1 / 14

By Final Year Chem. Engg. (Roll no:11-20) U.I.C.T,Mumbai-400019

Packed Column Extractor. By Final Year Chem. Engg. (Roll no:11-20) U.I.C.T,Mumbai-400019. An Overview. Same packing as used in G-L operations Advantage of using Packings Material of Packing to be used. Tower design. The Problem statement : Flow rate of organic stream= 1m 3 /hr

aldon
Télécharger la présentation

By Final Year Chem. Engg. (Roll no:11-20) U.I.C.T,Mumbai-400019

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Packed Column Extractor By Final Year Chem. Engg. (Roll no:11-20) U.I.C.T,Mumbai-400019

  2. An Overview • Same packing as used in G-L operations • Advantage of using Packings • Material of Packing to be used

  3. Tower design • The Problem statement: Flow rate of organic stream= 1m3/hr Impurity in inlet organic stream=10000ppm Desired impurity in exit stream=500ppm

  4. Approach • We assume that the dispersed and the continuous phases are in plug flow • We find the minimum value of Ud/Uc • Then we assume different values for Ud/Uc • Calculate hold up at flooding using • εf={[(Ud/Uc)2+8 (Ud/Uc)]0.5-3(Ud/Uc)}/[4*(1- Ud/Uc)] • Assume a certain percentage of flooding hold up as operating hold up and hence calculate ε • We find the terminal velocity of a drop using • eUo=C(a*ρc/(e3*g*Δ ρ)) (-0.5) formula given in the book by Degallson and Laddha where C=0.637 • Then we calculate Ud using the correlation for slip velocity

  5. Find diameter of the column: • D=((Qc/Uc)*(4/3.142))0.5 • Drop diameter is found using • d=1.6(γ/(ρc-ρd)g) 0.5 • Size of Raschig rings were taken as 1”,0.75” and 0.5”. • Overall Height of transfer unit: Koc.a=0.06* φ*(1- φ)/[(a*ρc/(g*e 3 *Δρ))0.5*(γ/(Δρ*g)) 0.5 *{(Sc)c 0.5 +(Sc)d 0.5 /m}] (This corellation is for packing size greater than drop size)

  6. [HTU oc]plug flow = Uc/Koc.a • [NTU oc]plug flow=(Cc1-Cc2)/(ΔC)LM • Height of column: Zt=[HTU oc]plug flow * [NTU oc]plug flow • Distributor design: We take nozzle velocity = 0.5*eUo Nozzle diameter value should be comparable with droplet diameter value. Hence, we take nozzle diameter= 6 mm. No of orifices = Qd/( Area of nozzle*Vn)

  7. Sample Calculations • Overall mass balance • Qo*ρo*(10,000-500)=Qa*ρa*(50,000-0) • 1*900*9500= Qa*1000*50000 • Thus Qa= .9*9500/50000 = 0.171 • Thus, the flow rate ratio (aq:org) or velocity ratio should be > 0.171

  8. Sample calculations- continued • Consider Raschig ring packings of size 1”. • Let the operating holdup=60% of holdup at flooding. • Let Ud/Uc=1.1. • εf = 0.3403 • Operating holdup = 0.2042 • Terminal velocity = 0.03191 m/s • By slip velocity relation, Ud= 0.00385 m/s • Uc= 0.0035 m/s • Column diameter: 0.318m • Drop diameter: 8 mm • Koc.a= 12.407 • HTU=3.327 ft=1.01m • NTU=3.429 • Column height=11.41 ft = 3.477 m • No of orifices for the distributor (orifice diameter=6 mm)= 70(approx)

  9. Observations • With increase in Ud/Uc, height of column decreases, diameter increases. • The diameter of the column increases when packings of smaller nominal diameter are used. • With increase in %flooding the total height of the column decreases.

  10. Thank You

More Related