diff options
Diffstat (limited to '389/CH3')
| -rwxr-xr-x | 389/CH3/EX3.1/Example3_1.sce | 36 | ||||
| -rwxr-xr-x | 389/CH3/EX3.2/Example3_2.sce | 32 | ||||
| -rwxr-xr-x | 389/CH3/EX3.3/Example3_3.sce | 25 | ||||
| -rwxr-xr-x | 389/CH3/EX3.4/Example3_4.sce | 41 | ||||
| -rwxr-xr-x | 389/CH3/EX3.5/Example3_5.sce | 55 | ||||
| -rwxr-xr-x | 389/CH3/EX3.6/Example3_6.sce | 57 | ||||
| -rwxr-xr-x | 389/CH3/EX3.7/Example3_7.sce | 68 | ||||
| -rwxr-xr-x | 389/CH3/EX3.8/Example3_8.sce | 48 |
8 files changed, 362 insertions, 0 deletions
diff --git a/389/CH3/EX3.1/Example3_1.sce b/389/CH3/EX3.1/Example3_1.sce new file mode 100755 index 000000000..82fa13885 --- /dev/null +++ b/389/CH3/EX3.1/Example3_1.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+
+// Illustration 3.1
+// Page: 53
+
+printf('Illustration 3.1 - Page: 53\n\n');
+
+// solution
+
+//****Data*****//
+// a = CO2 b = H2O
+Ca0 = 0;//[kmol/cubic m]
+Cai = 0.0336;//[kmol/cubic m]
+Dab = 1.96*10^(-9);// [square m/s]
+//*******//
+
+density = 998;// [kg/cubic m]
+viscosity = 8.94*10^(-4);//[kg/m.s]
+rate = 0.05;//[kg/m.s] mass flow rate of liquid
+L = 1;//[m]
+g = 9.81;//[m/square s]
+// From Eqn. 3.10
+del = ((3*viscosity*rate)/((density^2)*g))^(1/3);// [m]
+Re = 4*rate/viscosity;
+// Flow comes out to be laminar
+// From Eqn. 3.19
+Kl_avg = ((6*Dab*rate)/(3.141*density*del*L))^(1/2);//[kmol/square m.s.(kmol/cubic m)]
+bulk_avg_velocity = rate/(density*del);//[m/s]
+// At the top: Cai-Ca = Cai_Ca0 = Cai
+//At the bottom: Cai-Cal
+// From Eqn. 3.21 & 3.22
+Cal = Cai*(1-(1/(exp(Kl_avg/(bulk_avg_velocity*del)))));// [kmol/cubic m]
+rate_absorption = bulk_avg_velocity*del*(Cal-Ca0);// [kmol/s].(m of width)
+printf('The rate of absorption is %e',rate_absorption);
+// The actual value may be substantially larger.
\ No newline at end of file diff --git a/389/CH3/EX3.2/Example3_2.sce b/389/CH3/EX3.2/Example3_2.sce new file mode 100755 index 000000000..08a57fc8b --- /dev/null +++ b/389/CH3/EX3.2/Example3_2.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// Illustration 3.2
+// Page: 56
+
+printf('Illustration 3.2 - Page: 56\n\n');
+
+// solution
+
+//***Data****//
+d = 0.025;// [m]
+avg_velocity = 3;// [m/s]
+viscosity = 8.937*10^(-4);// [kg/m.s]
+density = 997;// [kg/m^3]
+//*********//
+
+kinematic_viscosity = viscosity/density;// [square m/s]
+Re = d*avg_velocity*density/viscosity;
+// Reynold's number comes out to be 83670
+// At this Reynold's number fanning factor = 0.0047
+f = 0.0047;
+L = 1;// [m]
+press_drop = 2*density*f*L*(avg_velocity^2)/(d);// [N/square m]
+P = 3.141*(d^2)*avg_velocity*press_drop/4;// [N.m/s] for 1m pipe
+m = 3.141*(d^2)*L*density/4;
+// From Eqn. 3.24
+Ld = ((kinematic_viscosity^3)*m/P)^(1/4);// [m]
+// From Eqn. 3.25
+Ud = (kinematic_viscosity*P/m)^(1/4);// [m/s]
+printf('Velocity of small eddies is %f m/s\n',Ud);
+printf('Length scale of small eddies is %e m',Ld);
\ No newline at end of file diff --git a/389/CH3/EX3.3/Example3_3.sce b/389/CH3/EX3.3/Example3_3.sce new file mode 100755 index 000000000..37eb6c59c --- /dev/null +++ b/389/CH3/EX3.3/Example3_3.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Illustration 3.3
+// Page: 69
+
+printf('Illustration 3.3 - Page: 69\n\n');
+
+// solution
+
+// Heat transfer analog to Eqn. 3.12
+// The Eqn. remains the same with the dimensionless conc. ratio replaced by ((tl-to)/(ti-to))
+
+// The dimensionless group:
+// eta = 2*Dab*L/(3*del^2*velocity);
+// eta = (2/3)*(Dab/(del*velocity))*(L/del);
+// Ped = Peclet no. for mass transfer
+// eta = (2/3)*(1/Ped)*(L/del);
+
+// For heat transfer is replaced by
+// Peh = Peclet no. for heat transfer
+// eta = (2/3)*(1/Peh)*(L/del);
+// eta = (2/3)*(alpha/(del*velocity))*(L/del);
+// eta = (2*alpha*L)/(3*del^2*velocity);
+printf('Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del^2*velocity)');
\ No newline at end of file diff --git a/389/CH3/EX3.4/Example3_4.sce b/389/CH3/EX3.4/Example3_4.sce new file mode 100755 index 000000000..ec1980073 --- /dev/null +++ b/389/CH3/EX3.4/Example3_4.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+
+// Illustration 3.4
+// Page: 69
+
+printf('Illustration 3.4 - Page: 69\n\n');
+
+// solution
+
+//***Data****//
+// a = UF6 b = air
+// The average heat transfer coefficient: Nu_avg = 0.43+0.532(Re^0.5)(Pr^0.31)
+// The analogus expression for mass transfer coefficient: Sh_avg = 0.43+0.532(Re^0.5)(Sc^0.31)
+d = 0.006;// [m]
+velocity = 3;// [m/s]
+surf_temp = 43;// [C]
+bulk_temp = 60;// [C]
+avg_temp = (surf_temp+bulk_temp)/2; //[C]
+density = 4.10;// [kg/cubic m]
+viscosity = 2.7*10^(-5);// [kg/m.s]
+Dab = 9.04*10^(-6);// [square m/s]
+press = 53.32;// [kN/square m]
+tot_press = 101.33;// [kN/square m]
+//******//
+
+avg_press = press/2; // [kN/square m]
+Xa = avg_press/tot_press;
+Xb = 1-Xa;
+Re = d*velocity*density/viscosity;
+Sc = viscosity/(density*Dab);
+Sh_avg = 0.43+(0.532*(2733^0.5)*(0.728^0.5));
+c = 273.2/(22.41*(273.2+avg_temp));// [kmol/cubic m]
+F_avg = Sh_avg*c*Dab/d;//[kmol/cubic m]
+Nb = 0;
+Ca1_by_C = press/tot_press;
+Ca2_by_C = 0;
+Flux_a = 1;
+// Using Eqn. 3.1
+Na = Flux_a*F_avg*log((Flux_a-Ca2_by_C)/(Flux_a-Ca1_by_C));//[kmol UF6/square m.s]
+printf('Rate of sublimation is %e kmol UF6/square m.s',Na);
\ No newline at end of file diff --git a/389/CH3/EX3.5/Example3_5.sce b/389/CH3/EX3.5/Example3_5.sce new file mode 100755 index 000000000..932326e6b --- /dev/null +++ b/389/CH3/EX3.5/Example3_5.sce @@ -0,0 +1,55 @@ +clear;
+clc;
+
+// Illustration 3.5
+// Page: 73
+
+printf('Illustration 3.5 - Page: 73\n\n');
+
+// solution
+
+//****Data****//
+velocity = 15;// [m/s]
+G = 21.3;// [kg/square m.s]
+//******//
+
+// Since the experimental data do not include the effects of changing Prandtl number.
+
+// Jh = (h/(Cp*density*viscosity)) = (h/Cp*G)*(Pr^(2/3)) = Shi(Re);
+
+// Shi(Re) must be compatible with 21.3*(G^0.6);
+// Let Shi(Re) = b*(Re^n);
+// Re = (l*G)/viscosity;
+
+// h = (Cp*G/(Pr^(2/3)))*b*(Re^n);
+// h = (Cp*G/(Pr^(2/3)))*b*((l*b/viscosity)^n) = 21.3*(G^0.6);
+
+n = 0.6-1;
+// b = 21.3*((Pr^(2/3))/Cp)*((l/viscosity)^(-n));
+
+// Using data for air at 38 C & 1 std atm.
+Cp1 = 1002;// [kJ/kg.K]
+viscosity1 = 1.85*10^(-5);//[kg/m.s]
+k1 = 0.0273;//[W/m.K]
+Pr1 = (Cp1*viscosity1)/k1;
+b_prime = 21.3*(Pr1^(2/3)/Cp1)*((1/viscosity1)^0.4);
+// b = b_prime*l^(0.4);
+// Jh = (h/(Cp*G))*Pr^(2/3) = b_prime*((l/Re)^(0.4)) = Shi(Re);
+
+// The heat mass transfer analogy will be used to estimate the mass transfer coefficient. (Jd = Jh)
+
+// Jd = (KG*Pbm*Mav*Sc^(2/3))/(density*viscosity) = Shi(Re) = b_prime*((l/Re)^0.4);
+
+// KG*Pbm = F = (b_prime*density*viscosity)/(Re^0.4*Mav*Sc^(2/3)) = (b_prime*(density*velocity)^0.6*(viscosity^0.4))/(Mav*Sc^(2/3));
+
+// For H2-H20, 38 C, 1std atm
+viscosity2 = 9*10^(-6);// [kg/m.s]
+density2 = 0.0794;// [kg/cubic m]
+Dab = 7.75*10^(-5);// [square m/s]
+Sc = viscosity2/(density2*Dab);
+
+// Assuming desity, Molecular weight and viscosity of the gas are essentially those of H2
+
+Mav = 2.02;// [kg/kmol]
+F = (b_prime*(density2*velocity)^0.6*(viscosity2^0.4))/(Mav*Sc^(2/3));// [kmol/square m.s]
+printf('The required mass transfer: %f kmol/square m.s',F);
\ No newline at end of file diff --git a/389/CH3/EX3.6/Example3_6.sce b/389/CH3/EX3.6/Example3_6.sce new file mode 100755 index 000000000..d6ba423e5 --- /dev/null +++ b/389/CH3/EX3.6/Example3_6.sce @@ -0,0 +1,57 @@ +clear;
+clc;
+
+// Illustration 3.6
+// Page: 77
+
+printf('Illustration 3.6 - Page: 77\n\n');
+
+// solution
+
+//***Data***//
+Dp = 0.0125;// [m]
+viscosity = 2.4*10^(-5);// [kg/m.s]
+Sc = 2;
+E = 0.3;
+Go = (2*10^(-3))/0.1;// molar superficial mass velocity [kmol/square m.s]
+//********//
+
+// a = CO b = Ni(CO)4
+// Nb = -(Na/4);
+Flux_a = 4/3;
+Ca2_by_C = 0;// At the metal interface
+// Ca1_by_C = Ya //mole fraction of CO in the bulk
+
+// Eqn. 3.1 becomes: Na = (4/3)*F*log((4/3)/((4/3)-Ya));
+
+// Let G = kmol gas/(square m bed cross section).s
+// a = specific metal surface
+// z = depth
+// Therefore, Na = -(diff(Ya*G))/(a*diff(z));// [kmol/((square m metal surface).s)];
+// For each kmol of CO consumed, (1/4)kmol Ni(CO)4 forms, representing a loss of (3/4) kmol per kmol of CO consumed.
+// The CO consumed through bed depth dz is therefore (Go-G)(4/3) kmol;
+// Ya = (Go-(Go-G)*(4/3))/G;
+// G = Go/(4-(3*Ya));
+// diff(YaG) = ((4*Go)/(4-3*Ya)^2)*diff(Ya);
+
+// Substituting in Eqn. 3.64
+// -(4*Go/((4-3*Ya)^2*a))*(diff(Ya)/diff(z)) = (4/3)*F*log(4/(4-3*Ya));
+
+// At depth z:
+// Mass velocity of CO = (Go-(Go-G)/(4/3))*28;
+// Mass velocity of Ni(CO)4 = ((Go-G)*(1/3))*170.7;
+// G_prime = 47.6*Go-19.6G; // total mass velocity [kg/square m.s]
+// Substituting G leads to:
+// G_prime = Go*(47.6-19.6*(4-3*Ya));// [kg/m.s]
+// Re = (Dp*G')/viscosity
+
+// With Go = 0.002 kmol/square m.s & Ya in the range 1-0.005, the range of Re is 292-444;
+// From table 3.3:
+// Jd = (F/G)*(Sc^(2/3)) = (2.06/E)*Re^(-0.575);
+// F = (2.06/E*(Sc)^(2/3))*(Go/(4-3*Ya))*Re^(-0.575);
+
+a = 6*(1-E)/Dp;
+
+// Result after arrangement:
+Z = integrate('-((4*Go)/((4-(3*Ya))^2*a))*(3/4)*(E*(Sc^(2/3))*(4-(3*Ya))/(2.06*Go)*(1/log(4/(4-(3*Ya)))))*(((Dp/viscosity)*(Go*(47.6-(19.6/(4-(3*Ya))))))^0.575)','Ya',1,0.005);// [m]
+printf('The bed depth required to reduce the CO content to 0.005 is %f m', Z);
\ No newline at end of file diff --git a/389/CH3/EX3.7/Example3_7.sce b/389/CH3/EX3.7/Example3_7.sce new file mode 100755 index 000000000..083394ecc --- /dev/null +++ b/389/CH3/EX3.7/Example3_7.sce @@ -0,0 +1,68 @@ +clear;
+clc;
+
+// Illustration 3.7
+// Page: 80
+
+printf('Illustration 3.7 - Page: 80\n\n');
+
+// solution
+
+//****Data*****//
+// a = water b = air
+out_dia = 0.0254;// [m]
+wall_thick = 0.00165;// [m]
+avg_velocity = 4.6;// [m/s]
+T1 = 66;// [C]
+P = 1;// [atm]
+Pa1 = 0.24;// [atm]
+k1 = 11400;// [W/(square m.K)]
+T2 = 24;// [C]
+k2 = 570;// [W/square m.K]
+k_Cu = 381;// [w/square m.K]
+//******//
+
+// For the metal tube
+int_dia = out_dia-(2*wall_thick);// [m]
+avg_dia = (out_dia+int_dia)/2;// [mm]
+Nb = 0;
+Flux_a = 1;
+Ya1 = 0.24;
+Yb1 = 1-Ya1;
+Mav = (Ya1*18.02)+(Yb1*29);// [kg/kmol]
+density = (Mav/22.41)*(273/(273+T1));// [kg/cubic m]
+viscosity = 1.75*10^(-5);// [kg/m.s]
+Cpa = 1880;// [J/kg.K]
+Cpmix = 1145;// [J/kg.K]
+Sc = 0.6;
+Pr = 0.75;
+G_prime = avg_velocity*density;// [kg/square m.s]
+G = G_prime/Mav;// [kmol/square m.s]
+Re = avg_dia*G_prime/viscosity;
+// From Table 3.3:
+// Jd = Std*Sc^(2/3) = (F/G)*Sc^(2/3) = 0.023*Re^(-0.17);
+Jd = 0.023*Re^(-0.17);
+F = (0.023*G)*(Re^(-0.17)/Sc^(2/3));
+
+// The heat transfer coeffecient in the absence of mass transfer will be estimated through Jd = Jh
+// Jh = Sth*Pr^(2/3) = (h/Cp*G_prime)*(Pr^(2/3)) = Jd
+h = Jd*Cpmix*G_prime/(Pr^(2/3));
+
+U = 1/((1/k1)+((wall_thick/k_Cu)*(int_dia/avg_dia))+((1/k2)*(int_dia/out_dia)));// W/square m.K
+
+// Using Eqn. 3.70 & 3.71 with Nb = 0
+// Qt = (Na*18.02*Cpa/1-exp(-(Na*18.02*Cpa/h)))*(T1-Ti)+(Lambda_a*Na);
+// Qt = 618*(Ti-T2);
+// Using Eqn. 3.67, with Nb = 0, Cai/C = pai, Ca1/C = Ya1 = 0.24;
+// Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1));
+
+// Solving above three Eqn. simultaneously:
+Ti = 42.2;// [C]
+pai = 0.0806;// [atm]
+Lambda_a = 43.4*10^6;// [J/kmol]
+Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1)));// [kmol/square m.s]
+Qt1 = 618*(Ti-T2);// [W/square m]
+Qt2 = ((Na*18.02*Cpa/(1-exp(-(Na*18.02*Cpa/h))))*(T1-Ti))+(Lambda_a*Na);// [W/square m]
+
+// since the value of Qt1 & Qt2 are relatively close
+printf('The local rate of condensation of water is %e kmol/square m.s',Na);
\ No newline at end of file diff --git a/389/CH3/EX3.8/Example3_8.sce b/389/CH3/EX3.8/Example3_8.sce new file mode 100755 index 000000000..717757b51 --- /dev/null +++ b/389/CH3/EX3.8/Example3_8.sce @@ -0,0 +1,48 @@ +clear;
+clc;
+
+// Illustration 3.8
+// Page: 81
+printf('Illustration 3.8 - Page: 81\n\n');
+printf('Illustration 3.8 (a)\n\n');
+
+// Solution (a)
+
+//***Data****//
+// a = water b = air
+Nb = 0;
+h = 1100;// [W/square m]
+//*****//
+
+Ma = 18.02;// [kg/kmol]
+Cpa = 2090;// [J/kg.K]
+T1 = 600;// [C]
+Ti = 260;// [C]
+// The positive dirn. is taken to be from the bulk gas to the surface.
+Has = 2.684*(10^6);// enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]
+Hai = 2.994*(10^6);// enthalpy of steam at 1 std atm, 260 C in [J/kg]
+
+// Radiation contributions to the heat transfer from the gas to the surface are negligible. Eqn. 3.70 reduces to
+Na = -((h/(Ma*Cpa))*log(1-((Cpa*(T1-Ti))/(Has-Hai))));// [kmol/square m.s]
+printf('The rate of steam flow reqd. is %f kmol/square m.s\n\n',Na);
+// negative sign indicates that the mass flux is into the gas
+
+printf('Illustration 3.8 (b)\n\n');
+
+// Solution (b)
+
+//***Data****//
+// a = water b = air
+h = 572;// [W/square m]
+T1 = 25;// [C]
+//******//
+
+Ti = 260;// [C]
+// The positive dirn. is taken to be from the bulk gas to the surface.
+Has = 1.047*10^(5);// enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]
+Hai = 2.994*(10^6);// enthalpy of steam at 1 std atm, 260 C in [J/kg]
+
+// Radiation contributions to the heat transfer from the gas to the surface are negligible. Eqn. 3.70 reduces to
+Na = -((h/(Ma*Cpa))*log(1-((Cpa*(T1-Ti))/(Has-Hai))));// [kmol/square m.s]
+printf('The rate of steam flow reqd. is %f kmol/square m.s',Na);
+// negative sign indicates that the mass flux is into
\ No newline at end of file |