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-rwxr-xr-x2471/CH2/EX2.1/Ex2_1.sce40
-rwxr-xr-x2471/CH2/EX2.2/Ex2_2.sce40
-rwxr-xr-x2471/CH2/EX2.3/Ex2_3.sce21
-rwxr-xr-x2471/CH2/EX2.4/Ex2_4.sce22
-rwxr-xr-x2471/CH2/EX2.5/Ex2_5.sce20
-rwxr-xr-x2471/CH2/EX2.6/Ex2_6.sce31
6 files changed, 174 insertions, 0 deletions
diff --git a/2471/CH2/EX2.1/Ex2_1.sce b/2471/CH2/EX2.1/Ex2_1.sce
new file mode 100755
index 000000000..697c46a67
--- /dev/null
+++ b/2471/CH2/EX2.1/Ex2_1.sce
@@ -0,0 +1,40 @@
+clear ;
+clc;
+// Example 2.1
+printf('Example 2.1\n\n');
+printf('Page No. 44\n\n');
+
+// given
+C= 35000;// cost of boiler
+C_grant=.25;// Capital grant available from goverment
+E= -(C-(C_grant*C));// Net expenditure
+Fs= 15250;// Fuel Saving
+r_i = 0.15;// interest
+r_t = 0.55;// tax
+
+a = [0 E Fs 0 E+Fs r_i*(E+Fs) 0 ]
+bal_1 = a(5)+a(6)-a(7)// Total Balance after 1st year
+
+c_all = 0.55;// capital allowance in 2nd year
+C_bal= (bal_1+0+Fs+(-(c_all*E)));// Cash Balance after 2nd year
+b = [bal_1 0 Fs -(c_all*E) C_bal r_i*C_bal r_t*(Fs+(r_i*C_bal))];
+bal_2 = b(5)+b(6)-b(7)//Total Balance after 2nd year
+
+c = [bal_2 0 Fs 0 bal_2+Fs r_i*(bal_2+Fs) r_t*(Fs+(r_i*(bal_2+Fs)))]
+bal_3= c(5)+c(6)-c(7)// Total Balance after 3rd year
+
+if(bal_2>0) then
+ disp('Pay back period is of two year')
+else
+ disp('Pay back period is of three year')
+end
+
+printf('Total saving at the end of second year is %3.0f Pound\n',bal_2);
+printf('Total saving at the end of third year is %3.0f Pound\n',bal_3);
+// Deviation in answer due to direct substitution
+
+
+
+
+
+
diff --git a/2471/CH2/EX2.2/Ex2_2.sce b/2471/CH2/EX2.2/Ex2_2.sce
new file mode 100755
index 000000000..19147c661
--- /dev/null
+++ b/2471/CH2/EX2.2/Ex2_2.sce
@@ -0,0 +1,40 @@
+clear ;
+clc;
+// Example 2.2
+printf('Example 2.2\n\n');
+printf('Page No. 45\n\n');
+
+// given
+C= 35000;// cost of boiler
+C_grant=0;// Capital grant available from goverment
+E= -(C-(C_grant*C));// Net expenditure
+Fs= 15250;// Fuel Saving
+r_i = 0.15;// interest
+r_t = 0.55;// tax
+
+a = [0 E Fs 0 E+Fs r_i*(E+Fs) 0 ]
+bal_1 = a(5)+a(6)-a(7)// Total Balance after 1st year
+
+c_all = 0.55;// capital allowance in 2nd year
+C_bal= (bal_1+0+Fs+(-(c_all*E)));// Cash Balance after 2nd year
+b = [bal_1 0 Fs -(c_all*E) C_bal r_i*C_bal r_t*(Fs+(r_i*C_bal))];
+bal_2 = b(5)+b(6)-b(7)//Total Balance after 2nd year
+
+c = [bal_2 0 Fs 0 bal_2+Fs r_i*(bal_2+Fs) r_t*(Fs+(r_i*(bal_2+Fs)))]
+bal_3= c(5)+c(6)-c(7)// Total Balance after 3rd year
+
+if(bal_2>0) then
+ disp('pay back period is of two year')
+else
+ disp('pay back period is of three year')
+end
+
+printf('Total saving at the end of second year is %3.2f Pound\n',bal_2);
+printf('Total saving at the end of third year is %3.2f Pound\n',bal_3);
+// Deviation in answer due to direct substitution
+
+
+
+
+
+
diff --git a/2471/CH2/EX2.3/Ex2_3.sce b/2471/CH2/EX2.3/Ex2_3.sce
new file mode 100755
index 000000000..db3dc5c36
--- /dev/null
+++ b/2471/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,21 @@
+clear ;
+clc;
+// Example 2.3
+printf('Example 2.3\n\n');
+printf('Page No. 46\n\n');
+
+// given
+F= 350*10^3;// fuel oils in gallons
+Ci= 5000;// cost of insulation of tanks
+
+As= 7500;//Annual Saving in Pound
+
+if(As> Ci) then
+disp("The investment has a pay-back period of less than 1 year");
+else
+disp("The investment has not a pay-back period of less than 1 year");
+end
+// Note- Since here pack back period is less than 1 year and the company is in profit so they can go with this fuel oil,
+// although it can be noted that there are more problems handling heavy fuels oils
+//and that the pay-back increases considerably the smaller the installation.
+//So the company can changeover from oil to coal as a fuel.
diff --git a/2471/CH2/EX2.4/Ex2_4.sce b/2471/CH2/EX2.4/Ex2_4.sce
new file mode 100755
index 000000000..c5db16ac0
--- /dev/null
+++ b/2471/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,22 @@
+clear ;
+clc;
+// Example 2.4
+printf('Example 2.4\n\n');
+printf('Page No. 47\n\n');
+
+// given
+F1= 500*10^3;// fuel oil in gallons
+F2= 500*10^3;// coal in gallons in Pound
+C1= 165*10^3;// cost of oil per year in Pound
+C2= 92*10^3;// cost of an equivalent of coal in Pound
+Ce= 100*10^3;// capital cost of extra handling eqiupment
+
+Cm= (Ce*0.2);// Maintenance , interest costs per year
+As= C1-C2;// Annual Saving in Pound
+printf('Annual Saving is %3.0f Pound\n',As)
+
+if((2*As)> Ce) then
+disp("Replacing an obsolete boiler plant is considerable");
+else
+disp("Replacing an obsolete boiler plant is not considerble");
+end
diff --git a/2471/CH2/EX2.5/Ex2_5.sce b/2471/CH2/EX2.5/Ex2_5.sce
new file mode 100755
index 000000000..8427acf5f
--- /dev/null
+++ b/2471/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,20 @@
+clear ;
+clc;
+// Example 2.5
+printf('Example 2.5\n\n');
+printf('Page No. 49\n\n');
+
+// given
+F= 10*10^3;// fuel oils in gallons
+Cs= 2200;// cost of maintaining tanks per year in Pound
+Ci= 1850;// cost of insulation of pipe in Pound
+
+As= (Cs*.85);//company saving is 85 per cent to the cost
+printf('Annual Saving on heating is %3.0f Pound\n',As)
+
+
+if(As> Ci) then
+disp("The investment has a pay-back period of less than 1 year");
+else
+disp("The investment has not a pay-back period of less than 1 year");
+end
diff --git a/2471/CH2/EX2.6/Ex2_6.sce b/2471/CH2/EX2.6/Ex2_6.sce
new file mode 100755
index 000000000..4af45b3fc
--- /dev/null
+++ b/2471/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,31 @@
+clear ;
+clc;
+// Example 2.6
+printf('Example 2.6\n\n');
+printf('Page No. 52\n\n');
+
+// given
+P1= 50;// Dry saturated steam pressure in bar
+P2= 0.5;// condenser pressure in bar
+
+//By using the steam tables saturation temperature is obtained at given pressures
+T1= 537//The saturation temperatue in K at 50 bar
+T2= 306//The saturation temperatue in K at 0.5 bar
+
+// For Carnot Cycle
+n=(1-(T2/T1))*100;
+printf('Efficiency percentage of Carnot Cycle is %3.0f \n',n)
+
+
+// For Rankine Cycle
+// By usins steam tables, the total heat and the sensibles heat and other remaining parameter has been calculated
+h1= 2794*10^3;//the total heat in dry steam at 50 bar in J/kg
+d= 0.655;// dryness fraction
+h2= 1725*10^3;// the entropy at state 2 in J/kg
+h3= 138*10^3;// the sensible heat at 0.5 bar in J/kg
+Vf= 1.03*10^-3;// volume of fluid im m^3,calculated from steam table
+W= (Vf*(P1-P2))*10^5;// pump work in J/kg
+E=(((h1-h2)-(W))/((h1-h3)-(W)))*100;
+printf('Efficiency percentage of Rankine Cycle is %3.0f \n',E)
+
+