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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /389/CH8/EX8.5/Example8_5.sce | |
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diff --git a/389/CH8/EX8.5/Example8_5.sce b/389/CH8/EX8.5/Example8_5.sce new file mode 100755 index 000000000..9dd6fdff6 --- /dev/null +++ b/389/CH8/EX8.5/Example8_5.sce @@ -0,0 +1,129 @@ +clear;
+clc;
+
+// Illustration 8.5
+// Page: 299
+
+printf('Illustration 8.5 - Page: 299\n\n');
+
+// solution
+
+//****Data****//
+// a = NH3 b = H2 c = N2 w = water
+P = 2;// [bars]
+Temp = 30;// [OC]
+L = 6.38;// [kg/s]
+W = 0.53;// [weir length,m]
+pitch = 12.5/1000;// [m]
+D = 0.75;// [Tower diameter,m]
+hW = 0.060;// [weir height,m]
+t = 0.5;// [tray spacing,m]
+//*******//
+
+// From Geometry of Tray Arrangement:
+At = 0.4418;// [Tower Cross section,square m]
+Ad = 0.0403;// [Downspout Cross section,square m]
+An = At-Ad;// [square m]
+Ao = 0.0393;// [perforation area,square m]
+Z = 0.5307;// [distance between downspouts,square m]
+z = (D+W)/2;// [average flow width,m]
+h1 = 0.04;// [weir crest,m]
+// From Eqn. 6.34
+Weff = W*(sqrt(((D/W)^2)-((((D/W)^2-1)^0.5)+((2*h1/D)*(D/W)))^2));// [m]
+q = Weff*(1.839*h1^(3/2));//[cubic m/s]
+// This is a recommended rate because it produces the liquid depth on the tray to 10 cm.
+Density_L = 996;// [kg/s]
+Mw = 18.02;// [kg/kmol]
+L1 = 6.38/Mw;// [kmol/s]
+Ma = 17.03;// [kg/kmol]
+Mb = 28.02;// [kg/kmol]
+Mc = 2.02;// [kg/kmol]
+MavG = (0.03*Ma)+(0.97*(1/4)*Mb)+(0.97*(3/4)*Mc);// [kg/kmol]
+Density_G = (MavG/22.41)*(P/0.986)*(273/(273+Temp));// [kg/cubic m]
+G = 0.893;// [kg/s]
+sigma = 68*10^(-3);// [N/m]
+abcissa = (L/G)*(Density_G/Density_L)^0.5;
+// From Table 6.2 (Pg169):
+alpha = 0.04893;
+beeta = 0.0302;
+// From Eqn. 6.30
+Cf = ((alpha*log10(1/abcissa))+beeta)*(sigma/0.02)^0.2;
+// From Eqn. 6.29
+Vf = Cf*((Density_L-Density_G)/Density_G)^(1/2);// [m/s]
+// 80% of flooding value:
+V = 0.8*Vf;// [m/s]
+G = 0.8*G;// [kg/s]
+G1 = G/MavG;// [kmol/s]
+Vo = V*An/Ao;// [m/s]
+l = 0.002;// [m]
+Do = 0.00475;// [m]
+// From Eqn. 6.37
+Co = 1.09*(Do/l)^0.25;
+viscosity_G = 1.13*10^(-5);// [kg/m.s]
+Reo = Do*Vo*Density_G/viscosity_G;
+// At Reynold's No. = Reo
+fr = 0.0082;
+g = 9.81;// [m/s^2]
+// From Eqn. 6.36
+deff('[y] = f36(hD)','y = (2*hD*g*Density_L/(Vo^2*Density_G))-(Co*(0.40*(1.25-(Ao/An))+(4*l*fr/Do)+(1-(Ao/An))^2))');
+hD = fsolve(1,f36);
+// From Eqn. 6.31;
+Aa = (Ao/0.907)*(pitch/Do)^2;// [square m]
+Va = V*An/Aa;// [m/s]
+// From Eqn. 6.38
+hL = 6.10*10^(-3)+(0.725*hW)-(0.238*hW*Va*(Density_G)^0.5)+(1.225*q/z);// [m]
+// From Eqn. 6.42
+hR = 6*sigma/(Density_L*Do*g);// m
+// From Eqn. 6.35
+hG = hD+hL+hR;// [m]
+Al = 0.025*W;// [square m]
+Ada = min(Al,Ad);
+// From Eqn. 6.43
+h2 = (3/(2*g))*(q/Ada)^2;// [m]
+// From Eqn.6.44
+h3 = hG+h2;
+// since hW+h1+h3 is essentially equal to t/2, flooding will not occur
+abcissa = (L/G)*(Density_G/Density_L)^0.5;
+V_by_Vf = V/Vf;
+// From Fig.6.17, V/Vf = 0.8 & abcissa = 0.239
+E = 0.009;
+
+// At the prevailing conditions:
+Dg = 2.296*10^(-5);// [square m/s]
+viscosity_G = 1.122*10^(-5);// [kg/m.s]
+ScG = viscosity_G/(Density_G*Dg)
+Dl = 2.421*10^(-9);// [square m/s]
+
+// From Henry's Law:
+m = 0.850;
+A = L1/(m*G1);
+
+// From Eqn. 6.61:
+NtG = (0.776+(4.57*hW)-(0.238*Va*Density_G^0.5)+(104.6*q/Z))/(ScG^0.5);
+// From Eqn. 6.64:
+thetha_L = hL*z*Z/q;// [s]
+// From Eqn. 6.62:
+NtL = 40000*(Dl^0.5)*((0.213*Va*Density_G^0.5)+0.15)*thetha_L;
+// From Eqn. 6.52:
+NtoG = 1/((1/NtG)+(1/(A*NtL)));
+// From Eqn. 6.51:
+EoG = 1-exp(-NtoG);
+// From Eqn. 6.63:
+DE = ((3.93*10^(-3))+(0.0171*Va)+(3.67*q/Z)+(0.1800*hW))^2;// [square m/s]
+// From Eqn. 6.59:
+Pe = Z^2/(DE*thetha_L);
+// From Eqn. 6.58:
+eta = (Pe/2)*((1+(4*m*G1*EoG/(L1*Pe)))^0.5-1);
+// From Eqn. 6.57:
+EMG = EoG*(((1-exp(-(eta+Pe)))/((eta+Pe)*(1+(eta+Pe)/eta)))+((exp(eta)-1)/(eta*(1+(eta/(eta+Pe))))));
+// From Eqn. 6.60:
+EMGE = EMG/((1+(EMG*(E/(1-E)))));
+// From Eqn. 8.16:
+EO = log(1+EMGE*((1/A)-1))/log(1/A);
+Np = 14*EO;
+yNpPlus1 = 0.03;
+x0 = 0;
+// From Eqn. 5.54(a):
+deff('[y] = f37(y1)','y = ((yNpPlus1-y1)/(yNpPlus1-m*x0))-(((A^(Np+1))-A)/((A^(Np+1))-1))');
+y1 = fsolve(0.03,f37);
+printf("Mole Fraction Of NH3 in effluent is %e",y1);
\ No newline at end of file |