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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3050/CH2 | |
download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip |
initial commit / add all books
Diffstat (limited to '3050/CH2')
-rwxr-xr-x | 3050/CH2/EX2.1/Ex2_1.sce | 13 | ||||
-rwxr-xr-x | 3050/CH2/EX2.2/Ex2_2.sce | 13 | ||||
-rwxr-xr-x | 3050/CH2/EX2.3/Ex2_3.sce | 13 | ||||
-rwxr-xr-x | 3050/CH2/EX2.4/Ex2_4.sce | 13 | ||||
-rwxr-xr-x | 3050/CH2/EX2.5/Ex2_5.sce | 19 | ||||
-rwxr-xr-x | 3050/CH2/EX2.6/Ex2_6.sce | 11 | ||||
-rwxr-xr-x | 3050/CH2/EX2.7/Ex2_7.sce | 13 | ||||
-rwxr-xr-x | 3050/CH2/EX2.8/Ex2_8.sce | 8 |
8 files changed, 103 insertions, 0 deletions
diff --git a/3050/CH2/EX2.1/Ex2_1.sce b/3050/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..b9a3d641d --- /dev/null +++ b/3050/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,13 @@ +//calculating GCV and NCV
+//Example 2.1
+clc
+clear
+C=60//percentage of Carbon in coal
+O=33//percentage of Oxygen in coal
+H=6//percentage of Hydrogen in coal
+S=0.5//percentage of Sulphur in coal
+N=0.5//percentage of Nitrogen in coal
+GCV=((8080*C)+(34500*(H-O/8))+(2240*S))/100//gross calorific value in kcal/kg
+NCV=(GCV-(0.09*H*587))//net calorific value in kcal/kg
+printf('Thus the higher calorific value of coal = %4.2f kcal/kg',GCV)
+printf('\n and the lower calorific value of coal = %4.2f kcal/kg',NCV)
diff --git a/3050/CH2/EX2.2/Ex2_2.sce b/3050/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..83d68f918 --- /dev/null +++ b/3050/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,13 @@ +//calculating GCV and NCV
+//Example 2.2
+clc
+clear
+C=90//percentage of Carbon in coal
+O=2//percentage of Oxygen in coal
+H=4//percentage of Hydrogen in coal
+S=2.5//percentage of Sulphur in coal
+N=1//percentage of Nitrogen in coal
+GCV=((8080*C)+(34500*(H-O/8))+(2240*S))/100//gross calorific value in kcal/kg
+NCV=(GCV-(0.09*H*587))//net calorific value in kcal/kg
+printf('Thus the gross calorific value of coal = %4.2f kcal/kg',GCV)
+printf('\n and the net calorific value of coal = %4.2f kcal/kg',NCV)
diff --git a/3050/CH2/EX2.3/Ex2_3.sce b/3050/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..4358a5822 --- /dev/null +++ b/3050/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,13 @@ +//calculating GCV and NCV
+//Example 2.3
+clc
+clear
+C=80//percentage of Carbon in coal
+O=3//percentage of Oxygen in coal
+H=7//percentage of Hydrogen in coal
+S=3.5//percentage of Sulphur in coal
+N=2.1//percentage of Nitrogen in coal
+GCV=((8080*C)+(34500*(H-O/8))+(2240*S))/100//gross calorific value in kcal/kg
+NCV=(GCV-(0.09*H*587))//net calorific value in kcal/kg
+printf('Thus the gross calorific value of coal = %4.0f kcal/kg',GCV)
+printf('\n and the net calorific value of coal = %4.0f kcal/kg',NCV)
diff --git a/3050/CH2/EX2.4/Ex2_4.sce b/3050/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..7f91fef7d --- /dev/null +++ b/3050/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,13 @@ +//calculating GCV and NCV
+//Example 2.4
+clc
+clear
+C=84//percentage of Carbon in coal
+O=8.4//percentage of Oxygen in coal
+H=5.5//percentage of Hydrogen in coal
+S=1.5//percentage of Sulphur in coal
+N=0.6//percentage of Nitrogen in coal
+GCV=((8080*C)+(34500*(H-O/8))+(2240*S))/100//gross calorific value in kcal/kg
+NCV=(GCV-(0.09*H*587))//net calorific value in kcal/kg
+printf('Thus the gross calorific value of coal = %4.0f kcal/kg',GCV)
+printf('\n and the net calorific value of coal = %4.0f kcal/kg',NCV)
diff --git a/3050/CH2/EX2.5/Ex2_5.sce b/3050/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..80e290db0 --- /dev/null +++ b/3050/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,19 @@ +//calculating proximate analysis
+//Example 2.5
+clc
+clear
+m1=1//mass of air dried coal in g
+m2=0.985//mass of dry coal residue after heating for 1hr in g
+m3=0.8//mass of residue after heating for 7min in g
+m4=0.1//mass of last residue
+Mm=m1-m2//mass of moisture in coal sample in g
+Mv=m2-m3//mass of volatile matter in g
+Ma=m4//mass of ash
+%m=Mm*100//percentage moisture
+%v=Mv*100//percentage of volatile matter
+%a=Ma*100//percentage of ash
+%c=100-(%m+%v+%a)//percentage of fixed carbon
+printf('Thus (i)percentage of moisture = %2.1f percent\n',%m)
+printf('(ii)percentage of volatile matter = %2.1f percent\n',%v)
+printf('(iii)percentage of ash = %2.0f percent\n',%a)
+printf('(iv)percentage of fixed carbon = %2.0f percent \n',%c)
diff --git a/3050/CH2/EX2.6/Ex2_6.sce b/3050/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..e25aa0f07 --- /dev/null +++ b/3050/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,11 @@ +//calculating percentage C and H
+//Example 2.6
+clc
+clear
+wt1=2.75//increase in wt of KOH tube in gm
+wt2=0.45//increase in wt of CaCl2 tube in gm
+wt=1//weight of coal sample in gm
+%c=(wt1*12*100)/(wt*44)//percentage of carbon
+%h=(wt2*2*100)/(wt*18)//percentage of hydrogen
+printf('Thus (i)Percentage of carbon = %2.0f percent',%c)
+printf('\n(ii)Percentage of hydrogen =%2.0f percent',%h)
diff --git a/3050/CH2/EX2.7/Ex2_7.sce b/3050/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..b860aa2fe --- /dev/null +++ b/3050/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,13 @@ +//calculating percentage S and N
+//Example 2.7
+clc
+clear
+wt1=2.6//weight of coal taken for quantitative analysis in gm
+wt=1.56//weight of coal sample taken in gm
+v=50-6.25//volume of H2SO4 used
+N=0.1//normality
+m=0.1755//wt of BaSO4 ppt. obtained
+%n=(v*N*1.4)/(wt)//percentage of nitrogen
+%su=(m*32*100)/(wt1*233)//percentage of sulphur
+printf('Thus (i)Percentage of nitrogen = %2.3f percent',%n)
+printf('\n(ii)Percentage of sulphur =%2.3f percent',%su)
diff --git a/3050/CH2/EX2.8/Ex2_8.sce b/3050/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..9aead48de --- /dev/null +++ b/3050/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,8 @@ +//calculating percentage S
+//Example 2.8
+clc
+clear
+wt=0.5//weight of coal taken for quantitative analysis in gm
+m=0.05//wt of BaSO4 ppt. obtained
+%su=(m*32*100)/(wt*233)//percentage of sulphur
+printf('Thus Percentage of sulphur =%2.3f percent',%su)
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