Mass Transfer - كلية الهندسة
Tikrit UniversityChemical Engineering DepartmentMass TransferThird YearByAssist. Prof. Dr. Ahmed Daham
Mass TransferTitle:Third YearAssist. Prof. Dr. Ahmed DahamMass TransferCourse Instructor: Assist. Prof. Dr. Ahmed Daham WiheebTextbook:1.Coulson J.M. & Richardson J.F., Chemical Engineering, Volume 1, six edition, ELBS,Pergamon Press. 2002.2.Coulson J.M. & Richardson J.F., Chemical Engineering, Volume 2, fifth edition, ELBS,Pergamon Press. 2002.References:1.2.3.4.5.Treybal R.E., Mass Transfer Operations, McGraw HillMcCabe W.L., Smith J.C. & Harriott P., Unit Operations in Chemical Engineering,McGraw Hill.Seader J.D.& Henley E.J., Separation Process Principles.Rousseau R.W., Handbook of Separation Process Technology, John WileyFoust A.S. et al, Principles of Unit Operations, John WileyCourse description:This course covers the fundamentals of the basic concepts of mass transportand understanding about diffusion theory, gas absorption, liquid-liquidextraction, leaching , distillation, humidification, drying and evaporation.Objectives:To give the students a application of diffusion theory to simple masstransfer problems. Analysis of chemical engineering unit operations involvingmass transfer. Design principles for mass transfer equipment. Solids/Liquidsseparation processes. liquids/Liquids separation processes. gases/Liquidsseparation processes.Grading:No.123456AssessmentHomework (HW), Quizzes(Q)Test 1Test 2Test 3Test 4Final Exam (F)Number11111% each10 %7.5 %7.5 %7.5 %7.5 %60 %% total107.57.57.57.560100Overall Total2Dates
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamSyllabus:NoTopicHours1Diffusion , flick's law, modes of diffusion42Multi-components mixture, correction of diffusivity43Diffusion in varying cross section area44Diffusivity coefficient in liquid and gas25Mass transfer theory26Mass transfer coefficient, wetted wall column47Absorption, equilibrium of gas and liquid48Packed tower69Tray tower610Calculation of tower diameter, stripping211Extraction , differential type412Completely immiscible413Party miscible414Leaching, batch leaching415Continuous leaching, constant under flow416Continuous leaching, variable under flow417Distillation , vapor-liquid equilibrium418Differential type219flash distillation220Continuous distillation (binary system)621Calculation the number of stages and reflux ration in continuous4distillation22Multi-feeds and side stream , lewis sorial method623Ponchon-savarit method424Batch distillation with constant reflux ratio6and with constant product composition25Multi-component distillation426Humidification , humidity227Cooling tower calculation228Height of cooling tower, carey method429Drying process630Evaporation43
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamChapter (10) in Volume (1)((Diffusion))The term diffusion (mass transfer) is used to denote the transference of a component ina mixture from a region where its concentration is high to a region where the concentration islower. Diffusion process can take place in a gas or vapour or in a liquid, and it can result fromthe random velocities of the molecules (molecular diffusion) or from the circulating or eddycurrents present in a turbulent fluid (eddy diffusion).Diffusion depends on:1.2.3.Driving force (ΔC), moles per unit volume (kmol/m3).The distance in the direction of transfer (Δz), meter (m).Diffusivity coefficient, unit area per unit time (m2/s).Fick's Law of diffusion:The rate of diffusion is governed by Fick's Law, first proposed by Fick in 1855 whichexpresses the mass transfer rate as a linear function of the molar concentration gradient. In amixture of two gases A and B, assumed ideal, Fick's Law for steady state diffusion may bewritten as:4
Mass TransferThird YearAssist. Prof. Dr. Ahmed Daham CA zJA αJA DABdCA dzFick ′ s first law of steady state diffusionWhere:JA :is the molecular diffusion flux of A , (moles per unit area per unit time)CA:is the concentration of A (moles of A per unit volume)kmolm3kmolm 2 .s.DAB: is known as the diffusivity or diffusion coefficient for A in B (unit area per unit time)m2sz:is distance in the direction of transfer (m).Diffusion with bulk of mass in motion:The Fick's first law of diffusion describes the mass transfer from the random movementof molecules of a stationary medium or a fluid in streamline flow. If circulating currents oreddies are present, then the molecular mechanism will be reinforced and the total masstransfer rate may be written as:Total diffusion Molecular diffusion Convection termConvection term Eddy diffusion Molar flux due to convectionConvection term Concentration * mass transfer velocity CA . VWhere:𝐤𝐦𝐨𝐥𝐦𝐚𝐬𝐬 𝐟𝐥𝐮𝐱𝐍𝐀 𝐍𝐁𝟐𝐦𝐦𝐚𝐬𝐬 𝐭𝐫𝐚𝐧𝐬𝐟𝐞𝐫 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐕 𝐦 .𝐬 ��𝐭𝐢𝐨𝐧𝐂𝐓𝐬𝟑𝐦Total diffusion NA JA CA . V𝐍𝐀 𝐃𝐀𝐁𝐝𝐂𝐀𝐂𝐀 𝐍𝐀 𝐍𝐁𝐝𝐳𝐂𝐓 (𝟏)Total diffusion equation in the form of concentration (normally used for liquids)5
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamThe total diffusion equation can be write in another forms:a.b.Partial pressure for gases.Mole fraction for gases and liquids.a.Total diffusion equation in the partial pressure form:If A and B are ideal gases in a mixture, the ideal gas law may be applied to each gasseparately and to the mixture:PV nRTP nVRTP CRTPA CA R TandPT CT R TPART1dCA dPRT ACA 𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐍𝐀 𝐍𝐁 (𝟐)Total diffusion equation in the form of partial pressure (normally used for gases)b.Total diffusion equation in the mole fraction form:XA PAPTorXA CACTPT XA PAandCT XA CAPT dXA dPAandCT dXA dCAThen:𝐍𝐀 𝐃𝐀𝐁 𝐏𝐓𝐑𝐓𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀 𝐍𝐁𝐝𝐳 (𝟑)Total diffusion equation in the form of mole fraction (used for gases and liquids)6
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamModes of diffusionThere are two modes of diffusion:Diffusion(1((2(Stagnant diffusionCounter diffusion𝐍𝐁 𝟎 In absorptionprocess(i(EquimolecularCounter diffusionUnequimolecularCounter diffusion𝐍𝐁 𝐍𝐀𝐍𝐁 𝐧 𝐍𝐀 In distillationcolumn1.(ii( In chemicalreactionStagnant diffusion (Mass transfer through a stationary second component):In several important processes, one component in a gaseous mixture will be transportedrelative to a fixed plane, such as a liquid interface, for example, and the other will undergo nonet movement. In gas absorption a soluble gas A is transferred to the liquid surface where itdissolves, whereas the insoluble gas B undergoes no net movement with respect to theinterface. Similarly, in evaporation from a free surface, the vapour moves away from thesurface but the air has no net movement. The mass transfer process therefore:𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐍𝐀 𝐍𝐁Since stagnant diffusion layer: (1)NB 07
Mass Transfer𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐍𝐀 𝟏 Third Year𝐝𝐏𝐀 𝐏𝐀 𝐍𝐀𝐝𝐳𝐏𝐓𝐏𝐀 𝐃𝐀𝐁 𝐏𝐓𝐑𝐓 𝐃𝐀𝐁 𝟏𝐑𝐓 𝐝𝐳𝐝𝐏𝐀𝐏𝟏 𝐏𝐀𝐓𝐍𝐀 𝐃𝐀𝐁 𝐏𝐓𝐑𝐓 𝐝𝐳𝐝𝐏𝐀𝐏𝐓 𝐏𝐀𝐍𝐀 𝐃𝐀𝐁𝐑𝐓 . (2)𝐝𝐏𝐀𝐝𝐳𝐍𝐀 𝐏𝐓𝐳𝟐 𝐳𝟏Assist. Prof. Dr. Ahmed Daham . . . . (3) . . . . (4)𝐥𝐧 . . . . (5)𝐏𝐓 𝐏𝐀𝟐𝐏𝐓 𝐏𝐀𝟏Example 10.1: Ammonia gas is diffusing at a constant rate through a layer of stagnant air1 mm thick. Conditions are such that the gas contains 50 percent by volume ammonia at oneboundary of the stagnant layer. The ammonia diffusing to the other boundary is quicklyabsorbed and the concentration is negligible at that plane. The temperature is 295 K and thepressure atmospheric, and under these conditions the diffusivity of ammonia in air is 0.18cm2/s. Estimate the rate of diffusion of ammonia through the layer.Solution:If the subscripts 1 and 2 refer to the two sides of the stagnant layer and the subscripts A and B referto ammonia and air respectively, then the rate of diffusion through a stagnant layer is given by:𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐏𝐓𝐳𝟐 𝐳𝟏𝐥𝐧𝐏𝐓 𝐏𝐀𝟐𝐏𝐓 𝐏𝐀𝟏Where:𝐏𝐓 101.3 kPa ,𝐏𝐀𝟐 0,𝐏𝐀𝟏 yA PT 0.5 101.3 50.65 kPa 𝐳 𝐳𝟐 𝐳𝟏 1 mm 1 10 3 m𝐑 8.314kJkmol . K,𝐓 298 KandDAB 0.188𝑐𝑚2𝑠 1.8 10 5𝑚2𝑠
Mass TransferThird YearAssist. Prof. Dr. Ahmed Daham1.8 10 5𝐍𝐀 101.3101.3 0kmol 4ln 5.153 108.314 295 1 10 3101.3 50.65m2 . s2.Counter diffusion:i.Equimolecular counter diffusion:When the mass transfer rates of the two components are equal and opposite the processis said to be one of equimolecular counter diffusion. Such a process occurs in the case of thebox with a movable partition. It occurs also in a distillation column when the molar latentheats of the two components are the same (λA λB) . At any point in the column a fallingstream of liquid is brought into contact with a rising stream of vapour with which it is not inequilibrium. The less volatile component is transferred from the vapour to the liquid and themore volatile component is transferred in the opposite direction. If the molar latent heats ofthe components are equal, the condensation of a given amount of less volatile componentreleases exactly the amount of latent heat required to volatilize the same molar quantity of themore volatile component. Thus at the interface, and consequently throughout the liquid andvapour phases, equimolecular counter diffusion is taking place (NB - NA).𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐍𝐀 𝐍𝐁 (1)Since equimolecular counter diffusion:NB - NA𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀𝐝𝐳 (3)𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀𝐝𝐳 (4)𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐏𝐀𝟐 𝐏𝐀𝟏𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐍𝐀 𝐍𝐀 (2) (5)𝐳𝟐 𝐳𝟏𝐏𝐀𝟏 𝐏𝐀𝟐𝐳 𝟐 𝐳𝟏9
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamDrift Factor:For stagnant diffusion:NA DABRTPT PA 2PTln zPT PA 1NA DABRTPT zPT PA 2 PT PA 1lnPT PA 2 PT PA 1From Dalton's Law of partial pressures:PT PA 2PT PA 1PT PA PBBy definition, PBm , the logarithmic mean of PB 1 and PB 2 , is given by:PT PA 2 PT PA 1PT PAln P P 2TA1NA D AB1PTRT zP BmPTWhere:P Bm PB 2 PB 1 PBmPBln P 2B1PA 1 PA 2is known as the drift factor.If the drift factor PTP Bm 1 ( this happen when the concentration of component Abeing transferred is low)Then,NA *D ABRTPA PA12z 2 z 1Thus the bulk flow enhances the mass transfer rate by a factorthe drift factor.10PTP Bm, known as
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamExample: In an air-carbon dioxide mixture at 298 K and 202.6 kPa, the concentration of CO2at two planes (3 mm) apart are 15 vol.% and 25 vol.%. The diffusivity of CO2 in air at 298 Kand 202.6 kPa is 8.2*10-6 m2/s. Calculate the rate of transfer of CO2 across the two planes,assuming:a.Equimolecular counter diffusion.b.Diffusion of CO2 through a stagnant air layer.Solution:PA 1 yA 1 . PT 0.15 202.6 30.39 kPaPA 2 yA 2 . PT 0.25 202.6 50.65 kPaa.Equimolecular counter diffusion.NA DABRTNA PA1 PA2z2 z18.2 10 68.314 298 (3 10 6 )b.Stagnant diffusion.NA DABRTkmolm2 . sPT PA 2PTln zPT PA 18.2 10 6NA 8.314 298ii.50.65 30.39 2.23 10 5202.6202.6 30.39ln 2.79 10 53 10 6202.6 50.65kmolm2 . sUnequimolecular counter diffusion:When the mass transfer rates of the two components are unequal and opposite, theprocess is said to be the unequimolecular diffusion, such a process occurs in a chemicalreaction.𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐍𝐀 𝐍𝐁Since unequimolecular counter diffusion: (1)NB - n NA11
Mass Transfer𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐍𝐀 𝟏 𝐍𝐀 𝐍𝐀 Third Year𝐝𝐏𝐀 𝐏𝐀 𝐝𝐳𝐏𝐓𝐏𝐀𝐏𝐓𝟏 𝐧 𝐃𝐀𝐁𝐑𝐓𝐏𝐓𝐝𝐳 𝐃𝐀𝐁𝐑𝐓𝟏 𝐳𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐏𝐓 𝐳Assist. Prof. Dr. Ahmed Daham𝐍𝐀 𝐧 𝐍𝐀 (2) 𝐃𝐀𝐁𝐑𝐓 (3) 𝐝𝐏𝐀𝐝𝐳𝐝𝐏𝐀𝐏𝟏 𝐏𝐀𝐓. . . . (4)𝟏 𝐧𝐝𝐏𝐀. . . . (5)𝐏𝐓 𝐏𝐀 𝟏 𝐧𝟏𝟏 𝐧𝐥𝐧𝐏𝐓 𝟏 𝐧 𝐏𝐀𝟐𝐏𝐓 𝟏 𝐧 𝐏𝐀𝟏Example: Species A in a gaseous mixture diffuses through a (3 mm) thick film and reaches acatalyst surface where the reaction A 3B takes place. If the partial pressure of A in thebulk of the gas is 8.5 kN/m2 and the diffusivity of A is 2*10-5 m2/s. Find the mole flux of A,given the pressure and temperature of the system are 101.3 kPa and 297 K, respectively.Solution:A 3B𝐧 𝐍𝐁𝟑 𝟑𝐍𝐀𝟏Given:DAB 2 10m2sT 297 K, 5, PT 101.3 kPaPA 1 8.5 kPaPA 2 0𝐍𝐀 𝐍𝐀 𝐃𝐀𝐁𝐑𝐓𝐏𝐓 𝐳2 10 58.314 297𝟏𝟏 𝐧𝐥𝐧101.33 10 3𝐏𝐓 𝟏 𝐧 𝐏𝐀𝟐𝐏𝐓 𝟏 𝐧 𝐏𝐀𝟏11 3ln101.3 2(0)kmol 2.12 10 5101.3 2(8.5)m2 . s12
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamMaxwell's Law for multicomponent mass transferThis argument can be applied to the diffusion of a constituent of a multicomponent gas.Considering the transfer of component A through a stationary gas consisting of componentsB, C, D, . etc, if the total partial pressure gradient can be regarded as being made up ofa series of terms each representing the contribution of the individual component gases. Themass transfer rate can be calculated from the previous equations using the effective diffusivityof A in the mixture (DAm).Calculation of the effective diffusivity of (A) in the mixture (DAm):Let A be the diffusing species through stagnant mixture of B, C, D . etc.𝐍𝐀 𝐃𝐀𝐦 𝐂𝐓𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀 𝐍𝐁 𝐍𝐂 𝐍𝐃𝐝𝐳 (𝟏)Where: DAm is the effective diffusivity of A in the mixture.Since stagnant diffusion layer of the mixture:𝐍𝐀 𝐃𝐀𝐦 𝐂𝐓NB NC ND 0𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀𝐝𝐳 (𝟐)𝐍𝐀 𝟏 𝐗 𝐀𝐝𝐗 𝐀 𝐂𝐓𝐃𝐀𝐦𝐝𝐳 (𝟑)Now consider binary system for diffusion of A in B.𝐍𝐀 𝐃𝐀𝐁 𝐂𝐓𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀 𝐍𝐁𝐝𝐳Since stagnant diffusion layer:𝐍𝐀 𝐃𝐀𝐁 𝐂𝐓NB 0𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀 𝐃𝐀𝐁 𝐂𝐓𝐝𝐗 𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀𝐝𝐗 𝐀 𝐂𝐓𝐃𝐀𝐁𝐝𝐳But:𝟏 𝐗𝐀 𝐗𝐁 𝐝𝐗 𝐀 𝐝𝐗 𝐁13
Mass TransferThird Year𝐍𝐀 𝐗 𝐁𝐝𝐗 𝐁 𝐂𝐓 𝐃𝐀𝐁𝐝𝐳Assist. Prof. Dr. Ahmed Daham . . (𝟒)Similarly for diffusion of A in C.𝐍𝐀 𝐃𝐀𝐂 𝐂𝐓𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀 𝐍𝐂𝐝𝐳Since stagnant diffusion layer:𝐍𝐀 𝐃𝐀𝐂 𝐂𝐓NC 0𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀 𝐃𝐀𝐂 𝐂𝐓𝐝𝐗 𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀𝐝𝐗 𝐀 𝐂𝐓𝐃𝐀𝐂𝐝𝐳But:𝟏 𝐗𝐀 𝐗𝐂 𝐝𝐗 𝐀 𝐝𝐗 𝐂𝐍𝐀 𝐗 𝐂𝐝𝐗 𝐂 𝐂𝐓 𝐃𝐀𝐂𝐝𝐳 . . (𝟓)Similarly for diffusion of A in D.𝐍𝐀 𝐃𝐀𝐃 𝐂𝐓𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀 𝐍𝐃𝐝𝐳Since stagnant diffusion layer:𝐍𝐀 𝐃𝐀𝐃 𝐂𝐓ND 0𝐝𝐗 𝐀 𝐗 𝐀 𝐍𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀 𝐃𝐀𝐃 𝐂𝐓𝐝𝐗 𝐀𝐝𝐳𝐍𝐀 𝟏 𝐗 𝐀𝐝𝐗 𝐀 𝐂𝐓𝐃𝐀𝐃𝐝𝐳But:𝟏 𝐗𝐀 𝐗𝐃𝐍𝐀 𝐗 𝐃𝐝𝐗 𝐃 𝐂𝐓 𝐃𝐀𝐃𝐝𝐳 𝐝𝐗 𝐀 𝐝𝐗 𝐃 . . (𝟔)14
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamNow adding Equations (4), (5) and ��� 𝐁 𝐗 𝐂 𝐗 𝐃 ) 𝐃𝐀𝐁 𝐃𝐀𝐂 𝐃𝐀𝐃𝐝𝐳𝐗𝐁 𝐗𝐂 𝐗𝐃 𝟏 𝐗𝐀But:𝐝(𝐗 𝐁 𝐗 𝐂 𝐗 𝐃 )𝐝𝐗 𝐀 𝐝𝐳𝐝𝐳𝐍𝐀 𝐗 𝐁𝐗𝐂𝐗𝐃𝐍𝐀 𝟏 𝐗 𝐀 𝐂𝐓 𝐃𝐀𝐁 𝐃𝐀𝐂 𝐃𝐀𝐃𝐂𝐓 𝐃𝐀𝐦𝟏 𝐗𝐀𝐗𝐁𝐗𝐂𝐗𝐃 r dilute mixture (low concentration of A), XA 0𝟏𝐗𝐁𝐗𝐂𝐗𝐃 ample: Nitrogen is diffusing under steady condition through a mixture of 2% N2, 20%C2H6 , 30% C2H4 and 48% C4H10 at 298 K and 100 kPa. The partial pressure of nitrogen attwo planes (1 mm ) apart are 13.3 & 6.67 kPa, respectively. Calculate the rate of N2 across thetwo planes. The diffusivity of N2 through C4H10 , C2H6 and C2H4 may be taken as 9.6*10-6m2/s , 14.8*10-6 m2/s and 16.3*10-6 m2/s, respectively.Solution:Since stagnant diffusion:NA DAmRTPT PA 2PTln zPT PA 11 yAyByCyD DAmDABDACDAD1 0.020.480.20.3 6 6DAm9.6 1014.8 1016.3 10 6DAm 1.22 10 51.22 10 5NA 8.314 298m2s100100 6.67kmolln 0.04920.001100 13.3m2 . s15
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamDiffusivities of gases and vapoursExperimental values of diffusivities are given in Table 10.2 for a number of gases andvapours in air at 298K and atmospheric pressure. The table also includes values of theSchmidt number Sc, the ratio of the kinematic viscosity (μ/A) to the diffusivity (D) for verylow concentrations of the diffusing gas or vapour. The importance of the Schmidt number inproblems involving mass transfer is discussed later.Experimental determination of diffusivitiesDiffusivities of vapours are most conveniently determined by the method developed byWinkelmann in which liquid is allowed to evaporate in a vertical glass tube over the top ofwhich a stream of vapour-free gas is passed, at a rate such that the vapour pressure ismaintained almost at zero (Figure 10.2). If the apparatus is maintained at a steadytemperature, there will be no eddy currents in the vertical tube and mass transfer will takeplace from the surface by molecular diffusion alone. The rate of evaporation can be followedby the rate of fall of the liquid surface, and since the concentration gradient is known, thediffusivity can then be calculated.16
Mass TransferNA DABRTNA NA Third YearPT PA 2PTln zPT PA 1D AB1PTRTzP BmD ABCTzC BmAssist. Prof. Dr. Ahmed Daham (1)PA 1 PA 2CA 1 CA 2Material balance on component A over dz:In out( in kg)(A * dz) ρL NA * A * M.wt * dtNA ρL dzMwt dtoutdz . . (2)inSubstitute Eq.(1) in to Eq.(2). To get:ρL dzDAB Mwt dtRTPT PA 2PTlnzPT PA 1PT PA 2DAB PTlnRT zPT PA 1t0ρLdt Mwt17z2z dzz1
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamρL . z2 2 z1 2 . RTPT PA 2. 2t . PT . lnPT PA 1DAB MwtWhere:ρLis the liquid density (kg/m3).Mwtis the molecular weight of liquid.PA 2 0 (always).NA DABzCTCBmCA 1 CA 2 . (3)Since PA 2 CA 2 0LetCA 1 CANA DABCAzCTCBm (4)Substitute Eq.(4) in to Eq.(2). To get:ρL dzC DAB AMwt dtzz2 2 z1 2 CTCBm2 Mwt . DAB . CAρLz2 z1 (z2 z1 2z1 ) 𝛒𝐋𝐌𝐰𝐭 . 𝐃𝐀𝐁 . 𝐂𝐀t2 Mwt . DAB . CAρL𝐭𝛒𝐋 𝐳𝟐 𝐳𝟏𝟐 𝐌𝐰𝐭 . 𝐃𝐀𝐁 . 𝐂𝐀 CTCBm𝐂𝐁𝐦𝐂𝐓𝐳𝟐 𝐳𝟏𝐂𝐁𝐦𝐳𝟏𝐂𝐓18CTCBmt
Mass TransferIf we drawThird Year𝐭𝐳𝟐 𝐳𝟏The slope s 𝐃𝐀𝐁 against 𝐳𝟐 𝐳𝟏ρL2 Mwt . DAB . CA𝛒𝐋𝟐 𝐌𝐰𝐭 . 𝐬𝐥𝐨𝐩𝐞 . 𝐂𝐀Assist. Prof. Dr. Ahmed Dahamthen:CBmCT𝐂𝐁𝐦𝐂𝐓Example 10.2: The diffusivity of the vapour of a volatile liquid in air can be convenientlydetermined by Winkelmann's method in which liquid is contained in a narrow diametervertical tube, maintained at a constant temperature, and an air stream is passed over the top ofthe tube sufficiently rapidly to ensure that the partial pressure of the vapour there remainsapproximately zero. On the assumption that the vapour is transferred from the surface of theliquid to the air stream by molecular diffusion alone, calculate the diffusivity of carbontetrachloride vapour in air at 321 K and atmospheric pressure from the experimental datagiven in Table 10.3.The vapour pressure of carbon tetrachloride at 321 K is 37.6 kN/m2 and the density of theliquid is 1540 kg/m3. The kilogram molecular volume may be taken as 22.4 m3.Solution:𝐭𝛒𝐋 𝐳𝟐 𝐳𝟏𝟐 𝐌𝐰𝐭 . 𝐃𝐀𝐁 . 𝐂𝐀 we draw𝐭𝐳𝟐 𝐳𝟏𝛒𝐋𝐌𝐰𝐭 . 𝐃𝐀𝐁 . 𝐂𝐀against𝐂𝐁𝐦𝐂𝐓𝐳𝟐 𝐳𝟏𝐂𝐁𝐦𝐳𝟏𝐂𝐓𝐳𝟐 𝐳𝟏19
Mass TransferThird Year𝛒𝐋𝟐 𝐌𝐰𝐭 . 𝐬𝐥𝐨𝐩𝐞 . 𝐂𝐀𝐃𝐀𝐁 122.4273321PT𝐂𝐁𝐦𝐂𝐓kss 3.1 107 22mmm𝐓𝐡𝐞 𝐬𝐥𝐨𝐩𝐞 𝐬 0.031𝐂𝐓 0.038101.3kmolm3or, CT 0.038RT8.314 321Mwt 154 kg/kmolAssist. Prof. Dr. Ahmed Dahamkmolm3𝛒𝐋 1540 kg/m3,𝐂𝐀 mole fraction 𝐂𝐓𝐂𝐀 PAPTCT 37.6101.30.038 0.0141kmolm3𝐂𝐁𝟏 𝐂𝐓 𝐂𝐀𝟏 0.038 0.0141 0.0239𝐂𝐁𝟐 𝐂𝐓 𝟎 0.03820
Mass Transfer𝐂𝐁𝐦 Third YearAssist. Prof. Dr. Ahmed Daham𝐂𝐁𝟐 𝐂𝐁𝟏 0.038 0.0239kmol �𝐀𝐁 15400.03032 154 3.1 107 (0.0141)0.038 9.12 10 6m2sExample: A small diameter tube closed at one end was filled with acetone to within 18 mmof the top and maintained at 290 K with a gentle stream of air blowing across the top. After15000 sec, the liquid level was fallen to 27.5 mm, the vapour pressure of acetone was 21.95kPa and atmospheric pressure was 99.75 kPa. Calculate the diffusivity of acetone in air.Given: the density of acetone is 790 kg/m3 and the molecular weight of acetone is 58kg/kmol.Solution:DAB MwtDAB ρL . z2 2 z1 2 . RTPT PA 2. 2t . PT . lnPT PA 1790 0.02752 0.0182 8.314 (290)m2 5 1.9 1099.75 0s58 . 15000 (99.75) . ln99.75 21.9521
Mass TransferThird YearAssist. Prof. Dr. Ahmed DahamPrediction of diffusivitiesThe diffusivity (DAB) for the transfer of one component (A) in another component (B)can be calculated empirically:1.Empirical correlation for calculation of gas diffusivity:The diffusi
Mass Transfer Third Year Assist. Prof. Dr. Ahmed Daham 9 1.8 10 5 8.314 295 101.3 1 10 3 ln 101.3 0 101 .3 50 65 5.153 10 4 kmol m2. s 2. Counter diffusion: i. Equimolecular counter diffusion: When the mass transfer rates of the two components are equal and opposite the process
Basic Heat and Mass Transfer complements Heat Transfer,whichispublished concurrently. Basic Heat and Mass Transfer was developed by omitting some of the more advanced heat transfer material fromHeat Transfer and adding a chapter on mass transfer. As a result, Basic Heat and Mass Transfer contains the following chapters and appendixes: 1.
Abstract—Recently, heat and mass transfer simulation is more and more important in various engineering fields. In order to analyze how heat and mass transfer in a thermal environment, heat and mass transfer simulation is needed. However, it is too much time-consuming to obtain numerical solutions to heat and mass transfer equations.
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Nov 01, 2014 · Chaldean Catholic Church Mass Schedule SATURDAY VIGIL: 6:00pm Ramsha, Evening Prayer 6:30pm Mass in English SUNDAY MASS: 8:00am Sapra, Morning Prayer 8:30am Holy Mass in Chaldean 10:00am Holy Mass in Arabic 11:30am Holy Mass in English 1:15pm Holy Mass in Chaldean 7:30pm Holy Mass in English DAILY MASS: MONDAY THRU FRIDAY
23.10 Radiant Heat Transfer Between Gray Surfaces 381 23.11 Radiation from Gases 388 23.12 The Radiation Heat-Transfer Coefficient 392 23.13 Closure 393 24. Fundamentals of Mass Transfer 398 24.1 Molecular Mass Transfer 399 24.2 The Diffusion Coefficient 407 24.3 Convective Mass Transfer 428 24.4 Closure 429 25.
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Chemical Reactions, and Stoichiometry Section 2.1: The Atomic Mass The atomic mass is the mass of 1 atom. Atoms are very small and their mass is a . molecular or ionic compound. The formula mass is expressed in a.m.u. Molar mass is the sum of atomic masses of all atoms in a mole of pure substance. The molar mass is expressed in g/mol.
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