path: root/Mass_Transfer_Operations_by_R_E_Treybal/5-Interphase_Mass_Transfer.ipynb

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 {
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	   "source": [
       "# Chapter 5: Interphase Mass Transfer"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 5.1: Local_overall_mass_transfer_coeffecient.sce"
		   ]
		  },
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"clear;\n",
"clc;\n",
"\n",
"// Illustration 5.1\n",
"// Page: 114\n",
"\n",
"printf('Illustration 5.1 - Page: 114\n\n');\n",
"\n",
"// solution\n",
"\n",
"//***Data***//\n",
"// a = NH3, b = H2O\n",
"d = 2.54*10^(-2);// [m]\n",
"Yag = 0.80;\n",
"Xal = 0.05;\n",
"T = 273+26.7;// [K]\n",
"Kl = 2.87*10^(-5);// [kmol/square m.s.(kmol/cubic m)]\n",
"Sh = 40;\n",
"Da = 2.297*10^(-5);// [square m.s]\n",
"P = 1.0133*10^(5);// [N/square m]\n",
"Xbm = 1.0;\n",
"//*********//\n",
"\n",
"Ma = 18;// [kg/kmol]\n",
"// Liquid:\n",
"// Because of large conc. of ammonia in gas F's rather than k's are used.\n",
"// Molecular weight of water and ammonia are nearly same.\n",
"// The density of the solution is practically that of water.\n",
"MolarDensity1 = 1000/Ma;// [kmol/cubic m]\n",
"// Kl is determined for dilute soln. where Xbm is practically 1.0\n",
"Fl = Kl*Xbm*MolarDensity1;// [kmol/square m.s]\n",
"Ma = 18;// [kg-/kmol]\n",
"// Gas:\n",
"MolarDensity2 = (1/22.41)*(273/(273+26.7));// [kmol/cubic m]\n",
"Fg = Sh*MolarDensity2*Da/d;// [kmol/square m.s]\n",
"\n",
"// Mass Transfer Flux\n",
"// Th eqb. distribuion data for NH3 from 'The Chemical Engineers Handbook' 5th Edt. p3-68:\n",
"// Data = [Xa,pa]\n",
"// Xa = NH3 mole fraction in gas phas\n",
"// pa = NH3 partial pressure in N/square m\n",
"Data = [0 0;0.05 7171;0.10 13652;0.25 59917;0.30 93220];\n",
"// Ya_star = mole fraction of NH3 in gas phase at eqb.\n",
"Ya_star = zeros(5);\n",
"for i = 1:5\n",
"    Ya_star(i) = (Data(i,2)/P);\n",
"end\n",
"// For transfer of only one component\n",
"Na_by_SummationN = 1.0;\n",
"Ya = zeros(5);\n",
"for i = 1:5\n",
"    Ya(i) = 1-((1-Yag)*(1-Xal)/(1-Data(i)));\n",
"end\n",
"scf(0);\n",
"plot(Data(:,1),Ya_star,Data(:,1),Ya);\n",
"xgrid();\n",
"xlabel('Xa = mole fraction of NH3 in liquid phase');\n",
"ylabel('Ya = mole fraction of NH3 in gas phase');\n",
"legend('equilibrium line','operating line');\n",
"title('Ya Vs Xa');\n",
"\n",
"// From intersection of operating line & Eqb. line\n",
"Xai = 0.274;\n",
"Yai = 0.732;\n",
"\n",
"// From Eqn.5.20\n",
"Na = Na_by_SummationN*Fg*log((Na_by_SummationN-Yai)/(Na_by_SummationN-Yag));// [kmol NH3 absorbed/square m.s]\n",
"printf('Local mass transfer flux for ammonia is %e kmol/square m.s',Na);"
   ]
   }
,
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		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 5.2: Stages_and_Mass_Transfer_Rates.sce"
		   ]
		  },
  {
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"clear;\n",
"clc;\n",
"\n",
"// Illustration 5.2\n",
"// Page: 130\n",
"\n",
"printf('Illustration 5.2 - Page: 130\n\n');\n",
"\n",
"// solution\n",
"\n",
"//****Data***//\n",
"// Eqb. data\n",
"// Data = [Wt% of moisture in the soap,Partial pressure of water in air(mm Hg)]\n",
"Data = [0 0;2.40 9.66;3.76 19.20;4.76 28.4;6.10 37.2;7.83 46.4;9.90 55.0; 12.63 63.2;15.40 71.9;19.02 79.5];\n",
"P = 760;// [mm Hg]\n",
"// Initial air\n",
"p1 = 12;// [mm Hg]\n",
"T = 273+75;// [K]\n",
"//******//\n",
"\n",
"// Y = kg water/kg dry air\n",
"// X = kg water/kg dry soap\n",
"// E = Air water phase\n",
"// R = Soap water phase\n",
"Y = zeros(10);\n",
"X = zeros(10);\n",
"for i = 1:10\n",
"    Y(i) = Data(i,2)/(P-Data(i,2))*(18.02/29);\n",
"    X(i) = Data(i,1)/(100-Data(i,1));\n",
"end\n",
"\n",
"printf('Illustration 5.2 (a)\n\n');\n",
"\n",
"// Soln. (a)\n",
"// First operation\n",
"Y1 = p1/(P-p1);// [kg water/kg dry soap]\n",
"// Initial Soap\n",
"S1 = 16.7/(100-16.7);// [kg water/kg dry soap]\n",
"// Final soap\n",
"S2 = 13/(100-13);// [kg water/kg dry soap]\n",
"Rs = 10*(1-0.167);// [kg dry soap]\n",
"// Using ideal gas law\n",
"Es = 10*((760-p1)/760)*(273/T)*(29/22.41);// [kg dry air]\n",
"slopeOperat = -Rs/Es;\n",
"\n",
"deff('[y] = f2(x)','y = slopeOperat*(x-S1)+Y1')\n",
"x = S1:-0.01:S2;\n",
"\n",
"// Second Operation\n",
"X1 = S2;\n",
"scf(1);\n",
"deff('[y] = f3(S)','y = slopeOperat*(S-X1)+Y1');\n",
"S = 0:0.01:S1;\n",
"plot(X,Y,x,f2,S,f3);\n",
"xlabel('kg water / kg dry soap');\n",
"ylabel('kg water / kg dry air');\n",
"legend('Equilibrium line','First Process','Second Process');\n",
"a = get('current_axes');\n",
"tight_limits = 'on';\n",
"a.data_bounds = [0 0;0.24 0.08];\n",
"xgrid();\n",
"title('Illustration 5.2(a)')\n",
"// Results for First Process\n",
"// The condition at abcissa S2 correspond to the end of first operation\n",
"printf('Conditions corresponding to First Operation \n')\n",
"printf('X = %f kg water/kg dry soap\n',S2);\n",
"printf('Y = %f kg water/kg dry air\n',f2(S2));\n",
"\n",
"// Results for Second Process\n",
"//  The point at which the line meets the equilibrium line corresponds to the final value\n",
"X2 = 0.103;\n",
"Y2 = (X2/(1+X2));\n",
"printf('Final moisture content of soap is %f %%\n\n',Y2*100);\n",
"\n",
"printf('Illustration 5.2 (b)\n\n');\n",
"\n",
"// Solution (b)\n",
"\n",
"Rs = 1*(1-0.167);// [kg dry soap/h]\n",
"// Entering soap\n",
"X1 = 0.20;// [kg water/kg dry soap]\n",
"// Leaving soap\n",
"x = 0.04;\n",
"X2 = x/(1-x);// [kg water/kg dry soap]\n",
"// Entering air\n",
"Y2 = 0.00996;// [from Illustration 5.2(a), kg water/kg dry air]\n",
"// The operating line of least slope giving rise to eqb. condition will indicate least amount of air usable.\n",
"// At X1 = 0.20; the eqb. condition:\n",
"Y1 = 0.0675;// [kg water/kg dry air]\n",
"scf(2);\n",
"deff('[y] = f4(x)','y = ((Y1-Y2)/(X1-X2))*(x-X1)+Y1');\n",
"x = X2:0.01:0.24;\n",
"plot(X,Y,x,f4);\n",
"xlabel('kg water / kg dry soap');\n",
"ylabel('kg water / kg dry air');\n",
"a = get('current_axes');\n",
"tight_limits = 'on';\n",
"a.data_bounds = [0 0;0.24 0.08];\n",
"xgrid();\n",
"title('Illustration 5.2(b)')\n",
"legend('Equilibrium line','Operating Line');\n",
"// By Eqn. 5.35\n",
"Es = Rs*(X1-X2)/(Y1-Y2);// [kg dry air/h]\n",
"Esv = (Es/29)*22.41*(P/(P-p1))*(T/273);// [cubic m/kg dry soap]\n",
"printf('Minimum amount of air required is %f cubic m/kg dry soap\n\n',Esv);\n",
"\n",
"printf('Illustration 5.2 (c)\n\n');\n",
"\n",
"// solution (c)\n",
"\n",
"Esnew = 1.30*Es;// [kg dry air/h]\n",
"Y1 = Rs*((X1-X2)/Esnew)+Y2;\n",
"scf(3);\n",
"deff('[y] = f5(x)','y = ((Y1-Y2)/(X1-X2))*(x-X1)+Y1');\n",
"x = X2:0.01:0.24;\n",
"plot(X,Y,x,f5);\n",
"xlabel('kg water / kg dry soap');\n",
"ylabel('kg water / kg dry air');\n",
"a = get('current_axes');\n",
"tight_limits = 'on';\n",
"a.data_bounds = [0 0;0.24 0.08];\n",
"xgrid();\n",
"title('Illustration 5.2(c)')\n",
"legend('Equilibrium line','Operating Line');\n",
"// with final coordinates X = X1 & y = Y1\n",
"// From figure, Total number of eqb . stages = 3\n",
"N = 3;\n",
"printf('Moisture content of air leaving the drier is %f kg water/kg dry air\n',Y1);\n",
"printf('Total number of eqb. stages = %d\n',N);"
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