diff options
Diffstat (limited to '2471')
115 files changed, 3307 insertions, 0 deletions
diff --git a/2471/CH1/EX1.1/Ex1_1.sce b/2471/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..1958fb9df --- /dev/null +++ b/2471/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,16 @@ + +clc; +// Example 1.1 +printf('Example 1.1\n\n'); +printf(' Page No. 08\n\n'); +// Solution + +// Given +m1= 40*10^3;// fuel oil in gallons per year +ga= 4.545*10^-3;// m^3 +m= m1*ga;// fuel oil in m^3 per year +Cv1= 175*10^3;// Btu per gallons +Bt= .2321*10^6;// J per m^3 +Cv= Cv1*Bt;// in J per year per m^3 +q=m*Cv;// in J per year +printf(' Heat available is %3.2e J per year\n',q) diff --git a/2471/CH1/EX1.10/Ex1_10.sce b/2471/CH1/EX1.10/Ex1_10.sce new file mode 100755 index 000000000..69380bdaf --- /dev/null +++ b/2471/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,57 @@ +clear ;
+clc;
+// Example 1.10
+printf('Example 1.10\n\n');
+printf('Page No. 19\n\n');
+// given
+
+//Monthly Energy Usage
+qunty = [15*10^3 4*10^3 90*10^3]
+cost = [4950 720 2250]// in Pound
+common_basis1 = [738 117 90]// in 10^6 Wh
+common_basis= [2655 421 324]// converted into 10^9 Joules
+unit_cost = cost ./ common_basis1// in Pound per 10^6 Wh
+p= 80;// production in tonnes
+EI = ((sum(common_basis))/p)*10^9;
+CI = sum(cost)/80;
+printf('Monthly energy index is %3.2e J per tonne \n',EI)
+printf('Monthly cost index is %.0f Pound per tonne \n\n',CI)//Deviation in answer is due to calculation error for sum of cost in the book
+
+// Boiler House Energy Audit
+qunty_b = [15000 10000]
+Com_basis_b_1 = [2655 36]// in 10^9 J
+Com_basis_b = [738 10]// in 10^6 Wh
+Cost_b = [4950 250]// in Pound
+b_output = 571*10^6;// in Wh
+EI_b = (b_output/(sum(Com_basis_b)*10^6));
+CI_b = (sum(Cost_b)/b_output)*10^3;// Pound converted into p
+printf('Energy index for boiler is %.3f \n',EI_b)
+printf('Cost index for boiler is %3.2e p per Wh\n \n',CI_b)
+
+//Power House Energy Audit
+P_gen = 200*10^6;// Power generated in Wh
+Com_basis_p_1 = [14.4 2055 -1000]// in 10^9 J
+Com_basis_p = [4.0 571 -278]// in 10^6 Wh
+Cost_p = [100 5196 -2530]// in Pound
+CI_p = (sum(Cost_p)/P_gen)*10^3;// Pound converted into p
+printf('Cost index for power house is %3.2e p per Wh\n\n',CI_p)//Deviation in answer is due to wrong calculation in the book
+
+//Space Heating Energy Audit
+deg_days = 260;// Number of degree-days
+Com_basis_s_1 = [36 100 105]// in 10^9 J
+Com_basis_s = [10.0 27.8 29.2]// in 10^6 Wh
+Cost_s = [250 253 179]// in Pound
+EI_s = ((sum(Com_basis_s)*10^6)/deg_days)
+CI_s = (sum(Cost_s)/deg_days)
+printf('Energy index for space heating is %3.2e Wh per degree-day\n',EI_s)
+printf('Cost index for space heating is %3.2f Pound per degree-day\n\n',CI_s)
+
+//Process Energy Audit
+T_pdt_output = 100;// in tonne
+Com_basis_pr_1 = [216 720 810 316]// in 10^9 J
+Com_basis_pr = [60 200 225 88]// in 10^6 Wh
+Cost_pr = [1500 2766 2047 540]// in Pound
+EI_pr = ((sum(Com_basis_pr)*10^6)/T_pdt_output);
+CI_pr = (sum(Cost_pr)/T_pdt_output);
+printf('Energy index for Process Energy Audit is %3.2e Wh per tonne \n',EI_pr)
+printf('Cost index for Process Energy Audit is %.2f Pound per tonne \n',CI_pr)
diff --git a/2471/CH1/EX1.2/Ex1_2.sce b/2471/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..2824a4c11 --- /dev/null +++ b/2471/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 1.2
+printf('Example 1.2\n\n');
+printf('Page No. 09\n\n');
+// Solution
+
+// Given
+Eo1= 1.775*10^9;// Annular energy consumption of oil in Btu
+Btu= 1055;// 1 Btu = 1055 Joules
+Eo= Eo1*Btu;// Annular energy consumption of oil in Joules
+Eg1= 5*10^3;// Annular energy consumption of gas in Therms
+Th= 1055*10^5;// 1 Th = 1055*10^3 Joules
+Eg= Eg1*Th;// Annular energy consumption of gas in Joules
+Ee1= 995*10^3;// Annular energy consumption of electricity in KWh
+KWh= 3.6*10^6;// 1 KWh = 3.6*10^6 Joules
+Ee= Ee1*KWh;// Annular energy consumption of electricity in Joules
+Et= ( Eo + Eg + Ee);// Total energy consumption
+P1= (Eo/Et)*100; // percentage of oil consumption
+P2= (Eg/Et)*100; // percentage of gas consumption
+P3= (Ee/Et)*100; // percentage of electricity consumption
+printf('percentage of oil consumption is %3.1f \n',P1)
+printf('percentage of gas consumption is %3.1f \n',P2)
+printf('percentage of electricity consumption is %3.1f \n',P3)
+
diff --git a/2471/CH1/EX1.3/Ex1_3.sce b/2471/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..e35a408b4 --- /dev/null +++ b/2471/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,22 @@ +clear ;
+clc;
+// Example 1.3
+printf('Example 1.3\n\n');
+printf('Page No. 10\n\n');
+// Solution
+
+// Given
+Et = 100*10^3;// total energy production in tonnes per annum
+Eo= 0.520*10^9;// oil consumption in Wh
+Eg= 0.146*10^9;// gas consumption in Wh
+Ee= 0.995*10^9;// electricity consumption in Wh
+Io= Eo/Et;
+Ig= Eg/Et;
+Ie= Ee/Et;
+Et1= Eo + Eg + Ee;// total energy consumption
+It= Et1/Et;
+printf('oil energy index is %3.0f Wh per tonne \n',Io)
+printf('gas energy index is %3.0f Wh per tonne \n',Ig)
+printf('electricity energy index is %3.0f Wh per tonne \n',Ie)
+printf('total energy index is %3.0f Wh per tonne ',It)
+
diff --git a/2471/CH1/EX1.4/Ex1_4.sce b/2471/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..5b515e949 --- /dev/null +++ b/2471/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,22 @@ +clear ;
+clc;
+// Example 1.4
+printf('Example 1.4\n\n');
+printf('Page No. 10\n\n');
+// Solution
+
+// Given
+mc= 1.5*10^3;// coke consumption in tonnes
+mg= 18*10^3;// gas consumption in therms
+me= 1*10^9;// electricity consumption in Wh
+Cc1= 72;// cost of coke in Pound per tonne
+Cg1= 0.20;// cost of gas in Pound per therm
+Ce1= 2.25*10^-5 ;// cost of electricity in Pound per Wh
+Cc= mc*Cc1;//in Pound
+Cg= mg*Cg1;//in Pound
+Ce= me*Ce1;//in Pound
+Ct= Cc+Cg+Ce;//in Pound
+printf('cost of coke consumption is %.0f Pound \n',Cc)
+printf('cost of gas consumption is %.0f Pound \n',Cg)
+printf('cost of electricity consumption is %.0f Pound \n',Ce)
+printf('total cost is %3.0f Pound \n',Ct)
diff --git a/2471/CH1/EX1.5/Ex1_5.sce b/2471/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..3b862ccdf --- /dev/null +++ b/2471/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,21 @@ +clear ;
+clc;
+// Example 1.5
+printf('Example 1.5\n\n');
+printf('Page No. 11\n\n');
+// Solution
+
+// Given
+Cc= 108.0*10^3;// cost of coke in Pound
+Cg= 3.6*10^3;// cost of gas in Pound
+Ce= 22.5*10^3;// cost of electricity in Pound
+Ct= Cc+Cg+Ce;// total cost of fuel in Pound
+E= 15*10^3;// total production in tonnes per year
+Ic= Cc/E;//Pound per tonne
+Ig= Cg/E;//Pound per tonne
+Ie= Ce/E;//Pound per tonne
+It= Ct/E;//Pound per tonne
+printf(' coke cost index is %3.2f Pound per tonne \n',Ic)
+printf(' gas cost index is %3.2f Pound per tonne\n',Ig)
+printf(' electricity cost index is %3.2f Pound per tonne\n',Ie)
+printf(' total cost index is %3.2f Pound per tonne\n',It)
diff --git a/2471/CH1/EX1.6/Ex1_6.sce b/2471/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..2cc062b72 --- /dev/null +++ b/2471/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,19 @@ +clear ; +clc; +// Example 1.6 +printf('Example 1.6\n\n'); +printf('Page No. 11\n\n'); +// Solution + +// Given +G1= 11.72*10^3;// hourly consumption of gas in therms +th= 34.13;// in Watts +G= G1*th;// hourly consumption of gas in Watts +O1= 4.32*10^9;// hourly consumption of oil in Joules +J= .278*10^-3;// in Watts +O= O1*J;// hourly consumption of oil in Watts +E= 500*10^3;// hourly consumption of electricity in Watts +// Pie Chart Representation : one input argument x=[G O E] +pie([G O E],["gas" "oil" "electricity"]);// Please see the graphics window +printf('The Pie chart is plotted in the figure'); + diff --git a/2471/CH1/EX1.7/Ex1_7.sce b/2471/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..1dc8ea9fe --- /dev/null +++ b/2471/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,17 @@ +close
+clear ;
+clc;
+// Example 1.7
+printf('Example 1.7\n\n');
+printf('Page No. 12\n\n');
+// Solution
+
+// Given
+O= 150*10^3;// energy consumption in office heating in Watts
+L= 120*10^3;// energy consumption in lighting in Watts
+B= 90*10^3;// energy consumption in boiler house in Watts
+P= 180*10^3;// energy consumption in process in Watts
+// Pie Chart Representation : one input argument x=[O L B P]
+pie([O L B P],["office heating" "lighting" "boiler heating" "process"]);// Please see the graphics window
+printf('The Pie chart is plotted in the figure');
+
diff --git a/2471/CH1/EX1.8/Ex1_8.sce b/2471/CH1/EX1.8/Ex1_8.sce new file mode 100755 index 000000000..a56f83edc --- /dev/null +++ b/2471/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+// Example 1.8
+printf('Example 1.8\n\n');
+printf('Page No. 16\n\n');
+// given
+
+qunty= [40 10000 400 90000]
+unit_price= [29 0.33 0.18 0.025]
+cost= (unit_price .* qunty)// in Pound
+common_basis= [310 492 11.7 90]// in 10^6 Wh
+per_unit_cost= (unit_price .* qunty) ./ common_basis// Pound per 10^6 Wh
+p= 150;// production in tonnes
+EI= sum(common_basis)*10^6/150
+CI= sum(unit_price .* qunty)/150
+printf('energy index is %3.2f Wh per tonne \n',EI)
+printf('cost index is %3.2f Wh per tonne \n',CI)
diff --git a/2471/CH1/EX1.9/Ex1_9.sce b/2471/CH1/EX1.9/Ex1_9.sce new file mode 100755 index 000000000..e7d6fd5d7 --- /dev/null +++ b/2471/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,24 @@ +clear ;
+clc;
+// Example 1.9
+printf('Example 1.9\n\n');
+printf('Page No. 17\n\n');
+//given
+
+p= [50, 55, 65, 50, 95, 90, 85, 80, 60, 90, 70, 110, 60, 105];// weakly production in tonnes
+s= [0.4, 0.35, 0.45, .31, 0.51,0.55, 0.45, 0.5, 0.4, 0.51, 0.4, 0.6, 0.45, 0.55];// weakly steam consumption in 10^6 kg
+coefs = regress(p,s);
+new_p = 0:120
+new_s = coefs(1) + coefs(2)*new_p;
+plot(p,s,'r*');
+set(gca(),"auto_clear","off")
+
+plot(new_p,new_s);// please see the corresponding graph in graphic window
+xtitle('weakly steam consumption-production','weakly output (tonnes)','steam consumption/week (10^6 kg)')
+l = legend([_('Given data'); _('Fitting function')],2);
+
+in= coefs(1)*10^6;// intercept of graph in kg/weak
+printf('At zero output the steam consumption is %3.0f in kg/weak \n',in)
+
+
+
diff --git a/2471/CH10/EX10.1/Ex10_1.sce b/2471/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..f074a6030 --- /dev/null +++ b/2471/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,30 @@ +clear ;
+clc;
+// Example 10.1
+printf('Example 10.1\n\n');
+printf('Page No. 293\n\n');
+
+// given
+m = 1;// mass flow rate of initial air mixture in kg/s
+T = 23.5;// Initial temperature in degree celcius
+m1 = 0.6;// Percentage of fresh air mixture
+T1 = 5;// Dry Bulb Temperature of fresh air in degree celcius
+w1 = 0.005;// Humidity of fresh air at temperature T1 in kg/kg
+m2 = 0.4;// Percentage of recirculated air mixture
+T2 = 25;// Dry Bulb Temperature of recirculated air in degree celcius
+w2 = 0.015;// Humidity of recirculated air at temperature T2 in kg/kg
+
+//In air conditioning => m1*w1 + m2*w2 = m*w
+w = (m1*w1 + m2*w2)/m;// in kg/kgs
+printf('The humidity of the air mixture is %.3f kg/kg \n',w)
+
+//The specific enthalpy in J/kg can be calculated by the formula => h = (1.005*10^3*T) +(w*((2.50*10^6)+(1.86*10^3*T))); where the T is the temperature and w is the humidity at temperature T
+h_f = (1.005*10^3*T1) +(w1*((2.50*10^6)+(1.86*10^3*T1)));// Specific enthalpy of fresh air in J/kg
+h_r = (1.005*10^3*T2) +(w2*((2.50*10^6)+(1.86*10^3*T2)));// Specific enthalpy of recirculated air in J/kg
+h_m = (1.005*10^3*T) +(w*((2.50*10^6)+(1.86*10^3*T)));// Specific enthalpy of final air mixture in J/kg
+
+h_t = (m1*h_f) + (m2*h_r);// Total enthalpy of initial air mixturein J/kg
+Q = m*(h_m - h_t);// in Watts
+printf('The load on the heater is% .3f W',Q)
+// Deviation in answer due to direct substitution and some approximation in answer in book
+
diff --git a/2471/CH10/EX10.2/Ex10_2.sce b/2471/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..3f53e615c --- /dev/null +++ b/2471/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+// Example 10.2
+printf('Example 10.2\n\n');
+printf('Page No. 298\n\n');
+
+// given
+m1 = 0.75;// Percentage of fresh air mixture
+T1 = 31;// Dry Bulb Temperature of fresh air in degree celcius
+w1 = 0.0140;// Humidity of fresh air at temperature T1 in kg/kg
+m2 = 0.75;// Percentage of recirculated air mixture
+T2 = 22;// Dry Bulb Temperature of recirculated air in degree celcius
+w2 = 0.0080;// Humidity of recirculated air at temperature T2 in kg/kg
+m = 1.50;// mass flow rate of final air mixture in kg/s
+T = 10;// Dew Point temperature in degree celcius
+
+//In air conditioning => m1*w1 + m2*w2 = m*w
+w = (m1*w1 + m2*w2)/m// in kg/kgs
+printf('The humidity of the air mixture is %.4f kg/kg \n',w)
+
+
+// from the psychrometric chart, at w = 0.011kg/kg, the dry bulb temperature is = 26.5 degree celcius also the humidity of saturated air at 10 degree celcius is 0.0075kg/kg
+T_w = 26.5;// Dry Bulb temerature in degree celcius
+w_10 = 0.0075;// humidity at temperatue T in kg/kg
+
+//the specific enthalpy in J/kg can be calculated by the formula => h = (1.005*10^3*T) +(w*((2.50*10^6)+(1.87*10^3*T))); where the T is the temperature and w is the humidity at temperature T
+
+h_a = (1.005*10^3*T_w) +(w*((2.50*10^6)+(1.88*10^3*T_w)));// Specific enthalpy of recirculated air in J/kg
+h_s = (1.005*10^3*T) +(w_10*((2.50*10^6)+(1.87*10^3*T)));// Specific enthalpy of saturated air at 10 degree celcius in J/kg
+
+Q = m*(h_a - h_s);// in Watts
+printf('The cooling load on the washer is %.0f W',Q)
+// Answer wrongly calculated in the book
diff --git a/2471/CH11/EX11.1/Ex11_1.sce b/2471/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..38b94026f --- /dev/null +++ b/2471/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,24 @@ +clear ;
+clc;
+// Example 11.1
+printf('Example 11.1\n\n');
+printf('Page No. 308\n\n');
+
+//given
+V = 205;// Flow rate in m^3
+T1 = 74;// in degree celcius
+T2 = 10;// in degree celcius
+m = 1000;// Steam in kg
+p = 950;// Density of steam in kg/m^3
+C = 85;// Cost in Pound per m^3
+C_V = 43.3*10^6;// Calorific value in J/kg
+Cp = 4.18*10^3;// heat capacity of water J/kg-K
+h = 2.33*10^6;// Heat of the steam in J/kg
+n = 0.65;// Average bolier efficiency
+
+S_cost = ((m*h*C)/(C_V*p*n));// Steam cost in Pound per 1000 kg
+E_save = V*m*(T1 - T2)*Cp;// Energy saving in J per day
+S_save = E_save/h;// in kg per day
+printf('the steam saving is %.0f kg per day \n',S_save)
+G_save = (S_cost*S_save)/m;// Pound per day
+printf('The gross saving is %.0f Pound per day per year',G_save)
diff --git a/2471/CH11/EX11.10/Ex11_10.sce b/2471/CH11/EX11.10/Ex11_10.sce new file mode 100755 index 000000000..f90606158 --- /dev/null +++ b/2471/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,31 @@ +clear ;
+clc;
+// Example 11.10
+printf('Example 11.10\n\n');
+printf('Page No. 324\n\n');
+
+//given
+Pr = 100;// Production in tonnes per day
+p = 10.2;// percentage of sulphur dioxide
+T1 = 900;//Burner temperature in degree celcius
+T2 = 425;//Required temperature in degree celcius
+P = 10;// Dry saturated steam pressure in bar
+T = 120;// Dry saturated steam temperature in degree celcius
+//At the given Temperature =T and Pressure P, the required heat Qr to geberate steam from feed water is calculated from the steam table.
+Qr = 2.27*10^6;// in J/kg
+
+Sp_1 = 1.14*10^3;// Specific heat of the inlet gas in J/kmol-K
+Sp_2 = 1.03*10^3;// Specific heat of the outlet gas in J/kmol-K
+pr_rate = 1.2;// production rate in kmol/s
+
+//In the calculation part, the book has taken percentage of sulphur dioxide p = 10.6 in the place of p = 10.2, so there exists a deviation in answer
+Q_in = ((Pr*pr_rate)/p) * Sp_1 * T1;// Heat content of the inlet gas in J/s
+Q_out = ((Pr*pr_rate)/p) * Sp_2 * T2;// Heat content of the outlet gas in J/s
+Qa = Q_in - Q_out;// Heat available for steam
+S = Qa/Qr;// in kg/s
+printf('The steam production is %.3f kg/s',S)//Deviation in answer is due to some wrong value substition as discussed above
+
+
+
+
+
diff --git a/2471/CH11/EX11.2/Ex11_2.sce b/2471/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..c424c259b --- /dev/null +++ b/2471/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,22 @@ +clear ;
+clc;
+// Example 11.2
+printf('Example 11.2\n\n');
+printf('Page No. 313\n\n');
+
+//given
+p1 = 10;//heat-sensitive liquor percen
+p2 = 50;//heat-sensitive liquor percent
+m = 0.28;// mass rate in kg/s
+t = 150;// time in h per week
+
+// This question does not contain any calculation part in it.
+I = [8250 1150 14850 16500];//Installation cost in Pound
+A = [69300 36800 23600 24600];// Annual steam cost in Pound
+A_S = [A(1)-A(1) A(1)-A(2) A(1)-A(3) A(1)-A(4)];// Annual savings in Pound
+
+printf('\t\t CAPITAL AND OPERATING COSTS OF EVAPORATION PLANT\n\n\t \t\tInstallation \t\t Annual \t\t Annual saving\n Type \t\t\t cost \t\t\tsteam cost \t\t(to single effect)\n \t \t\t (Pound) \t\t (Pound) \t\t (Pound)\n\nSingle effect \t\t %.0f \t\t\t %.0f \t\t\ %.0f \nDouble effect \t\t %.0f \t\t\t %.0f \t\t\ %.0f \nTriple effect + \n(vapour compression) \t %.0f \t\t %.0f \t\t\ %.0f \nTriple effect \t\t %.0f \t\t %.0f \t\t\ %.0f \n\n\n',I(1),A(1),A_S(1),I(2),A(2),A_S(2),I(3),A(3),A_S(3),I(4),A(4),A_S(4))
+
+
+printf(' The results enable the return on investment to be assessed by one of the standard economic procedures and the final selsction made.')
+
diff --git a/2471/CH11/EX11.3/Ex11_3.sce b/2471/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..d19aa63bf --- /dev/null +++ b/2471/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,40 @@ +clear ;
+clc;
+// Example 11.3
+printf('Example 11.3\n\n');
+printf('Page No. 314\n\n');
+
+//given
+f = 1;// feed of sodium hydroxide in kg
+v = 0.5;// produed vapour in kg
+A = 30;// in m^2
+T1 = 95;// Temperature of boiling solution in deg C
+U = 3*10^3;// heat transfer coefficent in W/m^2-K
+m = 1;// feed rate in kg/s
+Tf = 70;// Feed temperature in deg C
+h_f = 260*10^3;// Enthalpy of feed in J/kg
+h_b = 355*10^3;// Enthalpy of boiling solution in J/kg
+h_v = 2.67*10^6;// Enthalpy of vapour in J/kg
+P1 = 0.6;// Pressure in vapour space in bar
+
+Q = (v*h_b) + (v*h_v) -(f*h_f);// in W
+printf('The total energy requirement is %.0f W \n',Q)
+
+// As Q = A*U*dT
+dT = Q/(U*A);// in degree celcius
+T2 = dT + T1;// in degree celcius
+//The temperature of the heating steam T2 corresponds to a pressure of 1.4 bar. Dry saturated steam at 1.4 bar has a total enthalpy of 2.69*10^6 J/kg
+//Assuming an isentropic compression of the vapour from 0.6 bar to 1.4 bar, the outlet enthalpy is 2.84*10^6 J/kg
+
+// from steam table
+P2 = 1.4// pressure in bar
+h_s = 2.69*10^6;// enthalpy of dry saturated steam in J/kg
+h_v2 = 2.84*10^6 ;// the outlet enthalpy of vapour in J/kg
+
+W = v*(h_v2 - h_s);// Work in W
+T_E = W + 60*10^3;// in W
+printf('The total energy consumption is %.0f W',T_E)
+
+
+
+
diff --git a/2471/CH11/EX11.4/Ex11_4.sce b/2471/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..4fd2252ae --- /dev/null +++ b/2471/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,14 @@ +clear ;
+clc;
+// Example 11.4
+printf('Example 11.4\n\n');
+printf('Page No. 316\n\n');
+
+//given
+Cm_S = 10000;// Company saving in Pound per annum
+S = Cm_S/12;// Saving in Pound per months
+Ca_C = 10500;// Capital cost in Pound
+Ins_C = 7500;// Installation cost in Pound
+T_C = Ca_C + Ins_C;// Total cost in Pound
+T = T_C/S;// pay-back time in months
+printf('The pay-back period was %.0f months\n',T)
diff --git a/2471/CH11/EX11.5/Ex11_5.sce b/2471/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..34e6ff9e3 --- /dev/null +++ b/2471/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,26 @@ +clear ;
+clc;
+// Example 11.5
+printf('Example 11.5\n\n');
+printf('Page No. 318\n\n');
+
+//From the heat balance:-
+//Heat recovered in the boiler = heat gained by the air = heat lost by the flue gases
+//=> Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f
+// As mass flow rate of air/flue gas is not given in the book
+//Assuming m_a = m_f = 2.273 kg/s & Cp_a = 1*10^3 J/kg-K
+
+m_a = 2.273;// in kg/s
+m_f = m_a;// in kg/s
+Cp_a = 1*10^3;// Specific heat capacity of air in J/kg-K
+T1_a = 20;// Entrance temperature of air in degree celcius
+T2_a = 130;// Exit temperature of air in degree celcius
+dT_a = T2_a - T1_a;//in K
+T1_f = 260;// Entrance temperature of flue gases in degree celcius
+T2_f = 155;// Entrance temperature of flue gases in degree celcius
+dT_f = T1_f - T2_f;//in K
+
+//From heat balance:- Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f
+Cp_f = ((m_a*Cp_a*dT_a)/(m_f*dT_f));// in J/kg-K
+Q = m_f*Cp_f*dT_f;// in W
+printf('The total heat recovered at full load if %3.2e W',Q)
diff --git a/2471/CH11/EX11.6/Ex11_6.sce b/2471/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..381e96311 --- /dev/null +++ b/2471/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,13 @@ +clear ;
+clc;
+// Example 11.6
+printf('Example 11.6\n\n');
+printf('Page No. 320\n\n');
+
+C = 10000;// Installation cost of the pump in Pound
+S = 3500;// Saving in Pound per annum
+T = C/S;// in year
+printf('The pay back time is %.0f year\n\n',T)
+
+// This question further does not contain any calculation part in it.
+printf('In a heat-pump system the work input to drive the compressor,W, produces a heat absorption capacity,Q2,\nand to balance the energy flow, a quantity of heat, Q1, must be dissipated.\nThus the energy equation is\n -> Q1 = W + Q2\nand the coeffient of performance is \nC.O.P. = Q1/W = Q1/(Q1 - Q2)\n Consequently the C.O.P. is always greater than unity.\nThe maximum theoretical value of the C.O.P. is that predicted by the Carnot in chapter 2,namely :\n -> (C.O.P.)max = T1/(T1 - T2)')
diff --git a/2471/CH11/EX11.7/Ex11_7.sce b/2471/CH11/EX11.7/Ex11_7.sce new file mode 100755 index 000000000..55a33ed1c --- /dev/null +++ b/2471/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+// Example 11.7
+printf('Example 11.7\n\n');
+printf('Page No. 320\n\n');
+
+//given
+T1 = 40;// in degree
+T2 = 0;// in degree celcius
+//As from carnot cycle, C.O.P = (T1/(T1 - T2)), where temperature are in degree celcius
+C_O_P1 = ((T1+273)/((T1+273) - (T2+273)));
+printf('C.O.P. is %.1f \n',C_O_P1)
+
+// A secondary fluid as hot water at 60 deg C is used
+T3 = 60;// Temperature of hot water in degree celcius
+C_O_P2 = ((T3+273)/((T3+273) - (T2+273)));
+printf('C.O.P. when secondary fluid is used is %.1f \n',C_O_P2)
diff --git a/2471/CH11/EX11.8/Ex11_8.sce b/2471/CH11/EX11.8/Ex11_8.sce new file mode 100755 index 000000000..3f8a232f3 --- /dev/null +++ b/2471/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,8 @@ +clear ;
+clc;
+// Example 11.8
+printf('Example 11.8\n\n');
+printf('Page No. 323\n\n');
+
+// This question does not contain any calculation part in it.
+printf('No calculation is required as not in shown in book')
diff --git a/2471/CH11/EX11.9/Ex11_9.sce b/2471/CH11/EX11.9/Ex11_9.sce new file mode 100755 index 000000000..070fdb24a --- /dev/null +++ b/2471/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,43 @@ +clear ;
+clc;
+// Example 11.9
+printf('Example 11.9\n\n');
+printf('Page No. 324\n\n');
+
+//given
+T1 = 273;// Measured temperature In degree celcius
+P = 1;// Measured pressure in bar
+T2 = 290;// initial temperature In degree celcius
+T3 = 1000;// Final temperature In degree celcius
+T4 = 1150;// Entering tempearture In degree celcius
+v1 = 7;// in m^3/s
+v2 = 8;// in m^s
+M = 22.7;// in kmol/m^3
+d = 0.1;// Diameter in m
+A = 0.01;// Surface area per regenerator channel in m^2
+u = 1;// maximum velocity in m/s
+Cp_1 = 34*10^3;// Heat capacity at T4 temperature in J/kmol-K
+Cp_2 = 32*10^3;// Heat capacity at outlet temperature in J/kmol-K
+Cp_m = 30*10^3;// Heat capacity at mean temperature in J/kmol-K
+
+m_c = v1/M;// Molal air flow rate in kmol/s
+H_c1 = Cp_m*(T3 - T1);// Enthalpy of air at 1000K in J/mol
+H_c2 = Cp_m*(T2 - T1);// Enthalpy of air at 290 in J/mol
+Q = (m_c*(H_c1 - H_c2))/10^6;// in 10^6 W
+printf('The heat transfer, Q is %.1f *10^6 W \n',Q)
+
+m_F = v2/M;// Molal flow rate of flue gas in kmol/s
+dH = (Q/m_F)*10^6;// enthaply chnage of the flue gas in J/kmol
+H_F1 = Cp_1*(T4 - T1);// Enthalpy of the flue gas at 1150 K in J/kmol
+H_F2 =H_F1 - dH;// Enthalpy at the exit temperature in J/kmol
+T_F2 = (H_F2/Cp_2) + T1;// in K
+printf('The exit tempearture of the flue gas is %.0f K \n',T_F2)
+S_R = v2/u;//cross sectional area of the regenerator in m^2
+N = S_R/A;
+printf('The number of channels required is %.0f \n',N)
+printf('Consequently for this regenerator a square layout could be achieved with 40 channels arranged horizontally and 20 channels vertically.')
+
+
+
+
+
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)
+
+
diff --git a/2471/CH3/EX3.1/Ex3_1.sce b/2471/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 000000000..84cf04b8c --- /dev/null +++ b/2471/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,15 @@ +clear ;
+clc;
+// Example 3.1
+printf('Example 3.1\n\n');
+printf('Page No. 58\n\n');
+
+// given
+P = 10000;// Principal Amount
+i = 0.15;// Interest Rate
+n = 4;//years
+I = P*i*n;// Simple Interest
+Ts= P+I;// The total repayment
+printf('The total repayment is %.0f Euro\n',Ts)
+
+
diff --git a/2471/CH3/EX3.10/Ex3_10.sce b/2471/CH3/EX3.10/Ex3_10.sce new file mode 100755 index 000000000..24dab07ee --- /dev/null +++ b/2471/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,28 @@ +clear ;
+clc;
+// Example 3.10
+printf('Example 3.10\n\n');
+printf('Page No. 71\n\n');
+
+// given
+C = 2500;// Cost of the project
+P = 1000;// Cash in flow
+r_r = 0.12;// Rate of return
+S = 0;// Zero salvage value
+n = 4;//years
+
+for j= 1:1:4 // as for four years
+ d_(j) = P*(1/(1+r_r)^j);
+ end
+
+
+P_v = d_(1)+d_(2)+d_(3)+d_(4);//Present value of cash inflow
+N = P_v-C;
+printf('Net present value is %.0f Pound\n',ceil(N))
+
+if(P_v>C) then
+ disp('The project may be undertaken')
+else
+ disp('The project may not be undertaken')
+ end
+
diff --git a/2471/CH3/EX3.11/Ex3_11.sce b/2471/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..a37c4ca3a --- /dev/null +++ b/2471/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,29 @@ +clear ;
+clc;
+// Example 3.11
+printf('Example 3.11\n\n');
+printf('Page No. 72\n\n');
+
+// given
+Cash_out = 80000;// Present value of cash outflow for both projects E and F
+r_r = .2;// Rate of return
+n = 5;// years
+
+d = [0.833 0.694 0.579 0.482 0.402]// Discount Factor for 20% of rate of return for 5 years
+Ce = [10000 20000 30000 40000 50000]// Cash flow for project E in Pound
+Pe = [8330 13880 17370 19280 20100]// Present value for project E in Pound
+
+Cf = [50000 40000 30000 20000 10000]// Cash flow for project F in Pound
+Pf = [41650 27760 17370 9640 4020]// Present value for project F in Pound
+
+Cash_inE = sum(Pe)//Present value of cash inflow in Pound
+Cash_inF = sum(Pf)//Present value of cash inflow in Pound
+
+Net_E = Cash_inE - Cash_out;// net present value for project E in Pound
+Net_F = Cash_inF - Cash_out;// net present value for project F in Pound
+
+if (Net_E>Net_F) then
+ disp('Project E is selected based on NPV')
+else
+ disp('Project F is selected based on NPV')
+end
diff --git a/2471/CH3/EX3.12/Ex3_12.sce b/2471/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 000000000..573b3b043 --- /dev/null +++ b/2471/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 3.12
+printf('Example 3.12\n\n');
+printf('Page No. 72\n\n');
+
+// given
+Cash_inG = 43000;// Present value of cash inflow for project G in Pound
+Cash_outG = 40000;// Present value of cash outflow for project G in Pound
+Net_G = Cash_inG - Cash_outG;// Net present value for G in Pound
+PI_G = (Cash_inG/Cash_outG);// Profitability index for G
+
+Cash_inH = 23000;// Present value of cash inflow for project H in Pound
+Cash_outH = 20000;// Present value of cash outflow for project H in Pound
+Net_H = Cash_inH - Cash_outH;// Net present value for H in Pound
+PI_H = (Cash_inH/Cash_outH);// Profitability index for H
+
+//The higher the profitability index the more desirable is the project.
+if (PI_G>PI_H) then
+ disp('Project G is more attractive than Project H')
+else
+ disp('Project H is more attractive than Project G')
+end
+
+
diff --git a/2471/CH3/EX3.13/Ex3_13.sce b/2471/CH3/EX3.13/Ex3_13.sce new file mode 100755 index 000000000..790a63c33 --- /dev/null +++ b/2471/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 3.13
+printf('Example 3.13\n\n');
+printf('Page No. 73\n\n');
+
+// given
+Cash_out = 80000;// Present value of cash outflow for project F in Pound
+n = 5;// years
+Cash_in= [50000 40000 30000 20000 10000]// Cashn in \flow for project F in Pound
+NPV = 0;//At the end of 5 years
+
+//Let the unknown rate for project F be rm.
+
+//The amount standing at the end of 5 years is\n => 0 = 80000*(1+rm)^5 - 50000*(1+rm)^4 - 40000*(1+rm)^3 - 30000*(1+rm)^2 - 20000*(1+rm)^1 - 10000
+// By taking (1+rm) = x\n =>8*x^5 - 5*x^4 - 4*x^3 - 3*x^2 - 2*x - 1 = 0\n\n')
+
+function y=fsol1(x)
+ y= 8*x^5 - 5*x^4 - 4*x^3 - 3*x^2 - 2*x - 1;
+endfunction
+[xres]=fsolve(100,fsol1);
+xres
+rm = (xres - 1)*100;
+printf('The value of rm for project F is %3.0f per cent\n',ceil(rm))
+
diff --git a/2471/CH3/EX3.14/Ex3_14.sce b/2471/CH3/EX3.14/Ex3_14.sce new file mode 100755 index 000000000..84b34e6bd --- /dev/null +++ b/2471/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,53 @@ +clear ;
+clc;
+// Example 3.14
+printf('Example 3.14\n\n');
+printf('Page No. 74\n\n');
+
+// given
+n = 5;//years
+C = 80000;// Cost of the project in Pound
+Cash_in = [10000 20000 30000 40000 50000]// Cash inflow in Pound
+r_d1 = 15;// Discount factor of 15%
+r_d2 = 18 ;// Discount factor of 18%
+r_d3 = 20;// Discount factor of 20%
+
+//At discount of 15%
+df_1 = [0.870 0.756 0.658 0.572 0.497]// Discount factor for every year
+PV_1 = [8700 15120 19740 22880 24850]// Present value
+Net_1 = sum (PV_1);// net present value
+
+
+//At discount of 18%
+df_2 = [0.847 0.718 0.609 0.516 0.437]// Discount factor for every year
+PV_2 = [8470 14360 18270 20640 21850]// Present value
+Net_2 = sum (PV_2);// net present value
+
+
+//At discount of 20%
+df_3 = [0.833 0.694 0.579 0.482 0.402]// Discount factor for every year
+PV_3 = [8330 13880 17370 19280 20100]// Present value
+Net_3 = sum (PV_3);// net present value
+
+// f = N.P.V. cash inflow - N.P.V. cash outflow
+//(1) By Numerical Method
+ff = 2*((sum (PV_2) - C)/(sum (PV_2) - sum(PV_3)));// in percentage
+f = 18 + ff;
+printf('the internal rate of return in percentage is %3.2f \n\n',f)// Deviation in answer due to direct substitution
+
+//(2) By Graphical Interpolation
+f_1 = (sum (PV_1) - C)/10^3;//At discount factor of 15%
+f_2 = (sum (PV_2) - C)/10^3;//At discount factor of 18%
+f_3 = (sum (PV_3) - C)/10^3;//At discount factor of 20%
+
+x = [f_1 f_2 f_3];
+y = [r_d1 r_d2 r_d3];
+plot(x,y,'r*');
+
+plot2d (x,y);// please see the corresponding graph in graphic window
+xtitle('Discount factor against f','f ( *10^3 Pound)','Discount factor(%)')
+regress(x,y)
+coefs = regress(x,y);
+printf('the internal rate of return in percentage is %3.1f \n',coefs(1))// Deviation in answer due to direct substitution
+
+
diff --git a/2471/CH3/EX3.15/Ex3_15.sce b/2471/CH3/EX3.15/Ex3_15.sce new file mode 100755 index 000000000..b1aeb453e --- /dev/null +++ b/2471/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 3.15
+printf('Example 3.15\n\n');
+printf('Page No. 77\n\n');
+
+// given
+i_t = [20 40 60 80 100];// Insulation thickness in mm
+f_c = [2.2 3.5 4.8 6.1 7.4];// Fixed costs in (10^3 Pound / year)
+h_c = [10.2 6.5 5.2 4.6 4.2];// Heat costs in (10^3 Pound / year)
+t_c = [12.4 10 10 10.7 11.6];// Total costs in (10^3 Pound / year)
+
+//(a) Graphical solution
+//Refer figure 3.8
+C_T = 9750;// Minimum total cost in Pound
+t = 47;// Corresponding thickness of insulation in mm
+printf('The most economic thickness of insulation is %.0f mm \n',t)
+
+//(b) Numerical solution
+// The cost due to heat losses,C1, and the fixed costs,C2, vary according to the equations;-
+// C1 = (a/x) + b and C2 = (c*x) + d
+// Substituting the values of C1 and C2 together with the corresponding insulation thickness values , the following equations are obtained :-
+// C1 = (150*10^3/x) + 2.7*10^3 and C2 = (65*x) + 0.9*10^3
+//And to obtain the total costs
+//CT = C1 + C2 = (150*10^3/x) + (65*x) + 3.6*10^3
+// Differentiate to optimise, and put dCT/dx equal to zero
+//dCT/dx =-((150*10^3)/x^2) + 65 = 0
+
+//Let y = dCT/dx
+function y=fsol1(x)
+ y = -((150*10^3)/x^2) + 65;
+endfunction
+[xres]=fsolve(50,fsol1);
+x = xres;
+printf('The optimum thickness of insulation is %.0f mm \n',x)
diff --git a/2471/CH3/EX3.16/Ex3_16.sce b/2471/CH3/EX3.16/Ex3_16.sce new file mode 100755 index 000000000..e01e8d867 --- /dev/null +++ b/2471/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,26 @@ +clear ;
+clc;
+// Example 3.16
+printf('Example 3.16\n\n');
+printf('Page No. 79\n\n');
+
+// given
+tb = [36*10^3 72*10^3 144*10^3 216*10^3]; //operating time in s
+U = [971 863 727 636];// Mean overall heat transfer rate in W/m^2-K
+A = 50;// area in m^2
+dT = 25;// temperature difference in degree celcius
+ts = 54*10^3;// Time in sec (h converted to sec)
+//As Q = U*A*dT
+for i = [1:1:4]
+ Q(i) = (U(i)*A*dT)/10^6;
+ Q_a(i) = ((tb(i)*Q(i)*10^6)/(tb(i) + ts))/10^6;
+ printf('the average heat transfer rate is %.3f *10^6 W \n',Q_a(i))
+end
+
+//Refer figure 3.9
+printf('\n')
+Q_max = 0.67*10^6;// Maximum value of Q in W
+T_opt = 33;// Time in h
+printf('The maximum value of Q obtained is %3.2e W \n',Q_max)
+printf('The most econnomic opertaing time for the heat exchanger to run is %.0f h ',T_opt)
+
diff --git a/2471/CH3/EX3.17/Ex3_17.sce b/2471/CH3/EX3.17/Ex3_17.sce new file mode 100755 index 000000000..bd38ed106 --- /dev/null +++ b/2471/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,43 @@ +clear ;
+clc;
+// Example 3.17
+printf('Example 3.17\n\n');
+printf('Page No. 80\n\n');
+
+// given
+// C_T = 7*x + (40000/(x*y)) + 6*y + 10
+//Differentiating C_T with respect to x and y:-
+//dC_T/dx = 7 - (40000/(x^2*y))
+//dC_T/dy = - (40000/(x*y^2)) + 6
+
+//For optimum conditions :- dC_T/dx = dC_T/dy = 0
+//dC_T/dx = 0 => 7 - (40000/(x^2*y)) = 0
+//=> y = 40000/(7*x^2).......(1)
+//dC_T/dy = 0 =>- (40000/(y^2*x)) +6 = 0
+//=> y = (40000/(6*x))^0.5.......(2)
+
+//From equation (1) and (2)
+//=> 40000/(7*x^2) - (40000/(6*x))^0.5 = 0
+
+function y=fsol1(x)
+ y = 40000/(7*x^2) - (40000/(6*x))^0.5 ;
+endfunction
+[xres]=fsolve(20,fsol1);
+x = xres;
+
+//from equation (1)
+y = 40000/(7*x^2);
+
+//a = d^2C_T/dx^2 = 80000/(x^3*y)
+//b = d^2C_T/dy^2 = 80000/(x*y^3)
+a = 80000/(x^3*y);
+b = 80000/(x*y^3);
+if a > 0
+ if b > 0
+//The optimum conditions must occur at a point of minimum cost- C_T_m
+C_T_m = 7*x + (40000/(x*y)) + 6*y + 10;// in Pound
+printf('The minimum cost is %.1f Pound',C_T_m)
+ end
+end
+
+
diff --git a/2471/CH3/EX3.2/Ex3_2.sce b/2471/CH3/EX3.2/Ex3_2.sce new file mode 100755 index 000000000..490faae4f --- /dev/null +++ b/2471/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,12 @@ +clear ;
+clc;
+// Example 3.2
+printf('Example 3.2\n\n');
+printf('Page No. 58\n\n');
+
+// given
+P = 10000;// Principal Amount in Pound
+i = 0.15;// Interest Rate
+n = 4;//years
+Tc = P*(1+i)^n;
+printf('The total repayment after adding compond interest is %.0f Pound\n',Tc)
diff --git a/2471/CH3/EX3.3/Ex3_3.sce b/2471/CH3/EX3.3/Ex3_3.sce new file mode 100755 index 000000000..48c81812a --- /dev/null +++ b/2471/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,13 @@ +clear ;
+clc;
+// Example 3.3
+printf('Example 3.3\n\n');
+//Page No. 59
+
+// given
+P = 60000;/// Principal Amount in Pound
+i = 0.18;// Interest Rate
+n = 10;//years
+R = P*((i*(1+i)^n)/((1+i)^n -1));//Rate of Capital Recovery
+printf('The annual investment required is %.1f Pound\n',R)
+
diff --git a/2471/CH3/EX3.4/Ex3_4.sce b/2471/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..81fb992bb --- /dev/null +++ b/2471/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,19 @@ +clear ;
+clc;
+// Example 3.4
+printf('Example 3.4\n\n');
+printf('Page No. 61\n\n');
+
+// given
+P = 100000;/// Principal Amount of boiler plant in Pound
+n = 10;// service life in years
+S = 0;//Zero Salvage value
+nT = (n*(n+1)/2);//sum of years
+for i = 0:9
+ d_(i+1) = ((P-S)/nT)*(n-i);
+end
+printf('The Annual depreciation for first year is %.0f Pound\n',d_(1))
+printf('The Annual depreciation for second year is %.0f Pound\n\n',d_(2))
+printf('The Annual depreciation for third year is %.0f Pound\n',d_(3))
+printf('The Annual depreciation for ten year is %.0f Pound\n',d_(10))
+// Deviation in answer due to some .approximation of values in the book
diff --git a/2471/CH3/EX3.5/Ex3_5.sce b/2471/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 000000000..d1a2435d9 --- /dev/null +++ b/2471/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+// Example 3.5
+printf('Example 3.5\n\n');
+printf('Page No. 62\n\n');
+
+// given
+P = 40000;/// Principal Amount of boiler plant in Pound
+nT = 10;// service life in years
+S = 4000;// Salvage value
+n = 6;// years after which Asset value has to be calculated
+
+//(a) Straight line method
+d = ((P-S)/nT);// Depreciation
+Aa = (d*(nT-n)) + S;
+printf('The Asset value at the end of six years using Straight line method is %.0f Pound\n',Aa)
+
+// (b) Declining balance technique
+f = 1-(S/P)^(1/nT);// Fixed fraction of the residual asset
+Ab = P*(1-f)^n;
+printf('The Asset value at the end of six years using Declining balance technique is %.0f Pound\n\n',Ab)
+
+// (c) Sum of the years digit
+sum_nT = (nT*(nT+1)/2);//sum of 10 years
+sum_n = 45;//sum after 6 years
+dc = ((sum_n/sum_nT)*(P-S));// Depreciation after 6 years
+Ac = P-dc;
+printf('The Asset value at the end of six years using Sum of the years digit is %.0f Pound\n',Ac)// Deviation in answer due to direct substitution
+
+//(d) Sinking Fund Method
+r_i = 0.06;// Rate of interest
+Ad = P-((P-S)*(((1+r_i)^n-1)/((1+r_i)^nT-1)));
+printf('The Asset value at the end of six years using Sinking Fund Method is %.0f Pound\n',Ad)// Deviation in answer due to direct substitution
diff --git a/2471/CH3/EX3.6/Ex3_6.sce b/2471/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 000000000..dfd29dc94 --- /dev/null +++ b/2471/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,30 @@ +clear ;
+clc;
+// Example 3.6
+printf('Example 3.6\n\n');
+printf('Page No. 67\n\n');
+
+// given
+P = 9000;// Capital Cost in Pound
+n = 5;// Project lifetime
+Less_dep = 8000;// Less Depreciation
+
+//For Project A
+d1 = [4500 3750 3000 1500 750 ]// Saving in every year (before depreciation)
+dT1 = sum (d1)
+Net_S1 = dT1- Less_dep;// Total Net Saving
+Avg1 = Net_S1/n;// Average net annual saving
+R_R1 = (Avg1/P)*100;
+
+//For Project
+d2 = [750 2250 4500 4500 1500 ]// Saving in every year (before depreciation)
+dT2 = sum (d2)
+Net_S2 = dT2- Less_dep;// Total Net Saving
+Avg2 = Net_S2/n;// Average net annual saving
+R_R2 = (Avg2/P)*100;
+
+printf('The percentage of Rate of Return on original investment for Project A is %3.1f \n',R_R1)
+printf('The percentage of Rate of Return on original investment for Project B is %3.1f \n',R_R2)
+
+
+
diff --git a/2471/CH3/EX3.7/Ex3_7.sce b/2471/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 000000000..54df1588e --- /dev/null +++ b/2471/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,19 @@ +clear ;
+clc;
+// Example 3.7
+printf('Example 3.7\n\n');
+printf('Page No. 68\n\n');
+
+// given
+Pc = 10000;// Capital cost for project C in Pound
+Pd = 10000;// Capital cost for project d in Pound
+nc = 3;// pay back period for C
+nd = 3;// pay back period for D
+Ca = [4500 3500 2000 2000 1000];// Annual Cash flow for C in Pound
+Cc = [4500 8000 10000 12000 13000]// Cumulative Cash flow for C in Pound
+Da = [1500 4000 4500 2200 1800 1000];// Annual Cash flow for D in Pound
+Dc = [1500 5500 10000 12200 14000 15000]// Cumulative Cash flow for D in Pound
+Ac = Cc(5)-Pc;// in Pound
+Ad = Dc(6)-Pd;// in Pound
+printf('Additional amount from C after the pay back time is %3.f Pound\n',Ac)
+printf('Additional amount from D after the pay back time is %3.f Pound\n',Ad)
diff --git a/2471/CH3/EX3.8/Ex3_8.sce b/2471/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 000000000..11ae55f77 --- /dev/null +++ b/2471/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,23 @@ +clear ;
+clc;
+// Example 3.8
+printf('Example 3.8\n\n');
+printf('Page No. 69\n\n');
+
+//Refer figure 3.6
+// given
+n = 5;//years
+C = 80000;// COst of the project in Pound
+S = 0;// Zero Salvage Value
+A_E = [10000 20000 30000 40000 50000]// Annual Net cash flow for project E in Pound
+C_E = [10000 30000 60000 100000 150000]// Cummulative Net cash flow for project E in Pound
+A_F = [50000 40000 30000 20000 10000]// Annual Net cash flow for project F in Pound
+C_F = [50000 90000 120000 140000 150000]// Cummulative Net cash flow for project F in Pound
+
+//From the figure 3.6 (intercept of x-axis)
+P_F = 1.75;// in years
+P_E = 3.5;// in years
+printf('The pay-back time of project F is %.2f \n',P_F)
+printf('The pay-back time of project E is %.1f \n\n',P_E)
+
+printf('As the pay-back time is less for project F,\nProject F would always be choosen in practice\nsince prediction of savings in the early years are more reliable than long-term predictions.')
diff --git a/2471/CH3/EX3.9/Ex3_9.sce b/2471/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 000000000..f39ecdb5c --- /dev/null +++ b/2471/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,31 @@ +clear ;
+clc;
+// Example 3.9
+printf('Example 3.9\n\n');
+printf('Page No. 70\n\n');
+
+// given
+P = 1;/// Principal Amount in Pound
+
+r_i = 0.1;// Compound interest rate
+for i = [1:1:4]
+ c = P*(1+r_i)^i;
+ printf('compound intrest after year %.0f is equal to %.2f Pound\n',i,c)
+end
+
+new_P = 1000*P;//in Pound
+new_c = 1000*c;// in Pound
+printf('The new amount at the compound interest after fourth year is %.0f Pound\n\n',new_c)
+
+// Discount rate
+r_d = 0.10;// Discount rate
+for j= 1:1:4
+ d = P*(1/(1+r_d)^j);
+ printf('The amount receivable at discount in year %.0f is %.3f Pound\n',j,d)
+end
+
+new_P1 = new_c;// in Pound
+new_d = new_P1*d;// in Pound
+printf('The new amount receivable at discount in fourth year is %.0f Pound\n',new_d)
+
+
diff --git a/2471/CH4/EX4.1/Ex4_1.sce b/2471/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..d12e67b49 --- /dev/null +++ b/2471/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,16 @@ +clear ;
+clc;
+// Example 4.1
+printf('Example 4.1\n\n');
+printf('Page No. 88\n\n');
+
+// given
+K = 45// Thermal Conductivity in W/m-K
+L = 5*10^-3;// thickness in metre
+T1 = 100;// in degree celcius
+T2 = 99.9;// in degree celcius
+A = 1;// Area in m^2
+
+//By Fourier law of conduction
+Q = ((K*A*(T1-T2))/L);// in Watts
+printf('The rate of conductive heat transfer is %.0f W \n',Q)
diff --git a/2471/CH4/EX4.10/Ex4_10.sce b/2471/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..031e02de9 --- /dev/null +++ b/2471/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 4.10
+printf('Example 4.10\n\n');
+printf('Page No. 106\n\n');
+
+// given
+T1 = 150;// Surface temperature in degree celcius
+T2 = 20;// Ambient temperature in degree celcius
+d = 0.100; //Outside diametr of pipe in m
+h = 10;// Outside film coefficient in W/m^2-K
+t = 25*10^-3;// thickness of insulation in m
+K = 0.040;// Thermal conductivity of insulation in W/m-K
+
+r2 = d/2;//in m
+r1 = r2+t;// in m
+Q = ((T1-T2)/((1/(2*%pi*r1*h))+(log(r1/r2)/(2*%pi*K))));// in W/m
+printf('The heat loss per unit length is %.0f W/m',Q)
diff --git a/2471/CH4/EX4.2/Ex4_2.sce b/2471/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..9b91875f2 --- /dev/null +++ b/2471/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+// Example 4.2
+printf('Example 4.2\n\n');
+printf('Page No. 89\n\n');
+// given
+K1 = 45// Thermal Conductivity of mild steel in W/m-K
+K2 = 0.040// Thermal Conductivity of insulaton in W/m-K
+L1 = 5*10^-3;// thickness of mild steel in metre
+L2 = 50*10^-3;// thickness of insulation in metre
+T1 = 100;// in degree celcius
+T2 = 25;// in degree celcius
+A = 1;// Area in m^2
+
+//By Fourier law of conduction
+Q = (((T1-T2)/((L1/(K1*A))+(L2/(K2*A)))))// in Watts
+printf('The rate of conductive heat transfer is %.0f W \n',Q)
diff --git a/2471/CH4/EX4.3/Ex4_3.sce b/2471/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..58b696245 --- /dev/null +++ b/2471/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,27 @@ +clear ;
+clc;
+// Example 4.3
+printf('Example 4.3\n\n');
+printf('Page No. 90\n\n');
+
+// given
+K1 = 26;// Thermal Conductivity of stainless steel in W/m-K
+K2 = 0.038;// Thermal Conductivity of insulaton in W/m-K
+L1 = 3*10^-3;// thickness of stainless steel in metre
+L2 = 40*10^-3;// thickness of insulation in metre
+T1 = 105;// in degree celcius
+T2 = 25;// in degree celcius
+L = 15;// Length of pipe in metre
+d1 = 50*10^-3;// Internal diameter of pipe in metre
+d2 = 56*10^-3;// External diameter of pipe in metre
+
+r1 = d1/2;// in metre
+r2 = d2/2;// in metre
+
+rm_p = ((r2-r1)/log(r2/r1));// logarithmic mean radius of pipe in m
+rm_i = (((r2+L2)-r2)/log((r2+L2)/r2));// logarithmic mean radius of insulation in m
+
+//By Fourier law of conduction
+Q = (((T1-T2)/((L1/(K1*2*%pi*rm_p))+(L2/(K2*2*%pi*rm_i)))));// in W/m
+Q_L = Q*L;
+printf('The rate of conductive heat transfer per 15 m length of pie is %3.2f W\n',Q_L)// Deviation in answer due to direct substitution
diff --git a/2471/CH4/EX4.4/Ex4_4.sce b/2471/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..c7848b2b2 --- /dev/null +++ b/2471/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,24 @@ +clear ;
+clc;
+// Example 4.4
+printf('Example 4.4\n\n');
+printf('Page No. 93\n\n');
+
+// given
+dH = 12*10^-3;// Outer diameter of pipe in m
+dC = 10*10^-3;// Inner diameter of pipe in m
+L = 1*10^-3;// im m
+h_H = 10*10^3;// Heat Transfer Coefficient on vapour side in W/m^2-K
+h_C = 4.5*10^3;// Heat Transfer Coefficient on vapour side in W/m^2-K
+K = 26;// Thermal Conductivity of metal in W/m-K
+dM = (dH + dC)/2;// mean diameter in m
+h_Hf = 6*10^3;// Fouling factor for hot side
+h_Cf = 6*10^3;// Fouling factor for cold side
+
+U = (1/h_H)+((L*dH)/(K*dM))+(dH/(dC*h_C));
+Uh = (1/U);// in W/m^2-K
+printf('The original heat transfer coefficient is %3.0f W/sq.m K \n',Uh )// Deviation in answer due to direct substitution
+
+u = (1/h_H)+(1/h_Hf)+((L*dH)/(K*dM))+(dH/(dC*h_C))+(dH/(dC*h_Cf));
+Uf = (1/u);// in W/m^2-K
+printf('The final heat transfer coefficient due to fouling is %3.0f W/m^2-K \n',ceil(Uf))
diff --git a/2471/CH4/EX4.5/Ex4_5.sce b/2471/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..641a3fbad --- /dev/null +++ b/2471/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,32 @@ +clear ;
+clc;
+// Example 4.5
+printf('Example 4.5\n\n');
+printf('Page No. 95\n\n');
+
+// given
+m_h = 1.05;// Mass flow rate of hot liquid in kg/s
+Thi = 130;// Inlet Temperature of hot liquid in degree celcius
+Tho = 30;// Outlet Temperature of hot fluid in degree celcius
+Cph = 2.45*10^3;// Specific heat capacity of hot liquid in J/kg-K
+
+m_c = 4.10;// Mass flow rate of cold liquid in kg/s
+Tci = 20;// Inlet Temperature of cold liquid in degree celcius
+Cpc = 4.18*10^3;// Specific heat capacity of cold liquid in J/kg-K
+
+A = 6.8;// Area of heat exchanger in m^2
+Q = m_h*Cph*(Thi-Tho);// in Watts
+
+//From heat balance
+// m_c*Cpc*(Tci-Tco)= m_h*Cph*(Thi-Tho)= UAlTm = Q
+Tco = ((Q/(m_c*Cpc))+Tci);
+printf(' The Outlet Temperature of cold fluid is %.0f degree celcius\n',Tco)
+// As counter flow heat exchanger
+T1 = Thi-Tco;
+T2 = Tho-Tci;
+Tm = ((T1-T2)/log(T1/T2));
+
+U = (Q/(A*Tm));
+printf('The overall heat transfer coefficient is %.0f W/sq.m K \n',U)// Deviation in answer due to direct substitution
+
+
diff --git a/2471/CH4/EX4.6/Ex4_6.sce b/2471/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..ec8eda231 --- /dev/null +++ b/2471/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,24 @@ +clear ;
+clc;
+// Example 4.6
+printf('Example 4.6\n\n');
+printf('Page No. 98\n\n');
+
+// given
+v = 1.23;// velocity in m/s
+d = 25*10^-3;// diameter in m
+p = 980;// density in kg/m^3
+u = 0.502*10^-3;// viscosity in Ns/m^2
+Cp = 3.76*10^3;// Specific heat capacity in J/kg-K
+K = 0.532;// Thermal conductivity in W/m-K
+
+Re = (d*v*p)/u;//Reynolds Number
+Pr = (Cp*u)/K;// Prandtl Number
+Re_d = (Re)^0.8;
+Pr_d = (Pr)^0.4;
+
+// By Dittus-Boelter Equation
+//Nu = 0.0232 * Re^0.8 Pr^0.4 = (hd)/K
+Nu = 0.0232 * Re_d * Pr_d;// Nusselt Number
+h = (Nu*K)/d;//W/m^2-K
+printf('The film heat transfer coefficient is %3.2f W/sq.m K\n',h)// Deviation in answer due to direct substitution
diff --git a/2471/CH4/EX4.7/Ex4_7.sce b/2471/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..afe284398 --- /dev/null +++ b/2471/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,32 @@ +clear ;
+clc;
+// Example 4.7
+printf('Example 4.7\n\n');
+printf('Page No. 99\n\n');
+
+// (a) without insulation
+// given
+d_a = 0.150;// Diameter of pipe in m
+T1_a = 60;// Surface temperature in degree celcius
+T2_a = 10;// Ambient temperature in degree celcius
+
+//For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25
+h_a = 1.41*((T1_a-T2_a)/d_a)^0.25;//W/m^2-K
+A_a = %pi * d_a;// Surface Area per unit length in m^2/m
+Q_a = h_a*A_a*(T1_a - T2_a);// in W/m
+printf('The heat loss per unit length without insulation is %.0f W/m \n',ceil(Q_a))
+
+// (b) with insulation
+// given
+d_b = 0.200;// Diameter of pipe in m
+T1_b = 20;// Surface temperature in degree celcius
+T2_b = 10;// Ambient temperature in degree celcius
+
+//For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25
+h_b = 1.41*((T1_b-T2_b)/d_b)^0.25;//W/m^2-K
+A_b = %pi * d_b;// Surface Area per unit length in m^2/m
+Q_b = h_b*A_b*(T1_b - T2_b);// in W/m
+printf('the heat loss per unit length with insulation is %.1f W/m',Q_b)
+// Deviation in answer due to direct substitution
+
+
diff --git a/2471/CH4/EX4.8/Ex4_8.sce b/2471/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..418ccce06 --- /dev/null +++ b/2471/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,16 @@ +clear ;
+clc;
+// Example 4.8
+printf('Example 4.8\n\n');
+printf('Page No. 103\n\n');
+
+// given
+d = 0.100;// Diameter of pipe in m
+T1 = 383;// Surface temperature in Kelvin
+T2 = 288;// Surrounding air temperature in Kelvin
+e = 0.9;// Emissivity of pipe
+A = %pi * d;// Surface Area per unit length in m^2/m
+
+// By Stefan-Blotzmann law, the radiative heat transfer rate is Q = 5.669*e*A*((T1/100)^4-(T2/100)^4)
+Q = 5.669*e*A*((T1/100)^4-(T2/100)^4);// in W/m
+printf('The radiative heat loss per unit length is %.0f W/sq.m',ceil(Q))
diff --git a/2471/CH4/EX4.9/Ex4_9.sce b/2471/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..92657673d --- /dev/null +++ b/2471/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,31 @@ +clear ;
+clc;
+// Example 4.9
+printf('Example 4.9\n\n');
+printf('Page No. 103\n\n');
+
+// given
+A = 1;// Area in m^2
+T1 = 423;// Surface temperature in Kelvin
+T2 = 293;// Surrounding air temperature in Kelvin
+T1_c = 150;// Surface temperature in degree celcius
+T2_c = 20;// Ambient temperature in degree celcius
+e = 0.9;// Emissivity of pipe
+
+//(a) Horizontal Pipe
+d = 0.100;// Diameter of pipe in m
+//For laminar flow in pipe,Q= (1.41*((T1-T2)/d)^0.25)*(T1-T2)
+Q_Ca = (1.41*((T1_c-T2_c)/d)^0.25)*(T1_c-T2_c);// Convective heat transfer rate in W/m^2
+// By Stefan-Blotzmann law, the radiative heat transfer rate is Q = 5.669*e*((T1/100)^4-(T2/100)^4)
+Q_Ra = 5.669*e*((T1/100)^4-(T2/100)^4);// in W/m^2
+Q_Ta = Q_Ra + Q_Ca;// IN W/m^2
+printf('The total heat loss from per square meter area is %.2f W/sq.m\n',Q_Ta)// Deviation in answer due to direct substitution
+
+
+//(b) Vertical Pipe
+//For turbulent flow in pipe,Q= (1.24*(T1-T2)^1.33)
+Q_Cb = (1.24*(T1-T2)^1.33);// Convective heat transfer rate in W/m^2
+// By Stefan-Blotzmann law, the radiative heat transfer rate is Q = 5.669*e*((T1/100)^4-(T2/100)^4)
+Q_Rb = 5.669*e*((T1/100)^4-(T2/100)^4);// in W/m^2
+Q_Tb = Q_Rb + Q_Cb;// IN W/m^2
+printf('The total heat loss from per square meter area is %.0f W/sq.m\n',floor(Q_Tb))
diff --git a/2471/CH5/EX5.1/Ex5_1.sce b/2471/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..1322e523f --- /dev/null +++ b/2471/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+// Example 5.1
+printf('Example 5.1\n\n');
+printf('Page No. 110\n\n');
+
+// given
+Q = 0.30*10^6;// Heat transfer rate in W/sq.m
+T1 = 540;// Mean gas temperature in degree celcius
+T2 = 207;// Steam temperature in degree celcius
+K_tube = 40;// Thermal conductivity of tube in W/m-K
+K_scale = 2.5 ;// Thermal conductivity of scale in W/m-K
+L_tube = 4*10^-3;// Length of tube in m
+
+// By Fourier equation and neglecting curvature effect, Q/A = [(T1- T2)/((L_tube/K_tube)+(L_scale/K_scale))]
+L_scale = K_scale*(((T1-T2)/Q)-(L_tube/K_tube));
+printf('The thickness of scale is %.4f m',L_scale)
diff --git a/2471/CH5/EX5.10/Ex5_10.sce b/2471/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 000000000..4cb35f3c9 --- /dev/null +++ b/2471/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,29 @@ +clear ;
+clc;
+// Example 5.10
+printf('Example 5.10\n\n');
+printf('Page No. 123\n\n');
+
+// given
+m = 0.45;// Mass flow rate in kg/s
+P = 2;// pressure in bar
+T1 = 60;// in degree celcius
+T2 = 250;// in degree celcius
+h_s = 2971*10^3;// Specific enthalpy of superheated steam in J/kg
+h_d = 2706*10^3;// Specific enthalpy of dry saturated steam in J/kg
+h_e = h_s - h_d;//excess Specific enthalpy in J/kg
+h = 251*10^3;// in J/kg
+V_s = 0.885;// specific volume of dry saturated steam at 2bar in m^3/kg
+
+h_r = h_d- h;// heat required to convert water at 60 deg C into dry saturated steam at 2 bar
+w = (h_e/h_r);// in kg/kg
+printf('the quantity of water requried is %.3f kg/kg\n\n',w)
+
+M = m*w;// in kg/s
+printf('the total mass flow rate of water required is %.3f kg/s \n\n',M)
+
+M_d = M + m;// mass flow rate of desuperheated steam in kg/s
+V = M_d*V_s;// in m^3/s
+printf('the total mass flow rate of desuperheated steam required is %.4f m^3/s \n',V)
+// Deviation in answer due to some approximation in answer in the book
+
diff --git a/2471/CH5/EX5.11/Ex5_11.sce b/2471/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..08ded0b32 --- /dev/null +++ b/2471/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 5.11
+printf('Example 5.11\n\n');
+printf('Page No. 130\n\n');
+
+// given
+T1 = 180;// in degree celcius
+T2 = 350;// in degree celcius
+m = 0.5;// Mass flow rate in kg/s
+
+
+//from steam table
+hL_180 = 302*10^3;// Specific enthalpy at 180 degree celcius in J/kg
+hL_350 = 690*10^3;// Specific enthalpy at 350 degree celcius in J/kg
+
+Q = m*(hL_350 - hL_180);// in W
+printf('The required heat is %.0f W',Q)
diff --git a/2471/CH5/EX5.12/Ex5_12.sce b/2471/CH5/EX5.12/Ex5_12.sce new file mode 100755 index 000000000..4e58b0d9f --- /dev/null +++ b/2471/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,19 @@ +clear ;
+clc;
+// Example 5.12
+printf('Example 5.12\n\n');
+printf('Page No. 130\n\n');
+
+// given
+T1 = 200;// in degree celcius
+T2 = 300;// in degree celcius
+m_l = 0.55;// Mass flow rate of liquid in kg/s
+P = 3; //pressure in bar
+Cp = 2.34*10^3;// Mean haet capacity in J/kg-K
+h = 272*10^3;// Latent heat of eutectic mixture at 3 bar
+
+Q = m_l*Cp*(T2 -T1);// in Watts
+m = Q/h;// in kg/s
+printf('The mass flow rate of dry saturated eutectic mixture is %.2f kg/s',m)
+
+
diff --git a/2471/CH5/EX5.13/Ex5_13.sce b/2471/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 000000000..91eecce20 --- /dev/null +++ b/2471/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,38 @@ +clear ;
+clc;
+// Example 5.13
+printf('Example 5.13\n\n');
+printf('Page No. 131\n\n');
+
+// given
+T = 300;// in degree celcius
+v = 2;// velocity in m/s
+d = 40*10^-3;// diameter in m
+
+// From the table 5.3 and 5.4 given in the book
+K_d = [2.80 2.65 2.55 2.75]// in W/m^2-k
+Re = [117*10^3 324*10^3 159*10^3 208*10^3]//Reynolds number
+Pr = [12 4.50 10.0 7.3]//Prandtl Number
+
+// By Dittus-Boelter Equation
+//Nu = 0.0232 * Re^0.8*Pr^0.3 = (hd)/K
+//h = 0.0232 * Re^0.8*Pr^0.3 *(K/d)
+
+h_T = 0.0232 * Re(1)^0.8*Pr(1)^0.3*K_d(1);// //W/m^2-K
+printf('The film heat transfer coefficient using Transcal N is %.0f W/sq.m K \n',h_T)// Deviation in answer due to direct substitution
+
+
+h_D = 0.0232 * Re(2)^0.8*Pr(2)^0.3*K_d(2);// //W/m^2-K
+printf('The film heat transfer coefficient using Dowtherm A is %.0f W/sq.m K \n\n',h_D)// Deviation in answer due to direct substitution
+
+
+h_M = 0.0232 * Re(3)^0.8*Pr(3)^0.3*K_d(3);// //W/m^2-K
+printf('The film heat transfer coefficient using Marlotherm S is %.0f W/sq.m K \n',h_M)// Deviation in answer due to direct substitution
+
+
+h_S = 0.0232 * Re(4)^0.8*Pr(4)^0.3*K_d(4);// //W/m^2-K
+printf('The film heat transfer coefficient using Santotherm 60 is %.0f W/sq.m K \n',h_S)// Deviation in answer due to direct substitution
+
+
+
+
diff --git a/2471/CH5/EX5.14/Ex5_14.sce b/2471/CH5/EX5.14/Ex5_14.sce new file mode 100755 index 000000000..ab4a15249 --- /dev/null +++ b/2471/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,45 @@ +clear ;
+clc;
+// Example 5.14
+printf('Example 5.14\n\n');
+printf('Page No. 137\n\n');
+
+// given
+T1 = 25;// Wet-bulb temperature in degree celcius
+T2 = 40;//Dry-bulb temperature in degree celcius
+
+//By using the humidity chart and steam tables for air-water mixtures at the given temperatures, the all following data can be obtained
+
+//(a) humidity
+w = 0.014;// in kg/kg
+printf('the required humidity is %.3f kg/kg \n',w)
+
+
+//(b) relative humidity
+R_H = 30;// in percentage
+printf('the required relative humidity in percentage is %.0f\n\n',R_H)
+
+//(c) the dew point
+T_w = 20;// in degree celcius
+printf('the required dew-point temperature is %.0f deg C\n',T_w)
+
+//(d) the humid heat
+Cpa = 1.006*10^3;// Heat Capacity of bone dry air in J/kg-K
+Cpwv = 1.89*10^3;// Heat Capacity of water vapour in J/kg-K
+S = Cpa + (w*Cpwv);//in J/kg-K
+printf('the humid heat is %.0f J/kg-K\n\n',S )
+
+//(e) the humid volume
+V_G = ((1/29)+(w/18))*22.41*((T2 + 273)/273);//in m^3/kg
+printf('the humid volume is %.3f m^3/kg \n',V_G)
+
+//(f) adiabatic process
+w_A = 0.020;// in kg/kg
+printf('the humidity of the mixture if saturated adiabatically is %.3f kg/kg \n\n',w_A)
+
+// (h) isothermal process
+w_i = 0.049;// in kg/kg
+printf('the humidity of the mixture if saturated isothermally is %.3f kg/kg \n',w_i)
+
+
+
diff --git a/2471/CH5/EX5.15/Ex5_15.sce b/2471/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 000000000..aeb2df810 --- /dev/null +++ b/2471/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 5.15
+printf('Example 5.15\n\n');
+printf('Page No. 137\n\n');
+
+// given
+T = 25;// Wet-bulb temperature in degree celcius
+T1 = 30;//Dry-bulb temperature in degree celcius
+V = 5;// Volumetric flow rate of initial air-water mixture in m^3/s
+T2 = 70;// Final Dry-bulb temperature in degree celcius
+
+//By using the humidity chart and steam tables for air-water mixtures at the given temperatures, the all following data can be obtained
+w = 0.018;// humidity at 25/30 degree celcius in kg/kg
+Cpa_1 = 1.00*10^3;// Heat Capacity of bone dry air at 30 degree celcius in J/kg-K
+Cpwv_1 = 1.88*10^3;// Heat Capacity of water vapour at 30 degree celcius in J/kg-K
+Cpa_2 = 1.008*10^3;// Heat Capacity of bone dry air at 70 degree celcius in J/kg-K
+Cpwv_2 = 1.93*10^3;// Heat Capacity of water vapour at 70 degree celcius in J/kg-K
+lo = 2.50*10^6;// Specifc Latent heat of vapourisation of water at 0 degree celcius in J/kg
+
+S_1 = Cpa_1 + (w*Cpwv_1);// the humid heat at 30 degree celcius in J/kg-K
+S_2 = Cpa_2 + (w*Cpwv_2);//the humid heat at 70 degree celcius in J/kg-K
+
+hG_1 = ((S_1*T1) + (w*lo));//the specific enthalpy at 30 degree celcius in J/kg
+hG_2 = ((S_2*T2) + (w*lo));//the specific enthalpy at 70 degree celcius in J/kg
+VG_1 = ((1/29)+(w/18))*22.41*((T1 + 273)/273);// Humid volume at 30 degree celcius in m^3/kg
+m = V/VG_1;// Mass flow rate in kg/s
+Q = m*(hG_2 - hG_1);// in Watts
+printf('The required heat is %3.2f W \n',Q)// Deviation in answer is due to some approximation in calculation in the book
+
+w_2 = w;// given in the question
+VG_2 = ((1/29)+(w_2/18))*22.41*((T2 + 273)/273);// Humid volume at 70 degree celcius in m^3/kg
+V_f = m*VG_2;;// in m^3/s
+printf( 'The volumetric flow rate of initial air-water mixture is %3.2f m^3/s',V_f)
+
diff --git a/2471/CH5/EX5.2/Ex5_2.sce b/2471/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..2423b1969 --- /dev/null +++ b/2471/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,43 @@ +clear ;
+clc;
+// Example 5.2
+printf('Example 5.2\n\n');
+printf('Page No. 113\n\n');
+
+// given
+T1 = 10;// in degree celcius
+T2 = 70;// in degree celcius
+d = 25*10^-3;// Inside diameter in m
+v = 1.5;// veocity in m/s
+
+Tm = (T1+T2)/2;// Arithmetic Mean temperature in degree celcius
+// At Tm, All physical properties of water is calculated by using steam table
+
+//(a)Heat absorbed by water
+p = 992;// Density of water in kg/m^3 At Tm
+A = (%pi*d^2)/4;// Area in m^2
+m = p*v*A;// Mass flow rate in kg/s
+h_70 = 293*10^3;// Specific enthalpy of water in J/kg at 70 degree celcius(from steam table)
+h_10 = 42*10^3;// Specific enthalpy of water in J/kg at 10 degree celcius(from steam table)
+Q = m*(h_70 - h_10);// in W
+printf(' Heat absorbed by water is %.0f W \n',Q)
+
+//(b) Film heat transfer
+//At Tm, the following properites of water are found by using steam table
+u = 650*10^-6;// viscosity in Ns/m
+Cp = 4180;//Specific heat in J/kg-s
+K = 0.632;// Thermal conductivity in W/m-s
+
+
+Re = (d*v*p)/u;//Reynolds Number // answer wrongly calculated in the text book
+Pr = (Cp*u)/K;// Prandtl Number
+Re_d = (Re)^0.8;
+Pr_d = (Pr)^0.4;
+
+// By Dittus-Boelter Equation
+//Nu = 0.0232 * Re^0.8 Pr^0.4 = (hd)/K
+Nu = 0.0232 * Re_d * Pr_d;// Nusselt Number
+h = (Nu*K)/d;//W/m^2-K
+printf('The film heat transfer coefficient is %.0f W/sq.m K\n',h)// Deviation in answer due to direct substitution and wrongly calculated in the text book
+
+
diff --git a/2471/CH5/EX5.3/Ex5_3.sce b/2471/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..b1199e2d4 --- /dev/null +++ b/2471/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,19 @@ +clear ;
+clc;
+// Example 5.3
+printf('Example 5.3\n\n');
+printf('Page No. 117\n\n');
+
+// given
+T1 = 25;// in degree celcius
+T2 = 212;// in degree celcius
+x = 0.96;// dryness fraction
+m = 1.25;// Mass flow rate in kg/s
+
+//from steam table
+hL_212 = 907*10^3;// Specific enthalpy at 212 degree celcius in J/kg
+hL_25 = 105*10^3;// Specific enthalpy at 25 degree celcius in J/kg
+l_212 = 1890*10^3;// Latent heat of vapourisation at 212 degree celcius in J/kg
+
+Q = m*((hL_212+(x*l_212))-hL_25);// in W
+printf('The required heat is %.0f W',Q)
diff --git a/2471/CH5/EX5.4/Ex5_4.sce b/2471/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..fd1b65cb4 --- /dev/null +++ b/2471/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,23 @@ +clear ;
+clc;
+// Example 5.4
+printf('Example 5.4\n\n');
+printf('Page No. 117\n\n');
+
+// given
+T = 25;// in degree celcius
+x = 0.96;// dryness fraction
+m = 3.15;// Mass flow rate in kg/s
+CV = 42.6*10^6;// Calorific value in J/kg
+P = 15;// Pressure in bar
+n = 0.8;// Efficiency
+
+//from steam table
+hL_1 = 843*10^3;// Specific enthalpy in J/kg
+hL_2 = 293*10^3;// Specific enthalpy in J/kg
+l_1 = 1946*10^3;// Latent heat of vapourisation at 70 degree celcius in J/kg
+
+Q = m*((hL_1+(x*l_1))-hL_2);// in W
+Q_Ac = Q/n// Actual heat required in Watts
+Oil = Q_Ac/CV;
+printf('The oil required is %.3f kg/s',Oil)
diff --git a/2471/CH5/EX5.5/Ex5_5.sce b/2471/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..8c30a645c --- /dev/null +++ b/2471/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,20 @@ +clear ;
+clc;
+// Example 5.5
+printf('Example 5.5\n\n');
+printf('Page No. 120\n\n');
+
+// given
+T1 = 134;// in degree celcius
+T2 = 100;// in degree celcius
+x = 0.96;// dryness fraction
+m = 0.75;// Mass flow rate in kg/s
+
+//from steam table
+hL_134 = 563*10^3;// Specific enthalpy at 134 degree celcius in J/kg
+hL_100 = 419*10^3;// Specific enthalpy at 100 degree celcius in J/kg
+l_134 = 2162*10^3;// Latent heat of vapourisation at 134 degree celcius in J/kg
+
+Q = m*((hL_134+(x*l_134))-hL_100);// in W
+printf('The required heat is %.0f W',Q)// Deviation in answer due to direct substitution and some approximation in answer in book
+
diff --git a/2471/CH5/EX5.6/Ex5_6.sce b/2471/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..3aff2922d --- /dev/null +++ b/2471/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 5.6
+printf('Example 5.6\n\n');
+printf('Page No. 120\n\n');
+
+// given
+x = 0.90;// dryness fraction
+m = 0.25;// Mass flow rate in kg/s
+P = 0.7;// pressure in bar
+T1 = 10;// in degree celcius
+
+//from steam table
+h_10= 42*10^3;// Specific enthalpy of water at 10 degree celcius in J/kg
+h_25 = 105*10^3;// Specific enthalpy of water at 25 degree celcius in J/kg
+h_30 = 126*10^3;// Specific enthalpy of water at 30 degree celcius in J/kg
+h_s = 2432*10^3;// Specific enthalpy of steam in J/kg
+
+//(a)T2 = 25;
+T2 = 25;// in degree celcius
+// By heat balance, heat transfered at 10 degree celcius = heat gained at 25 degree celcius; "(m*h_s)+(h_10*y)= (m*h_25)+(h_25*y)"; where 'y' is the quqntity of water to be used at 25 degree celcius in kg/s
+y = (m*(h_s-h_25)/(h_25-h_10));
+printf('the quantity of water to be used at 25 degree celcius is %.2f kg/s \n',y)
+
+
+//(b)T2 = 30;
+T2 = 30;// in degree celcius
+// By heat balance, heat transfered at 10 degree celcius = heat gained at 30 degree celcius; "(m*h_s)+(h_10*y)= (m*h_30)+(h_30*y)"; where 'z' is the quqntity of water to be used at 30 degree celcius in kg/s
+z = (m*(h_s-h_30)/(h_30-h_10));
+printf('the quantity of water to be used at 30 degree celcius is %.2f kg/s \n',z)
+
+
+
+
+
diff --git a/2471/CH5/EX5.7/Ex5_7.sce b/2471/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..a97fc2aa1 --- /dev/null +++ b/2471/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 5.7
+printf('Example 5.7\n\n');
+printf('Page No. 121\n\n');
+
+// given
+x = 0.97;// dryness fraction
+m = 4.0;// Mass flow rate in kg/s
+v = 40;// velocity in m/s
+P = 10;// pressure in bar
+
+//from steam table
+Sp_vol = 0.194;// specific volume at 10 bar dry steam in m^3/kg
+
+Q = Sp_vol*x*m// Volumetric flow rate of steam in m^3/s
+d = sqrt((Q*m)/(v*%pi));
+printf('the required diameter of pipe is %.3f m',d)
diff --git a/2471/CH5/EX5.8/Ex5_8.sce b/2471/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 000000000..91e7b44c4 --- /dev/null +++ b/2471/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,17 @@ +clear ;
+clc;
+// Example 5.8
+printf('Example 5.8\n\n');
+printf('Page No. 122\n\n');
+
+// given
+T1 = 25;// in degree celcius
+T2 = 450;// in degree celcius
+m = 7.5;// Mass flow rate in kg/s
+
+//from steam table
+hL_450 = 3303*10^3;// Specific enthalpy at 450 degree celcius in J/kg
+hL_25 = 105*10^3;// Specific enthalpy at 25 degree celcius in J/kg
+
+Q = m*(hL_450 - hL_25);// in W
+printf('The required heat is %.0f W',Q)// Deviation in answer due to direct substitution and some approximation in answer in book
diff --git a/2471/CH5/EX5.9/Ex5_9.sce b/2471/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..34f05e95f --- /dev/null +++ b/2471/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,27 @@ +clear ;
+clc;
+// Example 5.9
+printf('Example 5.9\n\n');
+printf('Page No. 122\n\n');
+
+// given
+P1 = 15;// Pressure at state 1 in bar
+P2 = 1.5;// Pressure at state 2 in bar
+T1 = 198;// in degree celcius
+
+// as the process is adiabatic; => Q = 0; => ehthalpy at state1 = enthalpy at state 2
+h_1 = 2789*10^3;// specific enthalpy at state 1 in J/kg
+h_2 = h_1;//specific enthalpy at state 2 in J/kg
+
+T3 = 150;// in degree celcius
+T4 = 200;// in degree celcius
+h_3 = 2773*10^3;// specific enthalpy at state 3 in J/kg
+h_4 = 2873*10^3;// specific enthalpy at state 4 in J/kg
+
+// Assuming a liner realtionship between temperature and enthalpy for the temperature range 150-200 degree celcius
+h = ((h_4 - h_3)/(T4 - T3));// specific enthalpy per degree celcius in J/kg-degC
+t = ((h_2 - h_3)/h);// in degree celcius
+T2 = T3 + t;// in degree celcius
+printf('the temperature of the final superheated steam at 1.5 bar is %.0f deg C',T2)
+
+
diff --git a/2471/CH6/EX6.1/Ex6_1.sce b/2471/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..d111dd953 --- /dev/null +++ b/2471/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,23 @@ +clear ;
+clc;
+// Example 6.1
+printf('Example 6.1\n\n');
+printf('Page No. 142\n\n');
+
+// given
+L = 2.5;// Length of tubes in metre
+Do = 10*10^-3;// Internal diameter of tubes in metre
+m = 3.46;// mass flow rate in kg/s
+Th = 120;// Temperature of condening steam in degree celcius
+Tl_i = 20;// Inlet temperature of liquid in degree celcius
+Tl_o = 80;// Outlet temperature of liquid in degree celcius
+Cp = 2.35*10^3;// Specific heat capacity of liquid in J/kg-K
+U = 950;// Overall heat transfer coefficent in W/m^2-K
+
+T1 = Th- Tl_i;// in degree celcius
+T2 = Th- Tl_o;// in degree celcius
+Tm = ((T2-T1)/log(T2/T1));// logarithmic mean temperature of pipe in degree celcius
+a = %pi*Do*L;//Surface area per tube in m^2
+A = ((m*Cp*(Tl_o - Tl_i))/(U*Tm));// in m^2
+N = A/a;
+printf('The number of tubes required is %3.0f',N)
diff --git a/2471/CH6/EX6.10/Ex6_10.sce b/2471/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 000000000..fde8f8679 --- /dev/null +++ b/2471/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,46 @@ +clear ;
+clc;
+// Example 6.10
+printf('Example 6.10\n\n');
+printf('Page No. 158\n\n');
+
+// given
+//for Boiler-1
+P_1 = 15;// Boiler pressure in bar
+Ts_1 = 300;// Steam temperature in degree celcius
+Tf_1 = 80;// Feed water temperature in degree celcius
+X_1 = 0;// Steam dryness fraction
+m_s1 = 9000;// steam rate in kg/h
+m_f1 = 700;// Gas rate in kg/h
+G_CV1 = 43.0*10^6;// In J/kg
+//from steam table,at P = 15 bar and at given temperatures
+h2_1 = 3039*10^3;//Specific enthalpy of steam in J/kg
+h1_1 = 335*10^3;//Specific enthalpy of feed steam in J/kg
+
+E_G1 = ((m_s1*(h2_1-h1_1)*100)/(m_f1*G_CV1));//
+printf('The gross efficiency percentage is %.0f \n',E_G1)
+Ee_1 = ((m_s1/m_f1)*(h2_1-h1_1))/(2257*10^3);
+printf('the equivalent evaporation for boiler-1 is %3.1f kg \n\n',Ee_1)
+
+//for Boiler-2
+P_2 = 10;// Boiler pressure in bar
+Ts_2 = 180;// Steam temperature in degree celcius
+Tf_2 = 60;// Feed water temperature in degree celcius
+X_2 = 0.96;// Steam dryness fraction
+m_s2 = 7000;// steam rate in kg/h
+m_f2 = 510;// Gas rate in kg/h
+G_CV2 = 43.0*10^6;// In J/kg
+//from steam table,AT 10 bar and at temperature T = Ts_2
+h2 = (763+(X_2*2013))*10^3;//Specific enthalpy of steam in J/kg
+//At temperature T = Tf_2
+h1 = 251*10^3;//Specific enthalpy of feed steam in J/kg
+
+E_G2 = ((m_s2*(h2-h1)*100)/(m_f2*G_CV2));//
+printf('The gross efficiency percentage is %.0f\n',E_G2)
+Ee_2 = ((m_s2/m_f2)*(h2-h1))/(2257*10^3);
+printf('the equivalent evaporation for boiler-2 is %3.1f kg',Ee_2)
+
+
+
+
+
diff --git a/2471/CH6/EX6.11/Ex6_11.sce b/2471/CH6/EX6.11/Ex6_11.sce new file mode 100755 index 000000000..f3dfafd79 --- /dev/null +++ b/2471/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,37 @@ +clear ;
+clc;
+// Example 6.11
+printf('Example 6.11\n\n');
+printf('Page No. 167\n\n');
+
+// given
+m = 10*10^3;// Production of boiler in kg/h
+X = 0.95;//Dryness fraction
+P = 10;//Pressure ib bar
+T_fw = 95;// Feed water temperature in degree celcius
+T_mf = 230;// Mean flue gae temperature in degree celcius
+T_mb = 25;// Mean boiler house temperature in degree celcius
+Coal_c = 900;// Coal consumption in kg/h
+A = 0.08;// Ash content in coal
+C_c = 0.15;//carbon content in coal
+CV_coal = 33.50*10^6;// Calorific value of coal in J
+M = 28;// Mass of flue gas per kg coal in kg
+Cp = 1.05*10^3;// Mean Specific heat capacity of the flue gas in J/kg-K
+CV_c = 34*10^6;// Calorific value of carbon in J/kg
+
+M_s = m/Coal_c;// Mass of steam produced per kg coal in kg
+H_w = (M_s*(763+(X*2013) - 398)*10^3)/10^6;// Heat absorbed by water per kg coal in 10^6 J(from steam table at given pressure and dryness fraction)
+H_f = (M*Cp*(T_mf - T_mb))/10^6;// Heat in flue gas in 10^6 J
+H_uc = (A*C_c*CV_c)/10^6;//Heat in unburnt carbon in 10^6 J
+h_sup = (CV_coal)/10^6;// Heat supplied by coal in 10^6 J
+un_acc = (h_sup - (H_w + H_f + H_uc));// unaccounted heat losses in 10^6 J
+a = (h_sup/h_sup)*100;
+b = (H_w/h_sup)*100;
+c = (H_f/h_sup)*100;
+d = (H_uc/h_sup)*100;
+e = (un_acc/h_sup)*100;
+T = b + c + d + e;
+printf(' THERMAL BALANCE SHEET :\n\t\t\t\t 10^6 J \t percentage \n\n Heat supplied by coal \t\t %.2f \t\t %.0f\n Heat absorbed by water \t %.1f\t\t %.1f\n Heat in flue gas \t\t %.2f \t\t %.0f\n Heat in unburnt carbon \t %.2f \t\t %.1f \n unaccounted heat losses \t %.2f \t\t %.1f\n TOTAL \t\t\t\t %.2f \t\t %.1f',h_sup,a,H_w,b,H_f,c,H_uc,d,un_acc,e,h_sup,T);
+
+
+
diff --git a/2471/CH6/EX6.12/Ex6_12.sce b/2471/CH6/EX6.12/Ex6_12.sce new file mode 100755 index 000000000..d1d62b85a --- /dev/null +++ b/2471/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,179 @@ +clear ;
+clc;
+// Example 6.12
+printf('Example 6.12\n\n');
+printf('Page No. 168\n\n');
+
+// given
+C_Rate = 2920;// Coal consumption rate in kg/h
+S_Rate = 22.5*10^3;// Steam consumption rate in kg/h
+Ps = 20;// Steam pressure in bar
+Ts = 350;// Steam Temperature in degree celcius
+Tf_in = 70;// Feed water temperature inlet economiser in degree celcius
+Tf_out = 110;// Feed water temperature outlet economiser in degree celcius
+Tm_b = 25;// Mean Boiler house temperature in degree celcius
+Tm_f = 260;// Mean exit flue gas temperature in degree celcius
+CO2_f = 15.8;// CO2 content of dry exit flue gas by volume
+CO_f = 0;// CO content of dry exit flue gas by volume
+C_ash = 0.025;// Carbon in ash in [%]
+G = 0.005;// Grit produced in [%]
+//Analysis of coal(as fired)
+M = 0.105;// Moisture [%]
+VM = 0.308;//Volatile matter [%]
+FC = 0.497;// FIxed carbon [%]
+Ash =0.09;// ASh [%]
+C = 0.66;// Carbon percentage - [%]
+H2 = 0.042;// Hydrogen percentage - [%]
+S = 0.015;// Sulphur percentage - [%]
+N2 = 0.012;// Nitrogen percentage - [%]
+O2 = 0.076;// Oxygen percentage - [%]
+H20 = 0.105;// Moisture percentage - [%]
+G_CV = 26.90;// Gross Calorific Value in 10^6 J/kg
+CV_C = 33.8*10^6;// Calorif Value of carbon in J/kg
+CV_G = 33.8*10^6;// Calorif Value of Grit in J/kg
+Ps_l = 20;// Pressure of steam leaving the boiler in bar
+
+//(a) Calculation of excess air usage
+//(a.1) Theoretical oxygen requirement
+F = 1;// Fuel feed required in kg
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_S = 32; //mol. weight of S
+w_N2 = 28; // mol. weight of N2
+w_O2 = 32; // mol. weight of O2
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_S = S /w_S;//kmol of S
+mol_N2 = N2 /w_N2;//kmol of N2
+mol_O2 = O2 /w_O2;//kmol of O2
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2;// in kmol
+N2_air = (O2_req*76.8)/23.2;// in kmol (considering air consists of 76.8% N2 and 23.2% O2 )
+printf('(a.1) \n')
+printf('Total number of kmol of O2 required per kg of fuel is %3.4f kmol \n',O2_req)
+printf('N2 associated with O2 is %3.4f kmol \n',N2_air)
+
+//(a.2) Theoretical CO2 content of dry flue gas
+T = C_req + S_req + mol_N2 + N2_air;// Total flue gas in kmol
+CO2 = (C_req/T)*100;// in [%]
+printf('(a.2) \n')
+printf('Theoretical CO2 content of dry flue gas in percentage is %3.1f \n',CO2)
+
+//(a.3)Excess air based on CO2 content
+Ex_air = ((CO2 - CO2_f)/CO2_f)*100;// in [%]
+printf('(a.3) \n')
+printf('Excess air based on CO2 content in percentage is %.0f \n\n',floor(Ex_air))
+
+
+//(b) Fuel gas components
+//(b.1) Composition per kg fuel
+w_CO2 = 44;// mol. weight of CO2
+w_SO2 = 64;// mol. weight of SO2
+// FOR DRY GAS
+CO2_d = C_req * w_CO2;// In kg/kg
+SO2_d = S_req * w_SO2;// In kg/kg
+N2_d = mol_N2 * w_N2;// N2 from fuel In kg/kg
+N2_air_d = N2_air * w_N2;// N2 from air In kg/kg
+T_N2 = N2_d + N2_air_d;// In kg/kg
+T_dry = CO2_d + SO2_d + T_N2;// In kg/kg
+printf('(b.1) \n')
+printf('Composition of dry gas \n')
+printf('CO2 %.3f \n',CO2_d)
+printf('SO2 %.2f \n',SO2_d)
+printf('N2 from fuel %.2f \n',N2_d)
+printf('N2 from air %.2f \n',N2_air_d)
+printf('Total dry air %.2f kg/kg \n\n',T_dry)
+
+//FOR WET GAS
+w_H2O = 18;// mol. weight of H2O
+H2O_f = M;// H2O from fuel
+H2O_H2 = mol_H2 * w_H2O;// H2O from H2
+T_H2O = H2O_f + H2O_H2;// in kg/kg
+printf('Composition of wet gas \n')
+printf('H2O from fuel %.3f \n',H2O_f)
+printf('H2O from H2 %.3f \n',H2O_H2)
+printf('Total H2O in wet gas %.3f kg/kg \n\n',T_H2O)
+
+//FOR DRY EXCESS AIR
+O2_dry_ex = O2_req * w_O2 *0.3;//in kg/kg
+N2_dry_ex = N2_air * w_N2 *0.3;//in kg/kg
+T_dry_ex = O2_dry_ex + N2_dry_ex;// in kg/kg
+printf('Composition of dry excess air \n')
+printf('O2 %.3f \n',O2_dry_ex)
+printf('N2 %.3f \n',N2_dry_ex)
+printf('Total dry excess air %.3f kg/kg \n\n',T_dry_ex)
+
+//(b.2) Enthalpy
+// From steam table or from the appendix C.2; at the given pressure and temperatures, the following specific heat capacity for different gases are obtained
+Cp_CO2_T1 = 1.04*10^3;// Specific heat Capacity of CO2 at temperature Tm_f in J/kg-K
+Cp_CO2_T2 = 0.85*10^3;// Specific heat Capacity of CO2 at temperature Tm_b in J/kg-K
+Cp_SO2_T1 = 0.73*10^3;// Specific heat Capacity of SO2 at temperature Tm_f in J/kg-K
+Cp_SO2_T2 = 0.62*10^3;// Specific heat Capacity of SO2 at temperature Tm_b in J/kg-K
+Cp_N2_T1 = 1.07*10^3;// Specific heat Capacity of N2 at temperature Tm_f in J/kg-K
+Cp_N2_T2 = 1.06*10^3;// Specific heat Capacity of N2 at temperature Tm_b in J/kg-K
+Cp_O2_T1 = 0.99*10^3;// Specific heat Capacity of O2 at temperature Tm_f in J/kg-K
+Cp_O2_T2 = 0.91*10^3;// Specific heat Capacity of O2 at temperature Tm_b in J/kg-K
+
+Cp_dry_T1 = ((CO2_d * Cp_CO2_T1) + (SO2_d * Cp_SO2_T1) + (T_N2 * Cp_N2_T1))/T_dry;// in J/kg-K
+Cp_dry_T2 = ((CO2_d * Cp_CO2_T2) + (SO2_d * Cp_SO2_T2) + (T_N2 * Cp_N2_T2))/T_dry;// in J/kg-K
+Cp_air_T1 = ((O2_dry_ex * Cp_O2_T1) + (N2_dry_ex * Cp_N2_T1))/T_dry_ex;// in J/kg-K
+Cp_air_T2 = ((O2_dry_ex * Cp_O2_T2) + (N2_dry_ex * Cp_N2_T2))/T_dry_ex;// in J/kg-K
+printf('(b.2) \n')
+printf('Specific heat Capacity of dry gas at 260 deg C is %.0f J/kg-K \n',Cp_dry_T1)
+printf('Specific heat Capacity of dry gas at 25 deg C is %.0f J/kg-K \n',Cp_dry_T2)
+printf('Specific heat Capacity of dry excess air at 260 deg C is %.0f J/kg-K \n',Cp_air_T1)
+printf('Specific heat Capacity of dry excess air at 25 deg C is %.0f J/kg-K \n\n',Cp_air_T2)
+
+// From Steam table or Appendix B.3, Enthalpy of superheated steam is obtained at 260 deg C and 1 bar
+E_s = 2995*10^3;//in J/kg-K
+
+//(c) Heat transferred to water
+E_w = S_Rate / C_Rate;// Evaporation of water per kg of fuel in kg
+E = (E_w*(461 - 293)*10^3)/10^6;// in 10^6 J
+B = (E_w*(2797 - 461)*10^3)/10^6;// in 10^6 J
+S = (E_w*(3139 - 2797)*10^3)/10^6;// in 10^6 J
+printf('(c) \n')
+printf('Heat to water in Economiser is %.1f *10^6 J \n',E)
+printf('Heat to water in Boiler is %.2f *10^6 J \n',B)
+printf('Heat to water in Superheater is %.2f *10^6 J \n\n',S)
+
+//(d) Heat loss in flue gas
+hl = 105*10^3;// Enthalpy of steam at 25 deg C (from steam table) in J/kg-K
+loss_dry = T_dry*((Tm_f*Cp_dry_T1) - (Tm_b*Cp_dry_T2))/10^6;// in 10^6 J
+loss_wet = T_H2O*(E_s - hl)/10^6;// in 10^6 J
+loss_ex_air = T_dry_ex*((Tm_f*Cp_air_T1) - (Tm_b*Cp_air_T2))/10^6;// in 10^6 J
+printf('(d) \n')
+printf('Heat loss in dry flue gas is %.2f *10^6 J \n',loss_dry)
+printf('Heat loss in wet flue gas is %.2f *10^6 J \n',loss_wet)
+printf('Heat loss in dry excess air is %.2f *10^6 J \n\n',loss_ex_air)
+
+//(e) Heat loss in combustile matter in ash
+loss_ash = (Ash * C_ash * CV_C)/10^6;// in 10^6 J
+printf('(e) Heat loss in combustile matter in ash is %.2f *10^6 J \n',loss_ash)
+
+//(f) Heat loss in grit
+loss_grit = (G * CV_G)/10^6;// in 10^6 J
+printf('(f) Heat loss in grit is %.2f *10^6 J \n\n',loss_grit)
+
+//(g) Radiation and unaccounted heat loss
+h_sup = G_CV;// Heat supplied by the coal in 10^6 J
+loss_rad = (h_sup - (E + B + S + loss_dry + loss_wet + loss_ex_air + loss_ash + loss_grit));// Radiation and unaccounted loss in 10^6 J
+a = (h_sup/h_sup)*100;
+b = (E/h_sup)*100;
+c = (B/h_sup)*100;
+d = (S/h_sup)*100;
+e = (loss_dry/h_sup)*100;
+f = (loss_wet/h_sup)*100;
+g = (loss_ex_air/h_sup)*100;
+h = (loss_ash/h_sup)*100;
+i = (loss_grit/h_sup)*100;
+j = (loss_rad/h_sup)*100;
+T = b + c + d + e + f + g + h + i + j;
+printf('(g) THERMAL BALANCE SHEET :\n\t\t\t\t 10^6 J \t percentage \n Heat supplied by coal \t\t %.2f \t\t %.0f\n Heat to loss in : economiser \t %.2f \t\t %.1f\n \t\t boiler \t %.2f \t\t %.0f\n \t\t superheater %.2f \t\t %.1f\n Heat loss in : dry flue gas %.2f \t\t %.1f\n \t\t wet flue gas %.2f \t\t %.1f\n \t dry eecess air %.2f \t\t %.1f\n Heat loss in ash \t\t %.2f \t\t %.1f\n Heat loss in grit \t\t %.2f \t\t %.1f\n Radiation and unaccounted loss %.1f \t\t %.1f\n TOTAL \t\t\t\t %.2f \t\t %.1f',h_sup,a,E,b,B,c,S,d,loss_dry,e,loss_wet,f,loss_ex_air,g,loss_ash,h,loss_grit,i,loss_rad,j,h_sup,T)
+
+
+
diff --git a/2471/CH6/EX6.13/Ex6_13.sce b/2471/CH6/EX6.13/Ex6_13.sce new file mode 100755 index 000000000..b03877cd9 --- /dev/null +++ b/2471/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,57 @@ +clear ;
+clc;
+// Example 6.13
+printf('Example 6.13\n\n');
+printf('Page No. 188\n\n');
+
+// given
+P = 1.5;// Pressure in bar
+T = 111;// Temperature in degree celcius
+m = 2;// mass flow rate of process liquid in kg/s
+Cp = 4.01*10^3;// Mean Specific heat capacity in J/kg_K
+Tl_i = 20;// Inlet temperature of liquid in degree celcius
+Tl_o = 90;// Outlet temperature of liquid in degree celcius
+Ps = 15;// Pressure of steam in bar
+X = 0.97;// Dryness fraction of steam
+Pa = 1.5;//Pressure after adiabatic expansion in bar
+Ta = 80;// Temperature of injecting condensate in degree celcius
+
+//(a)
+Q = m*Cp*(Tl_o - Tl_i);// in W
+L = 2227*10^3;// Latent heat of 1.5 bar steam in J/kg
+m_a = Q/L;
+printf('(a) Mass flow rate of 1.5 bar steam is %.3f kg/s \n',m_a)
+
+//(b)
+//from steam table, Specific enthalpy of 0.97 dry 15 bar absolute steam
+h = ((843+(X*1946))*10^3);// in J/kg
+//the balance for the desuperheater,when y is the mass flow rate(kg/s) of condensate at 80 deg C is,on the basis of 1kg/s of superheated steam: => (1*2731*10^3)+(335*10^3*y)=(1+y)*2693*10^3
+y = (((2731-2693)*10^3)/((2693-335)*10^3))// in kg/s
+m_b = m_a/(1+y);// in kg/s
+printf('(b) Mass flow rate of 15 bar steam is %.3f kg/s \n',m_b)
+
+//(c)
+m_c = y*m_b;//in kg/s
+printf('(c) Mass flow rate of condensateis %.3f kg/s\n',m_c)
+
+//(d)
+v = 30;// steam velocity in m/s
+//from steam table
+V = 1.16;// Specific volum of 1.5 bar saturated steam in m^3/kg
+V_d = V*m_a;// in m^3/s
+d = ((V_d*4)/(v*%pi))^0.5;// im m
+printf('(d) The vapour main diameter is %3.2f m \n',d)
+
+//(e)
+l = 2.5;// Length of tubes in m
+d_i = 10*10^-3;// Internal Diameter of tube in m
+U = 1500;// Overall heat transfer coefficent in W/m^2-K
+
+a = %pi*d_i*l;// in m^2
+T1 = T - Tl_i;// in degree celcius
+T2 = T - Tl_o;// in degree celcius
+Tm = ((T2-T1)/log(T2/T1));// logarithmic mean temperature of pipe in degree celcius
+A = Q/(U*Tm);// in m^2
+N = A/a;
+printf('(e) The number of tubes required is %3.0f \n',N)
+
diff --git a/2471/CH6/EX6.2/Ex6_2.sce b/2471/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..a4605e6df --- /dev/null +++ b/2471/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 6.2
+printf('Example 6.2\n\n');
+printf('Page No. 142\n\n');
+
+// given
+v = 1.50;// velocity in m/s
+N_t = 100;// Number of tubes
+Do = 10*10^-3;// Internal diameter of tubes in metre
+m = 3.46;// mass flow rate in kg/s
+p = 1180;// density in kg/m^3
+
+A = (N_t*%pi*Do^2)/4;// otal cross-sectional area in m^2
+V = m/p;//Volumetric flow rate in m^3/s
+Fv = V/A;// Fluid velocity in m/s
+N_p = v/Fv;
+printf('the number of passes is %.0f',N_p)
diff --git a/2471/CH6/EX6.3/Ex6_3.sce b/2471/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..7308da1ce --- /dev/null +++ b/2471/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,24 @@ +clear ;
+clc;
+// Example 6.3
+printf('Example 6.3\n\n');
+printf('Page No. 144\n\n');
+
+// given
+Th_i = 130;//Inlet temperature of hot liquid in degree celcius
+Th_o = 90;// Outlet temperature of hot liquid in degree celcius
+Tc_i = 20;// Inlet temperature of cold liquid in degree celcius
+Tc_o = 50;// Outlet temperature of cold liquid in degree celcius
+
+//For Couter-current flow
+T1 = Th_i - Tc_o;
+T2 = Th_o - Tc_i;
+Tm_1 = ((T2-T1)/log(T2/T1));
+printf('The logarithmic mean temperature difference for counter-current flow is %.0f degree celcius \n',Tm_1)
+
+
+//For Co-current flow
+T3 = Th_i - Tc_i;
+T4 = Th_o - Tc_o;
+Tm_2 = ((T3-T4)/log(T3/T4));
+printf('The logarithmic mean temperature difference for co-current flow is %.0f degree celcius \n',Tm_2)
diff --git a/2471/CH6/EX6.4/Ex6_4.sce b/2471/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..aa883d858 --- /dev/null +++ b/2471/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,49 @@ +clear ;
+clc;
+// Example 6.4
+printf('Example 6.4\n\n');
+printf('Page No. 147\n\n');
+
+// given
+F = 1;// Fuel feed required in kg
+//By ultimate analysis of feed
+C = 0.86;// Carbon percentage - [%]
+H2 = 0.05;// Hydrogen percentage - [%]
+S = 0.001;// Sulphur percentage - [%]
+O2 = 0.08;// Oxygen percentage - [%]
+
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_O2 = 32; // mol. weight of O2
+w_S = 32; //mol. weight of S
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_O2 = O2 /w_O2;//kmol of O2
+mol_S = S /w_S;//kmol of S
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2;// in kmol
+printf('Total number of kmol of O2 required per kg of fuel is %3.3f kmol \n',O2_req)
+m_O2 = O2_req*w_O2;// Mass of O2 required per kg of fuel
+printf('Mass of O2 required per kg of fuel is %3.1f kg \n\n',m_O2)
+//Calculation of air
+m_air = m_O2/0.232;// in kg
+printf('Mass of air required per kg of fuel is %3.1f kg \n',m_air')
+//Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
+R = 8310;//Universal gas constant in J/kmol-K
+T = (273+20);// in K
+P = 1.013*10^5;// in N/m^2
+n = 1;// 1 kmol of air
+V_kmol = (n*R*T)/P;// In m^3/kmol
+M_air = 29;// Mol. weight of air
+V_kg = V_kmol/M_air;// in m^3/kg
+V_air = m_air*V_kg;// in m^3
+printf('Volume of air required is %3.1f m^3 \n',V_air')
+//Deviation in answer is due to some approximation in calculation in the book
+
+
+
+
diff --git a/2471/CH6/EX6.5/Ex6_5.sce b/2471/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..865d480c1 --- /dev/null +++ b/2471/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,51 @@ +clear ;
+clc;
+// Example 6.5
+printf('Example 6.5\n\n');
+printf('Page No. 148\n\n');
+
+// given
+F = 1;// Weight of coal in kg
+//By analysis of coal in weight basis
+C = 0.74;// Carbon percentage - [%]
+H2 = 0.05;// Hydrogen percentage - [%]
+S = 0.01;// Sulphur percentage - [%]
+N2 = 0.001;// Nitrogen percentage - [%]
+O2 = 0.05;// Oxygen percentage - [%]
+H20 = 0.09;// Moisture percentage - [%]
+Ash = 0.05;// Ash percentage - [%]
+
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_O2 = 32; // mol. weight of O2
+w_S = 32; //mol. weight of S
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_O2 = O2 /w_O2;//kmol of O2
+mol_S = S /w_S;//kmol of S
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2;// Total number of kmol of O2 required per kg of fuel in kmol
+m_O2 = O2_req*w_O2;// Mass of O2 required per kg of fuel
+printf('Mass of O2 required per kg of fuel is %3.2f kg \n',m_O2)
+//Calculation of air
+m_air = m_O2/0.232;// in kg
+printf('Mass of air required per kg of fuel is %3.1f kg \n',m_air')
+//Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
+R = 8310;//Universal gas constant in J/kmol-K
+T = (273+0);// in K
+P = 1.013*10^5;// in N/m^2
+n = 1;// 1 kmol of air
+V_kmol = (n*R*T)/P;// In m^3/kmol
+M_air = 29;// Mol. weight of air
+V_kg = V_kmol/M_air;// in m^3/kg
+V_air = m_air*V_kg;// in m^3
+printf('Volume of air required is %3.1f m^3\n',V_air')
+
+
+
+
+
diff --git a/2471/CH6/EX6.6/Ex6_6.sce b/2471/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..e5c4db907 --- /dev/null +++ b/2471/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,52 @@ +clear ;
+clc;
+// Example 6.6
+printf('Example 6.6\n\n');
+printf('Page No. 149\n\n');
+
+// given
+F = 1;// Fuel feed in kg
+C = 0.86;// Mass of Carbon in kg
+H2 = 0.05;// Mass of Hydrogen in kg
+S = 0.01;// Mass of Sulphur in kg
+O2 = 0.08;// Mass of Oxygen in kg
+
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_O2 = 32; // mol. weight of O2
+w_S = 32; //mol. weight of S
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_O2 = O2 /w_O2;//kmol of O2
+mol_S = S /w_S;//kmol of S
+//By kmol of product
+CO2 = mol_C*1;// CO2 formed by the reaction C + O2 -> CO2
+H2O = mol_H2*1;// H2O formed by the reaction H2 + (1/2)O2 -> H2O
+SO2 = mol_S*1;// SO2 formed by the reaction S + O2 -> SO2
+Pdt = CO2 + H2O + SO2;// Total kmol of combustion products in kmol
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2// Total number of kmol of O2 required per kg of fuel in kmol
+
+N2 = (O2_req*79)/21;// in kmol (considering air consists of 79% N2 and 21% O2 by moles)
+Wet_pdts = Pdt + N2;// Wet combustion products in kmol
+
+//Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
+R = 8310;//Universal gas constant in J/kmol-K
+T = (273+0);// in K
+P = 1.013*10^5;// in N/m^2
+n_wet = Wet_pdts;// in kmol
+V_wet = (n_wet*R*T)/P;// In m^3
+n_dry = n_wet - H2O;//in kmol
+V_dry = (n_dry*R*T)/P;// In m^3
+
+printf('Volume of wet flue gas is %3.2f m^3 \n',V_wet)
+printf('Volume of dry flue gas is %3.2f m^3',V_dry)
+
+
+
+
+
diff --git a/2471/CH6/EX6.7/Ex6_7.sce b/2471/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..b97e66a3b --- /dev/null +++ b/2471/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,72 @@ +clear ;
+clc;
+// Example 6.7
+printf('Example 6.7\n\n');
+printf('Page No. 150\n\n');
+
+// given
+F = 1;// Weight of fuel in kg
+e = 0.5;// excess air percentage
+C = 0.74;// Mass of Carbon in kg
+H2 = 0.05;// Mass of Hydrogen in kg
+S = 0.01;// Mass of Sulphur in kg
+N2 = 0.001;//Mass of Nitrogen in kg
+O2 = 0.05;// Mass of Oxygen in kg
+H2O = 0.09;// Mass of Moisture in kg
+Ash = 0.05;// Mass of Ash in kg
+
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_O2 = 32; // mol. weight of O2
+w_S = 32; //mol. weight of S
+w_N2 = 28;// mol. weight of N2
+w_H20 = 18;// mol. weight of H2O
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_O2 = O2 /w_O2;//kmol of O2
+mol_S = S /w_S;//kmol of S
+mol_N2 = N2 /w_N2;//kmol of N2
+mol_H2O = H2O /w_H20;//kmol of H20
+
+//By kmol of product
+CO2 = mol_C*1;// CO2 formed by the reaction C + O2 -> CO2
+H2O_air = mol_H2*1;// H2O formed by the reaction H2 + (1/2)O2 -> H2O
+SO2 = mol_S*1;// SO2 formed by the reaction S + O2 -> SO2
+Pdt = CO2 + H2O_air + SO2 + mol_N2 + mol_H2O;// Total kmol of combustion products in kmol
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2;// Total number of kmol of O2 required per kg of fuel in kmol
+
+Ex_O2 = O2_req*e;// Amount of excess oxygen in kmol
+
+N2_air = (O2_req*(1+e)*79)/21;// in kmol (considering air consists of 79% N2 and 21% O2 by moles)
+N2_flue = mol_N2 + N2_air;// Total N2 in flue gas in kmol
+H2O_flue = mol_H2O+ H2O_air;// Total H2O in flue gas in kmol
+
+T_wet = CO2 + H2O_flue + SO2 + Ex_O2 + N2_flue;//Total components of flue gas on a wet basis in kmol
+T_dry = CO2 + SO2 + Ex_O2 + N2_flue;//Total components of flue gas on a dry basis in kmol
+H2O_dry = 0;
+C_wet = ((CO2 / T_wet)*100);// in percentage
+H_wet = ((H2O_flue/T_wet)*100);// in percentage
+S_wet = ((SO2/T_wet)*100);// in percentage
+N_wet = ((N2_flue/T_wet)*100);// in percentage
+O_wet = ((Ex_O2/T_wet)*100);// in percentage
+
+C_dry = ((CO2 / T_dry)*100);// in percentage
+H_dry = ((H2O_dry/T_dry)*100);// in percentage
+S_dry = ((SO2/T_dry)*100);// in percentage
+N_dry = ((N2_flue/T_dry)*100);// in percentage
+O_dry = ((Ex_O2/T_dry)*100);// in percentage
+T1 = C_wet + H_wet + S_wet + N_wet + O_wet;// in percentage
+T2 = C_dry + S_dry + N_dry + O_dry;// in percentage
+printf('\t\t kmol \t\t percent composition by volume\n Component \t Wet \t Dry \t\t Wet \t Dry \n CO2 \t %.4f %.4f \t\t %.1f \t %.1f \n H2O \t %.4f %.0f \t\t\t %.1f \t\t %.1f \n SO2 \t %.4f %.4f \t\t %.1f \t\t %.1f \n N2 \t\t %.4f %.4f \t\t %.1f \t %.1f \n O2 \t\t %.4f %.4f \t\t %.1f \t\t %.1f \n TOTAL \t %.4f %.4f \t\t %.0f \t\t %.0f',CO2,CO2,C_wet,C_dry,H2O_flue, H2O_dry,H_wet,H_dry,SO2,SO2,S_wet,S_dry,N2_flue, N2_flue,N_wet,N_dry,Ex_O2,Ex_O2,O_wet,O_dry,T_wet,T_dry,T1,T2)
+//Deviation in answes is due to some calculation approxiamation in the book.
+
+
+
+
+
+
diff --git a/2471/CH6/EX6.8/Ex6_8.sce b/2471/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..01467a716 --- /dev/null +++ b/2471/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,20 @@ +clear ;
+clc;
+// Example 6.8
+printf('Example 6.8\n\n');
+printf('Page No. 156\n\n');
+
+// given
+H = 0.05;// Hydrogen percentage - [%]
+O = 0.08;// Oxygen percentage - [%]
+C = 0.86;// Carbon percentage - [%]
+S = 0.001;// Sulphur percentage - [%]
+
+G_CV = ((33.9*C)+143*(H-(O/8))+(9.1*S))*10^6;
+printf('The gross calorific value is %3.2e J/kg \n',G_CV)
+
+
+N_CV = ((33.9*C)+121*(H-(O/8))+(9.1*S))*10^6;
+printf('The net calorific value is %3.1e J/kg',N_CV)
+
+
diff --git a/2471/CH6/EX6.9/Ex6_9.sce b/2471/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 000000000..72f78c3f3 --- /dev/null +++ b/2471/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,30 @@ +clear ;
+clc;
+// Example 6.9
+printf('Example 6.9\n\n');
+printf('Page No. 157\n\n');
+
+// given
+P = 10;// Boiler pressure in bar
+Ts = 180;// Steam temperature in degree celcius
+Tf = 80;// Feed water temperature in degree celcius
+X = 0.95;// Steam dryness fraction
+m_s = 4100;// steam rate in kg/h
+m_f = 238;// Gas rate in kg/h
+G_CV = 53.5*10^6;// In J/kg
+N_CV = 48*10^6;//in J/kg
+
+//from steam table,AT 10 bar and at temperature T = Ts
+h2 = (763+(X*2013))*10^3;//Specific enthalpy of steam in J/kg
+//At temperature T = Tf
+h1 = 335*10^3;//Specific enthalpy of feed steam in J/kg
+
+E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));//
+printf('The gross efficiency percentage is %.0f \n',E_G)
+
+
+E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));//
+printf('The net efficiency percentage is %.0f',E_N)
+
+
+
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.')
+
+
+
+
diff --git a/2471/CH8/EX8.1/Ex8_1.sce b/2471/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..b21184e1d --- /dev/null +++ b/2471/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,12 @@ +clear ;
+clc;
+// Example 8.1
+printf('Example 8.1\n\n');
+printf('Page No. 222\n\n');
+
+// given
+V = 240;// Voltage in Volts
+I = 8;// Current in Amps
+//By ohm's law-> V = I*R
+R = V/I;// In ohms
+printf('The resistance of the given circuit is %.0f ohms',R)
diff --git a/2471/CH8/EX8.10/Ex8_10.sce b/2471/CH8/EX8.10/Ex8_10.sce new file mode 100755 index 000000000..d1bfd24e4 --- /dev/null +++ b/2471/CH8/EX8.10/Ex8_10.sce @@ -0,0 +1,13 @@ +clear ;
+clc;
+// Example 8.10
+printf('Example 8.10\n\n');
+printf('Page No. 240\n\n');
+
+P = 100;// Power in 10^3 W
+C = 5;// Charge in Euro per 10^3 per month
+PF = [1.0 0.9 0.8 0.7 0.6 0.5];// Power factor
+for i = [1:1:6]
+ VA = (PF(i)\P)*C
+ printf('Charge per month for power factor %.1f is %.0f Euro \n', PF(i),VA)
+end
diff --git a/2471/CH8/EX8.11/Ex8_11.sce b/2471/CH8/EX8.11/Ex8_11.sce new file mode 100755 index 000000000..b43db5a15 --- /dev/null +++ b/2471/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 8.11
+printf('Example 8.11\n\n');
+printf('Page No. 240\n\n');
+
+// This question doesnot contain any calculation part.
+//given
+P_F_1 = 0.7;// Initial power factor
+P_F_2 = 0.95;// Final power factor
+//Refer Figure 8.10
+red_I = 26;//reduction in current in per cent
+printf('The reduction in current is %.0f per cent \n',red_I)
+P_F_3 = 1.0;// Increased power factor
+// From figure 8.10
+Save = 4;// per cent
+printf('Increase in power factor from 0.95-1.0 only increases saving further by a %.0f per cent',Save)
+
diff --git a/2471/CH8/EX8.12/Ex8_12.sce b/2471/CH8/EX8.12/Ex8_12.sce new file mode 100755 index 000000000..fa6636688 --- /dev/null +++ b/2471/CH8/EX8.12/Ex8_12.sce @@ -0,0 +1,18 @@ +clear ;
+clc;
+// Example 8.12
+printf('Example 8.12\n\n');
+printf('Page No. 240\n\n');
+
+// This question doesnot contain any calculation part.
+//given
+C = 10000;// Installation cost of capacitors in Pound
+P_F_1 = 0.84;// Initial power factor
+P_F_2 = 0.97;// Final power factor
+//Refer Figure 8.10
+red_dem = 14;//reduction in maximum demand in per cent
+T = 9;// pay-back time in months
+
+printf('The reduction in maximum demand is %.0f per cent \n',red_dem)
+printf('The pay-back time was %.0f months',T)
+// This question does not contain any calculation part.
diff --git a/2471/CH8/EX8.13/Ex8_13.sce b/2471/CH8/EX8.13/Ex8_13.sce new file mode 100755 index 000000000..7f13eb13c --- /dev/null +++ b/2471/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,21 @@ +clear ;
+clc;
+// Example 8.13
+printf('Example 8.13\n\n');
+printf('Page No. 244\n\n');
+
+//given
+T1 = 21;// in degree celcius
+t1 = 8;// time in h per day
+c = 3.5;// cost in p per unit
+C1 = 38;// Total cost in Pound per 10^3 W
+
+T2 = 16;// in degree celcius
+t2 = 8;// time in h per day
+C2 = 27;// Total cost in Pound per 10^3 W
+
+Save = C1 - C2;// Saving in Pound per 10^3 W
+Save_deg = Save/(T1 - T2);// Total Saving in Pound per 10^3 W for each degree drop
+Save_per = (Save_deg/C1)*100;// Saving in percent
+printf('For each degree drop, an energy saving of %.0f per cent is achieved',floor(Save_per))
+
diff --git a/2471/CH8/EX8.14/Ex8_14.sce b/2471/CH8/EX8.14/Ex8_14.sce new file mode 100755 index 000000000..4c3789410 --- /dev/null +++ b/2471/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+// Example 8.14
+printf('Example 8.14\n\n');
+printf('Page No. 245\n\n');
+
+// This question doesnot contain any calculation part.
+//refer Table 8.6
+O_1 = 1750;
+O_2 = 0;
+O_3 = 2;
+O_4 = 150;
+O_5 = 1900;
+O_6 = 0;
+I_1 = 580;
+I_2 = 1658;
+I_3 = 0.5;
+I_4 = 40;
+I_5 = 1698;
+I_6 = 11;
+D_1 = 300;
+D_2 = 869;
+D_3 = 0.5;
+D_4 = 40;
+D_5 = 900;
+D_6 = 37;
+printf('\t ENERGY COSTS FOR HEATING STEEL BILLETS\n\n Components(10^3 W/tonne) \t (1)Oil fired \t (2)Induction \t (3)Direct resistance\n Fuel(electricity) \t\t %.0f \t \t %.0f \t \t%.0f\n Electricity(as prime energy) %.0f \t\t \t %.0f \t \t%.0f\n Metal loss(percent)\t\t %.0f \t\t \t %.1f \t \t%.1f\n Metal loss (as energy)\t\t %.0f \t\t \t %.0f \t \t%.0f\n Total energy needs\t\t %.0f \t\t \t %.0f \t \t%.0f\n Energy saving to(1)\t\t %.0f \t\t \t %.0f \t\t \t%.0f\n',O_1,I_1,D_1,O_2,I_2,D_2,O_3,I_3,D_3,O_4,I_4,D_4,O_5,I_5,D_5,O_6,I_6,D_6)
+
+
+
+
+
+
diff --git a/2471/CH8/EX8.15/Ex8_15.sce b/2471/CH8/EX8.15/Ex8_15.sce new file mode 100755 index 000000000..2e9beccdb --- /dev/null +++ b/2471/CH8/EX8.15/Ex8_15.sce @@ -0,0 +1,23 @@ +clear ;
+clc;
+// Example 8.15
+printf('Example 8.15\n\n');
+printf('Page No. 247\n\n');
+
+// This question doesnot contain any calculation part.
+//refer Table 8.7
+El = 35;// Percentage of electricity produced from primary fuel
+En_1 = 50;// Endothermic gas (m^3)
+En_2 = 100;// Endothermic gas (m^3)
+En_3 = 200;// Endothermic gas (m^3)
+G_1 = 97;// Gas use (10^3 Wh)
+G_2 = 194;// Gas use (10^3 Wh)
+G_3 = 386;// Gas use (10^3 Wh)
+El_1 = 24;// Electricity use (10^3 Wh)
+El_2 = 48;// Electricity use (10^3 Wh)
+El_3 = 95;// Electricity use (10^3 Wh)
+P_1 = 69;// Primary energy (10^3 Wh)
+P_2 = 137;// Primary energy (10^3 Wh)
+P_3 = 271;// Primary energy (10^3 Wh)
+printf(' USE OF ELECTRICITY AND GAS FOR HEATING ENDOTHERMIC GAS GENERATORS\n\n Endothermic gas (m^3) \t %.0f \t %.0f \t %.0f \n Gas use (10^3 Wh) \t %.0f \t %.0f \t %.0f \n Electricity use (10^3 Wh) \t %.0f \t %.0f \t %.0f \n Primary energy (10^3 Wh) \t %.0f \t %.0f \t %.0f \n',En_1,En_2,En_3,G_1,G_2,G_3,El_1,El_2,El_3,P_1,P_2,P_3)
+
diff --git a/2471/CH8/EX8.2/Ex8_2.sce b/2471/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..41902bca3 --- /dev/null +++ b/2471/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,23 @@ +clear ;
+clc;
+// Example 8.2
+printf('Example 8.2\n\n');
+printf('Page No. 223\n\n');
+
+// given
+V1 = 100;// In Volts
+V2 = 50;// In Volts
+R1 = 8;// Resistance in ohm
+R2 = 5;// Resistance in ohm
+R3 = 10;// Resistance in ohm
+R4 = 50;// Resistance in ohm
+//By refering figure 8.3, and applying kirchoff's current law and kirchoff's voltage law in the given circuit diagram, we get following equations:
+// I1 = I2 + I3
+//V1 - R1*I1 - V2 - R3*I3 = 0
+//V2 - R4*I3 + R3*I3 - R2*I2 = 0
+A = [1 -1 -1;8 0 10;0 55 -10];
+b = [0;50;50];
+x = A\b
+printf('The currents in I1 is %.1f A \n',x(1))
+printf('The currents in I2 is %.1f A \n',x(2))
+printf('The currents in I3 is %.1f A \n ',x(3))
diff --git a/2471/CH8/EX8.3/Ex8_3.sce b/2471/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..39218f66f --- /dev/null +++ b/2471/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 8.3
+printf('Example 8.3\n\n');
+printf('Page No. 226\n\n');
+
+// given
+R = 6;// Resistance in ohm
+Xc = 16;// Capacitive resistance in ohm
+Xl = 24;// Inductive resistance in ohm
+Z = ((R^2) + (Xc - Xl)^2)^0.5;// Impedance in ohm
+P_f = R/Z;// Power factor = cos(x) = 0.6
+x = acos(0.6);
+y = sqrt(1 - P_f^2);//y = sin (x)
+V = 200;// in Volts(sin wave voltage = ((200*2^1.5)*sinwt)
+I = V/Z;// Current in Amperes
+P = I^2 * R;// in W
+Q = V * I * y;// in VAR
+S = V * I;// in VA
+printf('The actual power is %.0f W \n',P)
+printf('The reactive power is %.0f VAR \n',Q)
+printf('The apparent power is %.0f VA \n',S)
+
+
+
diff --git a/2471/CH8/EX8.4/Ex8_4.sce b/2471/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..be86bbd63 --- /dev/null +++ b/2471/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,15 @@ +clear ;
+clc;
+// Example 8.4
+printf('Example 8.4\n\n');
+printf('Page No. 232\n\n');
+
+//given
+pump_1 = 100*10^3;// Required pump in W
+T_1 = 8;// Pump Operating time of each day
+Inc_op = 0.5;// Increased output per cent
+pump_ex = 50*10^3;// Extra pump requried in W
+
+// This question doesnot contain any calculation part.
+printf('there is no computational part in the problem')
+
diff --git a/2471/CH8/EX8.5/Ex8_5.sce b/2471/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..3e4f70609 --- /dev/null +++ b/2471/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,13 @@ +clear ;
+clc;
+// Example 8.5
+printf('Example 8.5\n\n');
+printf('Page No. 232\n\n');
+
+//given
+P = 600*10^3;// Power demand of pump in W
+T = 8;// Operating time in hour per day
+red = 1.00/10^3;// off-peak reduction in Pound per 10^3 W month
+M_save = P*red;// Monthly saving Pound per month
+A_save = M_save*12;// Annual saving in Pound per year
+printf('Annual saving is %.0f Pound per year',A_save)
diff --git a/2471/CH8/EX8.6/Ex8_6.sce b/2471/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..109a92b22 --- /dev/null +++ b/2471/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,14 @@ +clear ;
+clc;
+// Example 8.6
+printf('Example 8.6\n\n');
+printf('Page No. 234\n\n');
+
+//given
+T_lamp = 12*10^3;// Output for the tungsten filament lamp in lm per 10^3 W
+F_tube = 63*10^3;// Output for the fluorescent tubes in lm per 10^3 W
+Save = F_tube - T_lamp;// in W
+printf('Energy saving is %.0f lm per 10^3 W \n',Save)
+
+Save_pcent = (Save/F_tube)*100;
+printf('Energy saving per cent is %.0f ',floor(Save_pcent))
diff --git a/2471/CH8/EX8.7/Ex8_7.sce b/2471/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..488930879 --- /dev/null +++ b/2471/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 8.7
+printf('Example 8.7\n\n');
+printf('Page No. 235\n\n');
+
+//given
+N = 40;// Number of lamps
+T1 = 15;// Operating time in h per day
+P = 500;// POwer from the lamps in W
+T2 = 300;// Total operating time in days per year
+C = 2.5/10^3;// Electricity cost in p per Wh
+
+An_Cost = N*P*T1*T2*C*10^-2;// In euro
+printf('The Annual Cost is %.0f Euro \n',An_Cost)
+
+//Improvement in light by installing glassfibre skylights
+T3 = 5;// Extra Time for natural lighting in h per day
+New_An_Cost = N*P*(T1-T3)*T2*C*10^-2;// In euro
+printf('The New Annual Cost is %.0f Euro \n',New_An_Cost)
+
+Save = An_Cost - New_An_Cost;// in euro
+printf('The annual saving for a pay-back period of 2.5 years is %.0f ',Save)
+
+
diff --git a/2471/CH8/EX8.8/Ex8_8.sce b/2471/CH8/EX8.8/Ex8_8.sce new file mode 100755 index 000000000..0e9b8ddd9 --- /dev/null +++ b/2471/CH8/EX8.8/Ex8_8.sce @@ -0,0 +1,10 @@ +clear ;
+clc;
+// Example 8.8
+printf('Example 8.8\n\n');
+printf('Page No. 236\n\n');
+
+// This question doesnot contain any calculation part.
+
+//By refering figure 8.7 which shows Poer factor-load curve for a motor with a capacitor and one without a capacitor.
+printf('there is no computational part in the problem')
diff --git a/2471/CH8/EX8.9/Ex8_9.sce b/2471/CH8/EX8.9/Ex8_9.sce new file mode 100755 index 000000000..d5810dea5 --- /dev/null +++ b/2471/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,16 @@ +clear ;
+clc;
+// Example 8.9
+printf('Example 8.9\n\n');
+printf('Page No. 238\n\n');
+
+// This question doesnot contain any calculation part.
+//given
+l = 500*10^3;// Load in VA
+P_F = 0.6;// Power Factor
+Req_P_F = 0.9;// Required power factor
+//Refer to figure 8.8
+BC = 2.5;// units
+C_rt = 250*10^3;// in VAR (obtained from figure 8.8)
+printf('The required condenser rating is %.0f \n',C_rt)
+
diff --git a/2471/CH9/EX9.1/Ex9_1.sce b/2471/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..49f4ea6ab --- /dev/null +++ b/2471/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,52 @@ +clear;
+clc;
+// Example 9.1
+printf('Example 9.1\n\n');
+printf('Page No. 252\n\n');
+
+//given
+a = 40;// in m
+b = 25;// in m
+c = 20;// in m
+d = 10;// in m
+e = 5;// in m
+f = 2;// in m
+g = 3;// in m
+h = 6;// in m
+
+//(1) Production Area
+T1 = 21;// Temperature difference in degree celcius
+T2 = -3;// Temperature difference in degree celcius
+U1 = 1.2;// heat transfer coefficent in W/m-K
+U2 = 5.6;// heat transfer coefficent in W/m-K
+U3 = 2.0;// heat transfer coefficent in W/m-K
+U4 = 0.7;// heat transfer coefficent in W/m-K
+U5 = 0.9;// heat transfer coefficent in W/m-K
+// As Q = U*A*T
+Q1 = (b*h)*U1*T1;// Heat loss in W. wall in W
+Q2 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;// Heat loss in N. wall in W
+Q3 = (c*f)*U2*T1;// Heat loss in N. window in W
+Q4 = (b*g)*U3*T2;// Heat loss in N. wall/internal in W
+Q5 = (b*g)*U1*T1;// Heat loss in E. wall/external in W
+Q6 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;// Heat loss in S. wall in W
+Q7 = (c*f)*U2*T1;// Heat loss in S. window in W
+Q8 = (b*a)*U4*T1;// Heat loss in roof in W
+Q9 = (b*a)*U5*T1;// Heat loss in floor in W
+T_Q_P = Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7 + Q8 + Q9;// in W
+
+//For Office surface
+T3 = 24;// Temperature difference in degree celcius
+T4 = 3;// Temperature difference in degree celcius
+// As Q = U*A*T
+Q_1 = (b*g)*U3*T4;// Heat loss in W. wall in W
+Q_2 = (d*g)*U1*T3;// Heat loss in N. wall in W
+Q_3 = (((b-(2*e))*g) +(e*f))*U1*T3;// Heat loss in E. Wall in W
+Q_4 = (e*f)*U2*T3;// Heat loss in E. window in W
+Q_5 = (e*f)*U2*T3;// Heat loss in E. window in W
+Q_6 = (d*g)*U1*T3;// Heat loss in S. wall in W
+Q_7 = (b*d)*U4*T3;// Heat loss in S. roof in W
+Q_8 = (b*d)*U5*T3;// Heat loss in floor in W
+T_Q_O = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 + Q_6 + Q_7 + Q_8; //in W
+
+T_Q = T_Q_P + T_Q_O;// in W
+printf('Total building fabric loss is %.0f W',T_Q)
diff --git a/2471/CH9/EX9.2/Ex9_2.sce b/2471/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..4719ebd5d --- /dev/null +++ b/2471/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 9.2
+printf('Example 9.2\n\n');
+printf('Page No. 255\n\n');
+
+//given
+L_Br = 0.105;// Length of brickwork in m
+L_Bl = 0.100;// Length of blockwork in m
+L_C = 0.05;// Length of cavity in m
+K_Br = 0.84;// Thermal conductivity of brickwork in W/m-K
+K_Bl = 0.22;// Thermal conductivity of blockwork in W/m-K
+K_C_in = 0.033;// Thermal conductivity of insulation in cavity in W/m-K
+R_Ex = 0.055;// Resistance of external surface in W/m^2-K
+
+//As R = L/K
+R_Br = (L_Br/K_Br);// Resistance of brickwork in W/m^2-K
+R_Bl = (L_Bl/K_Bl);// Resistance of blockwork in W/m^2-K
+R_C = 0.18;// Resistance of cavity in W/m^2-K
+
+//Without insulation of cavity
+R_T = 0.938;// Total Resistance in W/m^2-K
+// Thermal transmittance - U = (1/R_T)
+U = (1/R_T);// in W/m^2-K
+printf('The U-value of external wall is %.2f W/sq.m K \n',U)
+
+//With insulation of cavity
+//As R = L/K
+R_C_in = (L_C/K_C_in);// Resistance of insulation in cavity in W/m^2-K
+In = R_C_in - R_C;// Net increase in W/m^2-K
+R_T_New = R_T + In;// New total resistance in W/m^2-K
+// Thermal transmittance - U = (1/R_T)
+U_New = (1/R_T_New);// in W/m^2-K
+printf('The new U-value is with foamed insulation %.3f W/sq.m K',U_New)
+
diff --git a/2471/CH9/EX9.3/Ex9_3.sce b/2471/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..88726abd9 --- /dev/null +++ b/2471/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 9.3
+printf('Example 9.3\n\n');
+printf('Page No. 256\n\n');
+
+//given
+N_1 = 1.5;// Ventilation rate in the production area (air changes per hour)
+N_2 = 1.0;// Ventilation rate in the office suite (air changes per hour)
+
+//From example 9.1
+V_P = 6000;// Voulme of production area in m^3
+V_O = 750;// Voulme of office suite in m^3
+T1 = 21;// Temperature difference in degree celcius
+T2 = -3;// Temperature difference in degree celcius
+T_P = 18;// Temperature difference in degree celcius
+F_loss = 74.4*10^3;// Total fabric loss in W
+
+// As Q_vent = 0.33 * N * V * (T1 - T2)
+Q_vent_P = 0.33 * N_1 * V_P * (T_P - T2);// Ventilation loss in production area in W
+Q_vent_O = 0.33 * N_2 * V_O * (T1 - T2);// Ventilation loss in office suite in W
+V_loss = Q_vent_P + Q_vent_O;// Total ventilation loss in W
+T_loss = F_loss + V_loss;// Total heat loss in W
+p = (V_loss/T_loss)*100;
+printf('percentage of ventilation loss is %.0f percent',p)
diff --git a/2471/CH9/EX9.4/Ex9_4.sce b/2471/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..642773d29 --- /dev/null +++ b/2471/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,59 @@ +clear ;
+clc;
+// Example 9.4
+printf('Example 9.4\n\n');
+printf('Page No. 260\n\n');
+
+//(a) Design loss
+T1 = 18;// Internal teemperature(specified as an Environmental temperature) in degree celcius
+//From example 9.1
+A = [150 200 40 75 75 200 40 1000 1000];// in m^2
+U = [1.2 1.2 5.6 2 1.2 1.2 5.6 0.7 0.9];// in W/m-K
+Qf = 58.3*10^3;// Fabric loss in production area in W
+T2 = -3;// in degree celcius
+s1 =0;
+s2 = 0;
+for i = [1:1:9]
+ s1 = s1+A(i);
+ s2 = s2+U(i)*A(i);
+end
+A_T = s1;// Total area in m^2
+UA_T = s2;// sum of U*A in W/m-K (answer wrongly calculated in the book)
+
+//From example 9.3
+N_1 = 1.5;// Ventilation rate in the production area (air changes per hour)
+V_P = 6000;// Voulme of production area in m^3
+
+//As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))
+C = 0.33*N_1*V_P*(1 + ((UA_T)/(4.8*A_T)));
+Q_vent = C * (T1 - T2);// in W
+T_Q1 = Qf + Q_vent;// in W
+printf('The total design loss is %.0f W \n',T_Q1) // (deviation in answer is due to error in calculation in the book)
+
+//(b) Reduced heat loss
+// The heat transfer coeffieint in this problem has been changed as U1
+U1 = [0.44 0.44 2.8 2 0.44 0.44 2.8 0.44 0.9];//in W/m^2-K
+T = [21 21 21 -3 21 21 21 21 21];// Temperature difference in degree celcius
+s3 = 0;
+s4 = 0;
+for i = [1:1:9]
+ s3 = s3+U1(i)*A(i);
+ s4 = s4+U1(i)*A(i)*T(i);
+end
+U1A_T = s3;// in W/m-k (answer wrongly calculated in the book)
+Q_loss = s4// in W
+
+
+//As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))
+C = 0.33*N_1*V_P*(1 + ((U1A_T)/(4.8*A_T)))
+Q_vent = C * (T1 - T2)// in W
+T_Q2 = Q_loss + Q_vent// in W
+
+Red = T_Q1 - T_Q2;// In W
+printf('The reduction in loss is %.0f W',Red) // (deviation in answer is due to error in calculation in the book)
+
+
+
+
+
+
diff --git a/2471/CH9/EX9.5/Ex9_5.sce b/2471/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..e4078b062 --- /dev/null +++ b/2471/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,38 @@ +clear ;
+clc;
+// Example 9.5
+printf('Example 9.5\n\n');
+printf('Page No. 265\n\n');
+
+//given
+T = 21;// Temperature difference in degree celcius
+Deg_d = 2186;// Total degree-days base(15.5 deg C) September_April
+T_D = 18;// Design Temperature in degree celcius
+T_O = 4;// base offset temperature in degree celcius
+T_b = T_D - T_O;// Base temperature in degree celcius
+
+// From Table 9.11 Correction factor for base tempratures other than 15.5 deg C is obtained. So for 14 deg c its 0.82
+C = 0.82;// Correction factor
+Do = Deg_d * C// Corrected degree-days
+
+//(a) Original construction
+//from example 9.4
+Q_d_1 = 133.7*10^3;// Design heat loss in W
+
+H_1 = Q_d_1/T;
+//As E = 24 * H * Do - E = Energy consumption in (Wh)
+E1 = (24*H_1 *Do)/10^6;// in 10^6 Wh ( from this step 'Do' is mistakely taken as 1972 inplace of 1792 in the solution of the book, so there is deviation in answer)
+E_1 = (E1 * 3600)*10^6;// in J
+printf('The total energy consumption in original construction is %.0e J \n',E_1)// Deviation in the answer is due to some calculation error as mentioned above
+
+//(b) Improved insulation
+//from example 9.4
+Q_d_2 = 104.4*10^3;// Design heat loss in W
+
+H_2 = Q_d_2/T;
+//As E = 24 * H * Do - E = Energy consumption in (Wh)
+E2 = (24*H_2 *Do)/10^6;// in 10^6 Wh ( from this step 'Do' is mistakely taken as 2972 inplace of 2792 in the solution of the book, so there is deviation in answer)
+E_2 = (E2 * 3600)*10^6;// in J
+printf('The total energy consumption in improved insulation is %.4e J \n',E_2)// Deviation in the answer is due to some calculation error as mentioned above
+
+
diff --git a/2471/CH9/EX9.6/Ex9_6.sce b/2471/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..2b3a80a06 --- /dev/null +++ b/2471/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 9.6
+printf('Example 9.6\n\n');
+printf('Page No. 268\n\n');
+
+//given
+U1 = 5.6;// Single glazing heat transfer coefficient in W/m^2_K
+U2 = 2.8;// Double glazing heat transfer coefficient in W/m^2_K
+Ti = 21;// Internal Temperature in degree celcius
+To = -1;// External Temperature in degree celcius
+R_H = 0.5;// Relative humidity
+Rs_i = 0.123;// Surface resistance in (W/m^2-K)^-1
+
+// At 21 Degree celcius and R.H. = 0.5, the dew point is 10.5 degree celcius
+Dew_pt = 10.5;// Dew point in degree celcius
+//As Ts_i = Ti - (Rs_i * U *(Ti - To))
+
+//(a) Single Glazing
+Ts_i_S = Ti - (Rs_i * U1 *(Ti - To));// in degree celcius
+printf('The internal surface temperature for single glazing is %.1f deg C \n',Ts_i_S)
+if (Dew_pt > Ts_i_S) then
+ disp('Surface condensation will occur since it is less than 10.5 deg C.')
+else
+ disp('No surface condensation is expected as it is greater than 10.5 deg C.')
+end
+
+//(b) Double Glazing
+Ts_i_D = Ti - (Rs_i * U2 *(Ti - To));// in degree celcius
+printf('The internal surface temperature for single glazing is %.1f deg C \n',Ts_i_D)
+if (Dew_pt > Ts_i_D) then
+ disp('Surface condensation will occur since it is less than 10.5 deg C.')
+else
+ disp('No surface condensation is expected since it is greater than 10.5 deg C.')
+end
diff --git a/2471/CH9/EX9.7/Ex9_7.sce b/2471/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..78cfb2b1e --- /dev/null +++ b/2471/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+// Example 9.7
+printf('Example 9.7\n\n');
+printf('Page No. 269\n\n');
+
+//given
+l_1 = 240;// existing length of solid brick in mm
+l_u = 25;// upgraded internal lining in mm
+l_e = 9.5;// Expanded polystyrenne in mm
+T_i = 20;// Internal temperature in degre celcius
+R_H_i = 50;// Internal Relative humidity in percent
+T_e = 0;// External temperature in degre celcius
+R_H_e = 90;// External Relative humidity in percent
+
+K = [0.123 0.059 0.714 0.286 0.055];// Thermal resistance in W/m^2-K
+V_r = [0.0 0.475 3.57 9.60 0.0];// Vapour Resistance in 10^9 N-s/kg
+
+//Refer Figure 9.3
+//From Figure 9.3, the tempeature, dew point, vapour pressure for different interface are obtained
+T = [18.01 17.06 5.51 0.89];// Temperature in degree celcius
+V_p = [1170 1148 986 550];//Vapour pressure in N/m^2
+D_P = [9.5 9.2 7.1 -1.5];// Dew point in degree celcius
+
+h = (T_i - T_e)/sum(K);// in W/m^2
+printf('The heat flow is %.2f W/m^2 \n',h)
+V_p_i = V_p(1);// Internal vapour pressure in N/m^2
+V_p_e = V_p(4);// External vapour pressure in N/m^2
+m = ((V_p_i - V_p_e)/sum(V_r))*10^-9;// in kg/s
+printf('The vapour mass flow is %.1e kg/s',m)
+
+
+
diff --git a/2471/CH9/EX9.8/Ex9_8.sce b/2471/CH9/EX9.8/Ex9_8.sce new file mode 100755 index 000000000..f6a2111a2 --- /dev/null +++ b/2471/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,41 @@ +clear ;
+clc;
+// Example 9.8
+printf('Example 9.8\n\n');
+printf('Page No. 275\n\n');
+
+//given
+A = 10;// in m^2
+S = 0.77;
+Sa = 0.54;
+//for South
+printf('\t\t\t SOUTH \n')
+I1 = [200 185 165 155 165 185 200];// in W-m^2
+I2 = [500 455 405 385 405 455 500];// in W-m^2
+for i = [1:1:7]
+A_G_S (i) = (A*I1(i)*S) + (A*I2(i)*Sa)
+end
+
+printf('The monthly peak cooling loads for the month March is %.0f W \n',A_G_S(1))
+printf('The monthly peak cooling loads for the month April is %.0f W \n',A_G_S(2))
+printf('The monthly peak cooling loads for the month May is %.0f W \n',A_G_S(3))
+printf('The monthly peak cooling loads for the month June is %.0f W \n',A_G_S(4))
+printf('The monthly peak cooling loads for the month July is %.0f W \n',A_G_S(5))
+printf('The monthly peak cooling loads for the month Aug. is %.0f W \n',A_G_S(6))
+printf('The monthly peak cooling loads for the month Sept. is %.0f W \n\n',A_G_S(7))
+
+ //For east
+ printf('\t\t\t EAST \n')
+I3 = [110 150 180 190 180 150 110];// in W-m^2
+I4 = [435 510 515 505 515 510 435];// in W-m^2
+for j = [1:1:7]
+A_G_E(j) = (A*I3(j)*S) + (A*I4(j)*Sa);
+end
+printf('The monthly peak cooling loads for the month March is %.0f W \n',A_G_E(1))
+printf('The monthly peak cooling loads for the month April is %.0f W \n',A_G_E(2))
+printf('The monthly peak cooling loads for the month May is %.0f W \n',A_G_E(3))
+printf('The monthly peak cooling loads for the month June is %.0f W \n',A_G_E(4))
+printf('The monthly peak cooling loads for the month July is %.0f W \n',A_G_E(5))
+printf('The monthly peak cooling loads for the month Aug. is %.0f W \n',A_G_E(6))
+printf('The monthly peak cooling loads for the month Sept. is %.0f W \n\n',A_G_E(7))
+
diff --git a/2471/CH9/EX9.9/Ex9_9.sce b/2471/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..893329a44 --- /dev/null +++ b/2471/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,43 @@ +clear ;
+clc;
+// Example 9.9
+printf('Example 9.9\n\n');
+printf('Page No. 277\n\n');
+
+//given
+A = 15;// glazing area in m^2
+l = 10;// Length of office in m
+h = 6;// height of office in m
+w = 3.5;// width of office in m
+Y_w = 4;// Admittance of wall in W/m^2-K
+Y_f = 3;// Admittance of floor in W/m^2-K
+Y_c = 3;// Admittance of ceiling in W/m^2-K
+N = 1.5;//Ventilation rate (air changes per hour)
+V = l*h*w;// Volume in m^3
+U_G = 5.6;// Transmittance in W/m^2-K
+
+//From table 9.18 and table 9.16
+To = 16.5;// External temperature of June in degree celcius
+T_O = 7.5;// Swing temperature in degre celcius
+I = 155; //Vertical S in W-m^2
+Is = 385;//Vertical S in W-m^2
+S = 0.77;// Solar gain factor
+Sa = 0.54;// Solar gain factor
+
+//As For the mean internal temperature -Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G)))
+Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G)));// in degree celcius
+printf('the mean internal temperature is %.1f deg C \n',Ti)
+
+A_G = (A*Is*Sa) + ((A*U_G) + (0.33*N*V))*T_O;// Swing in gain in W
+Net_A = 2*((w*h) + (l*w)) - A;// Net wall area in m^2
+A_f = l*h;// floor area in m^2
+A_c = l*h;//ceiling area in m^2
+A_Y_w = Net_A * Y_w;// Wall AY in W/K
+A_Y_f = A_f * Y_f;// Floor AY in W/K
+A_Y_c = A_c * Y_c;// ceiling AY in W/K
+A_Y_wi = 84;// Window AY in W/K
+Net_AY = A_Y_w + A_Y_f + A_Y_c + A_Y_wi// in W/K
+Ti_s = ((A_G)/((0.33*N*V) + (Net_AY)))// Internal Temperature swing in deg C
+T_p = Ti + Ti_s;// in deg C
+printf('Peak internal temperature is %.1f deg C',T_p) // Deviation in the answer is due to some calculation approximation in the book
+
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