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Chapter 28 & 30

U O T P. Chapter 28 & 30. 對流質量傳送係數 Convective Mass-Transfer Coefficient. 單 元 操 作. Kc ’ - Mass Transfer Coefficient. c iy. c A1. c A. c A2. z 1. z. z 2. 流動相. 單 元 操 作. kc ’ - Equimolar counter-diffusion. y. y Ai. x Ai. x. interface. Example Evaporizing A.

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Chapter 28 & 30

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  1. UOTP Chapter 28 & 30 對流質量傳送係數 Convective Mass-Transfer Coefficient

  2. 單元操作 Kc’-Mass Transfer Coefficient ciy cA1 cA cA2 z1 z z2 流動相

  3. 單元操作 kc’-Equimolar counter-diffusion y yAi xAi x interface

  4. Example Evaporizing A • The liquid A completely wets the surface, witch is a blotting paper. From the surface A is vaporizing though B.PA is 0.2 atm. (ky’= 6.78*10-5kgmole/m2-s) • a) Caculate =? NA ky u B:2atm PA2 =0 2 PA1=0.2atm 1 A 293K

  5. kc of A diffusion through stagnant

  6. Dimension Analysis (momentum, heat, mass ) • Momentum transfer 摩擦力 τw=Fw/A =f (ρ, u, D, μ) 無因次群= f = f( NRe )=f( ) • Heat transfer • Mass transfer （NRe , NSc）

  7. Schmidt Number NSC

  8. Example 28-2

  9. vx Incompressible flow (牛頓流體) Blasius’s solution 連續方程式 動量方程式 Navier-Stokes equation

  10. Analog in heat transfer 能量方程式 Fourier方程式

  11. Analog in mass transfer

  12. UOTP hx &kc over a plate (Laminar Flow) Re=xVρ/μ

  13. hx &kc over a plate (Laminar Flow) Pohlhausen 提出

  14. Example-28-3

  15. 動量、能量及質量輸送之類比 • (1) Reynolds Analog　　適用：Pr≈1,Sc≈1

  16. (2) Chilton-Colburn Analog

  17. Examp 28-4

  18. Examp 28-5

  19. Npr & Nsc

  20. kc’ in flow parallel to flate plate Gas Nre<200000 NSh＝0.664NReL0.5 NSc1/3 …(28-21) gas JD＝NShL/(NRELNSc1/3) ….(30-1) = 0.664NReL0.5 …..(30-2) NA Gas Nre>200000 NSh＝0.365NRe0.5 NSc1/3 JD＝NShL/(NRELNSc1/3) …..(30-1) = 0.365NReL0.5 …..(30-3) For 0.6<Sc<2500, 0.6<pr<100 J=Cf /2 Liquid or solid

  21. Example-30-1

  22. kc’ in flow through ball • 流過圓球外 u=0 • u>0 for gas u>0 NSh＝2.0＋C NRem NSc1/3

  23. kc’ in flow through ball • 流過圓球外 • u>0 for gas 1<Nre<48000 • u>0 for liquid 2<Nre<2000 • u>0 for liquid 2000<Nre<17000 NSh＝2.0＋0. 552NRe0.53 NSc1/3 …(30-9) NSh＝2.0＋0. 95NRe0.5 NSc1/3 u>0 NSh＝2.0＋0. 347NRe0.62 NSc1/3

  24. Example-30-2

  25. Examp • Air u=0.305 m/s • Mass transfer from C10H8 spheres to air • Dp=25.4mm ,T=45℃,DAB=6.92*10-6m2/s • PA0=0.555mmHg

  26. 單元操作 比較 管外流體熱傳（強制流動） • 因次分析 許多實驗式可查 • 垂直流過水平管外(適用液體＆NRe=1~104之氣體） NNu x NPr–0.3 =0.35+ 0.56NRe0.52 • 流過一球體 NNu = 2.0+ 0. 6 NRe0.50 NPr1/3

  27. kc’ in flow outside tube • 垂直流過水平管外 NSh NSc-0.3＝0.35＋0.56NRe0.52

  28. vx CA CA0 CAi r2 A kc’ in laminar flow inside tube • 流經圓管內 L

  29. 2. NSh＝0.023NRe0.8NSc1/3（ ）0.14 kc’ in turbulent flow inside tube • 流經圓管內 or vx CA CA0 CAi r2 A

  30. kc’ in laminar flow in wet-wall tube x • 流經濕壁塔 y u NAx falling film CA flow inside tube δ

  31. Example-30-4

  32. kc’ in flow through packed tower • 流過填充塔 • bed of spheres CA2 CA1

  33. kc’ in flow through packed tower • 流過填充塔 • bed of spheres Gas 10<Nre<10000 , liq 10<Nre<1500 v’ 氣體在空管之速度 JD＝ JH =(0.4548/ε)NRe-0.4069 liq 0.0016<Nre<55 165<Nsc<70000 JD＝ JH =(1.09/ε)NRe -2 /3 liq 55<Nre<1500 165<Nsc<10690 JD＝ JH =(0.25/ε)NRe –0.31

  34. kc’ in flow through packed tower • 流過填充塔 • bed of non-spheres Raschig Ring CA2 Lessing Ring CA1

  35. NA in flow through fluidize bed • 流過流體化床 Gas , liq 10<Nre<4000 v’ 氣體在空管之速度 JD＝ JH =(0.4548/ε)NRe-0.4069 liq 1<Nre<10 JD＝ JH =(1.1068/ε)NRe –0.72

  36. Examp 28-o1 • Dtower =0.0667m • C6H5COOH spheres Dp=6.375mm • Water 26.1℃ Q=5.514*10-7m3/s • Particles ε=0.435 surface area A=0.01198m2 • v’ =(5.514*10-7m3/s)/((π/4)Dtower 2) • 求kc • CA2

  37. NA to small particles(1) • Dp<0.6mm(600μm) • Gas bubbles or liq drops Dp>2.5mm Gas bubbles in liq Or solid particles in liq cA1 Diffusion free fall or rise by gravity

  38. Examp 28-o2 • Dp =100μm of air bubble 1 atm 37℃ • The solubity of O2 from air in water =2.26 kgmole/m3 • Dv of O2 in water is =3.2510-9m2/s • 求NA =kL (cAi -cA) =kc (cA - cAi)

  39. NA to small particles(2) • In high turbulent mixer Turbulent force larger than gravity

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