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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 /905/CH4 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '905/CH4')
| -rwxr-xr-x | 905/CH4/EX4.2/4_2.sce | 45 | ||||
| -rwxr-xr-x | 905/CH4/EX4.3/4_3.sce | 166 | ||||
| -rwxr-xr-x | 905/CH4/EX4.4/4_4.sce | 157 | ||||
| -rwxr-xr-x | 905/CH4/EX4.5/4_5.sce | 120 | ||||
| -rwxr-xr-x | 905/CH4/EX4.6/4_6.sce | 64 | ||||
| -rwxr-xr-x | 905/CH4/EX4.7/4_7.sce | 54 | ||||
| -rwxr-xr-x | 905/CH4/EX4.8/4_8.sce | 40 | ||||
| -rwxr-xr-x | 905/CH4/EX4.9/4_9.sce | 62 |
8 files changed, 708 insertions, 0 deletions
diff --git a/905/CH4/EX4.2/4_2.sce b/905/CH4/EX4.2/4_2.sce new file mode 100755 index 000000000..774da1f87 --- /dev/null +++ b/905/CH4/EX4.2/4_2.sce @@ -0,0 +1,45 @@ +clear;
+clc;
+
+// Illustration 4.2
+// Page: 227
+
+printf('Illustration 4.2 - Page: 227\n\n');
+
+// solution
+//*****Data*****//
+u = 3*10^-6; // [Kinematic viscosity, square m/s]
+v = 0.01; // [Superficial liquid velocity, m/s]
+g = 9.8; // [square m/s]
+//*****//
+// From table 4.1
+// For metal pall rings
+a_pr = 112.6; // [ square m/cubic m]
+e_pr = 0.951;
+Ch_pr = 0.784;
+// For Hiflow rings
+a_hr = 92.3; // [square m/cubic m]
+e_hr = 0.977;
+Ch_hr = 0.876;
+
+// Renoylds and Froude's number for metal pall rings
+Rel_pr = v/(u*a_pr);
+Frl_pr = v^2*a_pr/g;
+// From equation 4.5 since Rel is greater than 5, for pall rings
+// ah/a = x_pr
+x_pr = 0.85*Ch_pr*Rel_pr^0.25*Frl_pr^0.1;
+// From equation 4.3
+hl_pr = (12*Frl_pr/Rel_pr)^(1/3)*(x_pr)^(2/3);
+
+
+// Renoylds and Froude's number for Hiflow rings
+Rel_hr = v/(u*a_hr);
+Frl_hr = v^2*a_hr/g;
+// From equation 4.5 since Rel is greater than 5, for pall rings
+// ah/a = x_pr
+x_hr = 0.85*Ch_hr*Rel_hr^0.25*Frl_hr^0.1;
+// From equation 4.3
+hl_hr = (12*Frl_hr/Rel_hr)^(1/3)*(x_hr)^(2/3);
+
+printf("The specific liquid holdup for Metal pall ring and Hiflow ring are %f cubic m holdup/cubic m packed bed and %f cubic m holdup/cubic m packed bed respectively\n\n",hl_pr,hl_hr);
+
\ No newline at end of file diff --git a/905/CH4/EX4.3/4_3.sce b/905/CH4/EX4.3/4_3.sce new file mode 100755 index 000000000..1c8ceb80c --- /dev/null +++ b/905/CH4/EX4.3/4_3.sce @@ -0,0 +1,166 @@ +clear;
+clc;
+
+// Illustration 4.3
+// Page: 233
+
+printf('Illustration 4.3 - Page: 233\n\n');
+
+// solution
+//*****Data*****//
+// a-ammonia b-air c-water
+P = 101.3; // [kPa]
+T = 293; // [K]
+R = 8.314;
+Vb = 20; // [kmole/h]
+xab = 0.05;
+Vc = 1500; // [kg/h]
+d = 0.9; // [ammonia absorbed]
+Ma = 17; // [gram/mole]
+Mb = 29; // [gram/mole]
+Mc = 18; // [gram/mole]
+g = 9.8; // [square m/s]
+//*****//
+
+// For Inlet gas
+Mg = (1-xab)*Mb+xab*Ma; // [gram/mole]
+V = Vb*Mg/3600; // [kg/h]
+rowg = P*Mg/(R*T); // [kg/cubic m]
+Qg = V/rowg; // [cubic m/s]
+
+// For exiting liquid
+b = Vb*xab*Ma*d; // [ammonia absorbed in kg/h]
+L = (Vc+b)/3600; // [kg/s]
+rowl = 1000; // [kg/cubic m]
+
+X = (L/V)*(sqrt(rowg/rowl));
+// From equation 4.8
+Yflood = exp(-(3.5021+1.028*log(X)+0.11093*(log(X))^2));
+
+
+printf('Illustration 4.3(a) - Page: 233\n\n');
+// Solution(a)
+// For 25-mm ceramic Raschig rings
+Fp = 179; // [square ft/cubic ft]
+ul = 0.001; // [Pa.s]
+// From equation 4.6
+Csflood = sqrt(Yflood/(ul^0.1*Fp)); // [m/s]
+// From equation 4.7
+vgf = Csflood/(sqrt(rowg/(rowl-rowg))); // [m/s]
+// From equation 4.9
+deltaPf = 93.9*(Fp)^0.7; // [Pa/m of packing]
+
+// For operation at 70% of the flooding velocity
+f = 0.7;
+// From equation 4.10
+vg = f*vgf; // [m/s]
+D = sqrt(4*Qg/(vg*%pi));
+
+// From Table 4.1, for 25 mm ceramic Raschig rings
+a_c = 190; // [square m/cubic m]
+Ch_c = 0.577;
+e_c = 0.68;
+Cp_c = 1.329;
+
+// From equation 4.13
+dp = 6*(1-e_c)/a_c; // [m]
+// From equation 4.12
+Kw = 1/(1+(2*dp/(3*D*(1-e_c))));
+
+// The viscosity of the gas phase is basically that of air at 293 K and 1 atm
+ug = 1.84*10^-5; // [kg/m.s]
+// From equation 4.15
+Reg = vg*rowg*dp*Kw/(ug*(1-e_c));
+// From equation 4.14
+sia_o = Cp_c*((64/Reg)+(1.8/(Reg^0.08)));
+
+// From equation 4.11
+// deltaP_o/z = T
+T = sia_o*a_c*rowg*vg^2/(2*Kw*e_c^3); // [Pa/m]
+
+// Now
+Gx = L/(%pi*D^2/4); // [kg/square m.s]
+Rel = Gx/(a_c*ul);
+Frl = Gx^2*a_c/(rowl^2*g);
+
+// From equation 4.5
+// ah/a = x_pr
+x = 0.85*Ch_c*Rel^0.25*Frl^0.1;
+// From equation 4.3
+hl = (12*Frl/Rel)^(1/3)*(x)^(2/3);
+
+// From equation 4.16
+// daltaP/deltaP_o = Y
+Y = (e_c/(e_c-hl))^1.5*exp(Rel/200);
+// Therefore
+// deltaP/z = H
+H = Y*T; // [Pa/m]
+
+printf("The superficial velocity is %f m/s\n",vgf);
+printf("The pressure drop at flooding is %f Pa/m\n",deltaPf);
+printf("The superficial velocity at 70 percent of flooding is %f m/s\n",vg);
+printf("The column inside diameter at 70 percent of flooding is %f m\n",D);
+printf("The pressure drop for operation at 70 percent of flooding is %f Pa/m\n\n",H);
+
+
+printf('Illustration 4.3(b) - Page: 236\n\n');
+// Solution (b)
+// Similarly for 25 mm metal Hiflow rings above quantities are determined
+Fp1 = 42; // [square ft/cubic ft]
+Csflood1 = sqrt(Yflood/(ul^0.1*Fp1)); // [m/s]
+vgf1 = Csflood1/(sqrt(rowg/(rowl-rowg))); // [m/s]
+// From equation 4.9
+deltaPf1 = 93.9*(Fp1)^0.7; // [Pa/m of packing]
+
+// For operation at 70% of the flooding velocity
+f = 0.7;
+// From equation 4.10
+vg1 = f*vgf1; // [m/s]
+D1 = sqrt(4*Qg/(vg1*%pi));
+
+// For Hiflow rings
+a_h = 202.9; // [square m/cubic m]
+e_h = 0.961;
+Ch_h = 0.799;
+Cp_h = 0.689;
+
+// From equation 4.13
+dp1 = 6*(1-e_h)/a_h; // [m]
+// From equation 4.12
+Kw1 = 1/(1+(2*dp1/(3*D1*(1-e_h))));
+
+// The viscosity of the gas phase is basically that of air at 293 K and 1 atm
+ug = 1.84*10^-5; // [kg/m.s]
+// From equation 4.15
+Reg1 = vg1*rowg*dp1*Kw1/(ug*(1-e_h));
+// From equation 4.14
+sia_o1 = Cp_h*((64/Reg1)+(1.8/(Reg1^0.08)));
+
+// From equation 4.11
+// deltaP_o/z = T
+T1 = sia_o1*a_h*rowg*vg1^2/(2*Kw1*e_h^3); // [Pa/m]
+
+// Now
+Gx1 = L/(%pi*D1^2/4); // [kg/square m.s]
+Rel1 = Gx1/(a_h*ul);
+Frl1 = Gx1^2*a_h/(rowl^2*g);
+
+// From equation 4.5
+// ah/a = x_pr
+x1 = 0.85*Ch_h*Rel1^0.25*Frl1^0.1;
+// From equation 4.3
+hl1 = (12*Frl1/Rel1)^(1/3)*(x1)^(2/3);
+
+// From equation 4.16
+// daltaP/deltaP_o = Y
+Y1 = (e_h/(e_h-hl1))^1.5*exp(Rel1/200);
+// Therefore
+// deltaP/z = H
+H1 = Y1*T1; // [Pa/m]
+
+
+printf("The superficial velocity is %f m/s\n",vgf1);
+printf("The pressure drop at flooding is %f Pa/m\n",deltaPf1);
+printf("The superficial velocity at 70 percent of flooding is %f m/s\n",vg1);
+printf("The column inside diameter at 70 percent of flooding is %f m\n",D1);
+printf("The pressure drop for operation at 70 percent of flooding is %f Pa/m\n\n",H1);
\ No newline at end of file diff --git a/905/CH4/EX4.4/4_4.sce b/905/CH4/EX4.4/4_4.sce new file mode 100755 index 000000000..42167a9cc --- /dev/null +++ b/905/CH4/EX4.4/4_4.sce @@ -0,0 +1,157 @@ +clear;
+clc;
+
+// Illustration 4.4
+// Page: 237
+
+printf('Illustration 4.4 - Page: 237\n\n');
+
+// solution
+//*****Data*****//
+// a-ethanol b- gas(CO2 rich vapor) c-liquid water
+P = 110; // [kPa]
+T = 303; // [K]
+R = 8.314;
+Vb = 180; // [kmole/h]
+xab = 0.02; // [molar composition of ethanol in gas]
+Vc = 151.5; // [kmole/h]
+d = 0.97; // [ethanol absorbed]
+Ma = 46; // [gram/mole]
+Mb = 44; // [gram/mole]
+Mc = 18; // [gram/mole]
+g = 9.8; // [square m/s]
+//*****//
+
+// For Inlet gas
+Mg = (1-xab)*Mb+xab*Ma; // [gram/mole]
+V = Vb*Mg/3600; // [kg/h]
+rowg = P*Mg/(R*T); // [kg/cubic m]
+Qg = V/rowg; // [cubic m/s]
+
+// For exiting liquid
+b = Vb*xab*Ma*d; // [ethanol absorbed in kg/h]
+L = (Vc*Mc+b)/3600; // [kg/s]
+rowl = 986; // [kg/cubic m]
+
+X = (L/V)*(sqrt(rowg/rowl));
+// From equation 4.8
+Yflood = exp(-(3.5021+1.028*log(X)+0.11093*(log(X))^2));
+
+printf('Illustration 4.4(a) - Page: 237\n\n');
+// Solution(a)
+
+// For 50 mm metal Hiflow rings
+Fp = 16; // [square ft/cubic ft]
+ul = 6.31*10^-4; // [Pa.s]
+// From equation 4.6
+Csflood = sqrt(Yflood/(ul^0.1*Fp)); // [m/s]
+// From equation 4.7
+vgf = Csflood/(sqrt(rowg/(rowl-rowg))); // [m/s]
+// From equation 4.9
+deltaPf = 93.9*(Fp)^0.7; // [Pa/m of packing]
+
+// For operation at 70% of the flooding velocity
+f = 0.7;
+// From equation 4.10
+vg = f*vgf; // [m/s]
+D = sqrt(4*Qg/(vg*%pi));
+
+// From Table 4.1, for 50 mm metal Hiflow rings
+a = 92.3; // [square m/cubic m]
+Ch = 0.876;
+e = 0.977;
+Cp = 0.421;
+
+// From equation 4.13
+dp = 6*(1-e)/a; // [m]
+
+// From equation 4.12
+Kw = 1/(1+(2*dp/(3*D*(1-e))));
+
+// The viscosity of the gas phase is basically that of air at 303 K and 110 kPa
+ug = 1.45*10^-5; // [kg/m.s]
+// From equation 4.15
+Reg = vg*rowg*dp*Kw/(ug*(1-e));
+// From equation 4.14
+sia_o = Cp*((64/Reg)+(1.8/(Reg^0.08)));
+
+// From equation 4.11
+// deltaP_o/z = I
+I = sia_o*a*rowg*vg^2/(2*Kw*e^3); // [Pa/m]
+
+// Now
+Gx = L/(%pi*D^2/4); // [kg/square m.s]
+Rel = Gx/(a*ul);
+Frl = Gx^2*a/(rowl^2*g);
+
+// From equation 4.5
+// ah/a = x
+x = 0.85*Ch*Rel^0.25*Frl^0.1;
+// From equation 4.3
+hl = (12*Frl/Rel)^(1/3)*(x)^(2/3);
+
+// From equation 4.16
+// daltaP/deltaP_o = Y
+Y = (e/(e-hl))^1.5*exp(Rel/200);
+// Therefore
+// deltaP/z = H
+H = Y*I; // [Pa/m]
+
+printf('Since the pressure drop is too high, we must increase the tower diameter to reduce the pressure drop.\n');
+// The resulting pressure drop is too high; therefore, we must increase the tower diameter to reduce the pressure drop. Appendix D presents a Mathcad computer
+// program designed to iterate automatically until the pressure drop criterion is satisfied.
+// From the Mathcad program we get
+D1 = 0.738; // [m]
+printf("The tower diameter for pressure drop of 300 Pa/m of packed height is %f m\n\n",D1);
+
+printf('Illustration 4.4(b) - Page: 241\n\n');
+// Solution(b)
+
+// For the tower diameter of D = 0.738 m, the following intermediate results were obtained from the computer program in Appendix D:
+vg1 = 2.68; // [m/s]
+vl1 = 0.00193; // [m/s]
+hl1 = 0.017;
+ah1 = 58.8; // [square m/cubic m]
+Reg1 = 21890;
+Rel1 = 32.6;
+Kw1 = 1/(1+(2*dp/(3*D1*(1-e))));
+
+
+f1 = vg1/vgf;
+printf("The fractional approach to flooding conditions is %f\n\n",f1);
+
+printf('Illustration 4.4(c) - Page: 242\n\n');
+// Solution(c)
+// For ethanol
+Vc_a = 167.1; // [cubic cm/mole]
+sigma_a = 4.53*10^-10; // [m]
+// E/k = M
+M_a = 362.6; // [K]
+
+// For carbon dioxide
+sigma_b = 3.94*10^-10; // [m]
+M_b = 195.2; // [K]
+
+// From equation 1.48
+Vb_a = 0.285*Vc_a^1.048; // [cubic cm/mole]
+
+e1 = (9.58/(Vb_a)-1.12);
+// From equation 1.53
+Dl = 1.25*10^-8*((Vb_a)^-0.19 - 0.292)*T^1.52*(ul*10^3)^e1; // [square cm/s]
+
+// From equation 1.49
+Dg = 0.085; // [square cm/s]
+
+// From Table 4.2, for 50 mm metal Hiflow rings
+Cl = 1.168
+Cv = 0.408;
+// From equation 4.17
+kl = 0.757*Cl*sqrt(Dl*a*vl1*10^-4/(e*hl1)); // [m/s]
+mtcl = kl*ah1; // [s^-1]
+
+Sc = ug/(rowg*Dg*10^-4);
+// From equation 4.18
+ky = 0.1304*Cv*(Dg*10^-4*P*1000/(R*T))*(Reg1/Kw1)^(3/4)*Sc^(2/3)*(a/(sqrt(e*(e-hl1)))); // [mole/square m.s]
+mtcg = ky*ah1*10^-3; // [kmole/cubic m.s]
+printf("The gas and liquid volumetric mass transfer coefficients are %e kmole/cubic m.s and %e s^-1 respectively.\n\n",mtcg,mtcl);
+
diff --git a/905/CH4/EX4.5/4_5.sce b/905/CH4/EX4.5/4_5.sce new file mode 100755 index 000000000..3e350f401 --- /dev/null +++ b/905/CH4/EX4.5/4_5.sce @@ -0,0 +1,120 @@ +clear;
+clc;
+
+// Illustration 4.5
+// Page: 245
+
+printf('Illustration 4.5 - Page: 245\n\n');
+
+// solution
+//*****Data*****//
+// a-chloroform b-water c-air
+T = 298; // [K]
+Dv = 1; // [vessel diameter, m]
+Vb = 10; // [kg/s]
+ca = 240*10^-6; // [gram/l]
+xr = 0.9; // [chloroform which is to be removed]
+m = 220;
+Ds = 0.5; // [diameter of sparger, m]
+no = 90; // [number of orifices]
+Do = 3*10^-3; // [diameter of orifice, m]
+nm = 0.6; // [mechanical efficiency]
+rowb = 1000; // [kg/cubic m]
+R = 8.314;
+Mc = 29; // [gram/mole]
+Mb = 18; // [gram/mole]
+g = 9.8; // [square m/s]
+//*****//
+
+Vair = 0.1; // [kg/s as calculated in chapter 3]
+mg = Vair/no; // [mass flow rate through each orifice, kg/s]
+ug = 1.8*10^-5; // [kg/m.s]
+Reo = 25940; // [Renoylds number]
+// From equ. 4.20
+dp = 0.0071*Reo^-0.05; // [m]
+
+// Since the water column height is not known, therefore an iterative procedure must be implemented.
+// Assuming column height, Z = 0.5 m
+Z = 0.5; // [m]
+// For Z = 0.5 m
+rowl = 1000; // [kg/cubic m]
+Ps = 101.3; // [kPa]
+Po = Ps + (1000*9.8*0.5/1000); // [kPa]
+Pavg = (Po+Ps)/2; // [kPa]
+rowg = Pavg*Mc/(R*T); // [kg/cubic m]
+
+area = %pi*Dv^2/4; // [square m]
+vg = Vair/(rowg*area); // [m/s]
+// In this case rowl = rowg and sigma = sigmaAW
+// From equation 4.22
+// Vg = vg
+// vg/vs = 0.182
+vs = vg/0.182; // [m/s]
+vl = -Vb/(rowl*area); // [negative because water flows downward, m/s]
+// From equ 4.21
+
+deff('[y] = f12(phig)','y = vs - (vg/phig)-(-vl/(1-phig))');
+phig = fsolve(0.1,f12);
+// Now in this case
+S = vl/(1-phig);
+// Value of 'S' comes out to be less than 0.15 m/s
+// Therefore
+dp = (dp^3*Po/Pavg)^(1/3); // [m]
+// From equ 4.23
+a = 6*phig/dp; // [m^-1]
+// Now we calculate diffusivity of chloroform
+Vba = 88.6; // [cubic cm/mole]
+u = 0.9*10^-3; // [Pa-s]
+e = (9.58/(Vba)-1.12);
+// From equation 1.53
+Dl = 1.25*10^-8*((Vba)^-0.19 - 0.292)*T^1.52*(u*10^3)^e; // [square cm/s]
+
+// And Schmidt number is
+Scl = 833; // [Schmidt Number]
+
+// Now we calculate dp*g^(1/3)/Dl^(2/3) = J
+J = dp*g^(1/3)/(Dl*10^-4)^(2/3)
+Reg = dp*vs*rowl/u; // [Gas bubble Renoylds number]
+// From equ 4.25
+Shl = 2 + 0.0187*Reg^0.779*Scl^0.546*J^0.116;
+
+// For dilute solution xbm = 1 or c = 55.5 kmole/cubic m
+// Then for Nb = 0
+c = 55.5; // [kmole/cubic m]
+kx = Shl*c*Dl*10^-4/dp; // [kmole/square m.s]
+mtc = kx*a; // [kmole/cubic m.s]
+
+L = Vb/Mb; // [kmole/s]
+Gmx = L/area; // [kmole/square m.s]
+V = Vair/Mc; // [kmole/s]
+A = L/(m*V); // [absorption factor]
+
+// From equ 4.28
+ // For, xin/xout = x = 10
+ x = 10;
+Z = (Gmx/(kx*a*(1-A)))*log(x*(1-A)+A);
+
+// With this new estimated Z ,we again calculate average pressure in the // column of water
+Po1 = 110.1; // [kPa]
+Pavg1 = 105.7; // [kPa]
+rowg1 = Pavg1*Mc/(R*T);
+// Now value of rowg1 obtained is very close to value used in the first // iteration. Therefore on three iteractions we achieve a value of 'Z'
+Z1 = 0.904; // [m]
+
+rowgo = Po1*Mc/(R*T); // [kg/cubic m]
+vo1 = 4*mg/(%pi*Do^2*rowgo); // [m/s]
+// Therefore, vo1^2/(2*gc) = F
+gc = 1;
+F = vo1^2/(2*gc); // [J/kg]
+// And R*T*log(Po/Ps)/Mc = G
+G = R*T*1000*log(Po1/Ps)/Mc; // [J/kg]
+Zs = 0
+// And (Z1-Zs)*g/gc = H
+H = (Z1-Zs)*g/gc; // [J/kg]
+// From equ 4.27
+W = F+G+H; // [J/kg]
+// Now the air compressor power is
+W1 = W*Vair*10^-3/nm; // [kW]
+
+printf("The depth of the water column required to achieve the specified 90percent removal efficiency is %f m\n\n",Z1);
+printf("The power required to operate the air compressor is %f kW\n\n",W1);
\ No newline at end of file diff --git a/905/CH4/EX4.6/4_6.sce b/905/CH4/EX4.6/4_6.sce new file mode 100755 index 000000000..8d5a2eb31 --- /dev/null +++ b/905/CH4/EX4.6/4_6.sce @@ -0,0 +1,64 @@ +clear;
+clc;
+
+// Illustration 4.6
+// Page: 255
+
+printf('Illustration 4.6 - Page: 255\n\n');
+
+// solution
+//*****Data*****//
+Ff = 0.9; // [foaming factor]
+sigma = 70; // [liquid surface tension, dyn/cm]
+Do = 5; // [mm]
+//From Example 4.4
+// X = 0.016;
+p = 15 // [pitch, mm]
+// From equ 4.35
+// Ah/Aa = A
+A = 0.907*(Do/p)^2; // [ratio of vapor hole area to tray active area]
+
+// Assume
+t = 0.5; // [m]
+// From equ 4.32
+alpha = 0.0744*t+0.01173;
+beeta = 0.0304*t+0.015;
+
+// Since X<0.1, therefore
+X = 0.1;
+// From equ 4.31
+Cf = alpha*log10(1/X) + beeta;
+// Since Ah/Aa > 0.1, therefore
+Fha = 1;
+Fst = (sigma/20)^0.2; // [surface tension factor]
+// From equ 4.30
+C = Fst*Ff*Fha*Cf;
+
+// From Example 4.4
+rowg = 1.923; // [kg/cubic m]
+rowl = 986; // [kg/cubic m]
+Qg = 1.145; // [cubic m/s]
+// From equation 4.29
+vgf = C*(sqrt((rowl-rowg)/rowg)); // [m/s]
+// Since X<0.1
+// Equ 4.34 recommends Ad/At = B = 0.1
+B = 0.1;
+// For an 80% approach to flooding, equation 4.33 yields
+f = 0.8;
+D = sqrt((4*Qg)/(f*vgf*%pi*(1-B))); // [m]
+// At this point, the assumed value of tray spacing ( t = 0.5 m) must be // checked against the recommended values of Table 4.3. Since the calculated
+// value of D < 1.0 m, t = 0.5 m is the recommended tray spacing, and no
+// further iteration is needed.
+
+deff('[y] = f14(Q)','y = B-((Q-sin(Q))/(2*%pi))');
+Q = fsolve(1.5,f14);
+Lw = D*sin(Q/2); // [m]
+rw = D/2*cos(Q/2); // [m]
+
+At = %pi/4*D^2; // [total cross sectional area, square m]
+Ad = B*At; // [Downcomer area, square m]
+Aa = At-2*Ad; // [ Active area over the tray, square m]
+Ah = 0.101*Aa; // [Total hole area, square m]
+
+printf('Summarizing, the details of the sieve-tray design are as follows:\n\n');
+printf(" Diameter = %f m\n Tray spacing = %f m\n Total cross-sectional area = %f square m\n Downcomer area = %f square m\n Active area over the tray = %f square m\n Weir length = %f m\n Distance from tray center to weir = %f m\n Total hole area = %f square m\n Hole arrangement: 5 mm diameter on an equilateral-triangular pitch 15 mm between hole centers, punched in stainless steel sheet metal 2 mm thick\n\n",D,t,At,Ad,Aa,Lw,rw,Ah);
\ No newline at end of file diff --git a/905/CH4/EX4.7/4_7.sce b/905/CH4/EX4.7/4_7.sce new file mode 100755 index 000000000..724bc9e4a --- /dev/null +++ b/905/CH4/EX4.7/4_7.sce @@ -0,0 +1,54 @@ +clear;
+clc;
+
+// Illustration 4.7
+// Page: 257
+
+printf('Illustration 4.7 - Page: 257\n\n');
+
+// solution//
+Do = 5; // [mm]
+g = 9.8; // [square m/s]
+hw = 50; // [mm]
+// From example 4.4
+Qg = 1.145; // [cubic m/s]
+// From example 4.6
+Ah = 0.062; // [square m]
+// Do/l = t = 5/2 = 2.5
+t = 2.5;
+// Ah/Aa = A = 0.101
+A = 0.101;
+rowg = 1.923; // [kg/cubic m]
+rowl = 986; // [kg/cubic m]
+roww = 995; // [kg/cubic m]
+
+vo = Qg/Ah; // [m/s]
+// From equation 4.39
+Co = 0.85032 - 0.04231*t + 0.0017954*t^2; // [for t>=1]
+// From equation 4.38
+hd = 0.0051*(vo/Co)^2*rowg*(roww/rowl)*(1-A^2); // [cm]
+
+// From example 4.6
+Aa = 0.615; // [square m]
+va = Qg/Aa; // [m/s]
+
+// From equation 4.41
+Ks = va*sqrt(rowg/(rowl-rowg)); // [m/s]
+phie = 0.274;
+
+// From equation 4.4
+ql = 0.000815; // [cubic m/s]
+
+// From example 4.6
+Lw = 0.719; // [m]
+Cl = 50.12 + 43.89*exp(-1.378*hw);
+sigma = 0.07; // [N/m]
+// From eqution 4.40
+hl = phie*(hw*10^-1+Cl*(ql/(Lw*phie))^(2/3));
+
+// From equation 4.42
+ho = 6*sigma/(g*rowl*Do*10^-3)*10^2; // [cm]
+// From equation 4.37
+ht = hd+hl+ho; // [cm of clear liquid/tray]
+deltaPg = ht*g*rowl*10^-2; // [Pa/tray]
+printf("The tray gas-pressure drop for the ethanol is %f Pa/tray\n\n",deltaPg);
\ No newline at end of file diff --git a/905/CH4/EX4.8/4_8.sce b/905/CH4/EX4.8/4_8.sce new file mode 100755 index 000000000..ef91a48d5 --- /dev/null +++ b/905/CH4/EX4.8/4_8.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// Illustration 4.8
+// Page: 259
+
+printf('Illustration 4.8 - Page: 259\n\n');
+
+// solution//
+// From Example 4.4, 4.6 and 4.7
+
+Do = 5*10^-3; // [m]
+rowg = 1.923; // [kg/cubic m]
+rowl = 986; // [kg/cubic m]
+g = 9.8; // [square m/s]
+hl = 0.0173; // [m]
+vo = 18.48; // [m/s]
+phie = 0.274;
+Ks = 0.082; // [m]
+A = 0.101; // [Ah/Aa]
+t = 0.5; // [m]
+
+Fr = sqrt(rowg*vo^2/(rowl*g*hl)); // [Froude number]
+if(Fr>=0.5)
+ printf('Weeping is not significant\n\n');
+else()
+ printf('Significant weeping occurs\n\n');
+ end
+// From above weeping is not a problem under this circumstances
+// From equation 4.47
+k = 0.5*(1-tanh(1.3*log(hl/Do)-0.15));
+
+// From equation 4.46
+h2q = (hl/phie) + 7.79*(1+6.9*(Do/hl)^1.85)*(Ks^2/(phie*g*A)); // [m]
+// From equation 4.45
+E = 0.00335*(h2q/t)^1.1*(rowl/rowg)^0.5*(hl/h2q)^k;
+// From Example 4.4, the gas mass flow rate is V = 2.202 kg/s
+V = 2.202; // [kg/s]
+Le = E*V; // [kg/s]
+printf("The entrainment flow rate for the ethanol absorber is %f m/s\n\n",Le);
\ No newline at end of file diff --git a/905/CH4/EX4.9/4_9.sce b/905/CH4/EX4.9/4_9.sce new file mode 100755 index 000000000..e17d68723 --- /dev/null +++ b/905/CH4/EX4.9/4_9.sce @@ -0,0 +1,62 @@ +clear;
+clc;
+
+// Illustration 4.9
+// Page: 264
+
+printf('Illustration 4.9 - Page: 264\n\n');
+
+// solution//
+// From examples 4.4, 4.6 and 4.7
+
+Do = 5*10^-3; // [m]
+Ml = 18.63; // [molecular weight of water, gram/mole]
+Mg = 44.04; // [molecular weight of carbon dioxide, gram/mole]
+rowg = 1.923; // [kg/cubic m]
+rowl = 986; // [kg/cubic m]
+vo = 18.48; // [m/s]
+hl = 0.0173; // [m]
+ug = 1.45*10^-5; // [kg/m.s]
+phie = 0.274;
+A = 0.101; // [Ah/Aa]
+Dg = 0.085; // [square cm/s]
+Dl = 1.91*10^-5; // [square cm/s]
+Aa = 0.614; // [square m]
+Qg = 1.145; // [cubic m/s]
+t = 0.5; // [m]
+h2q = 0.391; // [m]
+rw = 0.34; // [m]
+ql = 0.000815; // [cubic m/s]
+g = 9.8; // [square m/s]
+G = 2.202/44.04; // [kg/s]
+L = 0.804/18.63; // [kg/s]
+
+Refe = rowg*vo*hl/(ug*phie);
+
+cg =rowg/Mg; // [kmole/cubic m]
+cl = rowl/Ml; // [kmole/cubic m]
+
+// For the low concentrations prevailing in the liquid phase, the ethanol- // water solution at 303 K obeys Henry's law, and the slope of the equilibriu// m curve is m = 0.57
+m = 0.57;
+// From equation 4.53
+a1 = 0.4136;
+a2 = 0.6074;
+a3 = -0.3195;
+Eog = 1-exp(-0.0029*Refe^a1*(hl/Do)^a2*A^a3/((sqrt(Dg*(1-phie)/(Dl*A)))*m*cg/cl+1));
+// From equation 4.62
+Deg = 0.01; // [square m/s]
+Peg = 4*Qg*rw^2/(Aa*Deg*(t-h2q)); // [Peclet number]
+// Since Peclet number is greater than 50, therefore vapor is unmixed
+// From equation 4.60
+Del = 0.1*sqrt(g*h2q^3); // [square m/s]
+// From equation 4.59
+Pel = 4*ql*rw^2/(Aa*hl*Del);
+N = (Pel+2)/2;
+lambda = m*G/L;
+// From equation 4.58
+Emg = ((1+lambda*Eog/N)^N -1)/lambda*(1-0.0335*lambda^1.073*Eog^2.518*Pel^0.175);
+// From example 4.8
+E = 0.05;
+// Substituting in equation 4.63
+Emge = Emg*(1-0.8*Eog*lambda^1.543*E/m);
+printf("The entrainment corrected Murphree tray efficiency for the ethanol is %f.\n\n",Emge);
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