initial commit / add all books

13 files changed, 405 insertions, 0 deletions

diff --git a/1943/CH4/EX4.1/Ex4_1.sce b/1943/CH4/EX4.1/Ex4_1.sce
new file mode 100755
index 000000000..7c59f44cd
--- /dev/null
+++ b/1943/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,39 @@
+
+clc

+clear

+//Input data

+C=84;//The mass of carbon present in the fuel in %

+H=10;//The mass of hydrogen present in the fuel in %

+S=3.2;//The mass of sulphur present in the fuel in %

+O=1.6;//The mass of oxygen present in the fuel in %

+I=1.2;//The mass of incombustible in the fuel in %

+X=15.72;//The flue gas of combined CO2 and SO2 by volume in %

+Og=1;//The flue gas of O2 by volume in %

+Y=100;//Let us consider the fuel oil in kg

+C1=12;//Molecular weight of Carbon 

+H1=2;//Molecular weight of hydrogen

+S1=32;//Molecular weight of sulphur

+O1=32;//Molecular weight of oxygen

+Co2=44;//Molecular weight of carbondioxide

+So2=64;//Molecular weight of sulphurdioxide

+N1=28;//Molecular weight of nitrogen 

+H2O=18;//Molecular weight of water

+

+//Calculations

+b=C/C1;//Equating coefficients of the carbon from equation

+g=H/H1;//Equating coefficients of the hydrogen from equation

+d=S/S1;//Equating coefficients of the sulphur from the equation

+e=(b+d)/(X/Og);//By volumetric analysis

+x=b+d+e+(g/2)-(O/O1);//Moles of oxygen are supplied for combustion

+f=3.76*x;//Equating coefficients of the nitrogen from equation

+Mo=x*O1;//Mass of oxygen supplied in kg

+Ma=Mo/0.232;//Mass of air supplied for 100 kg of fuel in kg

+Wa=Ma/100;//Mass of air supplied for 1 kg fuel in kg

+Wrh={(11.5*C)+(34.5*[(H)-(O/8)])+(4.3*S)}/100;//Theoretical air required per kg of fuel in kg

+E=[(Wa-Wrh)/Wrh]*100;//Percentage of excess air in %

+D=(b*Co2)+(d*So2)+(e*O1)+(f*N1);//Mass of dry flue gas formed for 100 kg fuel in kg

+dfg=D/100;//Mass of dry flue gas formed per kg of fuel in kg

+Mw=(g*H2O)/100;//Mass of water vapour formed per kg of fuel

+

+//Output

+printf('(a) Mass of air supplied WA = %3.2f kg \n (b)Percentage excess air supplied = %3.2f percentage \n (c)mass of dry flue gas formed = %3.2f kg \n (d) Mass of water vapour formed = %3.2f kg',Wa,E,dfg,Mw)

diff --git a/1943/CH4/EX4.10/Ex4_10.sce b/1943/CH4/EX4.10/Ex4_10.sce
new file mode 100755
index 000000000..62f265176
--- /dev/null
+++ b/1943/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,20 @@
+
+clc

+clear

+//Input data

+tg=180;//The gas temperature in degree centigrade

+p=250;//The draught produced by the ID fan in mm

+nf=0.52;//The efficiency of the fan

+Va=0.858;//Volume flow rate of air in m^3/kg

+g=9.812;//gravitational force constant in m/s^2

+Wf=10;//Coal rate in t/h

+Wa=12.9;//Actual air required per kg fuel in kg air/kg fuel

+ta=30;//Room temperature in degree centigrade

+

+//Calculations

+Wfg=[(Wf+(Wa*10))*1000]/3600;//Fuel gas required in kg/s

+Vfg=[Va*(tg+273)]/(ta+273);//Volume flow rate of fuel gas in m^3/kg

+ID=((Wfg*Vfg*p*g)/(nf))/1000;//ID fan motor capacity in kW

+

+//Output

+printf('The motor capacity of the ID fan is %3.2f kW ',ID)

diff --git a/1943/CH4/EX4.11/Ex4_11.sce b/1943/CH4/EX4.11/Ex4_11.sce
new file mode 100755
index 000000000..0004d7fb9
--- /dev/null
+++ b/1943/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,39 @@
+
+clc

+clear

+//Input data

+CO2=13.2;//The volume of carbondioxide present in the partial analysis of dry flue gas in %

+O2=3.2;//The volume of oxygen present in the partial analysis of dry flue gas in %

+C=88;//The mass of carbon present in the coal according to coal analysis on mass basis in %

+H=4.4;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %

+A=7.6;//The mass of ash present in the coal according to coal analysis on mass basis in %

+M=0;//Moisture present in the fuel was nil

+Mc=12;//Molecular weight of the carbon 

+Mh=2;//Molecular weight of the hydrogen

+Mo=32;//Molecular weight of the oxygen

+Mho=18;//Molecular weight of water

+p=101.325;//Atmospheric pressure in kPa

+

+//Calculations

+c=C/Mc;//Equating coefficients of the carbon from the equation 

+g=H/Mh;//Equating coefficients of the hydrogen from the equation 

+x=(CO2/100)/(O2/100);//From dry fuel gas analysis (dfg) 

+d=[[(CO2/100)*(47.5)]-7.333]/[[(CO2/100)*(3.032)]-1];//Coefficient of the carbonmonoxide in the equations product side

+b=c-d;//Coefficient of the carbondioxide in the equation product side

+a=10.21-(0.742*d);//Coefficient of the oxygen in the reactant side of the equation 

+e=b/x;//Coefficient of the oxygen in the product side of the equation 

+f=3.76*a;//Equating coefficients of the nitrogen from the equation 

+ma=(a*Mo)/0.232;//Mass of air supplied for 100 kg coal in kg

+ma1=ma/100;//Mass of air supplied per kg coal in kg

+T=b+d+e+f;//Total number of moles of dry flue gas (dfg)

+CO21=(b/T)*100;//Carbondioxide by volume in percentage

+O21=(e/T)*100;//Oxygen by volume in percentage

+CO1=(d/T)*100;//Carbonmonoxide by volume in percentage

+N21=(f/T)*100;//Nitrogen by volume in percentage

+Mwv=(g*Mho)/100;//Mass of watervapour formed per kg coal in kg

+Mf=(g)/(b+d+e+f+g);//Mole fraction of water vapour in flue gas

+P=Mf*p;//Partial pressure of water vapour in kPa

+D=32.9;//Dew point temperature from steam tables in degree centigrade

+

+//Output

+printf('(a)The complete volumetric composition of the dry flue gas is \n Carbondioxide by volume = %3.2f percentage \n Oxygen by volume = %3.2f percentage \n Carbonmonoxide by volume = %3.2f percentage \n Nitrogen by volume = %3.2f percentage \n (b) The actual amount of air supplied per kg coal = %3.2f kg \n (c) Mass of water vapour formed per kg coal = %3.2f kg \n (d) The dew point temperature of the flue gas = %3.2f degree centigrade ',CO21,O21,CO1,N21,ma1,Mwv,D)

diff --git a/1943/CH4/EX4.12/Ex4_12.sce b/1943/CH4/EX4.12/Ex4_12.sce
new file mode 100755
index 000000000..87adcf53a
--- /dev/null
+++ b/1943/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,24 @@
+
+clc

+clear

+//Input data

+H=200;//Height of the stack in m

+D=4;//Diameter of the stack in m

+m=1000;//Mass flow rate of gas in kg/s

+Ts=100;//Stack exit gas temperature in degree centigrade

+Ta=5;//Ambient air temperature in degree centigrade

+Vw=50;//Wind velocity in Km/h

+Cp=1.005;//Specific heat of the gas in kJ/kgK

+pi=3.142;//Mathematical constant the value of pi

+

+//Calculations

+Vw1=(50*1000)/(60*60);//Wind velocity in m/s

+Qe=m*Cp*(Ts-Ta);//Heat emission from plume in kW

+Qe1=Qe/1000;//Heat emission from the plume in MW

+p=(101.325)/(0.287*373);//Density of the gas in kg/m^3

+A=(pi*D^2)/4;//Area of the stack in m^2

+Vs=m/(p*A);//Stack gas exict velocity in m/s

+H1=[(2.62*(Qe1^(1/2))*1000)/Vw1]-[(0.029*Vs*D)/Vw1];//The height of the gas plume in m

+

+//Output

+printf('The height of the gas plume is H = %3.1f m ',H1)

diff --git a/1943/CH4/EX4.13/Ex4_13.sce b/1943/CH4/EX4.13/Ex4_13.sce
new file mode 100755
index 000000000..b6bd4e83b
--- /dev/null
+++ b/1943/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,47 @@
+
+clc

+clear

+//Input data

+CV=20;//Calorific value of the fuel in MJ/kg

+C=65;//The amount of carbon present in the fuel according to gravimetric analysis in %

+H=25;//The amount of hydrogen present in the fuel according to gravimetric analysis in %

+O=10;//The amount of oxygen present in the fuel according to gravimetric analysis in %

+p1=1;//Pressure at the inlet of the compressor in bar

+t1=27;//Temperature at the inlet of the compressor in degree centigrade

+p2=4;//The pressure which compressor compresses it isentropically in bar

+Re=78;//The regenerator effectiveness in %

+CO2=6;//The amount of carbondioxide according to the analysis of dry exhaust gas in %

+CO=1.5;//The amount of carbonmonoxide according to the analysis of dry exhaust gas in %

+Cp=1.005;//Specific heat capacity of the air in kJ/kgK

+i=1.44;//Isentropic index for the air

+Cp1=1.15;//Specific heat capacity of the air in kJ/kgK

+i1=1.33;//Isentropic index for the combustion products

+Mc=12;//Molecular weight of the carbon 

+Mh=2;//Molecular weight of the hydrogen

+Mo=32;//Molecular weight of the oxygen

+Mho=18;//Molecular weight of water

+T0=288;//Datum temperature in K (Assumed)

+

+//Calculations

+h=(C/100)/(Mc);//Equating coefficients of the carbon from the equation 

+e=(H/100)/Mh;//Equating coefficients of the hydrogen from the equation 

+y=(CO/100)/(CO2/100);//From dry exhaust gas analysis for solving

+a=h/(1+y);//The coefficient of the carbondioxide in the product side of the equation 

+b=h-a;//The coefficient of the carbonmonoxide in the product side of the equation 

+z=b/(CO/100);//The sum of coefficients of the product side of the equation 

+x=z-(b/2)+(e/2);//Mol of air supplied in kmol

+wa=x*28.96;//Air supplied in kg/kg fuel

+wf=1;//Assuming 1 kg of fuel supplied

+T2=(t1+273)*(p2/p1)^((i-1)/i);//Temperature at the outlet of the compressor in K

+T3=[[(wa*Cp*(T2-T0))+(wf*CV*1000)]/[(wa+wf)*(Cp1)]]+T0;//Maximum temperature of the cycle in K

+T4=T3/[(4)^((i1-1)/i1)];//Temperature at point of the cycle in K

+T5=[(Re/100)*(T4-T2)]+T2;//Temperature at point of the cycle in K

+Wc=wa*Cp*(T2-(t1+273));//Work done by the compressor in kW

+Wt=23.54*Cp1*(T3-T4);//Work done by the turbine in kW

+Q1=23.54*Cp1*(T3-T5);//Total work done by the system in kW

+nc=(Wt-Wc)/Q1;//Efficiency of the cycle 

+nc1=nc*100;//Efficiency of the cycle in %

+spc=3600/(Wt-Wc);//Specific fuel consumption in kg/kWh

+

+//Output 

+printf('(a) The maximum temperature of the cycle T3 = %3.0f K \n (b)Thermal efficiency of the plant = %3.3f or %3.2f percentage\n (c) Specific fuel consumption = %3.3f kg/kWh ',T3,nc,nc1,spc)

diff --git a/1943/CH4/EX4.2/Ex4_2.sce b/1943/CH4/EX4.2/Ex4_2.sce
new file mode 100755
index 000000000..d54483c56
--- /dev/null
+++ b/1943/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,18 @@
+
+clc 

+clear

+//Input data

+CO2=11.5;//Percentage of carbondioxide present in combustion in %

+O2=2.7;//Percentage of oxygen present in the combustion in %

+CO=0.7;//Percentage of carbonmonoxide present in the combuston in %

+

+//Calculations 

+a=85.1/3.76;//Equating moles for nitrogen from the equation 

+x=(CO2+CO)/3;//Equating moles for carbon from the equation 

+b=[a-CO2-(CO/2)-O2]*2;//Equating moles for oxygen from the equation 

+y=a/x;//Moles of oxygen supplied for one mole of propane gas

+z=5;//Theoretically 5 moles of oxygen are required for reacting

+E=[(y-z)/z]*100;//The excess of air supplied in %

+

+//Output

+printf('The percentage excess air used is = %3.1f percentage',E)

diff --git a/1943/CH4/EX4.3/Ex4_3.sce b/1943/CH4/EX4.3/Ex4_3.sce
new file mode 100755
index 000000000..aba54e17b
--- /dev/null
+++ b/1943/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,23 @@
+
+clc

+clear

+//Input data

+CO2=12.1;//The amount of carbondioxide released from the combustion in %

+O2=3.8;//The amount of oxygen released from the combustion in %

+CO=0.9;//The amount of carbonmonoxide released from the combustion in %

+MO=32;//Molecular weight of Oxygen

+

+//Calculations

+a=83.2/3.76;//Equating moles for nitrogen from the equation 

+b=(2*a)-(2*CO2)-(2*O2)-CO;//Equating moles for oxygen from the equation

+x=CO2+CO;//Equating moles for carbon from the equation

+y=2*b;//Equating moles for hydrogen from the equation

+z=18.75;//Moles of Oxygen from the stoichiometric equation 

+z1=a;//Moles of Oxygen from the combustion equation 

+E=[(z1-z)/z]*100;//Percentage of excess air in%

+A=(a*MO)/0.232;//Actual air supplied per mole of C13H23

+Mc=179;//Molecular weight of C13H23

+Af=A/Mc;//Air fuel ratio during the test

+

+//Output

+printf('(a) The air fuel ratio during the test = %3.2f \n (b) The excess or deficiency of air used = %3.0f Percentage of excess air used',Af,E)

diff --git a/1943/CH4/EX4.4/Ex4_4.sce b/1943/CH4/EX4.4/Ex4_4.sce
new file mode 100755
index 000000000..d4098f1d9
--- /dev/null
+++ b/1943/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,51 @@
+
+clc

+clear

+//Input data

+C=61;//The mass of carbon present in the coal according to coal analysis on mass basis in %

+H=4;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %

+O=3;//The mass of oxygen present in the coal according to coal analysis on mass basis in %

+N=2;//The mass of nitrogen present in the coal according to coal analysis on mass basis in %

+S=1;//The mass of sulphur present in the coal according to coal analysis on mass basis in %

+M=4;//The mass of moisture present in the coal according to coal analysis on mass basis in %

+A=25;//The mass of ash present in the coal according to coal analysis on mass basis in %

+HHV=24.3;//The high heating value of the coal i.e energy released by complete combustion of 1 kg fuel in MJ/kg

+CO2=12;//The amount of carbondioxide by volume according to dry flue gas analysis in %

+CO=1.5;//The amount of carbonmonoxide by volume according to dry flue gas analysis in %

+O2=7;//The amount of oxygen by volume according to dry flue gas analysis in %

+N2=79.5;//The amount of nitrogen by volume according to dry flue gas analysis in %

+Te=170;//Exhaust gas temperature in degree centigrade

+L=0.03;//Energy loss other than dry exhaust loss and incomplete combustion is 3% of HHV

+R=150;//Steam generation rate in t/h

+Po=100;//Steam condition at boiler outlet in bar

+To=500;//Steam condition at boiler outlet in degree centigrade

+Ti=160;//Feed water inlet temperature in degree centigrade

+HCO2=33083;//Heat of reaction in kJ/kg carbon

+HCO=9500;//Heat of reaction in kJ/kg carbon

+cp=1.05;//Heat capacity of dry flue gas (dfg) in kJ/kgK

+Ta=30;//The ambient temperature of air in degree centigrade

+Mc=44;//Molecular weight of Carbondioxide

+Mco=28;//Molecular weight of carbonmonoxide

+Mo=32;//Molecular weight of oxygen

+Mn=28;//Molecular weight of nitrogen

+Mx=12;//Molecular weight of carbon

+h1=3373.7;//Enthalpy at 100 bar and 500 degree centigrade in kJ/kg

+hf=675.55;//Enthalpy at 160 degree centigrade in kJ/kg

+hg=2724.7;//Enthalpy at 100 bar in kJ/kg

+

+//Calculations

+Mdfg=([(C/100)*[(Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(N2/100))]]/[Mx*[(CO2/100)+(CO/100)]]);//Mass of dry flue gas produced per kg of fuel in kg

+Ed=Mdfg*cp*(Te-Ta);//Energy loss due to dry exhaust gas in kJ/kg fuel

+//Since Mdfg is 11.73kg through sciab calculation, there is a variation in Ed value and Ei value

+Ei=[(Mdfg)*(HCO2-HCO)*(Mx/Mco)]*[(Mco*(CO/100))/[(Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(O2/100))+(Mco*(N2/100))]];//Energy loss due to incomplete combustion in kJ/kg fuel

+El=L*HHV;//Energy loss other than dry exhaust loss and incomplete combustion loss in MJ/kg fuel

+TEl=(Ed/1000)+(Ei/1000)+El;//Total energy loss in MJ/kg fuel

+Be=[(HHV-TEl)/(HHV)]*100;//Boiler efficiency in %

+Wf=([(R*1000)*(h1-hf)]/((Be/100)*HHV*1000))/3600;//The fuel burning rate in kg/s

+Wth=(11.5*(C/100))+(34.5*[(H/100)-(O/800)])+(4.3*(S/100));//Thearetical air required per kg of fuel in kg

+WA=[[(3.04*(N2/100)*(C/100))]/[(CO2/100)+(CO/100)]]-[(N/100)*(1/0.768)];//Actual air supplied per kg of fuel in kg

+per=[(WA-Wth)/Wth]*100;//Percentage excess air used in %

+pea=[(h1-hg)/(h1-hf)]*100;//Percentage of energy absorbed in the superheater

+

+//Output

+printf('(a)The amount of dry flue gas produced per kg fuel = %3.2f kg \n (b)The dry exhaust loss = %3.1f kJ/kg fuel and incomplete combustion loss per kg fuel = %3.2f kJ/kg fuel \n (c)The boiler efficiency = %3.2f percentage \n (d) THe fuel burning rate = %3.3f kg/s \n (e)The percentage of excess air used = %3.2f percentage \n (f) The percentage of energy absorbed in the superheater = %3.2f percentage',Mdfg,Ed,Ei,Be,Wf,per,pea)

diff --git a/1943/CH4/EX4.5/Ex4_5.sce b/1943/CH4/EX4.5/Ex4_5.sce
new file mode 100755
index 000000000..d809fa7d9
--- /dev/null
+++ b/1943/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,34 @@
+
+clc

+clear

+//Input data

+C=83.7;//The amount of carbon present in the fuel oil according to ultimate analysis of a fuel oil in %

+H=12.7;//The amount of hydrogen present in the fuel oil according to ultimate analysis of a fuel oil in %

+O=1.2;//The amount of oxygen present in the fuel oil according to ultimate analysis of a fuel oil in %

+N=1.7;//The amount of nitrogen present in the fuel oil according to ultimate analysis of a fuel oil in %

+S=0.7;//The amount of sulphur present in the fuel oil according to ultimate analysis of a fuel oil in %

+td=27;//The dry bulb temperature of combustion air in degree centigrade

+tw=21;//The wet bulb temperature of combustion air in degree centigrade

+E=0.3;//Excess air and assuming complete combustion in %

+t=200;//Temperature to find total volume of combustion products in degree centigrade

+p=1.013;//Pressure to find total volume of combustion procucts in bar

+

+//Calculations

+Wth=(11.5*(C/100))+[34.5*((H/100)-(O/100)*(1/8))]+(4.3*(S/100));//Theoretical air required per kg of fuel in kg

+WA=(1+E)*Wth;//Actual air required per kg of fuel in kg/kg fuel

+sh=0.0132;//Specific humidity at DBT and WBT in kg moisture/kg dry air

+W=WA*sh;//Water vapour entering with air per kg fuel in kg vap/kg fuel

+Tw=(9*(H/100))+WA;//Total water vapour formed per kg fuel in kg

+CO2=(44/12)*(C/100);//mass of carbondioxide gas per kg of fuel

+O2=0.232*E*Wth;//Mass of oxygen gas per kg of fuel

+N2=0.768*(1+E)*Wth+(N/100);//Mass of nitrogen gas per kg of fuel

+SO2=(64/32)*(S/100);//Mass of nitrogen gas per kg of fuel

+H2O=1.383;//Mass of water per kg of fuel

+M=(CO2/44)+(O2/32)+(N2/28)+(SO2/64)+(H2O/18);//Moles of combustion gases formed per kg fuel

+VG=M*22.4*[(273+t)/273]*(1.013/1.013);//Volume of flue gases at 200 degree centigrade and 1.013 bar per kg fuel

+CO21=((CO2/44)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas CO2 by volume

+O21=((O2/32)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas O2 by volume

+N21=((N2/28)/[(CO2/44)+(O2/32)+(N2/28)])*100;//Composition of dry flue gas N2 by volume

+

+//Output

+printf('(a)The total volume of combustion products at 200 degee centigrade and 1.013 bar = %3.2f m^3 \n (b)The dry flue gas analysis based on carbondioxide,oxygen and nitrogen is \n Carbondioxide = %3.2f percent \n Oxygen = %3.2f percent \n Nitrogen = %3.2f percent',VG,CO21,O21,N21)

diff --git a/1943/CH4/EX4.6/Ex4_6.sce b/1943/CH4/EX4.6/Ex4_6.sce
new file mode 100755
index 000000000..dfacec0b8
--- /dev/null
+++ b/1943/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,33 @@
+
+clc

+clear

+//Input data

+C2H6=22.6;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %

+CH4=73.6;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %

+CO2=2.4;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %

+N2=1.4;//The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %

+E=0.25;//Assuming combustion air to be dry and in excess

+t=260;//The temperature for the total gas volume for complete combustion in degree centigrade

+p=1.013;//The pressure for the total gas volume for complete combustion in bar

+Mch=30;//Molecular weight of C2H6

+Mc=16;//Molecular weight of CH4

+Mco=44;//Molecular weight of CO2

+Mn=28;//Molecular weight of N2

+Mo=32;//Molecular weight of O2

+Mh=18;//Molecular weight of H2O

+

+//Calculations

+x=100;//Assuming 100 moles of fuel gas 

+Mf=[(C2H6/100)*Mch]+[(CH4/100)*Mc]+[(N2/100)*Mn]+[(CO2/100)*Mco];//Molecular weight of fuel gas

+Ma=[(226.3*(Mo+(3.76*Mn))*(1+E))]/28.96;//Moles of air supplied

+Mc=1440;//Moles of combustion gas from the equation 

+Mr=x+Ma+Mc;//Total reaction molecules

+Mwc=[(121.2*Mco)+(215*Mh)+(56.6*Mo)+(1065.4*Mn)]/Mc;//Molecular weight of combustion gas in kg/kgmol

+Mt=Mc/(x*20);//Total number of moles of combustion gas per kg fuel gas

+VG=Mt*22.4*[(273+t)/273];//Volume of combustion products per kg fuel gas

+CO21=(121.2/(121.2+56.6+1065.4))*100;//Gas analysis of CO2 by volume

+O21=(56.6/1243.2)*100;//Gas analysis of O2 by volume

+N2=(1065.4/1243.2)*100;//Gas analysis of N2 by volume

+

+//Output

+printf('(a)The molecular weight of the combustion products M = %3.2f kg/kg mol \n (b) The total gas volume for complete combustion at 260 degree centigrade and 1.013 bar is %3.2f m^3/kg fuel \n (c)The dry flue gas analysis on \n carbondioxide = %3.1f percent \n oxygen = %3.1f percent \n nitrogen = %3.1f percent ',Mwc,VG,CO21,O21,N2)

diff --git a/1943/CH4/EX4.7/Ex4_7.sce b/1943/CH4/EX4.7/Ex4_7.sce
new file mode 100755
index 000000000..cc1cdd3d6
--- /dev/null
+++ b/1943/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,19 @@
+
+clc

+clear

+//Input data

+CO21=9.7;//Carbondioxide gas analysis before the air preheater 

+CO22=9.2;//Carbondioxide gas analysis after the air preheater

+O21=4.0;//Oxygen gas analysis before the air preheater

+O22=4.9;//Oxygen gas analysis after the air preheater

+N21=86.3;//Nitrogen gas analysis before the air preheater

+N22=85.9;//Nitrogen gas analysis after the air preheater

+C=72;//The coal used shows the carbon percentage by mass in %

+

+//Calculations

+W1=[(3.04)*(N21/100)*(C/100)]/[(CO21/100)];//Before air preheater in kg air/kg fuel

+W2=[(3.04)*(N22/100)*(C/100)]/[(CO22/100)];//After air preheater in kg air/kg fuel

+A=W2-W1;//Air leakage in kg air/kg fuel

+

+//Output

+printf('The air leakage into the air preheater per kg of coal fired is %3.0f kg air/kg fuel',A)

diff --git a/1943/CH4/EX4.8/Ex4_8.sce b/1943/CH4/EX4.8/Ex4_8.sce
new file mode 100755
index 000000000..65f3383a4
--- /dev/null
+++ b/1943/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,31 @@
+
+clc

+clear

+//Input data

+n=6;//Total lancashire boilers in a textile factory

+Ws=6;//Each boiler supplying steam in t/h

+p=16;//Pressure at which steam is supplied in bar

+t=250;//Temperature at which steam is supplied in degree centigrade

+CV=43960;//Calorific value of the fuel oil in kJ/kg

+no=75;//Overall efficiency of the boiler in %

+a=16;//The amount of air required for efficient burning of the fuel inkg

+H=20;//Drought of water gauge required at the base of chimney in mm

+tf=320;//The flue gases leave the boiler in degree centigrade

+ts=300;//The average temperature of the gases in the stack in degree centigrade

+ta=30;//The atmospheric temperature in degree centigrade

+R=0.287;//Real Gas constant in kJ/kgK

+h1=2919.2;//enthalpy at the entrance of the boiler in kJ/kgK

+hf=125.8;//Enthalpy at the feed in kJ/kgK

+pi=3.1412;//Mathematical constant

+g=9.81;//gravitational fore constant in m/s^2

+P=1.013;//Atmospheric pressure in bar

+

+//Calculations

+H1=[(H*R*(273+ta)*(273+ts))]/[P*100*[(273+ts)-(273+ta)]];//The draught produced in m

+Wf=[[(Ws*1000)*6*(h1-hf)]/((no/100)*CV)]/3600;//Air fuel ratio in kg/s

+Wa=a*Wf;//Actual air fuel ratio in kg/s

+Wfg=17*Wf;//Air fuel ratio in kg/s

+D=[[(Wfg*R*(273+ts)*(4/pi))]/[(101.3)*(2*g*H1)^(1/2)]]^(1/2);//Diameter at its base in m

+

+//Output

+printf('(a)The height of the stack H = %3.2f m \n (b)The diameter at its base D = %3.2f m ',H1,D)

diff --git a/1943/CH4/EX4.9/Ex4_9.sce b/1943/CH4/EX4.9/Ex4_9.sce
new file mode 100755
index 000000000..83570f703
--- /dev/null
+++ b/1943/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,27 @@
+
+clc

+clear

+//Input data

+Wf=10;//Coal rate in t/h

+C=78;//The mass of carbon present in the coal according to coal analysis on mass basis in %

+H=3;//The mass of hydrogen present in the coal according to coal analysis on mass basis in %

+O=3;//The mass of oxygen present in the coal according to coal analysis on mass basis in %

+S=1;//The mass of sulphur present in the coal according to coal analysis on mass basis in %

+M=7;//The mass of moisture present in the coal according to coal analysis on mass basis in %

+A=8;//The mass of ash present in the coal according to coal analysis on mass basis in %

+E=0.3;//Excess air in percentage

+p=180;//Plenum chamber pressure in mm water gauge

+nm=0.6;//Mechanical efficiency of the fan

+ta=30;//Room temperature in degree centigrade

+R=0.287;//Real gas constant

+P=101.325;//Atmospheric pressure in kPa

+g=9.812;//gravitational force constant m/s^2

+

+//Calculations

+Wth=(11.5*(C/100))+(34.5*[(H/100)-(O/(8*100))])+(4.3*(S/100));//Theoretical air required per kg fuel in kg air/kg fuel

+WA=Wth*(1+0.3);//Actual air required per kg fuel in kg air/kg fuel

+Va=(R*(273+ta))/P;//Volume flow rate of air in m^3/kg

+FD=((WA*Wf*1000*Va*p*g)/(3600*nm))/1000;//FD fan motor capacity in kW

+

+//Output

+printf('The required motor capacity needed for the FD fan is %3.2f kW ',FD)