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ChE 131: Transport Processes Outline 1.Class Policies 2.Introduction 3.Review Course Assessment 3 Long Exams Final Exam 3 Machine Problems Classwork 60% 20% 15% 5% Policies to Remember Submit 12 sheets of colored pad paper at least the day before an exam. Get an official excuse slip from the College if you miss an exam and you have a valid excuse. No exemptions will be given for the final exam. Policies to Remember Quizzes may be given from time to time. All quizzes shall be written in bluebooks. No makeup shall be given to missed quizzes. Outline 1.Class Policies 2.Introduction 3.Review Transport Phenomena What exactly are "transport phenomena"? Transport phenomena are really just a fancy way that Chemical Engineers group together three areas of study that have certain ideas in common. These three areas of study are: • Fluid mechanics • Heat transfer • Mass transfer Transport processes Transport Processes Momentum Transport – transfer of momentum which occurs in moving media (fluid flow, sedimentation, mixing, filtration, etc.) Heat Transport – transfer of energy from one region to another (drying, evaporation, distillation) Mass Transport – transfer of mass of various chemical species from one phase to another distinct phase (distillation, absorption, adsorption, etc.) Why Study Transport Phenomena? Why Study Transport Phenomena? Why Study Transport Phenomena? Chemical Engineering Thermodynamics Transport Phenomena PROCESS EQUIPMENT DESIGN Chemical Reaction Kinetics Materials Science Process Economics Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR Levels of Analysis MACROSCOPIC Use of macroscopic balances MICROSCOPIC Overall assessment of a system MOLECULAR Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR Small region/volume element is selected Use of equations of change Velocity, temperature, pressure and concentration profiles are determined Levels of Analysis MACROSCOPIC MICROSCOPIC MOLECULAR Molecular structure and intermolecular forces become significant Complex molecules, extreme T and P, chemically reacting systems Review LET’S REVIEW!!! Dimensional Analysis Check the dimensional consistency of the following empirical equation for heat transfer between a flowing fluid and the surface of a sphere: 1 G 0.5 h 2.0kD 0.6D 0.5 h – heat transfer coefficient (W/m2-K) D – diameter of sphere (m) k – thermal conductivity of fluid (W/m-K) G – mass velocity of fluid (kg/m2-s) μ – viscosity (kg/m-s) cp – heat capacity (J/kg-K) 0.17 0.33 0.67 p c k Dimensional Analysis We use the following convention: Energy unit – E Mass unit – M Length unit – L Time unit – t Temperature unit – T Dimensional Analysis E /t E 2 2 L T L t T E /t E L T L t T For the heat transfer coefficient: For thermal conductivity: For diameter: L For viscosity: M Lt Dimensional Analysis M For mass velocity: 2 L t E For heat capacity: M T Combining: E E 1 2 L t T L t T L 1 0.5 L 0.5 0.17 0.17 0.33 0.67 M L t E E 0.5 0.17 0.33 0.33 0.67 0.67 0.67 L t M M T L t T Dimensional Analysis E E 1 2 L t T L t T L 0.5 0.17 0.17 0.33 0.67 L t E E 1 M 0.5 0.5 0.17 0.33 0.33 0.67 0.67 0.67 L L t M M T L t T Simplifying: E E 2 2 L t T L t T M 0.5 0.17 0.33 t L 0.17 0.5 1 0.67 0.5 0.17 0.67 T E 0.33 0.67 0.33 0.67 Dimensional Analysis Simplifying: E E 2 2 L t T L t T 0.5 0.17 0.33 0.17 0.5 1 0.67 0.33 0.67 M L E E E E 2 2 2 L t T L t T L t T t 0.5 0.17 0.67 T 0.33 0.67 Material Balance An evaporator is fed continuously with 25 metric tons/h of a solution consisting of 10% NaOH, 10% NaCl, and 80% H2O. During evaporation, water is boiled off, and salt precipitates as crystals, which are settled and removed from the remaining liquor. The concentrated liquor leaving the evaporator contains 50% NaOH, 2% NaCl, and 48% H2O. Calculate the MT of water evaporated per hour, the MT of salt precipitated per hour, and MT of liquor produced per hour. Material Balance 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H2O EVAPORATOR H (water) 1.0 H2O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H2O C (crystals) 1.0 NaCl NaOH bal: 0.10(25) = 0.5M + 0C + 0H NaCl bal: 0.10(25) = 0.02M + 1.0C + 0H H2O bal: 0.80(25) = 0.48M + 0C + 1.0H Material Balance 25 MT/h 0.1 NaOH 0.1 NaCl 0.8 H2O EVAPORATOR H (water) 1.0 H2O M (mother liquor) 0.5 NaOH 0.02 NaCl 0.48 H2O C (crystals) 1.0 NaCl H = water evaporated per hour = 17.6 MT/h C = salt precipitated per hour = 2.4 MT/h M = liquor produced per hour = 5 MT/h Material Balance Dry gas containing 75% air and 25% NH3 vapor enters the bottom of a cylindrical packed absorption tower that is 2 ft in diameter. Nozzles in the top of the tower distribute water over the packing. A solution of NH3 in H2O is drawn at the bottom of the column, and scrubbed gas leaves the top. The gas enters at 80°F and 760 mm Hg. It leaves at 60°F and 730 mm Hg. The leaving gas contains, on the solute-free basis, 1.0% NH3. If the entering gas flows through the empty bottom of the column at velocity (upward) of 1.5 ft/s, how many ft3 of entering gas are treated per hour? How many pounds of NH3 are absorbed per hour? Material Balance G (scrubbed gas) 0.01 NH3 (solute-free) W (water) 1.0 H2O SCRUBBER D (dry gas) 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3 Volume of gas entering = velocity diameter of tower ft ft 3 2 3600 s = 1.5 2 ft s 4 1h 16964.6 h Material Balance Convert solute-free basis percentage to mass fraction: 0.01 NH3 , solute-free = xNH3 1 xNH3 xNH3 0.0099 x air 0.9901 We now rewrite our diagram: Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O SCRUBBER D (dry gas) 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3 Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior, Material Balance Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior and a basis of 1 hour: P V TSTP n2 nSTP PSTP VSTP T 760 mm Hg 16964.6 ft 3 460 32R n2 1 lbmol 3 760 mm Hg 359 ft 460 32R n2 42.35 lbmol Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3 Air balance: 0.75(42.35) = 0.9901G G = amount of dry gas = 32.08 lbmol dry gas Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 S (water + ammonia) x H2O y NH3 NH3 balance: 0.25(42.35) = 0.0099(32.08) + xS xS = amount of NH3 absorbed = 10.27 lbmol NH3 Material Balance G (scrubbed gas) 0.0099 NH3 0.9901 air W (water) 1.0 H2O SCRUBBER D (dry gas) = 42.35 lbmol 0.75 air 0.25 NH3 Pounds of NH3 absorbed: S (water + ammonia) x H2O y NH3 17 lb NH3 lb NH3 10.27 lb mol NH3 174.58 hr lb mol NH3 Energy Balance Air is flowing steadily through a horizontal heated tube. The air enters at 40°F and at a velocity of 50 ft/s. It leaves the tube at 140°F and 75 ft/s. The average specific heat of air is 0.24 Btu/lb-°F. How many Btu’s per pound of air are transferred through the wall of the tube? Energy Balance Energy Balance: v H g z Q Ws 2 2 Simplifying: v 2 cp T Q 2 Energy Balance Energy Balance: v 2 cp T Q 2 Btu 0.24 140 40F lbm F 752 502 2 Q 24.06 Btu lbm 1 lbf ft 2 Btu Q 2 s 32.174 lb ft 778.2 lbf ft m 2 s Differential Equation Solve the following differential equation: dy y tan( x) cos( x) dx Differential Equation dy y tan( x) cos( x) dx This equation follows the form: dy yP( x) Q( x) dx whose solution is: y e P ( x ) dx P ( x ) dx Q( x)e dx C Differential Equation dy y tan( x) cos( x) dx sin( x) P ( x)dx tan( x)dx cos( x) dx Let u = cos(x), du = sin(x)dx sin( x) 1 1 1 cos( x) dx u du ln(u) ln u ln cos( x) lnsec( x) Differential Equation dy y tan( x) cos( x) dx y e P ( x ) dx P ( x ) dx Q( x)e dx C y e ln(sec( x )) cos( x )eln(sec( x )) dx C y eln(cos( x )) cos( x ) sec( x )dx C y cos( x ) dx C y x cos( x ) C