diff options
Diffstat (limited to '858/CH3')
30 files changed, 750 insertions, 0 deletions
diff --git a/858/CH3/EX3.1/example_1.sce b/858/CH3/EX3.1/example_1.sce new file mode 100755 index 000000000..df337ce67 --- /dev/null +++ b/858/CH3/EX3.1/example_1.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 3.1 page number 90\n\n")
+
+//to find the coal consumption
+w_C = 0.6; //amount of carbon in coal
+N2_content = 40 //in m3 per 100m3 air
+
+air_consumed = N2_content/0.79;
+weight_air = air_consumed*(28.8/22.4);
+O2_content = air_consumed*32*(0.21/22.4); //in kg
+
+H2_content = 20 //in m3
+
+steam_consumed = H2_content*(18/22.4);
+
+C_consumption1 = (12/18)*steam_consumed; //in reaction 1
+C_consumption2 = (24/32)*O2_content; //in reaction 2
+
+total_consumption = C_consumption1+C_consumption2;
+coal_consumption = total_consumption/w_C;
+
+printf("coal consumption = %f kg",coal_consumption)
diff --git a/858/CH3/EX3.10/example_10.sce b/858/CH3/EX3.10/example_10.sce new file mode 100755 index 000000000..06a1e7a48 --- /dev/null +++ b/858/CH3/EX3.10/example_10.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.10 page number 98\n\n")
+
+//to find the amount of water evaporated
+xf = 0.15;
+xl = (114.7)/(114.7+1000);
+xc = 1;
+
+K2Cr2O7_feed = 1000*0.15; //in kg
+
+n = 0.8;
+C = n*K2Cr2O7_feed;
+V = (K2Cr2O7_feed-120 - 880*0.103)/(-0.103);
+
+printf("amount of water evaporated = %f kg",V)
diff --git a/858/CH3/EX3.11/example_11.sce b/858/CH3/EX3.11/example_11.sce new file mode 100755 index 000000000..bbd396719 --- /dev/null +++ b/858/CH3/EX3.11/example_11.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 3.10 page number 98\n\n")
+
+//to find the yield of crystals
+
+xc = 106/286;
+xf = 0.25;
+xl = 27.5/127.5;
+
+water_present = 100*(1-xf); //in kg
+V = 0.15*75; //in kg
+C = (100*xf - 88.7*xl)/(xc-xl);
+Na2CO3_feed = 25/xc;
+
+yield = (C/Na2CO3_feed)*100;
+
+printf("yield = %f ",yield)
diff --git a/858/CH3/EX3.12/example_12.sce b/858/CH3/EX3.12/example_12.sce new file mode 100755 index 000000000..acb84650c --- /dev/null +++ b/858/CH3/EX3.12/example_12.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.12 page number 99\n\n")
+
+//to find the fraction of air recirculated
+
+r = 50 //weight of dry air passing through drier
+w1 = 1.60 //in kg per kg dry solid
+w2 = 0.1 //in kg/kg dry solid
+H0 = 0.016 //in kg water vapor/kg dry air
+H2 = 0.055 //in kg water vapor/kg dry air
+
+y = 1 - (w1-w2)/(r*(H2-H0));
+printf("fraction of air recirculated = %f",y)
+
+H1 = H2 - (w1-w2)/r;
+printf("\n\nhumidity of air entering the drier = %f kg water vapor/kg kg dry air",H1)
+
+//check
+H11 = H2*y+H0*(1-y);
+if H1 == H11 then printf("\n\nfraction of air recirculated = %f \n verified",y)
+end
diff --git a/858/CH3/EX3.13/example_13.sce b/858/CH3/EX3.13/example_13.sce new file mode 100755 index 000000000..6b2c194a0 --- /dev/null +++ b/858/CH3/EX3.13/example_13.sce @@ -0,0 +1,28 @@ +clc
+clear
+printf("example 3.13 page number 100\n\n")
+
+//to find the volumetric flow rate and fraction of air passing through the cooler
+
+//basis 60m3/h of conditioned air at 25 degree C and 60% RH
+
+Hf = 0.012;
+Hi = 0.033;
+H1 = 0.0075;
+
+water_vapor = Hf/18; //in kmol of water vapor
+dry_air = 1/28.9; //in kmol
+total_mass = water_vapor+dry_air;
+
+volume = 22.4*(298/273)*total_mass;
+weight = 60/volume;
+printf("weight of dry air handled per hr = %f kg",weight)
+
+//part 1
+inlet_watervapor = 0.033/18; //in kmol of water vapor
+volume_inlet = 22.4*(308/273)*(inlet_watervapor+dry_air);
+printf("\n\nvolumetric flow rate of inlet air = %f cubic meter",volume_inlet*weight)
+
+//part 2
+y = (Hf - Hi)/(H1 - Hi);
+printf("\n\nfraction of inlet air passing through cooler = %f",y)
diff --git a/858/CH3/EX3.14/example_14.sce b/858/CH3/EX3.14/example_14.sce new file mode 100755 index 000000000..439a75167 --- /dev/null +++ b/858/CH3/EX3.14/example_14.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.14 page number 102\n\n")
+
+//to find the fraction of purged recycle and total yield
+
+//x- moles of N2 and H2 recycled; y - moles of N2 H2 purged
+
+Ar_freshfeed = 0.2;
+//argon in fresh feed is equal to argon in purge
+
+y = 0.2/0.0633; //argon in purge = 0.0633y
+x = (0.79*100 - y)/(1-0.79);
+printf("y = %f kmol\nx = %f kmol",y,x)
+
+//part 1
+fraction = y/x;
+printf("\n\nfration of recycle that is purged = %f",fraction)
+
+//part 2
+yield = 0.105*(100+x);
+printf("\n\noverall yield of ammonia = %f kmol",yield)
diff --git a/858/CH3/EX3.15/example_15.sce b/858/CH3/EX3.15/example_15.sce new file mode 100755 index 000000000..c83ccebf4 --- /dev/null +++ b/858/CH3/EX3.15/example_15.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 3.15 page number 107\n\n")
+
+//to find change in enthalpy
+H0_CH4 = -74.9 //in kJ
+H0_CO2 = -393.5 //in kJ
+H0_H2O = -241.8 //in kJ
+
+delta_H0 = H0_CO2+2*H0_H2O-H0_CH4;
+printf("change in enthalpy = %f kJ",delta_H0)
diff --git a/858/CH3/EX3.16/example_16.sce b/858/CH3/EX3.16/example_16.sce new file mode 100755 index 000000000..372f5cac7 --- /dev/null +++ b/858/CH3/EX3.16/example_16.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.16 page number 107\n\n")
+
+//to compare the enthalpy change in two reactions
+
+H0_glucose = -1273 //in kJ
+H0_ethanol = -277.6 //in kJ
+H0_CO2 = -393.5 //in kJ
+H0_H2O = -285.8 //in kJ
+
+//for reaction 1
+delta_H1 = 2*H0_ethanol+2*H0_CO2-H0_glucose;
+printf("enthalpy change in reaction 1 = %f KJ",delta_H1)
+
+//for reaction 2
+delta_H2 = 6*H0_H2O+6*H0_CO2-H0_glucose;
+printf("\n\nenthalpy change in reaction 2 = %f kJ",delta_H2)
+
+if delta_H1>delta_H2 then disp ("reaction 2 supplies more energy")
+ else disp ("reaction 1 supplies more energy")
+end
diff --git a/858/CH3/EX3.17/example_17.sce b/858/CH3/EX3.17/example_17.sce new file mode 100755 index 000000000..18eb0c538 --- /dev/null +++ b/858/CH3/EX3.17/example_17.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 3.17 page number 108\n\n")
+
+//to find enthalpy of formation of CuSO4.5H2O
+
+delta_H2 = 11.7 //in kJ/mol
+m_CuSO4 = 16 //in gm
+m_H2O = 384 //in gm
+
+delta_H3 = -((m_CuSO4+m_H2O)*4.18*3.95*159.6)/(16*10^3)
+delta_H1 = delta_H3 - delta_H2;
+
+printf("enthalpy of formation = %f kJ/mol",delta_H1)
diff --git a/858/CH3/EX3.18/example_18.sce b/858/CH3/EX3.18/example_18.sce new file mode 100755 index 000000000..37db2eb07 --- /dev/null +++ b/858/CH3/EX3.18/example_18.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 3.18 page number 108\n\n")
+
+//to find the temperature of combustion
+
+H_combustion = 1560000 //in kJ/kmol
+H0_CO2 = 54.56 //in kJ/kmol
+H0_O2 = 35.2 //in kJ/kmol
+H0_steam = 43.38 //in kJ/kmol
+H0_N2 = 33.32 //in kJ/kmol
+
+t = H_combustion/(2*H0_CO2+3*H0_steam+0.875*H0_O2+16.46*H0_N2);
+
+printf("theoritical temperature of combustion = %f degree C",t)
diff --git a/858/CH3/EX3.19/example_19.sce b/858/CH3/EX3.19/example_19.sce new file mode 100755 index 000000000..59cde39aa --- /dev/null +++ b/858/CH3/EX3.19/example_19.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 3.19 page number 109\n\n")
+
+//to find the heat of reaction and consumption of coke
+
+H_NaCl = 410.9 //in MJ/kmol
+H_H2SO4 = 811.3 //in MJ/kmol
+H_Na2SO4 = 1384 //in MJ/kmol
+H_HCl = 92.3 //in MJ/kmol
+
+Q = H_Na2SO4 + 2*H_HCl -2*H_NaCl-H_H2SO4;
+printf("heat of reaction = %f MJ\n\n",Q)
+
+heat_required = 64.5*(500/73);
+coke_consumption = heat_required/19
+printf("amount of coke oven gas consumed = %f cubic meter",coke_consumption)
diff --git a/858/CH3/EX3.2/example_2.sce b/858/CH3/EX3.2/example_2.sce new file mode 100755 index 000000000..11d7be1d1 --- /dev/null +++ b/858/CH3/EX3.2/example_2.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.2 page number 91\n\n")
+
+//to find amount of ammonia and air consumed
+
+NH3_required = (17/63)*1000; //NH3 required for 1 ton of nitric acid
+NO_consumption = 0.96;
+HNO3_consumption = 0.92;
+NH3_consumed = NH3_required/(NO_consumption*HNO3_consumption);
+volume_NH3 = NH3_consumed*(22.4/17);
+printf("volume of ammonia consumed= %f cubic metre/h",volume_NH3)
+
+NH3_content = 11 //% by volume
+air_consumption = volume_NH3*((100-11)/11);
+printf("\n\nvolume of air consumed = %f cubic metre/h",air_consumption)
diff --git a/858/CH3/EX3.20/example_20.sce b/858/CH3/EX3.20/example_20.sce new file mode 100755 index 000000000..804cafb66 --- /dev/null +++ b/858/CH3/EX3.20/example_20.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.20 page number 109\n\n")
+
+//to find the rate of heat flow
+
+cp_water = 146.5 //in kj/kg
+cp_steam = 3040 //in kJ/kg
+d = 0.102 //in m
+u = 1.5 //in m/s
+density = 1000 //in kg/m3
+
+m = (3.14/4)*d^2*u*density;
+Q = m*(cp_steam-cp_water);
+
+printf("rate of heat flow = %f kW",Q)
diff --git a/858/CH3/EX3.21/example_21.sce b/858/CH3/EX3.21/example_21.sce new file mode 100755 index 000000000..67fc2f755 --- /dev/null +++ b/858/CH3/EX3.21/example_21.sce @@ -0,0 +1,4 @@ +clc
+//EXAMPLE 3.21
+//To find the calorific value of coal
+disp('this is a theoritical problem.Refer the book for solution')
diff --git a/858/CH3/EX3.22/example_22.sce b/858/CH3/EX3.22/example_22.sce new file mode 100755 index 000000000..893cbe949 --- /dev/null +++ b/858/CH3/EX3.22/example_22.sce @@ -0,0 +1,48 @@ +clc
+clear
+printf("example 3.22 page number 110\n\n")
+
+//to find the amount of air required for combustion and composition of flue gas
+wt_C = 0.75 //in kg
+wt_H2 = 0.05 //in kg
+wt_O2 = 0.12 //in kg
+wt_N2 = 0.03 //in kg
+wt_S = 0.01 //in kg
+wt_ash = 0.04 //in kg
+
+O2_C = wt_C*(32/12); //in kg
+O2_H2 = wt_H2*(16/2); //in kg
+O2_S = wt_S*(32/32); //in kg
+O2_required = O2_C+O2_H2+O2_S;
+
+oxygen_supplied = O2_required - wt_O2;
+air_needed = oxygen_supplied/0.23;
+printf("amount of air required = %f kg",air_needed)
+
+volume = (22.4/28.8)*air_needed;
+printf("\n\nvolume of air needed = %f cubic meter",volume)
+
+air_supplied = 1.20*air_needed;
+N2_supplied = air_supplied*0.77;
+total_N2 = N2_supplied+wt_N2;
+
+O2_fluegas = air_supplied*0.23 - oxygen_supplied;
+
+wt_CO2 = wt_C+O2_C;
+wt_SO2 = wt_S+O2_S;
+
+moles_CO2 = wt_CO2/44;
+moles_SO2 = wt_SO2/64;
+moles_N2 = total_N2/28;
+moles_O2 = O2_fluegas/32;
+total_moles = moles_CO2+moles_SO2+moles_N2+moles_O2;
+
+x_CO2 = moles_CO2/total_moles;
+x_SO2 = moles_SO2/total_moles;
+x_N2 = moles_N2/total_moles;
+x_O2 = moles_O2/total_moles;
+
+printf("\n\nCO2 = %f",x_CO2*100)
+printf("\n\nSO2 = %f",x_SO2*100)
+printf("\n\nN2 = %f",x_N2*100)
+printf("\n\nO2 = %f",x_O2*100)
diff --git a/858/CH3/EX3.23/example_23.sce b/858/CH3/EX3.23/example_23.sce new file mode 100755 index 000000000..8dbd593fe --- /dev/null +++ b/858/CH3/EX3.23/example_23.sce @@ -0,0 +1,73 @@ +clc
+clear
+printf("example 3.23 page number 110\n\n")
+
+//to find the composition of flue gas
+
+C = 0.8 //in kg
+H2 = 0.05 //in kg
+S = 0.005 //in kg
+ash = 0.145 //in kg
+
+//required oxygen in kg
+C_O2 = C*(32/12);
+H2_O2 = H2*(16/2);
+S_O2 = S*(32/32);
+O2_supplied = C_O2+S_O2+H2_O2;
+printf("amount of O2 supplied = %f kg\n\n",O2_supplied)
+
+wt_air = O2_supplied*(100/23);
+wt_airsupplied = 1.25*wt_air;
+printf("amount of air supplied = %f kg\n\n",wt_airsupplied)
+
+//flue gas composition
+m_N2 = wt_airsupplied*0.77; //in kg
+mole_N2 = m_N2/28;
+
+m_O2 = (wt_airsupplied-wt_air)*0.23; //in kg
+mole_O2 = m_O2/32;
+
+m_CO2 = C*(44/12); //in kg
+mole_CO2 = m_CO2/44;
+
+m_H2O = H2*(18/2); //in kg
+mole_H2O = m_H2O/18;
+
+m_SO2 = S*(64/32); //in kg
+mole_SO2 = m_SO2/64;
+
+m = m_N2+m_O2+m_CO2+m_H2O+m_SO2
+
+//percent by weight
+w_N2 = m_N2/m;
+printf("percentage of N2 by weight = %f\n\n",w_N2*100)
+
+w_O2 = m_O2/m;
+printf("percentage of O2 by weight = %f\n\n",w_O2*100)
+
+w_CO2 = m_CO2/m;
+printf("percentage of CO2 by weight = %f\n\n",w_CO2*100)
+
+w_H2O = m_H2O/m;
+printf("percentage of H2O by weight = %f\n\n",w_H2O*100)
+
+w_SO2 = m_SO2/m;
+printf("percentage of SO2 by weight = %f\n\n",w_SO2*100)
+
+m1 = mole_N2+mole_O2+mole_CO2+mole_H2O+mole_SO2
+
+//percent by mole
+x_N2 = mole_N2/m1;
+printf("percentage of N2 by mole = %f\n\n",x_N2*100)
+
+x_O2 = mole_O2/m1;
+printf("percentage of O2 by mole = %f\n\n",x_O2*100)
+
+x_CO2 = mole_CO2/m1;
+printf("percentage of CO2 by mole = %f\n\n",x_CO2*100)
+
+x_H2O = mole_H2O/m1;
+printf("percentage of H2O by mole = %f\n\n",x_H2O*100)
+
+x_SO2 = mole_SO2/m1;
+printf("percentage of SO2 by mole = %f\n\n",x_SO2*100)
diff --git a/858/CH3/EX3.24/example_24.sce b/858/CH3/EX3.24/example_24.sce new file mode 100755 index 000000000..d336acfef --- /dev/null +++ b/858/CH3/EX3.24/example_24.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 3.24 page number 112\n\n")
+
+//to find volumetric composition of flue glass
+
+wt_H2 = 0.15;
+wt_C = 0.85;
+O2_H2 = wt_H2*(16/2);
+O2_C = wt_C*(32/12);
+
+total_O2 = O2_H2+O2_C;
+
+wt_air = total_O2/0.23;
+
+air_supplied = 1.15*(wt_air);
+N2_supplied = 0.77*air_supplied/28;
+O2_supplied = 0.23*(air_supplied-wt_air)/32;
+moles_CO2 = 0.85/12;
+
+printf("moles of CO2 = %f kmol\n\n",moles_CO2)
+printf("moles of N2 = %f kmol \n\n",N2_supplied)
+printf("moles of O2 = %f kmol\n\n",O2_supplied)
+
+total_moles = N2_supplied+O2_supplied+moles_CO2;
+
+printf("percentage of CO2 = %f\n\n",(moles_CO2/total_moles)*100)
+printf("percentage of N2 = %f\n\n",(N2_supplied/total_moles)*100)
+printf("percentage of O2 = %f",(O2_supplied/total_moles)*100)
diff --git a/858/CH3/EX3.25/example_25.sce b/858/CH3/EX3.25/example_25.sce new file mode 100755 index 000000000..4882ba72b --- /dev/null +++ b/858/CH3/EX3.25/example_25.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 3.25 page number 113\n\n")
+
+//to find the excess air supplied
+
+N2 = 80.5 //in m3
+air_supplied = N2/0.79 //in m3
+volume_O2 = air_supplied*0.21; //in m3
+O2_fluegas = 6.1 //in m3
+
+O2_used = volume_O2 - O2_fluegas;
+excess_air_supplied = (O2_fluegas/O2_used)*100;
+
+printf("percentage of excess air supplied = %f",excess_air_supplied)
diff --git a/858/CH3/EX3.26/example_26.sce b/858/CH3/EX3.26/example_26.sce new file mode 100755 index 000000000..f5fdcf056 --- /dev/null +++ b/858/CH3/EX3.26/example_26.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 3.26 page number 114\n\n")
+
+//to find the outlet temperature of water
+
+q_NTP = 10*(200/101.3)*(273/313);
+m_CO2 = 44*(q_NTP/22.4);
+s_CO2 = 0.85 //in kJ/kg K
+
+Q = m_CO2*s_CO2*(40-20) //Q = ms*delta_T
+
+d0 = 0.023 //in mm
+A0 = (3.14/4)*d0^2;
+di = 0.035 //in mm
+Ai = (3.14/4)*di^2;
+
+A_annular = Ai-A0;
+u = 0.15 //in m/s
+m_water = A_annular*(u*3600)*1000 //in kg/hr
+
+s_water = 4.19 //in kJ/kg K
+t = 15+(Q/(m_water*s_water));
+
+printf("exit water temperature = %f degree C",t)
diff --git a/858/CH3/EX3.27/example_27.sce b/858/CH3/EX3.27/example_27.sce new file mode 100755 index 000000000..e99d03ca8 --- /dev/null +++ b/858/CH3/EX3.27/example_27.sce @@ -0,0 +1,38 @@ +clc
+clear
+printf("example 3.27 page number 114\n\n")
+
+//to find the area of heating surface
+F = 1000 //in kg
+xF = 0.01
+
+solid_feed = F*xF;
+water_feed = F - solid_feed;
+
+tF = 40 //in degree C
+hF = 167.5 //in kJ/kg
+xL = 0.02;
+
+solid_liquor = 10 //in kg
+L = solid_liquor/xL;
+tL = 100 //in degree C
+hL = 418.6 //in kJ/kg
+
+V = F -L;
+
+tv = 100 //in degree C
+Hv = 2675 //in kJ/kg
+ts = 108.4 //in degree C
+Hs = 2690 //in kJ/kg
+tc = 108.4 //in degree C
+hc = 454 //in kJ/kg
+
+//applying heat balance
+S = (F*hF-V*Hv-L*hL)/(hc-Hs);
+printf("weight of steam required = %f kg/hr",S)
+
+Q = S*(Hs-hc);
+U = 1.4 //in kW/m2K
+delta_t = ts-tL;
+A = 383.2/(U*delta_t);
+printf("\n\narea of heating surface = %f square meter",A)
diff --git a/858/CH3/EX3.28/example_28.sce b/858/CH3/EX3.28/example_28.sce new file mode 100755 index 000000000..26b253515 --- /dev/null +++ b/858/CH3/EX3.28/example_28.sce @@ -0,0 +1,32 @@ +clc
+clear
+printf("example 3.28 page number 115\n\n")
+
+//to find the top and bottom product,condenser duty,heat input to rebpoiler
+hF = 171 //in kJ/kg
+hD = 67 //in kJ/kg
+hL = hD;
+
+hW = 200 //in kJ/kg
+H = 540 //in kJ/kg
+
+disp('part 1')
+F = 1000 //in kg/h
+xF = 0.40
+xW = 0.02;
+xD = 0.97;
+D = F*(xF-xW)/(xD-xW);
+W = F-D;
+
+printf("bottom product = %f kg/hr",W)
+printf("\ntop product = %f kg/hr\n\n",D)
+
+disp('part 2')
+L = 3.5*D;
+V = L+D;
+Qc = V*H-L*hL-D*hD;
+printf("condenser duty = %f KJ/hr\n\n",Qc)
+
+disp('part 3')
+Qr = Qc - 24200;
+printf("rate of heat input to reboiler = %f kJ/hr",Qr)
diff --git a/858/CH3/EX3.29/example_29.sce b/858/CH3/EX3.29/example_29.sce new file mode 100755 index 000000000..9315a25b5 --- /dev/null +++ b/858/CH3/EX3.29/example_29.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 3.29 page number 117\n\n")
+
+//to find the rate of crystal formation, cooling water rate, required area
+
+F = 1000; //in kg
+V = 0.05*F; //in kg
+xF = 0.48;
+xL = 75/(100+75);
+xC = 1;
+C = (F*xF-950*xL)/(1-0.429);
+printf ("rate of crystal formation = %f kg",C)
+
+L = F-C-V;
+
+//cooling water
+W = (F*2.97*(85-35)+126.9*75.2-V*2414)/(4.19*11);
+printf("\n\nrate of cooling water = %f kg",W)
+
+delta_T1 = 56;
+delta_T2 = 17;
+delta_Tm = (delta_T1-delta_T2)/(log(delta_T1/delta_T2))
+U = 125;
+
+A=(F*2.97*(85-35)+126.9*75.2-V*2414)/(U*delta_Tm*3.6);
+printf("\n\narea = %f square meter",A)
diff --git a/858/CH3/EX3.3/example_3.sce b/858/CH3/EX3.3/example_3.sce new file mode 100755 index 000000000..780428bac --- /dev/null +++ b/858/CH3/EX3.3/example_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 3.3 page number 91\n\n")
+
+//to find the consumption of NaCl and H2SO4 in HCl consumption
+
+HCl_production = 500 //required to be produced in kg
+NaCl_required = (117/73)*HCl_production;
+yield = 0.92;
+purity_NaCl= 0.96;
+
+actual_NaCl = NaCl_required/(purity_NaCl*yield);
+printf("amount of NaCl required = %f kg",actual_NaCl)
+
+purity_H2SO4 = 0.93;
+H2SO4_consumption = (98/73)*(HCl_production/(yield*purity_H2SO4));
+printf("\n\namount of H2SO4 consumed = %f kg",H2SO4_consumption)
+
+Na2SO4_produced = (142/73)*HCl_production;
+printf("\n\namount of Na2SO4 produced = %f kg",Na2SO4_produced)
diff --git a/858/CH3/EX3.30/example_30.sce b/858/CH3/EX3.30/example_30.sce new file mode 100755 index 000000000..5ae626691 --- /dev/null +++ b/858/CH3/EX3.30/example_30.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 3.30 page number 118\n\n")
+
+//to find the heat of combustion
+
+delta_n = 10-12; //mole per mole napthanlene
+
+//basis 1g
+moles_napthalene = (1/128);
+
+disp('part 1')
+Qv = 40.28 //in kJ
+Qp = Qv-(delta_n*moles_napthalene*8.3144*298/1000);
+printf("heat of combustion = %f kJ\n\n",Qp)
+
+disp('part 2')
+delta_H = 44.05 //in kJ/gmol
+water_formed = 4/128; //in g mol
+Qp1 = Qp - (delta_H*water_formed);
+printf("heat of combustion = %f kJ",Qp1)
diff --git a/858/CH3/EX3.4/example_4.sce b/858/CH3/EX3.4/example_4.sce new file mode 100755 index 000000000..1a0fb8293 --- /dev/null +++ b/858/CH3/EX3.4/example_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 3.4 page number 92\n\n")
+
+//to find the period of service
+
+C2H2_produced = (1/64)*0.86; //in kmol
+volume_C2H2 = C2H2_produced*22.4*1000; //in l
+
+//assuming ideal behaviour,
+volume = (100/101.3)*(273/(273+30));
+time = (volume_C2H2/volume)*(1/60);
+printf("time of service = %f hr",time)
+
+
+
+
diff --git a/858/CH3/EX3.5/example_5.sce b/858/CH3/EX3.5/example_5.sce new file mode 100755 index 000000000..6f284f6a0 --- /dev/null +++ b/858/CH3/EX3.5/example_5.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 3.5 page number 92\n\n")
+
+//to find the screen effectiveness
+
+xv = 0.88;
+xf = 0.46;
+xl = 0.32;
+F= 100 //in kg
+
+L = (F*(xf-xv))/(xl-xv);
+V = F-L;
+printf("L = %f Kg \nV = %f Kg",L,V)
+Eo = (V*xv)/(F*xf);
+
+printf(" \n\neffectiveness based on oversized partices = %f \n\n",Eo)
+Eu = (L*(1-xl))/(F*(1-xf));
+
+printf("effectiveness based on undersized partices = %f",Eu)
+E=Eu*Eo;
+
+printf("\n\noverall effectiveness = %f",E)
diff --git a/858/CH3/EX3.6/example_6.sce b/858/CH3/EX3.6/example_6.sce new file mode 100755 index 000000000..d6085e1ef --- /dev/null +++ b/858/CH3/EX3.6/example_6.sce @@ -0,0 +1,41 @@ +clc
+clear
+printf("example 3.6 page number 94\n\n")
+
+//to find the flow rate and concentration
+
+G1 = 3600 //in m3/h
+P = 106.6 //in kPa
+T = 40 //in degree C
+q = G1*(P/101.3)*(273/((273+T))); //in m3/s
+m = q/22.4; //in kmol/h
+y1 = 0.02;
+Y1 = y1/(1-y1);
+
+printf("mole ratio of benzene = %f kmol benzene/kmol dry gas",Y1)
+
+Gs = m*(1-y1);
+printf("\n\nmoles of benzene free gas = %f kmol drygas/h",Gs)
+
+//for 95% removal
+Y2 = Y1*(1-0.95);
+printf("\n\nfinal mole ratio of benzene = %f kmol benzene/kmol dry gas",Y2)
+
+x2 = 0.002
+X2 = 0.002/(1-0.002);
+
+//at equilibrium y* = 0.2406X
+//part 1
+//for oil rate to be minimum the wash oil leaving the absorber must be in equilibrium with the entering gas
+
+y1 = 0.02;
+x1 = y1/(0.2406);
+X1 = x1/(1-x1);
+min_Ls = Gs*((Y1-Y2)/(X1-X2));
+printf("\n\nminimum Ls required = %f kg/h",min_Ls*260)
+
+//for 1.5 times of the minimum
+Ls = 1.5*min_Ls;
+printf("\n\nflow rate of wash oil = %f kg/h",Ls*260)
+X1 = X2 + (Gs*((Y1-Y2)/Ls));
+printf("\n\nconcentration of benzene in wash oil = %f kmol benzene/kmol wash oil",X1)
diff --git a/858/CH3/EX3.7/example_7.sce b/858/CH3/EX3.7/example_7.sce new file mode 100755 index 000000000..2438d9d76 --- /dev/null +++ b/858/CH3/EX3.7/example_7.sce @@ -0,0 +1,49 @@ +clc
+clear
+printf("example 3.7 page number 95\n\n")
+
+//to find the extraction of nicotine
+xf = 0.01
+Xf = xf/(1-xf);
+Feed = 100 //feed in kg
+c_nicotine = Feed*Xf; //nicotine conc in feed
+c_water = Feed*(1-Xf) //water conc in feed
+
+//part 1
+function[f] = F1(x)
+ funcprot(0)
+ f = (x/150)-0.9*((1-x)/99);
+endfunction
+
+//initial guess
+x = 10;
+y = fsolve(x,F1);
+printf("amount of nicotine removed N = %f kg",y)
+//part 2
+function[f] = F1(x)
+ f = (x/50)-0.9*((1-x)/99);
+endfunction
+
+//initial guess
+x = 10;
+N1 = fsolve(x,F1);
+printf("\n\namount of nicotine removed in stage 1, N1 = %f kg",N1)
+function[f] = F1(x,N1)
+ f = (x/50)-0.9*((1-x-N1)/99);
+endfunction
+
+//initial guess
+x = 10;
+N2 = fsolve(x,F1);
+printf("\n\namount of nicotine removed in stage 2, N2 = %f kg",N2)
+function[f] = F1(x,N1,N2)
+ f = (x/50)-0.9*((1-x-N2-N1)/99);
+endfunction
+
+//initial guess
+x = 10;
+N3 = fsolve(x,F1);
+
+printf("\n\namount of nicotine removed in stage 3, N3 = %f kg",N3)
+N = N1+N2+N3;
+printf("\n\ntotal amount of nicotine removed = %f kg",N)
diff --git a/858/CH3/EX3.8/example_8.sce b/858/CH3/EX3.8/example_8.sce new file mode 100755 index 000000000..f6948ef03 --- /dev/null +++ b/858/CH3/EX3.8/example_8.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 3.8 page number 96\n\n")
+
+//to find the amount of water in residue
+
+vp_water = 31.06 //in kPa
+vp_benzene = 72.92 //in kPa
+
+P = vp_water +vp_benzene;
+x_benzene = vp_benzene/P;
+x_water = vp_water/P;
+
+initial_water = 50/18; //in kmol of water
+initial_benzene = 50/78 //in kmol of benzene
+water_evaporated = initial_benzene*(x_water/x_benzene);
+water_left = (initial_water - water_evaporated);
+
+printf("amount of water left in residue = %f kg",water_left*18)
diff --git a/858/CH3/EX3.9/example_9.sce b/858/CH3/EX3.9/example_9.sce new file mode 100755 index 000000000..4df8514ed --- /dev/null +++ b/858/CH3/EX3.9/example_9.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 3.9 page number 97\n\n")
+
+//to find the vapor content of dimethylanaline
+po_D = 4.93 //in kPa
+po_W = 96.3 //in kPa
+n = 0.75 //vaporization efficiency
+
+P = n*po_D+po_W;
+printf("P = %f kPa",P)
+
+x_water = 96.3/100;
+x_dimethylanaline = 1-x_water;
+wt_dimethylanaline = (x_dimethylanaline*121)/(x_dimethylanaline*121+x_water*18);
+printf("\n\nweight of dimethylanaline in water = %f",wt_dimethylanaline*100)
+
+//part 1
+n = 0.8;
+po_D = 32 //in kPa
+actual_vp = n*po_D;
+p_water = 100 - actual_vp;
+steam_required = (p_water*18)/(actual_vp*121);
+printf("\n\namount of steam required = %f kg steam/kg dimethylanaline",steam_required)
+
+//part 2
+x_water = p_water/100;
+wt_water = x_water*18/(x_water*18+(1-x_water)*121);
+printf("\n\nweight of water vapor = %f \nweight of dimethylanaline =%f",wt_water*100,100*(1-wt_water))
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