initial commit / add all books

author: priyanka 2015-06-24 15:03:17 +0530
committer: priyanka 2015-06-24 15:03:17 +0530
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tree: ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH10
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diff --git a/647/CH10/EX10.1/Example10_1.sce b/647/CH10/EX10.1/Example10_1.sce
new file mode 100755
index 000000000..3c7437b8b
--- /dev/null
+++ b/647/CH10/EX10.1/Example10_1.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 10.1
+// Page: 390
+
+printf("Example: 10.1 - Page: 390\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.1 on page number 390 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.1 on page 390 of the book.");
+\ No newline at end of file
diff --git a/647/CH10/EX10.10/Example10_10.sce b/647/CH10/EX10.10/Example10_10.sce
new file mode 100755
index 000000000..1a0644d94
--- /dev/null
+++ b/647/CH10/EX10.10/Example10_10.sce
@@ -0,0 +1,31 @@
+clear;
+clc;
+
+// Example: 10.10
+// Page: 412
+
+printf("Example: 10.10 - Page: 412\n\n");
+
+// Solution
+
+//*****Data******//
+P = 101.3;// [kPa]
+P1sat = 100.59;// [kPa]
+P2sat = 99.27;// [kPa]
+x1 = 0.532;
+//****************//
+
+x2 = 1 - x1;
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+
+// For solution containing 10 mol percent benzene:
+x1 = 0.10;
+x2 = 1 - x1;
+gama1 = exp(A/(1 + (A*x1/(B*x2))^2));
+gama2 = exp(B/(1 + (B*x2/(A*x1))^2));
+printf("Activity Coeffecient\n");
+printf("gama1 = %.3f\n",gama1);
+printf("gama2 = %.3f\n",gama2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.11/Example10_11.sce b/647/CH10/EX10.11/Example10_11.sce
new file mode 100755
index 000000000..ccb833502
--- /dev/null
+++ b/647/CH10/EX10.11/Example10_11.sce
@@ -0,0 +1,34 @@
+clear;
+clc;
+
+// Example: 10.11
+// Page: 413
+
+printf("Example: 10.11 - Page: 413\n\n");
+
+// Solution
+
+//*****Data******//
+P = 760;// [mm of Hg]
+P1sat = 995;// [mm of Hg]
+P2sat = 885;// [mm of Hg]
+x1 = 0.335;
+T = 64.6;// [OC]
+//****************//
+
+x2 = 1 - x1;
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+
+// For solution containing 11.1 mol percent acetone:
+x1 = 0.111;
+x2 = 1 - x1;
+gama1 = exp((A*x2^2)/(x2 + (A*x1/(B))^2));
+gama2 = exp((B*x1^2)/(x1 + (B*x2/(A))^2));
+y1 = 1/(1 + (gama2*x2*P2sat/(gama1*x1*P1sat)));
+y2 = 1 - y1;
+printf("Equilibrium Composition\n");
+printf("Acetone Composition = %.2f %%\n",y1*100);
+printf("Chloform composition = %.2f %%\n",y2*100);
+\ No newline at end of file
diff --git a/647/CH10/EX10.12/Example10_12.sce b/647/CH10/EX10.12/Example10_12.sce
new file mode 100755
index 000000000..522e83e4e
--- /dev/null
+++ b/647/CH10/EX10.12/Example10_12.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 10.12
+// Page: 414
+
+printf("Example: 10.12 - Page: 414\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.12 on page number 414 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.12 on page 414 of the book.");
+\ No newline at end of file
diff --git a/647/CH10/EX10.13/Example10_13.sce b/647/CH10/EX10.13/Example10_13.sce
new file mode 100755
index 000000000..90b1811dc
--- /dev/null
+++ b/647/CH10/EX10.13/Example10_13.sce
@@ -0,0 +1,54 @@
+clear;
+clc;
+
+// Example: 10.13
+// Page: 418
+
+printf("Example: 10.13 - Page: 418\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: iso - butanol
+// 2: iso - propanol
+T = 50 + 273;// [K]
+x1 = 0.3;
+V1 = 65.2;// [cubic cm/mol]
+V2 = 15.34;// [cubic cm/mol]
+// For Wilson equation:
+a12 = 300.55;// [cal/mol]
+a21 = 1520.32;// [cal/mol]
+// For NTRL equation:
+b12 = 685.21;// [cal/mol]
+b21 = 1210.21;// [cal/mol]
+alpha = 0.552;
+R = 2;// [cal/mol K]
+//******************//
+
+x2 = 1 - x1;
+// A: Estimation of activity coeffecient using Wilson equation:
+// From Eqn. 10.65:
+A12 = (V2/V1)*exp(-a12/(R*T));
+// From Eqn. 10.66:
+A21 = (V1/V2)*exp(-a21/(R*T));
+// From Eqn. 10.67:
+gama1 = exp(-log(x1 + A12*x2) + x2*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))));
+// From Eqn. 10.68:
+gama2 = exp(-log(x2 + A21*x1) - x1*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))));
+printf("Wilson equation\n");
+printf("Activity Coeffecient of iso - butanol is %.3f\n",gama1);
+printf("Activity Coeffecient of iso - propanol is %.3f\n",gama2);
+printf("\n");
+
+// A: Estimation of activity coeffecient using NTRL equation:
+t12 = b12/(R*T);
+t21 = b21/(R*T);
+G12 = exp(-alpha*t12);
+G21 = exp(-alpha*t21);
+// From Eqn. 10.70:
+gama1 = exp((x2^2)*(t21*(G21/(x1 + x2*G21))^2 + (t12*G12/(x2 + x1*G12)^2)));
+// From Eqn. 10.71:
+gama2 = exp((x1^2)*(t12*(G12/(x2 + x1*G12))^2 + (t21*G21/(x1 + x2*G21)^2)));
+printf("NTRL equation\n");
+printf("Activity Coeffecient of iso - butanol is %.3f\n",gama1);
+printf("Activity Coeffecient of iso - propanol is %.3f\n",gama2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.14/Example10_14.sce b/647/CH10/EX10.14/Example10_14.sce
new file mode 100755
index 000000000..dc9b31eb6
--- /dev/null
+++ b/647/CH10/EX10.14/Example10_14.sce
@@ -0,0 +1,35 @@
+clear;
+clc;
+
+// Example: 10.14
+// Page: 426
+
+printf("Example: 10.14 - Page: 426\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.4;// [mole fraction of ethane in vapour phase]
+x2 = 0.6;// [mole fraction of propane in vapour phase]
+P = 1.5;// [MPa]
+//***************//
+
+// Assume T = 10 OC
+T = 10;// [OC]
+// From Fig. 10.14 (Pg 426):
+K1 = 1.8;
+K2 = 0.5;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+// Since y1 + y2 > 1, so we assume another value of T = 9 OC.
+T = 9;// [OC]
+// From Fig. 10.14 (Pg 426):
+K1 = 1.75;
+K2 = 0.5;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+// Since y1 + y2 = 1. Therefore:
+printf("Bubble Temperature is %d OC\n",T);
+printf("Composition of the vapour bubble:\n y1 = %.2f\n y2 = %.2f",y1,y2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.15/Example10_15.sce b/647/CH10/EX10.15/Example10_15.sce
new file mode 100755
index 000000000..361763b84
--- /dev/null
+++ b/647/CH10/EX10.15/Example10_15.sce
@@ -0,0 +1,40 @@
+clear;
+clc;
+
+// Example: 10.15
+// Page: 428
+
+printf("Example: 10.15 - Page: 428\n\n");
+
+// Solution
+
+//*****Data******//
+y1 = 0.20;// [mole fraction of methane in vapour phase]
+y2 = 0.30;// [mole fraction of ethane in vapour phase]
+y3 = 0.50;// [mole fraction of propane in vapour phase]
+T = 30;// [OC]
+//*************//
+
+// Assume P = 2.0 MPa
+P = 2.0;// [MPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 8.5;
+K2 = 2.0;
+K3 = 0.68;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 2.15 MPa at 30 OC.
+P = 2.15;// [MPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 8.1;
+K2 = 1.82;
+K3 = 0.62;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Dew Pressure is %.2f MPa\n",P);
+printf("Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f",x1,x2,x3);
+\ No newline at end of file
diff --git a/647/CH10/EX10.16/Example10_16.sce b/647/CH10/EX10.16/Example10_16.sce
new file mode 100755
index 000000000..ee0025c9c
--- /dev/null
+++ b/647/CH10/EX10.16/Example10_16.sce
@@ -0,0 +1,94 @@
+clear;
+clc;
+
+// Example: 10.16
+// Page: 429
+
+printf("Example: 10.16 - Page: 429\n\n");
+
+// Solution
+
+// Dew point Pressure
+//*****Data******//
+y1 = 0.10;// [mole fraction of methane in vapour phase]
+y2 = 0.20;// [mole fraction of ethane in vapour phase]
+y3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 690 kPa
+P = 690;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 20.0;
+K2 = 3.25;
+K3 = 0.92;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 10135 kPa at 10 OC.
+P = 10135;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 13.20;
+K2 = 2.25;
+K3 = 0.65;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 870 kPa at 10 OC.
+P = 870;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 16.0;
+K2 = 2.65;
+K3 = 0.762;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Dew Pressure is %d kPa\n",P);
+printf("Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n",x1,x2,x3);
+printf("\n");
+
+// Bubble point Pressure
+//*****Data******//
+x1 = 0.10;// [mole fraction of methane in vapour phase]
+x2 = 0.20;// [mole fraction of ethane in vapour phase]
+x3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 2622 kPa
+P = 2622;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.60;
+K2 = 1.11;
+K3 = 0.335;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 2760 kPa at 10 OC.
+P = 2760;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.25;
+K2 = 1.07;
+K3 = 0.32;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 2656 kPa at 10 OC.
+P = 2656;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.49;
+K2 = 1.10;
+K3 = 0.33;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Bubble Pressure is %d kPa\n",P);
+printf("Composition of the vapour bubble:\n y1 = %.4f\n y2 = %.4f\n y3 = %.4f",y1,y2,y3);
+\ No newline at end of file
diff --git a/647/CH10/EX10.17/Example10_17.sce b/647/CH10/EX10.17/Example10_17.sce
new file mode 100755
index 000000000..443de93ad
--- /dev/null
+++ b/647/CH10/EX10.17/Example10_17.sce
@@ -0,0 +1,50 @@
+clear;
+clc;
+
+// Example: 10.17
+// Page: 432
+
+printf("Example: 10.17 - Page: 432\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: acetone 2: acetonitrile 3: nitromethane
+z1 = 0.45;
+z2 = 0.35;
+z3 = 0.20;
+P1sat = 195.75;// [kPa]
+P2sat = 97.84;// [kPa]
+P3sat = 50.32;// [kPa]
+//***************//
+
+// Bubble Point Calculation:
+Pbubble = z1*+P1sat + z2*P2sat +z3*P3sat;// [kPa]
+
+// Dew Point Calculation:
+Pdew = 1/((z1/P1sat) + (z2/P2sat) + (z3/P3sat));// [kPa]
+K1 = P1sat/Pdew;
+K2 = P2sat/Pdew;
+K3 = P3sat/Pdew;
+// Overall Material balance:
+// For 1 mol of the feed.
+// L + V = 1......................................... (1)
+// F*zi = L*xi + V*yi ............................... (2)
+// zi = (1 - V)*xi + V*yi ........................... (3)
+// Substituting xi = yi/K in eqn. (3)
+// yi = zi*Ki/(1 + V*(Ki - 1))
+// Since, Sum(yi) = 1.
+deff('[y] = f(V)','y = (z1*K1/(1 + V*(K1 - 1))) + (z2*K2/(1 + V*(K2 - 1))) + (z3*K3/(1 + V*(K3 - 1))) - 1');
+V = fsolve(0.8,f);
+L = 1 - V;
+y1 = z1*K1/(1 + V*(K1 - 1));
+y2 = z2*K2/(1 + V*(K2 - 1));
+y3 = z3*K3/(1 + V*(K3 - 1));
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+printf(" L = %e mol\n",L);
+printf(" V = %e mol\n",V);
+printf(" y1 = %.4f\n y2 = %.4f\n y3 = %.4f\n",y1,y2,y3);
+printf(" x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n",x1,x2,x3);
+\ No newline at end of file
diff --git a/647/CH10/EX10.18/Example10_18.sce b/647/CH10/EX10.18/Example10_18.sce
new file mode 100755
index 000000000..72db8defc
--- /dev/null
+++ b/647/CH10/EX10.18/Example10_18.sce
@@ -0,0 +1,46 @@
+clear;
+clc;
+
+// Example: 10.18
+// Page: 433
+
+printf("Example: 10.18 - Page: 433\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: Benzene 2: Toulene
+z1 = 0.81;
+Temp = 60;// [OC]
+P = 70;// [kPa]
+// Antonine Constants:
+A1 = 14.2321;
+B1 = 2773.61;
+C1 = 220.13;
+A2 = 15.0198;
+B2 = 3102.64;
+C2 = 220.02;
+//******************//
+
+deff('[P1] = f1(T)','P1 = exp(A1 - B1/(T + C1))');
+P1sat = f1(Temp);// [kPa]
+deff('[P2] = f2(T)','P2 = exp(A2 - B2/(T + C2))');
+P2sat = f2(Temp);// [kPa]
+// P = x1*P1sat + x2*P2sat;
+// x2 = 1 - x1;
+deff('[y] = f3(x1)','[y] = P - (x1*P1sat + (1 - x1)*P2sat)');
+x1 = fsolve(7,f3);
+y1 = x1*P1sat/P;
+x2 = 1 - x1;
+y2 = 1 - y1;
+
+// Basis: 1 mol of feed stream.
+F = 1;// [mol]
+// F*zi = L*xi + V*yi = L*xi + (1 - L)*yi
+deff('[y] = f4(L)','[y] = F*z1 - (L*x1 + (1 - L)*y1)');
+L = fsolve(7,f4);// [mol]
+V = 1 - L;// [mol]
+printf(" L = %.4f mol\n",L);
+printf(" V = %.4f mol\n",V);
+printf(" y1 = %.4f\n y2 = %.4f\n",y1,y2);
+printf(" x1 = %.4f\n x2 = %.4f\n",x1,x2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.19/Example10_19.sce b/647/CH10/EX10.19/Example10_19.sce
new file mode 100755
index 000000000..ecfe4e404
--- /dev/null
+++ b/647/CH10/EX10.19/Example10_19.sce
@@ -0,0 +1,45 @@
+clear;
+clc;
+
+// Example: 10.19
+// Page: 413
+
+printf("Example: 10.11 - Page: 436\n\n");
+
+// Solution
+
+//*****Data******//
+// (1): acetone (2): carbon tetrachloride
+T = 45;// [OC]
+// Data = [P (torr), x1, y1]
+Data = [315.32 0.0556 0.2165;339.70 0.0903 0.2910;397.77 0.2152 0.4495;422.46 0.2929 0.5137; 448.88 0.3970 0.5832;463.92 0.4769 0.6309;472.84 0.5300 0.6621;485.16 0.6047 0.7081;498.07 0.7128 0.7718;513.20 0.9636 0.9636];
+//*************//
+
+// From the standard data (Pg 531):
+// For Acetone:
+A1 = 14.2342;
+B1 = 2691.46;
+C1 = 230.00;
+// For carbon tetrachloride:
+A2 = 13.6816;
+B2 = 2355.82;
+C2 = 220.58;
+P1sat = exp(A1 - B1/(T + C1));// [kPa]
+P2sat = exp(A2 - B2/(T + C2));// [kPa]
+P1sat = P1sat*760/101.325;// [torr]
+P2sat = P2sat*760/101.325;// [torr]
+P = Data(:,1);
+x1 = Data(:,2);
+y1 = Data(:,3);
+x2 = 1 - x1;
+y2 = 1 - y1;
+gama1 = (y1.*P./x1)/P1sat;
+gama2 = (y2.*P./x2)/P2sat;
+Value = log(gama1./gama2);
+scf(2);
+plot(x1,Value);
+xgrid();
+xlabel("x1");
+ylabel("ln(y1/y2)");
+// Since the whole area is above X - axis:
+printf("The data is not consistent thermodynamically\n");
+\ No newline at end of file
diff --git a/647/CH10/EX10.2/Example10_2.sce b/647/CH10/EX10.2/Example10_2.sce
new file mode 100755
index 000000000..5d26bb565
--- /dev/null
+++ b/647/CH10/EX10.2/Example10_2.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 10.2
+// Page: 399
+
+printf("Example: 10.2 - Page: 399\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.2 on page number 399 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.2 on page 399 of the book.");
+\ No newline at end of file
diff --git a/647/CH10/EX10.3/Example10_3.sce b/647/CH10/EX10.3/Example10_3.sce
new file mode 100755
index 000000000..558d52926
--- /dev/null
+++ b/647/CH10/EX10.3/Example10_3.sce
@@ -0,0 +1,25 @@
+clear;
+clc;
+
+// Example: 10.3
+// Page: 400
+
+printf("Example: 10.3 - Page: 400\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.6;// [mole fraction of ethylene]
+x2 = 0.4;// [mole fraction of propylene]
+T = 423;// [K]
+P1_sat = 15.2;// [vapour pressure of ethylene, atm]
+P2_sat = 9.8;// [vapour pressure of propylene, atm]
+//**************//
+
+P = x1*P1_sat + x2*P2_sat;// [atm]
+printf("The total pressure is %.2f atm\n",P);
+// In vapour phase:
+y1 = x1*P1_sat/P;// [mole fraction of ethylene]
+y2 = x2*P2_sat/P;// [mole fraction of propylene]
+printf("Mole fraction of ethylene in vapour phase is %.1f\n",y1);
+printf("Mole fraction of propylene in the vapour phase is %.1f\n",y2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.4/Example10_4.sce b/647/CH10/EX10.4/Example10_4.sce
new file mode 100755
index 000000000..82a3f8a92
--- /dev/null
+++ b/647/CH10/EX10.4/Example10_4.sce
@@ -0,0 +1,31 @@
+clear;
+clc;
+
+// Example: 10.4
+// Page: 400
+
+printf("Example: 10.4 - Page: 400\n\n");
+
+// Solution
+
+//*****Data******//
+Temp = 77;// [OC]
+P = 75;// [kPa]
+deff('[P1] = f1(T)','P1 = exp(14.35 - 2942/(T + 220))');
+deff('[P2] = f2(T)','P2 = exp(14.25 - 2960/(T + 210))');
+//*************//
+
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+deff('[y] = f3(x1)','y = P - (x1*P1sat) - (1 - x1)*P2sat');
+x1 = fsolve(7,f3);
+x2 = 1 - x1;
+printf("In Liquid phase\n");
+printf("The mole fraction of X is %.3f\n",x1);
+printf("The mole fraction of Y is %.3f\n",x2);
+
+y1 = x1*P1sat/P;
+y2 = 1 - y1;
+printf("In Vapour phase\n");
+printf("The mole fraction of X is %.3f\n",y1);
+printf("The mole fraction of Y is %.3f\n",y2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.5/Example10_5.sce b/647/CH10/EX10.5/Example10_5.sce
new file mode 100755
index 000000000..2cfcf33f2
--- /dev/null
+++ b/647/CH10/EX10.5/Example10_5.sce
@@ -0,0 +1,58 @@
+clear;
+clc;
+
+// Example: 10.5
+// Page: 401
+
+printf("Example: 10.5 - Page: 401\n\n");
+
+// Solution
+
+//*****Data******//
+deff('[P1] = f1(T)','P1 = exp(14.3916 - 2795/(T + 230))');
+deff('[P2] = f2(T)','P2 = exp(14.2724 - 2945.47/(T + 224))');
+deff('[P3] = f3(T)','P3 = exp(14.2043 - 2972.64/(T + 209))');
+//*************//
+
+// Solution (i)
+
+//*****Data******//
+Temp = 75;// [OC]
+P = 75;// [kPa]
+x1 = 0.30;
+x2 = 0.40;
+//*************//
+
+x3 = 1 - (x1 + x2);
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+P3sat = f3(Temp);// [kPa]
+P = x1*P1sat + x2*P2sat + x3*P3sat;// [kPa]
+y1 = x1*P1sat/P;
+y2 = x2*P2sat/P;
+y3 = x3*P3sat/P;
+printf("Solution (i)\n");
+printf("The mole fraction of acetone is %.3f\n",y1);
+printf("The mole fraction of acetonitrile is %.3f\n",y2);
+printf("The mole fraction of nitromethane is %.3f\n",y3);
+
+// Solution (ii)
+
+//*****Data*****//
+Temp = 80;// [OC]
+y1 = 0.45;
+y2 = 0.35;
+//**************//
+
+y3 = 1 - (y1 + y2);
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+P3sat = f3(Temp);// [kPa]
+P = 1/((y1/P1sat) + (y2/P2sat) + (y3/P3sat));// [kPa]
+x1 = y1*P/P1sat;
+x2 = y2*P/P2sat;
+x3 = y3*P/P3sat;
+printf("Solution (ii)\n");
+printf("The mole fraction of acetone is %.3f\n",x1);
+printf("The mole fraction of acetonitrile is %.3f\n",x2);
+printf("The mole fraction of nitromethane is %.3f\n",x3);
+\ No newline at end of file
diff --git a/647/CH10/EX10.6/Example10_6.sce b/647/CH10/EX10.6/Example10_6.sce
new file mode 100755
index 000000000..4e401790a
--- /dev/null
+++ b/647/CH10/EX10.6/Example10_6.sce
@@ -0,0 +1,30 @@
+clear;
+clc;
+
+// Example: 10.6
+// Page: 403
+
+printf("Example: 10.6 - Page: 403\n\n");
+
+// Solution
+
+//*****Data******//
+deff('[P1] = f1(T)','P1 = exp(16.5915 - 3643.31/(T - 33.424))');
+deff('[P2] = f2(T)','P2 = exp(14.2532 - 2665.54/(T - 53.424))');
+deff('[A] = f3(T)','A = 2.771 - 0.00523*T');
+Temp = 318.15;// [K]
+x1 = 0.25;
+//**************//
+
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+A = f3(Temp);
+x2 = 1 - x1;
+gama1 = exp(A*x2^2);
+gama2 = exp(A*x1^2);
+P = x1*gama1*P1sat + x2*gama2*P2sat;
+y1 = x1*gama1*P1sat/P;
+y2 = x2*gama2*P2sat/P;
+printf("In Vapour phase\n");
+printf("The mole fraction of methanol is %.3f\n",y1);
+printf("The mole fraction of methyl acetate is %.3f\n",y2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.7/Example10_7.sce b/647/CH10/EX10.7/Example10_7.sce
new file mode 100755
index 000000000..d61fb697f
--- /dev/null
+++ b/647/CH10/EX10.7/Example10_7.sce
@@ -0,0 +1,34 @@
+clear;
+clc;
+
+// Example: 10.7
+// Page: 408
+
+printf("Example: 10.7 - Page: 408\n\n");
+
+// Solution
+
+//*****Data******//
+Temp = 30;// [OC]
+A = 0.625;
+//**************//
+
+P1sat = exp(13.71 - 3800/Temp);// [kPa]
+P2sat = exp(14.01 - 3800/Temp);// [kPa]
+// At azeotropic point:
+// P = gama1*P1sat + gama2*P2sat
+// gama1/gama2 = P2sat/P1sat
+// log(gama1) - log(gama2) = log(P2sat) - log(P1sat)
+// Val = log(gama1) - gama2
+Val = log(P2sat) - log(P1sat);
+// log(gama1) = (A*x2^2)
+// log(gama2) = (A*x1^2)
+// A(x2^2 - x1^2) = 0.625*(x2^2 - x1^2)..................... (1)
+// x1 + x2 = 1............................................. (2)
+// On simplifying, we get:
+// A*(1 - (2*x1)) = Val
+x1 = (1/2)*(1 - Val/A);
+x2 = 1 - x1;
+printf("Azeotropic Composition\n");
+printf("The mole fraction of component 1 is %.3f\n",x1);
+printf("The mole fraction of component 2 is %.3f\n",x2);
+\ No newline at end of file
diff --git a/647/CH10/EX10.8/Example10_8.sce b/647/CH10/EX10.8/Example10_8.sce
new file mode 100755
index 000000000..a3b7043b4
--- /dev/null
+++ b/647/CH10/EX10.8/Example10_8.sce
@@ -0,0 +1,13 @@
+clear;
+clc;
+
+// Example: 10.8
+// Page: 410
+
+printf("Example: 10.8 - Page: 410\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 10.8 on page number 410 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 10.8 on page 410 of the book.");
+\ No newline at end of file
diff --git a/647/CH10/EX10.9/Example10_9.sce b/647/CH10/EX10.9/Example10_9.sce
new file mode 100755
index 000000000..610d9271a
--- /dev/null
+++ b/647/CH10/EX10.9/Example10_9.sce
@@ -0,0 +1,25 @@
+clear;
+clc;
+
+// Example: 10.9
+// Page: 412
+
+printf("Example: 10.9 - Page: 412\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.42;
+x2 = 0.58;
+P = 760;// [mm of Hg]
+P1sat = 786;// [mm of Hg]
+P2sat = 551;// [mm of Hg]
+//***************//
+
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+printf("Van Laar Constants\n");
+printf("A = %.3f\n",A);
+printf("B = %.3f\n",B);
+\ No newline at end of file
author	priyanka	2015-06-24 15:03:17 +0530
committer	priyanka	2015-06-24 15:03:17 +0530
commit	b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
tree	ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH10
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