Mass Transfer 1phenomenon of one-dimensional conduction.pptx

mass transfer phenomenon

1. TRANSPORT PHENOMENA: 3 rd section MASS TRANSFER Diah Susanti, Ph.D Undergraduate program Materials and metallurgical engineering department Institut teknologi sepuluh nopember (its) April 19, 2020

2. SOURCE: “Transport Phenomena: A Unified Approach”, Robert S. Brodkey and Harry C. Hershey, Mc. Graw -Hill, 1988 Chapter 2.

3. The Analogy General rate equation: (RATE) = DRIVING FORCE / RESISTANCE

4. Mass Transfer (Mass Diffusion ): Fick's Law (1) at T, P is constant The minus sign indicates that the direction of the flux is from the high to low concentration side, while the direction of the concentration increase is from the low to high side ( J A /A ) is the molar flux, which is the number of mole A transferred per unit of time per area ( kmol m -2 s -1 ) (J A ) is the molar transfer rate that is the number of mole A transferred per unit time (kmols -1 ) A is the area where there is no volume flow (stationary) (m 2 ) Subscript x shows that the direction of the transfer is only in the x direction D is proportionality constant = diffusivity coefficient (m 2 s -1 ) specific for each material C A is concentration ( kmol m -3 ) x is the distance (m) So that : Rate = J A Driving force = C A Resistance = x/(DA)   N 2 CO 2 x = 0 C A = 0 x = L C A L Direction of flux Direction of concentration increase

5. Fick’s law: (2) in which + + k (3) i , j, k are the unit vectors in the x, y, z directions, respectively, and  (del) is an operator which may operate on any scalar. Mass transfer is more complex than heat or momentum transfers because at least two species are needed. If there is one particular molecule in a mixture, the molecule must diffuse through other molecules. For example, the second component is called B. There are two possibilities for B; diffuse or not. If it does not participate to diffuse, then it is called A diffuses on a stationary film ( diffusion through stagnant film). The mass transfer equation for B in the x direction is: (4) For 3-dimensional directions: (5) The diffusion coefficients for A and B will be the same if they are ideal gases. In reality, the two are not the same. Therefore it can be written with D AB and D BA . D AB is the diffusion coefficient of component A which diffuses through A + B D BA is the diffusion coefficient of component B which diffuses through A + B  

6. NOTATION The picture on the left shows gas fluxes A and B through the room between two stationary walls. Therefore, the calculation of flux A or B is relative to the stationary coordinates, not the coordinates that moves. Thus we can use the N A /A and N B /A notation to replace the J A /A and J B /A notations used to express molar flux relative to volumetric velocity. wall wall 1 Fluxes 2 (J A /A) x (J B /A) x x 2 Concentration Profile C A,1 C B,1 C B,2 C A,2

7. In the picture above it can be seen that gas A diffuses from left to right, and therefore the molecules must be replaced by gas molecules B so that the pressure is constant. Because at each position x the number of molecules is constant, there is no volume flow. The diffusion is called ‘ equimolar counter diffusion ’. For ‘ equimolar counter diffusion ’ N A = - N B (minus sign shows the opposite direction) ( N A /A) x = -( N B /A) x (6) Because there is no volumetric flow, so: ( N A /A) x =(J A /A) x =-D AB ( C A /x) (7) ( N B /A) x =(J B /A) x =-D BA ( C B /x ) (8) The total concentration is constant because pressure is kept constant: C T = C A + C B = constant (9) ( C T /x) = ( C A /x) + ( C B /x) = 0 (10) Hence, ( C A /x) = - ( C B /x) (11) Therefore, D AB = D BA (12) This equation (11) is restricted to the binary diffusion of ideal gas at constant T and P . Equimolar Counter Diffusion

8. Mass flux Mass flux is defined as mass per area per time. It is written mathematically as : ( j A /A) x = [(J A /A) x ]M A (13) ( n A /A) x = [(N A /A) x ]M A (14) Mass flux unit is kgm -2 s -1 ,while the unit of mass flowrate is kgs -1 . The driving force is the mass concentration = (C A )M A =  A (15) where M A is molecular weight of A (kg kmol -1 ) and  A is the density of A (kg m -3 ) So that the Fick’s law could be written as: ( j A /A) x =-D (  A /x) (16)

9. PARTIAL PRESSURE Ideal gas equation: PV = nRT (17) So that C A = = (18) where is the partial pressure of A, y A is the mole fraction of A. (19) Hence ) at T,P constant (20)  

10. Mass transfer example problems Example 2.6. Two gas streams of CO 2 and air are flowing in a channel. The channel is divided into equal volumes by a piece of iron 4 cm thick as shown in the figure below. At the A-A plane there is a hole 1.2 cm in diameter drilled in the iron so that CO 2 diffuses from left to right and air from right to left. At the plane A-A, the two gases are at a pressure of 2 atm and a temperature of 20 o C. Upstream of the hole, both gases are pure. Under the conditions given, the concentration of CO 2 is 0.083 kmol m -3 , i.e the concentration of CO 2 on the left at the point A . At the right end of the hole, the concentration of CO 2 is considered 0 because air flows quickly. The diffusion coefficient of CO 2 in the air is 1.56x10 -3 m 2 s -1 . Please find: a). CO 2 Molar Flux b). How many pounds of CO 2 flow through the hole for 1 hour -A---------A- iron iron x air CO 2 4 cm d=1,2 cm

11. Answer: Known: P = 2 atm , T = 20 C, C A0 = 0.083 kmol m -3 . C Ax = 0 kmol m -3 . D AB = 1.56x10 -3 m 2 s -1 . L = 4 cm = 4x10 -2 m = 0.04 m, d = 1.2 cm = 0.012 m  A = 1/4 d 2 = 1/4 (0.012) 2 = 1.131x10 -4 m 2 . Asked: a). CO 2 Molar Flux b). How many pounds of CO 2 flow through the hole for 1 hour Answer: The CO 2 diffuses through the hole in the iron at steady-state if the flow rate of CO 2 and air are constant. Since both gases are at the same temperature and pressure, eq. (6) ( N A /A) x =(J A /A) x =-D AB ( C A /x) (6) At steady state, the molar flux ( N A /A) x must be constant in the hole throughout its length because all the CO 2 entering from the left exits into the air stream. Hence eq. (6) rearranges to: =- ( 6) = = 3.237X10 -3 kmol m -2 s -1 . So, the CO 2 molar flux is 3.237X10 -3 kmol m -2 s -1  the molar flux with respect to the stationary coordinates (the apparatus)  

12. b). The mass flux : ( n A /A) x = [(N A /A) x ]M A ( 14) M A = M CO2 = 44 kg mol -1 . The mass flow rate: ( n A ) x = [(N A /A) x ] AM A = 3.237X10 -3 kmol m -2 s -1 x1.131x10 -4 m 2 x 44 kg mol -1 =1.61x10 -5 kg/s. Converted to lbm /hour: 1.61x10 -5 kg/s x 3600 s/hour x 1 lbm /0.4539 kg = 0.128 lbm /hour. The mass flow rate is positive because the CO 2 flows from left to right with a mass flow rate of 0.128 lbm /hour .

13. Example 2.7. Air and CO 2 mix together in a simple T pipe. Air at pressures of 3 atm and 30  C enters at the T end with a flow rate of 2 kmol min -1 . CO 2 at 3 atm and temperature at 30  C enters at the other end of T pipe with a flow rate of 4 kmol min -1 . The two gases flow out from the middle T, still at 3 atm and 30 o C. a. Calculate the incoming CO 2 concentration (kg m -3 ) b. Calculate the concentration of N 2 that enters the air stream ( lbmol ft -3 ) c. Calculate the concentration of CO 2 out ( mol cm -3 ) d. Calculate the concentration of O 2 out ( kmol m -3 ) Air N 2 = 79% O 2 = 21% 3 atm , 30 o C 2 kmol min -1 CO 2 3 atm, 30 o C 4 kmol min -1 Air & CO 2 3 atm , 30 o C

14. ANSWER: Known: Air at P = 3 atm and T = 30 C = 303 K, N air = 2 kmol /min. CO 2 at P = 3 atm and T = 30 C = 303 K, N CO2 = 4 kmol /min. The two gas streams exit still at 3 atm and 303 K. R (gas constant) = 8.2057x10 -2 atm m 3 kmol -1 K -1 = 0.082057 atm m 3 kmol -1 K -1 Asked: a . Calculate the incoming CO 2 concentration (kg m -3 ) b. Calculate the concentration of N 2 that enters the air stream ( lbmol ft -3 ) c. Calculate the concentration of CO 2 out ( mol cm -3 ) d. Calculate the concentration of O 2 out ( kmol m -3 )

15. Answer: a). C CO2 entering = C A = = ( 18) C CO2 = = = 0.121 kmol m -3 . To convert into kg m -3 , we should multiply with M CO2 . So, concentration of CO 2 = 0.121 kmol m -3 x 44 kg/ mol = 5.31 kg m -3 . b). C N2 entering = C N2 = = = = 0.0953 kmol m -3 Converted to lbmol ft -3 : 0.0953 x x = 5.945x10 -3 lbmol /ft 3 . So the concentration of N 2 entering is 5.945x10 -3 lbmol /ft 3 ( Note: Air is 79% nitrogen and 21% oxygen, so y N2 = 0.79 and y O2 = 0.21). c). The exiting gases are comprised of CO 2 =4 kmol , N 2 =0.79x2 kmol =1.58 kmol and O 2 = 0.21x 2 kmol = 0.42 kmol . Total mole is 6 kmol . Mole fraction of gases: y CO2 = 4/6 = 0.667, y N2 = 1.58/6 = 0.263, and y O2 = 0.42/6 = 0.07. So the concentration of CO 2 in the exit stream is . d). The concentration of O 2 exiting is: . Hence the concentration of O 2 is .  

16. Example 2.8. A tank containing 15% CO 2 in the air is connected to a second tank that contains only air. The connecting pipe has a diameter of 5 cm and a length of 30 cm. Both tanks are at a pressure of 1 atm and a temperature of 298.15 K. The volume of each tank is very large compared to the volume of the pipe, so changes in concentration in the tank can be ignored for a very long time after the experiment begins. The diffusion coefficient of CO 2 in the air at 1 atm and 25  C is 0.164 x 10 -4 m 2 s -1 . Calculate the initial mass transfer rate of CO 2 . Does the air transfer? 15% CO 2 air 30 cm 5 cm 1 2

17. Known: y CO2 in air = 0.15. Pipe d = 5 cm, L = 30 cm. The pressures and temperatures of both tanks are 1 atm and 298.15 K. D CO2 in air at T and P is 0.164 x 10 -4 m 2 s -1 x = 30 – 0 = 30 cm = 0.3 m. A = 1/4 d 2 = 1/4 ( 0.05) 2 = 1.96x10 -3 m 2 . R = 8.314x10 3 Nm kmol -1 K -1 . Asked: Calculate the initial mass transfer rate of CO 2 . Does the air transfer? Answer: Assume that both gases are ideal. The diffusion transfer is equimolar counter diffusion. So, ( N A /A) x =(J A /A) x =-D AB ( C A /x) (7) ( N A /A) x = ) N A = A x 1.96x10 -3 m 2 = + 6.57x10 -10 kmol /s. The plus sign indicates that the diffusion is from 1 to 2 ( to the right). In order to keep the pressure 1 atm , the air should flow from right to left (in the opposite direction) with a flow rate is 6.57x10 -10 kmol /s.