8 files changed, 195 insertions, 0 deletions
diff --git a/2471/CH7/EX7.1/Ex7_1.sce b/2471/CH7/EX7.1/Ex7_1.sce
new file mode 100755
index 000000000..5ef1f1b2e
--- /dev/null
+++ b/2471/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,15 @@
+clear ;
+clc;
+// Example 7.1
+printf('Example 7.1\n\n');
+printf('Page No. 201\n\n');
+
+// given
+C = 220*10^3;//Original annual cost of fuel in Pound
+O_E = 73;// Original Efficiency
+Fl_i = 20;// Initial Flue loss 
+Fl_f = 18.7;// Final Flue loss 
+N_E = O_E + (Fl_i - Fl_f);// New Efficiency
+F_save = C*((N_E-O_E)/N_E);
+printf('Fuel saving is %.0f Pound',F_save)
+//Deviation in answer is due to some wrong calculation the book, instead of new efficiency in the denominator in line 13, the book has taken original efficiency
diff --git a/2471/CH7/EX7.2/Ex7_2.sce b/2471/CH7/EX7.2/Ex7_2.sce
new file mode 100755
index 000000000..06c7cdb0e
--- /dev/null
+++ b/2471/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,39 @@
+clear ;
+clc;
+// Example 7.2
+printf('Example 7.2\n\n');
+printf('Page No. 201\n\n');
+
+//From Example 2.1
+
+// given
+C= 35000;// cost of boiler  in Pound
+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\n',bal_3);
+// Deviation in answer due to direct substitution
+
+
+printf('The data in example 2.1 indicated that:- \n Saving could be made by replacing exising oil-fired burners by new burners requiring considerably less atomising steam.\n The financial saving are 15.25*10^3 Pound per year for an insulation and capital cost of 35*10^3 Pound.')
diff --git a/2471/CH7/EX7.3/Ex7_3.sce b/2471/CH7/EX7.3/Ex7_3.sce
new file mode 100755
index 000000000..7a4339ac7
--- /dev/null
+++ b/2471/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,15 @@
+clear;
+clc;
+// Example 7.3
+printf('Example 7.3\n\n');
+printf('Page No. 203\n\n');
+
+// given
+C = 250*10^3;//Original annual cost of fuel in Pound
+O_E = 71.5;// Original Efficiency
+Fl_i = 20;// Initial Flue loss 
+Fl_f = 17.5;// Final Flue loss 
+N_E = O_E + (Fl_i - Fl_f);// New Efficiency
+F_save = C*((N_E-O_E)/N_E);// in Pound
+printf('Fuel saving is %.0f Pound per year',F_save)
+//Deviation in answer is due to some calculation approximation the book
diff --git a/2471/CH7/EX7.4/Ex7_4.sce b/2471/CH7/EX7.4/Ex7_4.sce
new file mode 100755
index 000000000..47c82fe52
--- /dev/null
+++ b/2471/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,10 @@
+clear;
+clc;
+// Example 7.4
+printf('Example 7.4\n\n');
+printf('Page No. 204\n\n');
+
+// This question doesnot contain any calculation part.
+//Refer figure 7.3, 7.4, 7.5
+T_max = 200;// Flue gas exit temperature in degree celcius
+printf(' The company investigate four alternative methods of heat abstraction using the flue gas.\n\n  System-1   The efficiency of the furnace without any air preheater is 79.2 per cent.\n  System-2   The efficiency of the furnace, with the air preheater only in the system operating as shown in figure 7.3, is increased to 86.6 per cent.\n  System-3   By the incorporation of the heat exchanger,the furnace efficiency is increased to 93.3 per cent using the arrangement shown in figure 7.4.\n  System-4   Using no preheating,finally achievied an overall thermal  efficiency of 93.7 per cent.\n  \t     The new air preheater scheme is shown in figure 7.5.\n\n The pay-back period in all instances is less than 3.5 years.')
diff --git a/2471/CH7/EX7.5/Ex7_5.sce b/2471/CH7/EX7.5/Ex7_5.sce
new file mode 100755
index 000000000..56b9c747f
--- /dev/null
+++ b/2471/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,37 @@
+clear ;
+clc;
+// Example 7.5
+printf('Example 7.5\n\n');
+printf('Page No. 205\n\n');
+
+//The temperature difference is not given the question.
+//Refer Table 7.1
+T1 = 1000;// Furnace operating temperature in degree celcius
+//T2 is  back calculated by the first condition given in table 7.1 and applying Fourier,s law of condition
+T2 = 997.9545;// in degree Celcius()
+dT = T1 - T2;// in degree celcius
+t = 120;// Continuous cycle time in h
+K1 = 44;// Thermal conductivity (W/m-K)
+K2 = 11;// Thermal conductivity (W/m-K)
+K3 = 4;// Thermal conductivity (W/m-K)
+x1 = 250*10^-3;// mm converted into m
+x2 = 50*10^-3;// mm converted into m
+dT = T1 - T2;// in K
+
+//By Fourier,s law of heat conduction- Q = (dT *K)/x in W/sq.m
+
+//For 250 mm firebrick
+Q1 = (dT *K1)/x1;// im W/sq.m
+printf('Energy losses by 250 mm firebrick is %.0f W/sq.m \n',Q1)
+
+//For 250 mm hot-face insulation
+Q2 = (dT *K2)/x1;// im W/sq.m
+printf('Energy losses by 250 mm hot-face insulation is %.0f W/sq.m \n',Q2) //Deviation in answer is due to assumption of T2 as its not mentioned in the question
+
+//For 250 mm hot-face insulation backed by 50 mm insulation
+//As the resistances are in series - R = (x1/K1)+ (x2/K2) and Q = dt/R in W/sq.m
+R = (x1/K2)+ (x2/K3);// in ohm
+Q3 = dT/R;// in W/sq.m
+printf('Energy losses by 250 mm hot-face insulation backed by 50 mm insulation is %.0f W/sq.m \n\n',Q3) //Deviation in answer is due to assumption of T2 as its not mentioned in the question
+
+
diff --git a/2471/CH7/EX7.6/Ex7_6.sce b/2471/CH7/EX7.6/Ex7_6.sce
new file mode 100755
index 000000000..b90f150bd
--- /dev/null
+++ b/2471/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,18 @@
+clear ;
+clc;
+// Example 7.6
+printf('Example 7.6\n\n');
+printf('Page No. 209\n\n');
+
+// given
+P = 150*10^3;// Power of compressor in W
+F_load = .78;// full load percentage of the time
+Re = .7;//  Heat Recovery
+T = 2200;//Compressor operating time in h/year
+C = 20*10^-6;// Energy cost in Pound/Wh
+
+H_Re = P*F_load*Re;// in W
+printf('Heat recovered is %.0f W \n',H_Re)
+E_save = H_Re*T*C;// in Pound/year
+printf('Economic Saving is %3.2f Pound per year',E_save)
+//Deviation in answer is due to some calculation approximation the book
diff --git a/2471/CH7/EX7.7/Ex7_7.sce b/2471/CH7/EX7.7/Ex7_7.sce
new file mode 100755
index 000000000..7dc53df3c
--- /dev/null
+++ b/2471/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,17 @@
+clear ;
+clc;
+// Example 7.7
+printf('Example 7.7\n\n');
+printf('Page No. 212\n\n');
+
+// given
+C_S = 1/10^3;// Cost of steam production in p/Wh
+P = 75*10^3;// Power required in W
+T = 4*10^3;// Production time in h/year
+C_T = 7*10^3;// Cost of turbine in Pound
+R_T = 4*10^3;// Annual running cost of turbine in W
+C_M = 1.5*10^3;// Cost of electric motor in Pound
+R_M = 14*10^3;// Running cost of electric motor in Pound
+C_M_A = 3.5/10^3;// Auunal running cost of electic motor in p/Wh
+Save_R = R_M - R_T;// in Pound per year
+printf('The saving in running costs would be %3.1e Pound per year',Save_R)
diff --git a/2471/CH7/EX7.8/Ex7_8.sce b/2471/CH7/EX7.8/Ex7_8.sce
new file mode 100755
index 000000000..8f23af7ad
--- /dev/null
+++ b/2471/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,44 @@
+clear ;
+clc;
+// Example 7.8
+printf('Example 7.8\n\n');
+printf('Page No. 214\n\n');
+
+// given
+m_s = 5.3;// Factory requirement of process steam in kg/s
+Pr_s_1 = 2.5;// Pressure of process steam at bar absolute
+E_load_1 = 1.10*10^3;// Electrical load requirement in W
+E_load_2 = 1.5*10^3;// Electrical load requirement in W
+m_e = 6.0;// Mass flow rate of generated electricity in kg/s
+Pr_e = 14;// Pressure of generated electricity at bar absolute
+T_heat = 2.790*10^6;// Total heat content in J/kg
+
+//The 14 bar absolute steam would undergo an adiabatic heat drop and the steam will be expanded
+h_drop = 306*10^3;// Adiabatic heat drop in J/kg
+Pr_2 = 2.5;// Expanded pressure at bar absolute
+Ex_stm = 0.11;// Exhaust steam percent
+Ef_T = 0.65;// Tubine efficiency
+R_h_drop = h_drop * Ef_T;// Real heat drop in J/kg
+P_T = m_e * R_h_drop;// Power generated by turbine in W
+Ef_G = 0.94;// Generator efficiency
+P_G = 1.13*10^6;// Output of generator in W
+
+//(a) Combined heat and power system
+Eq_Eva = 8;// Equivalent evaporation of steam per kg coal in kg
+C_req = m_e/Eq_Eva;// in kg/s
+printf('Coal Required is %.2f kg/s\n',C_req)
+printf('If the plant operates on a 140-h week for 50 weeks per annum the coal consumption is 18.9*0^6 kg per year.\nAt an average price of, for example, 35 Pound per tonne, the aanual cost is 660*10^3 Pound.\n\n')
+
+//(b) Coal required for process steam
+// for low pressure steam
+Eq_Eva_2 = 8.25;// Equivalent evaporation of steam per kg coal in kg
+Coal_req = m_s/Eq_Eva_2;// in kg/s
+printf('Coal Required is %.3f kg/s\n\n',Coal_req)
+printf('Assuming similar operating conditions for the plant the total coal consumption is 16.2*10^6 kg per year,\nand the annual cost is 556*10^6 Pound.\n')
+
+//(c)Electrical Power
+printf('The cost of 1.15*10^6 W of electricity for the same period of time is,assuming a cost of 23 Pound per 10^6 Wh,177*10^3Pound.\nThe coal equivalent to generate 1.15*10^6 W of power for the grid would be about 5.0*10^6 kg per year.\n\nThe C.H.P. unit saves a coal equivalent of 2.3*10^6kg per year,\nover the system generating process steam and utilizing grid electricty.\nThe economic savings are 83*1063 Pound per year illustrating the benefits of a C.H.P. syatem in this case.')
+
+
+
+