From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 905/CH4/EX4.2/4_2.sce | 45 ++++++++++++++ 905/CH4/EX4.3/4_3.sce | 166 ++++++++++++++++++++++++++++++++++++++++++++++++++ 905/CH4/EX4.4/4_4.sce | 157 +++++++++++++++++++++++++++++++++++++++++++++++ 905/CH4/EX4.5/4_5.sce | 120 ++++++++++++++++++++++++++++++++++++ 905/CH4/EX4.6/4_6.sce | 64 +++++++++++++++++++ 905/CH4/EX4.7/4_7.sce | 54 ++++++++++++++++ 905/CH4/EX4.8/4_8.sce | 40 ++++++++++++ 905/CH4/EX4.9/4_9.sce | 62 +++++++++++++++++++ 8 files changed, 708 insertions(+) create mode 100755 905/CH4/EX4.2/4_2.sce create mode 100755 905/CH4/EX4.3/4_3.sce create mode 100755 905/CH4/EX4.4/4_4.sce create mode 100755 905/CH4/EX4.5/4_5.sce create mode 100755 905/CH4/EX4.6/4_6.sce create mode 100755 905/CH4/EX4.7/4_7.sce create mode 100755 905/CH4/EX4.8/4_8.sce create mode 100755 905/CH4/EX4.9/4_9.sce (limited to '905/CH4') 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); -- cgit