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+clear;

+clc;

+

+// Illustration 6.13

+// Page: 380

+printf('Illustration 6.13 -  Page: 380\n\n');

+

+// solution

+//*****Data*****//

+// A-benzene  B-toluene  C-1,2,3-trimethylbenzene

+// From example 6.10

+Nmin = 4.32; // [stages]

+// From example 6.11

+Rmin = 0.717; // [minimum reflux ratio]

+// For R = 1

+R = 1;

+X = (R-Rmin)/(R+1);

+// From equ 6.88

+Y = 1-exp((1+54.4*X)/(11+117.2*X)*(X-1)/sqrt(X));

+// Fro equ 6.86

+N = (Y+Nmin)/(1-Y);

+// From example 6.10 99.7% of the LNK (benzene) is recovered in the distillate// , 95% of the light key is in the distillate, and 95% of the heavy key is in// the bottoms

+

+// For a basis of 100 mol of feed, the material balances for three components // are

+// For distillate

+nAd = 39.88; // [LNK, moles of benzene]

+nBd = 28.5; // [LK, moles of toluene]

+nCd = 1.50; // [HK, moles of 1,2,3-trimethylbenzene]

+nTd = nAd+nBd+nCd; // [total number of moles]

+xAd = nAd/nTd;

+xBd = nBd/(nTd);

+xCd = nCd/(nTd);

+

+// For bottoms

+nAb = 0.12;

+nBb = 1.50;

+nCb = 28.50;

+nTb = nAb+nBb+nCb;

+xAb = nAb/nTb;

+xBb = nBb/nTb;

+xCb = nCb/nTb;

+

+D = nTd;

+W = nTb;

+// From problem statement

+Zlk = 0.3;

+Zhk = Zlk;

+// Substituting in equation 6.89

+// T = Nr/Ns

+T = (Zhk/Zlk*W/D*(xBb/xCd)^2)^0.206;

+

+// Solution of simultaneous equation 

+function[f]=H(e)

+    f(1) = e(1)-e(2)*T;

+    f(2) = e(1)+e(2)-N;

+        funcprot(0);

+endfunction

+

+// Initial guess

+e = [5 4];

+y = fsolve(e,H);

+Nr = y(1); // [number of stages in rectifying section]

+Ns = y(2); // [number of stages in stripping section]

+disp(Ns,Nr);

+printf('Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed.');