6 files changed, 229 insertions, 0 deletions
diff --git a/2465/CH18/EX18.1/Example_1.sce b/2465/CH18/EX18.1/Example_1.sce
new file mode 100644
index 000000000..d93962d69
--- /dev/null
+++ b/2465/CH18/EX18.1/Example_1.sce
@@ -0,0 +1,27 @@
+//Chapter-18,Example 1,Page 404
+clc();
+close();
+
+H=6
+
+W= 2200    //water equivalent of bomb calorimeter
+
+w= 550    //weight of water taken
+
+del_t = 2.42    //rise in temperature
+
+m= 0.92    //weight of coal burnt
+
+L =580    //latent heat of steam
+
+fuse = 10   //fuse correction
+
+acid =50    //acid correction
+
+HCV=((W+w)*(del_t)-(acid+fuse))/m
+
+NCV=HCV-(0.09*H*L)
+
+printf("HCV = %.2f cal/g",HCV)
+
+printf("\n NCV = %.2f cal/g",NCV)
diff --git a/2465/CH18/EX18.2/Example_2.sce b/2465/CH18/EX18.2/Example_2.sce
new file mode 100644
index 000000000..76386308f
--- /dev/null
+++ b/2465/CH18/EX18.2/Example_2.sce
@@ -0,0 +1,35 @@
+//Chapter-18,Example 2,Page 405
+clc();
+close();
+
+W1 = 2.5     //weight of coal
+
+W2 = 2.415   //weight of coal after heating at 110 C
+
+W_res= 1.528  //weight of residue
+
+W_ash= 0.245    //weight of ash
+
+Mois= W1-W2    //moisture in sample
+
+per_M=Mois*100/W1
+
+printf("the percentage of moisture is %.2f ",per_M )
+
+VCM=W2-W_res  //amount of VCM in sample
+
+per_VCM=VCM*100/W1
+
+printf("\n the percentage of VCM is %.2f ",per_VCM )
+
+per_ash=W_ash*100/W1
+
+printf("\n the percentage of ash is %.2f",per_ash )
+
+Fix_C= W_res-W_ash  //fixed carbon
+
+per_fix=Fix_C*100/W1
+
+printf("\n the percentage of fixed carbon is %.2f",per_fix )
+
+//mistake in textbook
diff --git a/2465/CH18/EX18.3/Example_3.sce b/2465/CH18/EX18.3/Example_3.sce
new file mode 100644
index 000000000..b3e14c5d3
--- /dev/null
+++ b/2465/CH18/EX18.3/Example_3.sce
@@ -0,0 +1,27 @@
+//Chapter-18,Example 3,Page 405
+clc();
+close();
+
+H=0.77
+
+W= 395    //water equivalent of bomb calorimeter
+
+w= 3500    //weight of water taken
+
+T1=26.5    //temperature
+
+T2=29.2    //temperature
+
+m= 0.83    //weight of fuel burnt
+
+L =587   //latent heat of steam
+
+HCV=((W+w)*(T2-T1))/m
+
+NCV=HCV-(0.09*H*L)
+
+printf("HCV = %.2f cal/g",HCV)
+
+printf("\n NCV = %.2f cal/g",NCV)
+
+//calculation mistake in textbook
diff --git a/2465/CH18/EX18.4/Example_4.sce b/2465/CH18/EX18.4/Example_4.sce
new file mode 100644
index 000000000..ba8dab90c
--- /dev/null
+++ b/2465/CH18/EX18.4/Example_4.sce
@@ -0,0 +1,17 @@
+//Chapter-18,Example 4,Page 406
+clc();
+close();
+
+V1= 25     //volume of H2SO4
+
+V2 =15     //volumeof NaOH
+
+v= V1*0.1-V2*0.1    //volume of H2SO4  consumed
+
+//100 cc H2SO4 ==17 g NH3 == 14 g N
+//1 cc H2SO4 = 14/1000 g N =0.014 g N
+//0.014 g N is present in 1 g coal
+
+N= 0.014*100
+
+printf("the percentage of nitrogen is %.2f ",N)
diff --git a/2465/CH18/EX18.5/Example_5.sce b/2465/CH18/EX18.5/Example_5.sce
new file mode 100644
index 000000000..811aae588
--- /dev/null
+++ b/2465/CH18/EX18.5/Example_5.sce
@@ -0,0 +1,86 @@
+//Chapter-18,Example 5,Page 406
+clc();
+close();
+
+
+H2 =0.24   //composition of H2
+
+CH4 =0.3   //composition of CH4
+
+CO =0.06   //composition of CO
+
+C2H6 =0.11   //composition of C2H6
+
+C2H4 =0.045   //composition of C2H4
+
+C4H8 =0.025   //composition of C4H8
+
+N2=0.12   //composition of N2
+
+CO2=0.08  //composition of CO2
+
+O2=0.02   //composition of O2
+
+//for reaction H2 + (1/2)O2 = H2O
+
+V1=H2*(1/2)   //volume of O2 required
+
+//for reaction CH4 + 2O2  = CO2 + 2H2O
+
+V2=CH4*2   //volume of O2 required
+vCO2_1=CH4*1   //volume of CO2
+
+//for reaction C2H6 + (7/2)O2 = 2CO2 +3H2O
+
+V3=C2H6*(7/2)   //volume of O2 required
+vCO2_2=C2H6*2   //volume of CO2
+
+//for reaction C2H4 + 3O2 = 2CO2 +2H2O
+
+V4=C2H4*3   //volume of O2 required
+vCO2_3=C2H4*2   //volume of CO2
+
+//for reaction C4H8 + 6O2 = 4CO2 +4H2O
+
+V5=C4H8*6   //volume of O2 required
+vCO2_4=C4H8*4   //volume of CO2
+
+//for reaction CO + (1/2)O2  = CO2
+
+V6=CO*(1/2)   //volume of O2 required
+vCO2_5=CO*1   //volume of CO2
+
+total_O2= V1+V2+V3+V4+V5+V6-O2  //total volume of oxygen
+
+//as air contains 21% of O2 by volume
+//when 40% excess
+
+V_air = total_O2*(100/21)*(140/100)   //volume of air 
+
+printf("the air to fuel ratio is %.3f",V_air)
+
+total_CO2 = vCO2_1+vCO2_2+vCO2_3+vCO2_4+vCO2_5+CO2  //total volume of CO2
+
+total_dry= total_CO2 +[N2+(79*V_air/100)]+[(V_air*21/100)-total_O2]
+
+printf("\n the total volume of dry products is %.4f cubicmeter ",total_dry)
+
+CO2_dry =total_CO2*100/total_dry
+
+N2_dry =[N2+(79*V_air/100)]*100/total_dry
+
+O2_dry =[(V_air*21/100)-total_O2]*100/total_dry
+
+printf("\n  Composition of products of combustion on dry basis")
+
+printf("\n  CO2 = %.3f",CO2_dry)
+
+printf("\n  N2 = %.3f",N2_dry)
+
+printf("\n  O2 = %.3f",O2_dry)
+
+//calculation mistake in textbook
+
+
+
+
diff --git a/2465/CH18/EX18.6/Example_6.sce b/2465/CH18/EX18.6/Example_6.sce
new file mode 100644
index 000000000..9441a7a58
--- /dev/null
+++ b/2465/CH18/EX18.6/Example_6.sce
@@ -0,0 +1,37 @@
+//Chapter-18,Example 6,Page 407
+clc();
+close();
+
+CO =0.46   //composition of CO
+
+CH4 =0.1   //composition of CH4
+
+H2 =0.4   //composition of H2
+
+C2H2 =0.02   //composition of C2H2
+
+N2=0.01   //composition of N2
+
+//for reaction CO + (1/2)O2  = CO2
+
+V1=CO*(1/2)   //volume of O2 required
+
+//for reaction CH4 + 2O2  = CO2 + 2H2O
+
+V2=CH4*2   //volume of O2 required
+
+//for reaction H2 + (1/2)O2 = H2O
+
+V3=H2*(1/2)   //volume of O2 required
+
+//for reaction C2H2 + (5/2)O2 = 2CO2 +H2O
+
+V4=C2H2*(5/2)   //volume of O2 required
+
+total_v= V1+V2+V3+V4
+
+//as air contains 21% of O2 by volume
+
+V_air = total_v*100/21   //volume of air 
+
+printf("the volume of air required is %.3f cubicmeter",V_air)