 1: 50 ℃

Ethanol Ethyl Acetate

Chapman and Engskog equation 1.32 x10-5 9.13 x10-6

Other empirical equation 1.4 x10-5 1.05 x10-5

Tert-butyl Alcohol

9.533 x10-6

1 x10-5

Chapman and Engskog equation 1.44 x10-5 9.95 x10-6 1.037 x10-5

Other empirical equation 1.53 x10-5 1.14 x10-5 1.09 x10-5

Ethanol

Chapman and Engskog equation 1.57x10-5

Other empirical equation 1.66 x10-5

Ethyl Acetate Tert-butyl Alcohol

1.08 x10-5 1.13 x10-5

1.23 x10-5 1.18 x10-5

@ 50 ℃: Liquid Ethanol

Z2 (cm) 8

Z1 (cm) 7.9

Ethyl Acetate

8.8

8.5

Tert-butyl Alcohol

8.5

8.4

@ 65 ℃: Liquid Ethanol

Z2 (cm) 8.8

Z1 (cm) 8.1

Ethyl Acetate Tert-butyl Alcohol

8.8 8.8

8.6 8.4

@ 80 ℃: Liquid Ethanol Ethyl Acetate Tert-butyl Alcohol

Z2 (cm) 8.1 8.6 8.4

Z1 (cm) 8.0 8.4 8.2

Liquid

Capillary tube method 4.645 x10-6 9.01 x10-6 2.957 x10-6

Trial 2: 65 ℃ Liquid Ethanol Ethyl Acetate Tert-butyl Alcohol

Capillary tube method 1.655 x10-5 5.72 x10-6 2.916 x10-6

Trial 3: 80 ℃ Liquid

Capillary tube method 1.058 x10-6 1.297 x10-6 1.34 x10-6

Analysis and Interpretation: The results from the experiment and computations for the different methods were compared. It is observed that trial 3 which is at 80°C has lower diffusivity than the other two lower temperatures. Conclusions and Recommendations: Temperature can greatly affect the rate of diffusion of the liquids. When the temperature was set to high values, the concentration gradient was also increased, proving that the rate of diffusion is directly proportional to the temperature. It is recommended that the other method was given. Also, using clean apparatus and uncontaminated chemicals is a must as it can greatly affect the results. Questions: 1. Define the following: a. Fick’s Law of Diffusion b. Equimolar Counter Diffusion c. Unicomponent Diffusion 2. A gas CH4 and He in a tube at 101.325 kPa pressure and 298 K. at one point the partial pressure of methane is PA1 = 60.79 kPa, and a point 0.02 m distance away, PA2 = 20.26 kPa. If the total pressure is constant throughout the tube, calculate the flux of methane at steady state for equimolar counter diffusion. 3. Ammonia gas is diffusing through N2 nondiffusing since it is insoluble in one boundary. The total pressure of NH3 at one point is 1.333 x 10-5 m2/s, at the other point 20 mm away it is 6.666 x 103 kPa. The DAB for the mixture at 101325 Pa and 298 K is 2.30 x 10-5 m2/s. a. Calculate the flux of NH3 in kgmols/s-m2. b. Do the same as (a) but assume N2 also diffuses; this is, both boundaries to both gases and flux is equimolar counter diffusion, in which case is the flux greater. 4. Mass transfer is occurring from a sphere of naphthalene having a radius of 10 mm. The sphere is in large volume of air at 52.6 ℃ and 1 atm abs pressure. The vapor pressure of naphthalene at 52.6 ℃ is 1.0 mmHg. The diffusivity of naphthalene in air at 0 ℃ is 5.16 x 10-6 m2/s. Calculate the rate of evaporation of naphthalene from the surface in kgmol/s- m2. Answers to Questions: 1. Fick’s law of diffusion states that the rate of diffusion of a chemical species at a location in a gas mixture( or liquid or solid dolution) is proportional to the concentration gradient of that species at that location. In equimolar counter-diffusion, the molar fluxes or A and B are equal, but opposite in direction, and the total pressure is constant throughout. Unicomponent diffusion is where one component (A) at one boundary (at the end of the diffusion path) can’t pass through the other component B. ( ) 2. ( ) ((

)

)

PB1 = P – PA1 = 1.013 x 105 – 1.333 x 105 = 87970 PB2 = P – PA2 = 1.013 x 105 – 6.666 x 105 = 94634 (

)

( (

) )(

( ( (

)(

)(

)( )(

)(

)(

)

) ) )

3. PA1 = 1.333x10^4 PA2 = 6.666 x10^3 PB1 x10= P-PA1 = 87970 Pa PB2 = P – PA2 = 94634 Pa PBD = PB2 – PB1 / ln(PB2/PB1) = 1.17x10^5 Pa NA = [(2.30x10^-5)(1.013x110^5)(1.333x10^4)]/[(8314)(298)(.02)(1.17x10^5)] NA = 5.36x10^-6 Kmol/ m^2(s) NA = Dab/RTZ (PA1-PA2) NA= (2.3x10^5/8314x298x0.02) (1.333x10^4 – 6.666x10^3) NA= 3.09x10^-6 Kmol/ m^2 s