By P Chattopadhyay
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Extra resources for Absorption & stripping
98) A. G s ' dY =A. dX A. k L , c' a. 1-1. kmollm3) Again, X = mole ratio of solute in liq phase c =-CT -c where, c = mols of solute per unit vol. of liq kmollm 3 cT =total mols of solute + solvent per unit vol. ofliq, kmol/m 3 C Ls' dX = - -X, c T ... 99) I+X = k Lc ' a. cT' [_X. _ X) 1____ , =k I+Xi . a. -X 1 ( (I+XJ(I+X) • dZ J dZ • ... a,c T ... 101) ... dX _ X - Xt ... -X 1 For dilute solutions, the Eqn. 100 simplifies to z f dZ ° Xb = f Ls . a,c T Xt Xi-X ... 3. , (-kL, )kG, y) is constant.
_ X) 1____ , =k I+Xi . a. -X 1 ( (I+XJ(I+X) • dZ J dZ • ... a,c T ... 101) ... dX _ X - Xt ... -X 1 For dilute solutions, the Eqn. 100 simplifies to z f dZ ° Xb = f Ls . a,c T Xt Xi-X ... 3. , (-kL, )kG, y) is constant. Under these circumstances we can transform Eqn. 93 to: Z fdZ = ° Yb Gs KG,p . a. P f (1 + Y) (1 + y*) . dY • Y Y- Y* ... 106) t Eqn. 1. 94 to : Z = [HTU]o,G' [NTU]o,G ... 107) Eqn. 97 becomes: f Z Yb °f Gs dY . dZ= --=---KG,p . a. P Y Y _ Y* ... 108) t Likewise Eqns. 104 transform to: ...
A,c T Xt Xi-X ... 3. , (-kL, )kG, y) is constant. Under these circumstances we can transform Eqn. 93 to: Z fdZ = ° Yb Gs KG,p . a. P f (1 + Y) (1 + y*) . dY • Y Y- Y* ... 106) t Eqn. 1. 94 to : Z = [HTU]o,G' [NTU]o,G ... 107) Eqn. 97 becomes: f Z Yb °f Gs dY . dZ= --=---KG,p . a. P Y Y _ Y* ... 108) t Likewise Eqns. 104 transform to: ... L Z fdZ- ° - ... 110) Xb f Ls dX KL,c·a,cT· Xt X*-X ... P The height of an overallliq transfer unit is: Ls [HTU] ° L = -----''''-, kL,c·a,cT ... 112) ... 45 Absorption The number of overall gas transfer units is : Yb [NTU]OG = • Y)(I+ y*) * .
Absorption & stripping by P Chattopadhyay