mass transfer final ppt

mass transfer final ppt - MASS TRANSFER Tuan Tran ( Leader)...

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Unformatted text preview: MASS TRANSFER Tuan Tran ( Leader) Gerry Bong GOALS Part 1: To determine the diffusivity from the data To compare the evaporation rate based on the height of the air column To calculate the diffusivity with no bulk convective flow Part 2:- To determine the evaporative flux from a water surface- Two different air velocities were used- To calculate the thermal and concentration boundary layer thicknesses and how they vary with air speed- To calculate the average mass transfer coefficient K m and how it scales with the velocity PART 1: DIFFUSION THROUGH STAGNANT FILM Experimental Procedure Fill the bath with water Turn on the AC circuit breaker and the heater circuit breaker Set over temperature to 80 C Press the “J” button to set the temperature 60 C MEASURE THE DIAMETER OF THE TUBE Method 1: Measure the height and volume of methanol used V= pi*r 2 *h , V= 40 µl and h= 10 mm Therefore, 40*10-3 cm 3 = pi*r 2 * 1(cm) r = 0.113 cm D = 0.226 cm MEASURE THE DIAMETER OF THE TUBE Method 2: Measure the height and mass of methanol Look up the literature value of the density of methanol V= m/ ρ , m= 0.035 g and ρ = 0.7918 g/ cm 3 V= 0.035/0.7918 = 0.442 cm 3 Thus, V= pi*r 2 *h r= 0.119 cm D= 0.238 cm Take the average of the values from the two methods The average radius of the tube is: r ave = 0.116 cm The average diameter of the tube is: D ave =0.232 cm SETTING UP THE EQUIPMENT Add methanol to the tube ~1 inch Adjust the ruler, the tube, and the sight glass Turn on the vacuum pump and check for air flowing Record the height of the methanol over time DATA ANALYSIS The data table consists of: time (min) z 2 (mm) is the height from the bottom of the tube to the air flow column the height of the top liquid layer and the bottom of the tube z 1 (mm) the height of liquid at time t z 2 – z 1 (mm) is the length of the diffusion path a Air flow z 1 z 2 Methanol DETERMINE D AB N A,z =y A (N A,z + N B,z ) – cD AB dy A /dz N B,z = 0 because B is insoluble in A N A.z =– cD AB / (1-y A ) dy A /dz Integrate by separation of variables we get N A,z = cD AB /z * (y B2-y B1 ) /y B,lm (1) Where z= z 2-z 1 and y B,lm = (y B2-y B1 )/ln(y B2 /y B1 ) But N A,z = ρ A,L /M A * dz/dt (2)...
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This note was uploaded on 03/11/2011 for the course CHEM ENG 154 taught by Professor Maboudian during the Spring '11 term at Berkeley.

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mass transfer final ppt - MASS TRANSFER Tuan Tran ( Leader)...

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