initial commit / add all books

13 files changed, 658 insertions, 0 deletions

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)

+

+

+