Modified the code

6 files changed, 0 insertions, 229 deletions

diff --git a/2465/CH18/EX18.1/Ex18_1.sce b/2465/CH18/EX18.1/Ex18_1.sce
deleted file mode 100644
index d93962d69..000000000
--- a/2465/CH18/EX18.1/Ex18_1.sce
+++ /dev/null
@@ -1,27 +0,0 @@
-//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/Ex18_2.sce b/2465/CH18/EX18.2/Ex18_2.sce
deleted file mode 100644
index 76386308f..000000000
--- a/2465/CH18/EX18.2/Ex18_2.sce
+++ /dev/null
@@ -1,35 +0,0 @@
-//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/Ex18_3.sce b/2465/CH18/EX18.3/Ex18_3.sce
deleted file mode 100644
index b3e14c5d3..000000000
--- a/2465/CH18/EX18.3/Ex18_3.sce
+++ /dev/null
@@ -1,27 +0,0 @@
-//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/Ex18_4.sce b/2465/CH18/EX18.4/Ex18_4.sce
deleted file mode 100644
index ba8dab90c..000000000
--- a/2465/CH18/EX18.4/Ex18_4.sce
+++ /dev/null
@@ -1,17 +0,0 @@
-//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/Ex18_5.sce b/2465/CH18/EX18.5/Ex18_5.sce
deleted file mode 100644
index 811aae588..000000000
--- a/2465/CH18/EX18.5/Ex18_5.sce
+++ /dev/null
@@ -1,86 +0,0 @@
-//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/Ex18_6.sce b/2465/CH18/EX18.6/Ex18_6.sce
deleted file mode 100644
index 9441a7a58..000000000
--- a/2465/CH18/EX18.6/Ex18_6.sce
+++ /dev/null
@@ -1,37 +0,0 @@
-//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)