diff options
Diffstat (limited to '905/CH8')
| -rwxr-xr-x | 905/CH8/EX8.1/8_1.sce | 46 | ||||
| -rwxr-xr-x | 905/CH8/EX8.2/8_2.sce | 34 | ||||
| -rwxr-xr-x | 905/CH8/EX8.3/8_3.sce | 44 | ||||
| -rwxr-xr-x | 905/CH8/EX8.4/8_4.sce | 43 | ||||
| -rwxr-xr-x | 905/CH8/EX8.5/8_5.sce | 23 | ||||
| -rwxr-xr-x | 905/CH8/EX8.6/8_6.sce | 84 | ||||
| -rwxr-xr-x | 905/CH8/EX8.7/8_7.sce | 85 | ||||
| -rwxr-xr-x | 905/CH8/EX8.8/8_8.sce | 96 |
8 files changed, 455 insertions, 0 deletions
diff --git a/905/CH8/EX8.1/8_1.sce b/905/CH8/EX8.1/8_1.sce new file mode 100755 index 000000000..12d23f8a6 --- /dev/null +++ b/905/CH8/EX8.1/8_1.sce @@ -0,0 +1,46 @@ +clear;
+clc;
+
+// Illustration 8.1
+// Page: 479
+
+printf('Illustration 8.1 - Page: 479\n\n');
+
+// solution
+// ****Data****//
+P_total = 1; // [bar]
+T1 = 320; // [K]
+T_c = 562.2; // [K]
+P_c = 48.9; // [bar]
+A = -6.983;
+B = 1.332;
+C = -2.629;
+D = -3.333;
+//*****//
+
+x1 = 1-(T1/T_c);
+deff('[y] = f12(P1)','y = log(P1/P_c)-(A*x1+B*x1^1.5+C*x1^3+D*x1^6)/(1-x1)');
+P1 = fsolve(.01,f12);// [bar]
+printf("Vapor pressure of benzene at 320 K is %f bar\n\n",P1);
+
+M_benzene = 78 // [gram/mole]
+printf('Illustration 8.1 (a)\n');
+
+// Solution (a)
+// For nitrogen
+M_nitrogen = 28; // [gram/mole]
+// From equation 8.2
+Y = P1/(P_total - P1); //[mole C6H6/ mole N2]
+Y_s1 = Y*(M_benzene/M_nitrogen); // [gram C6H6/gram N2]
+
+printf("Absolute humidity of mixture of benzene and nitrogen is %f gram C6H6/gram N2\n\n",Y_s1);
+
+printf('Illustration 8.1 (b)\n');
+// Solution (b)
+// For carbon dioxide
+M_carbondioxide = 44; // [gram/mole]
+// From equation 8.2
+Y = P1/(P_total - P1); //[mole C6H6/ mole C02]
+Y_s2 = Y*(M_benzene/M_carbondioxide); // [gram C6H6/gram CO2]
+
+printf("Absolute humidity of mixture of benzene and carbon dioxide is %f gram C6H6/gram CO2\n",Y_s2);
\ No newline at end of file diff --git a/905/CH8/EX8.2/8_2.sce b/905/CH8/EX8.2/8_2.sce new file mode 100755 index 000000000..60f3ee90d --- /dev/null +++ b/905/CH8/EX8.2/8_2.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Illustration 8.2
+// Page: 480
+
+printf('Illustration 8.2 - Page: 480\n\n');
+
+// solution
+// A - water vapor B - air
+// REference state is air
+
+// ****Data****//
+T_ref = 273; // [Reference temperature, K]
+T = 303; // [K]
+P_total = 1; // [atm]
+P_A = 4.24; // [Vapor pressure of water at 303K, kPa]
+M_A = 18; // [gram/mole]
+M_B = 29; // [gram/mole]
+C_A = 1.884; // [kJ/kg.K]
+C_B = 1.005; // [kJ/kg.K]
+lambda = 2502.3; // [Latent heat of Vaporization at 273K, kJ/kg]
+//*****//
+
+P_total = P_total*101.325; // [kPa]
+
+// From equation 8.2
+Y_s = P_A/(P_total - P_A)*(M_A/M_B); //[kg H2O/ kg dry air]
+printf("Absolute humidity of mixture of water vapor and air is %f kg H2O/kg dry air\n\n",Y_s);
+
+// From equation 8.3
+H_s = C_B*(T-T_ref) + Y_s*(C_A*(T-T_ref) + lambda); // [kJ/kg dry air]
+
+printf("Enthalpy per unit mass of dry air of a saturated mixture at 303 K and 1 atm is %f kJ/kg dry air\n",H_s);
\ No newline at end of file diff --git a/905/CH8/EX8.3/8_3.sce b/905/CH8/EX8.3/8_3.sce new file mode 100755 index 000000000..63c1bbace --- /dev/null +++ b/905/CH8/EX8.3/8_3.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+// Illustration 8.2
+// Page: 482
+
+printf('Illustration 8.3 - Page: 482\n\n');
+
+// solution
+// A - water vapor B - air
+//*****Data*****
+T = 328; // [dry bulb temperature, K]
+P_total = 1; // [atm]
+H = 30; // [relative humidity, %]
+//*****//
+P_vapA = 15.73; // [vapor pressure of water, kPa]
+P_total = P_total*101.325; // [kPa]
+M_A = 18; // [gram/mole]
+M_B = 29; // [gram/mole]
+
+P_A = (H/100)*P_vapA;// [partial pressure of A,kPa]
+
+printf('Illustration 8.3 (a)\n\n');
+// At dew point partial pressure is equal to vapor pressure
+// Using Antonnie equation we can find dew point temperature
+
+printf("Dew point temperature is 304.5 K\n")
+
+// From equation 8.1
+Y_s = P_A/(P_total-P_A)*(M_A/M_B);
+printf("Absolute humidity of air-water mixture at 328 K is %f kg H2O/kg dry air\n\n",Y_s);
+
+printf('Illustration 8.3 (b)\n\n');
+
+//soluton (b)
+T_ref = 273; // [K]
+C_A = 1.884; // [kJ/kg.K]
+C_B = 1.005; // [kJ/kg.K]
+lambda = 2502.3; // [Latent heat of Vaporization at 273 K, kJ/kg]
+
+// From equation 8.3
+H_s = C_B*(T-T_ref) + Y_s*(C_A*(T-T_ref) + lambda);
+
+printf("Enthalpy per unit mass of dry air of a saturated mixture relative to 273 K is %f kJ/kg dry air\n",H_s);
\ No newline at end of file diff --git a/905/CH8/EX8.4/8_4.sce b/905/CH8/EX8.4/8_4.sce new file mode 100755 index 000000000..5bb40e52a --- /dev/null +++ b/905/CH8/EX8.4/8_4.sce @@ -0,0 +1,43 @@ +clear;
+clc;
+
+// Illustration 8.4
+// Page: 484
+
+printf('Illustration 8.4 - Page: 484\n\n');
+
+// Solution
+// a - water vapor b - air
+//*****Data*****
+T_G1 = 356; // [K]
+P_total = 101.325; // [kPa]
+Y_1 = .03; // [kg water/kg dry air]
+//*****//
+
+C_pa = 1.884; // [kJ/kg.K]
+C_pb = 1.005; // [kJ/kg.K]
+
+C_s1 = C_pb + Y_1*C_pa;// [kJ/kg.K]
+
+T_1 = 373.15; // [K]
+T_c = 647.1; // [K]
+M_a = 18.02; // [gram/mole]
+M_b = 28.97; // [gram/mole]
+lambda_1 = 2256; // [Latent Heat of Vaporizarion at T_1, kJ/kg]
+
+// Using equation 8.10
+// T_as = T_G1- (Y_as - Y_l)*lambda_as/C_s1
+// where lambda_2 = lambda_1*((1-T_as/T_c)/(1-T_1/T_c))^.38
+// Y_as = P_a/(P_total-P_a)*M_a/M_b
+// and P_a = exp(16.3872-(3885.7/(T_as-42.98))) - Antoine equation for component 'a'
+
+deff('[y] = f12(T_as)',' y = T_as - T_G1 + ((exp(16.3872 - (3885.7/(T_as - 42.98)))/(P_total - (exp(16.3872 - (3885.7/(T_as - 42.98))))))*(M_a/M_b) - Y_1)*(lambda_1*((1-T_as/T_c)/(1-T_1/T_c))^.38/C_s1)');
+T_as = fsolve(310,f12); // [K]
+printf("Adiabatic Saturation Temperature is %f K\n",T_as);
+
+// Now using equation 8.2
+
+P_a = exp(16.3872-(3885.7/(T_as-42.98))); // [kPa]
+Y_as = P_a/(P_total-P_a)*M_a/M_b; // [kg water/kg dry air]
+
+printf("Absolute humidity is %f kg water/kg dry air\n",Y_as);
\ No newline at end of file diff --git a/905/CH8/EX8.5/8_5.sce b/905/CH8/EX8.5/8_5.sce new file mode 100755 index 000000000..9534f6d30 --- /dev/null +++ b/905/CH8/EX8.5/8_5.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Illustration 8.5
+// Page: 487
+
+printf('Illustration 8.5 - Page: 487\n\n');
+
+// Solution
+//*****Data*****//
+T_w = 320; // [K]
+T_g = 340; // [K]
+lambda_w = 2413; // [Latent Heat of Vaporization at 320K, kJ/kg]
+Y_w1 = 0.073; // [kg water/kg dry air]
+//*****//
+A = 0.95; // [For air water system,A, kJ/kg.K]
+
+// here A = hg/ky, psychrometric ratio
+// Air-water mixture is saturated at 320K and 1 atm
+// Using equation 8.15
+
+Y_w2 = Y_w1 - ((T_g-T_w)*A/lambda_w); // [kg water/kg dry air]
+printf("Absolute humidity of air-water mixture at 340 K and 1 atm is %f kg water/kg dry air\n ",Y_w2);
\ No newline at end of file diff --git a/905/CH8/EX8.6/8_6.sce b/905/CH8/EX8.6/8_6.sce new file mode 100755 index 000000000..71cadc66b --- /dev/null +++ b/905/CH8/EX8.6/8_6.sce @@ -0,0 +1,84 @@ +clear;
+clc;
+
+// Illustration 8.6
+// Page: 487
+
+printf('Illustration 8.6 - Page: 487\n\n');
+
+// a - toluene b - air
+//*****Data*****
+T_G1 = 333; // [K]
+P_total = 101.325; // [kPa]
+Y_1 = 0.05; // [kg vapor/kg dry air]
+//*****//
+
+C_pa = 1.256; // [kJ/kg.K]
+C_pb = 1.005; // [kJ/kg.K]
+
+C_s1 = C_pb + Y_1*C_pa
+
+T_1 = 383.8; // [K]
+T_c = 591.8; // [K]
+M_a = 92; // [gram/mole]
+M_b = 28.97; // [gram/mole]
+lambda_1 = 33.18*1000/92; // [Latent heat of vaporization at T_1, kJ/kg]
+
+// Constants of antoine equation
+A = 13.9320;
+B = 3057; // [K]
+C = -55.52; // [K]
+
+printf('Illustration 8.6 (a)\n');
+
+// Solution (a)
+
+// Using equation 8.10
+// T_as = T_G1 - (Y_as - Y_l)*lambda_as/C_s1
+// where lambda_2 = lambda_1*((1-T_as/T_c)/(1-T_1/T_c))^.38
+// Y_as = P_a/(P_total-P_a)*M_a/M_b
+// P_a = exp(A-B/(T+c))
+
+deff('[y] = f12(T_as)',' y = T_as - T_G1 + ((exp(13.9320 - (3057/(T_as - 55.52)))/(P_total - (exp(13.9320 - (3057/(T_as - 55.52))))))*(M_a/M_b) - Y_1)*(lambda_1*((1-T_as/T_c)/(1-T_1/T_c))^.38/C_s1)');
+T_as = fsolve(273,f12); // [K]
+printf("Adiabatic Saturation Temperature is %f K\n",T_as);
+
+// Now using equation 8.2
+
+P_a = exp(13.9320-(3057/(T_as-55.52))); // [kPa]
+Y_as = P_a/(P_total-P_a)*M_a/M_b; // [kg vapor/kg dry air]
+
+printf("Absolute humidity is %f kg vapor/kg dry air\n\n",Y_as);
+
+printf('Illustration 8.6 (b)\n');
+
+// Solution (b)
+
+// Thermodynamic properties of mixture of toluene and air
+row = 1.06; // [kg/cubic m]
+u = 19.5*10^-6; // [P]
+Pr = 0.7;
+Dab = 0.1; //[From Wilke-Lee equation, square cm/s]
+Sc = u/(row*Dab*10^-4);
+
+// Using equation 8.16
+
+A_1 = C_s1*(Sc/Pr)^0.567; // [kJ/kg.K]
+// here A_1 = hg/ky, psychrometric ratio
+
+// Using equation 8.15
+// T_w = T_G1 - (Y_w-Y_1)*lambda_w/(hg/ky)
+// where lambda_w = lambda_1*((1-T_w/T_c)/(1-T_1/T_c))^.38
+// Y_w = P_a/(P_total-P_a)*M_a/M_b
+// P_a = exp(A-B/(T+c))
+
+deff('[z] = f15(T_w)',' z = T_w - T_G1 + ((exp(13.9320 - (3057/(T_w - 55.52)))/(P_total - (exp(13.9320 - (3057/(T_w - 55.52))))))*(M_a/M_b) - Y_1)*(lambda_1*((1-T_w/T_c)/(1-T_1/T_c))^.38/A_1)');
+T_w = fsolve(273,f15); // [K]
+printf("Wet bulb Temperature is %f K\n",T_w);
+
+// Now using equation 8.2
+
+P_a = exp(13.9320-(3057/(T_w-55.52))); // [kPa]
+Y_w = P_a/(P_total-P_a)*M_a/M_b; // [kg vapor/kg dry air]
+
+printf("Absolute humidity is %f kg vapor/kg dry air\n",Y_w);
diff --git a/905/CH8/EX8.7/8_7.sce b/905/CH8/EX8.7/8_7.sce new file mode 100755 index 000000000..352a827c5 --- /dev/null +++ b/905/CH8/EX8.7/8_7.sce @@ -0,0 +1,85 @@ +clear;
+clc;
+
+// Illustration 8.7
+// Page: 493
+
+printf('Illustration 8.7 - Page: 493\n\n');
+
+
+// solution
+
+//****Data****//
+L_min = 2.27;// [kg/square m.s]
+G_min = 2;// [kg/square m.s]
+L2_prime = 15;// [kg/s]
+Templ2 = 318;// [K]
+Tempg1 = 303;// [Entering air dry bulb, K]
+Tempw1 = 297;// [ Entering air wet bulb, K]
+Kya = 0.90;// [kg/cubic m.s]
+//*******//
+
+H1_prime = 72.5;// [kJ/kg dry air]
+Y1_prime = 0.0190;// [kg water/kg dry air]
+Templ1 = 302;// [K]
+Cal = 4.187;// [kJ/kg]
+
+// Equilibrium Data:
+// Data = [Temp.(K),H_star(kJ/kg)]
+Data_star = [302 100;305.5 114;308 129.8;310.5 147;313 166.8;315.5 191;318 216];
+
+// The operating line for least slope:
+H2_star = 210;// [kJ/kg]
+Data_minSlope = [Templ1 H1_prime;Templ2 H2_star];
+deff('[y] = f14(Gmin)','y = ((L2_prime*Cal)/Gmin)-((H2_star-H1_prime)/(Templ2-Templ1))');
+Gmin = fsolve(2,f14);// [kg/s]
+Gs = 1.5*Gmin;// [kg/s]
+
+// For the Operating Line:
+y = deff('[y] = f15(H2)','y = ((H2-H1_prime)/(Templ2-Templ1))-((L2_prime*Cal)/Gs)');
+H2 = fsolve(2,f15);// [kJ/kg dry air]
+Data_opline = [Templ1 H1_prime;Templ2 H2];
+
+scf(4);
+plot(Data_star(:,1),Data_star(:,2),Data_minSlope(:,1),Data_minSlope(:,2),Data_opline(:,1),Data_opline(:,2));
+xgrid();
+legend('Equilibrium line','Minimum Flow Rate Line','Operating Line');
+xlabel("Liquid Temperature, K");
+ylabel("Enthalphy Of Air Water vapour, kJ / kg dry air");
+
+// Tower cross section Area:
+Al = L2_prime/L_min;// [square m]
+Ag = Gs/G_min;// [square m]
+A = min(Al,Ag);// [square m]
+printf("Cross sectional is %f square m\n",A);
+
+// Data from operating line:
+// Data1 = [Temp.(K),H_prime(kJ/kg)]
+Data1 = [302 72.5;305.5 92;308 106.5;310.5 121;313 135.5;315.5 149.5;318 164.2];
+
+// Driving Force:
+Data2 = zeros(7,2);
+// Data2 = [Temp[K],driving Force]
+for i = 1:7
+ Data2(i,1) = Data1(i,1);
+ Data2(i,2) = 1/(Data_star(i,2)-Data1(i,2));
+end
+
+// The data for operating line as abcissa is plotted against driving force;
+Area = 3.28;
+// From Eqn. 7.54
+deff('[y] = f16(Z)','y = Area-(Kya*Z/G_min)');
+Z = fsolve(2,f16);
+printf("The height of tower is %f m\n",Z);
+NtoG = 3.28;
+HtoG = G_min/Kya;// [m]
+
+// Make up water
+// Assuming the outlet air is essentially saturated:
+Y2_prime = 0.048;// [kg water/kg dry air]
+H2 = 164.2; // [kJ/kg dry air]
+// This corresponds to an exit-air temperature of 312.8 K
+
+// Approximate rate of evaporation
+R = Gs*(Y2_prime-Y1_prime);
+printf("Rate of evaporation is %f kg/s\n",R);
\ No newline at end of file diff --git a/905/CH8/EX8.8/8_8.sce b/905/CH8/EX8.8/8_8.sce new file mode 100755 index 000000000..6da0b6257 --- /dev/null +++ b/905/CH8/EX8.8/8_8.sce @@ -0,0 +1,96 @@ +clear;
+clc;
+
+// Illustration 8.8
+// Page: 495
+
+printf('Illustration 8.8 - Page: 495\n\n');
+
+// solution (a)
+printf('Illustration 8.8 (a) - Page: 495\n\n');
+
+// a - water vapor b - air
+//****Data****//
+T_L2 = 314; // [inlet water temperature, K]
+T_L1 = 303; // [outlet water temperature, K]
+T_d = 306; // [dry bulb temperature ,K]
+T_w1 = 298; // [wet bulb temperature, K]
+Z = 3; // [packed tower depth, m]
+G_x = 3; // [mass velocity, kg/square m.s]
+G_s =2.7; // [mass velocity, kg/square m.s]
+//*****//
+
+T_o = 273; // [reference temperature, K]
+C_al = 4.187; // [kJ/kg.K]
+C_pb = 1.005; // [kJ/kg.K]
+C_pa = 1.884; // [kJ/kg.K]
+P_total = 101.325; // [kPa]
+lambda_0 = 2502.3; // [kJ/kg]
+M_a = 18.02; // [gram/mole]
+M_b = 28.97; // [gram/mole]
+
+// Equilibrium Data:
+// Data = [Temp.(K),H_eqm(kJ/kg)],[H_eqm - Equilibrium gas enthalpy]
+Data_eqm = [273 9.48;283 29.36;293 57.8;303 99.75;313 166.79;323 275.58;333 461.5];
+
+scf(4);
+plot(Data_eqm(:,1),Data_eqm(:,2));
+xgrid();
+legend('Equilibrium line');
+xlabel("Liquid Temperature, K");
+ylabel("Enthalphy Of Air Water vapour, kJ / kg dry air");
+
+P_a = exp(16.3872-(3885.7/(T_w1-42.98))); // [kPa]
+Y_m1 = P_a/(P_total-P_a)*(M_a/M_b); // [kg water/kg dry air]
+H_g1 = C_pb*(T_w1-T_o) + Y_m1*(C_pa*(T_w1-T_o)+lambda_0); // [Enthalpy of saturated mixture, kJ/kg dry air]
+
+// From overall energy balance
+H_g2 = H_g1 + G_x*C_al*(T_L2-T_L1)/G_s; // [Enthalpy of exit air, kJ/kg]
+
+// For calculation of mass transfer unit, Ntog
+// Data1 = [T_L1 H_g1,.....,T_L2 H_g2]
+Data1 = zeros(10,2);
+deltaT = (T_L2-T_L1)/9;
+for i = 1:10
+ Data1(i,1) = T_L1 + (i-1)*deltaT;
+ Data1(i,2) = H_g1 + G_x*C_al*(Data1(i,1)-T_L1)/G_s;
+end
+
+// Data for enthalpy of exit air at different temperature varying from T_L1 to T_L2, operating line
+Data1 = [303 76.17;304.22 81.85;305.44 87.53;306.67 93.22;307.89 98.91;309.11 104.59;310.33 110.28;311.56 115.96;312.78 121.65;314 127.35];
+
+// Data of equilibrium gas enthalpy at different temperature varying from T_L1 to T_L2 from the above equilibrium graph
+Data2 = [303 100;304.22 107.93;305.44 116.12;306.67 124.35;307.89 132.54;309.11 140.71;310.33 148.89;311.56 157.14;312.78 165.31;314 177.67];
+
+// Driving force
+Data3 = zeros(10,2);
+// Data3 =[Equilibrium gas enthalpy, driving force]
+for i = 1:10
+ Data3(i,1) = Data1(i,2);
+ Data3(i,2) = 1/(Data2(i,2)-Data1(i,2));
+end
+
+// The data for Equilibrium gas enthalpy as abcissa is plotted against driving force
+Area = 1.642;
+N_tog = 1.642;
+H_tog = Z/N_tog; // [m]
+
+// Overall volumetric mass-transfer coefficient, K_ya
+K_ya = G_s/H_tog;
+printf("Overall volumetric mass-transfer coefficient is %f kg/cubic m.s\n\n",K_ya);
+
+// Solution (b)
+printf('Illustration 8.8 (b) - Page: 495\n\n');
+
+T_w2 = 288; // [New entering-air wet-bulb temperature, K]
+P_a2 = exp(16.3872-(3885.7/(T_w2-42.98))); // [kPa]
+Y_m2 = P_a2/(P_total-P_a2)*(M_a/M_b); // [kg water/kg dry air]
+H_g11 = C_pb*(T_w2-T_o) + Y_m2*(C_pa*(T_w2-T_o)+lambda_0); // [Enthalpy of saturated mixture, kJ/kg dry air]
+
+// the change in water temperature through the tower must remain the same as in part (a), namely T_L2b-T_L1b = 11K
+// Since N_tog is a function of both water temperatures(T_L1',T_L2'), this provides the second relation needed to calculate the values of T_L2b and T_L1b
+// The two equations are solved simultaneously by trial and error method, from above we get T_L1' = 297K
+T_L1b = 297; // [K]
+T_L2b = T_L1b + 11; //[K]
+S = T_L1b - T_w2; // [wet bulb temperature approach, K]
+printf("The outlet water temperature and wet bulb temperature approach is %f K and %f K respectively ",T_L1b,S);
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