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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1775 | |
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initial commit / add all books
Diffstat (limited to '1775')
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diff --git a/1775/CH1/EX1.1/Chapter1_Example1.sce b/1775/CH1/EX1.1/Chapter1_Example1.sce new file mode 100755 index 000000000..12b509a48 --- /dev/null +++ b/1775/CH1/EX1.1/Chapter1_Example1.sce @@ -0,0 +1,24 @@ +//Chapter-1, Illustration 1, Page 15
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.91;//Percentage composition of Carbon
+H=0.03;//Percentage composition of Hydrogen
+O=0.02;//Percentage composition of Oxygen
+N=0.008;//Percentage composition of Nitrogen
+S=0.008;//Percentage composition of Sulphur
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)))+(4.3*S);//Mass of air per kg of coal in kg
+
+//OUTPUT
+mprintf('Minimum mass of air per kg of coal is %3.2f kg',m)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.10/Chapter1_Example10.sce b/1775/CH1/EX1.10/Chapter1_Example10.sce new file mode 100755 index 000000000..79466d5e7 --- /dev/null +++ b/1775/CH1/EX1.10/Chapter1_Example10.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 10, Page 23
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+H2=0.27;//Percentage composition of H2 by volume
+CO2=0.18;//Percentage composition of CO2 by volume
+CO=0.125;//Percentage composition of CO by volume
+CH4=0.025;//Percentage composition of CH4 by volume
+N2=0.4;//Percentage composition of N2 by volume
+
+//CALCULATIONS
+v=(2.38*(H2+CO))+(9.52*CH4);//Volume of air required for complete combustion in (m^3)
+
+//OUTPUT
+mprintf('Volume of air required for complete combustion is %3.3f (m^3)',v)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.11/Chapter1_Example11.sce b/1775/CH1/EX1.11/Chapter1_Example11.sce new file mode 100755 index 000000000..329cb1ff6 --- /dev/null +++ b/1775/CH1/EX1.11/Chapter1_Example11.sce @@ -0,0 +1,33 @@ +//Chapter-1, Illustration 11, Page 24
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+H2=0.5;//Percentage composition of H2 by volume
+CO2=0.1;//Percentage composition of CO2 by volume
+CO=0.05;//Percentage composition of CO by volume
+CH4=0.25;//Percentage composition of CH4 by volume
+N2=0.1;//Percentage composition of N2 by volume
+pCO2=8;//Percentage volumetric analysis of CO2
+pO2=6;//Percentage volumetric analysis of O2
+pN2=86;//Percentage volumetric analysis of N2
+
+
+//CALCULATIONS
+v=(2.38*(H2+CO))+(9.52*CH4);//Volume of air required for complete combustion in (m^3)
+vN2=v*0.79;//Volume of nitrogen in the air in m^3
+a=CO+CH4+CO2;//CO2 formed per m^3 of fuel gas burnt
+b=vN2+N2;//N2 formed per m^3 of fuel gas burnt
+vt=a+b;//Total volume of dry flue gas formed in m^3
+ve=(pO2*vt)/(21-pO2);//Excess air supplied in m^3
+V=v+ve;//Total quantity of air supplied in m^3
+
+//OUTPUT
+mprintf('Air-fuel ratio by volume is %3.3f:1',V)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.12/Chapter1_Example12.sce b/1775/CH1/EX1.12/Chapter1_Example12.sce new file mode 100755 index 000000000..d3d2ed7af --- /dev/null +++ b/1775/CH1/EX1.12/Chapter1_Example12.sce @@ -0,0 +1,40 @@ +//Chapter-1, Illustration 12, Page 24
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+H2=0.14;//Percentage composition of H2 by volume
+CO2=0.05;//Percentage composition of CO2 by volume
+CO=0.22;//Percentage composition of CO by volume
+CH4=0.02;//Percentage composition of CH4 by volume
+O2=0.02;//Percentage composition of O2 by volume
+N2=0.55;//Percentage composition of N2 by volume
+e=0.4;//Excess air supplied
+//CALCULATIONS
+v=(2.38*(H2+CO))+(9.52*CH4)-(4.76*O2);//Volume of air required for complete combustion in (m^3)
+ve=v*e;//Volume of excess air supplied in m^3
+vtN2=v-(v*0.21);//Volume of N2 in theoretical air in m^3
+veN2=ve-(ve*0.21);//Volume of N2 in excess air in m^3
+vt=vtN2+veN2;//Total volume of N2 in air supplied in m^3
+vCO2=CO+CH4+CO2;//CO2 formed per m^3 of fuel gas
+vN2=vt+N2;//N2 formed per m^3 of fuel gas
+veO2=ve*0.21;//Volume of excess O2 per m^3 of fuel gas
+vT=vCO2+vN2+veO2;//Total volume of dry combustion products
+pCO2=(vCO2*100)/vT;//Percentage volume of CO2
+pN2=(vN2*100)/vT;//Percentage volume of N2
+pO2=(veO2*100)/vT;//Percentage volume of O2
+
+//OUTPUT
+mprintf('Volume of air required for complete combustion is %3.3f (m^3) \n Volume of CO2 per m^3 of gas fuel is %3.2f m^3/m^3 of gas fuel \n Volume of N2 per m^3 of gas fuel is %3.3f m^3/m^3 of gas fuel \n Volume of excess O2 per m^3 of gas fuel is %3.2f m^3/m^3 of gas fuel \n Total volume of dry combustion products is %3.3f m^3/m^3 of gas fuel \n Percentage volume of CO2 is %3.1f percent \n Percentage volume of N2 is %3.2f percent \n Percentage volume of O2 is %3.2f percent',v,vCO2,vN2,veO2,vT,pCO2,pN2,pO2)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.2/Chapter1_Example2.sce b/1775/CH1/EX1.2/Chapter1_Example2.sce new file mode 100755 index 000000000..e9298ab9f --- /dev/null +++ b/1775/CH1/EX1.2/Chapter1_Example2.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 2, Page 16
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.86;//Percentage composition of Carbon
+H=0.12;//Percentage composition of Hydrogen
+O=0.01;//Percentage composition of Oxygen
+S=0.01;//Percentage composition of Sulphur
+v=0.773;//Specific volume of air at N.T.P in (m^3)/kg
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)))+(4.3*S);//Theoretical mass of air per kg of coal in kg
+vth=m*v;//Theoretical volume of air at N.T.P per kg fuel in (m^3)/kg of fuel
+
+//OUTPUT
+mprintf('Theoretical volume of air at N.T.P per kg fuel is %3.2f (m^3)/kg of fuel',vth)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.3/Chapter1_Example3.sce b/1775/CH1/EX1.3/Chapter1_Example3.sce new file mode 100755 index 000000000..6b98548ab --- /dev/null +++ b/1775/CH1/EX1.3/Chapter1_Example3.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 3, Page 16
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.78;//Percentage composition of Carbon
+H=0.06;//Percentage composition of Hydrogen
+O=0.078;//Percentage composition of Oxygen
+N=0.012;//Percentage composition of Nitrogen
+S=0.03;//Percentage composition of Sulphur
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)))+(4.3*S);//Minimum quantity of air required in kg
+mt=((11*C)/3)+(9*H)+(2*S)+(8.32+N);//Total mass of products of combustion in kg
+
+//OUTPUT
+mprintf('Minimum quantity of air required for complete combustion is %3.2f kg \n Total mass of products of combustion is %3.3f kg',m,mt)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.4/Chapter1_Example4.sce b/1775/CH1/EX1.4/Chapter1_Example4.sce new file mode 100755 index 000000000..0e1b512ac --- /dev/null +++ b/1775/CH1/EX1.4/Chapter1_Example4.sce @@ -0,0 +1,43 @@ +//Chapter-1, Illustration 4, Page 17
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.84;//Percentage composition of Carbon
+H=0.09;//Percentage composition of Hydrogen
+CO2=0.0875;//Volumetric composition of CO2
+CO=0.0225;//Volumetric composition of CO
+O2=0.08;//Volumetric composition of Oxygen
+N2=0.81;//Volumetric composition of Nitrogen
+M1=44;//Molecular mass of CO2
+M2=28;//Molecular mass of CO
+M3=32;//Molecular mass of O2
+M4=28;//Molecular mass of N2
+
+//CALCULATIONS
+c1=CO2*M1;//Proportional mass of CO2
+c2=CO*M2;//Proportional mass of CO
+c3=O2*M3;//Proportional mass of O2
+c4=N2*M4;//Proportional mass of N2
+c=c1+c2+c3+c4;//Total proportional mass of constituents
+m1=c1/c;//Mass of CO2 per kg of flue gas in kg
+m2=c2/c;//Mass of CO per kg of flue gas in kg
+m3=c3/c;//Mass of O2 per kg of flue gas in kg
+m4=c4/c;//Mass of N2 per kg of flue gas in kg
+d1=m1*100;//Mass analysis of CO2
+d2=m2*100;//Mass analysis of CO
+d3=m3*100;//Mass analysis of O2
+d4=m4*100;//Mass analysis of N2
+m=((3*m1)/11)+((3*m2)/7);//Mass of carbon in kg
+md=C/m;//Mass of dry flue gas in kg
+
+//OUTPUT
+mprintf('Mass of dry flue gases per kg of coal burnt is %3.1f kg',md)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.5/Chapter1_Example5.sce b/1775/CH1/EX1.5/Chapter1_Example5.sce new file mode 100755 index 000000000..4891b92d1 --- /dev/null +++ b/1775/CH1/EX1.5/Chapter1_Example5.sce @@ -0,0 +1,48 @@ +//Chapter-1, Illustration 5, Page 17
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.624;//Percentage composition of Carbon
+H=0.042;//Percentage composition of Hydrogen
+O=0.045;//Percentage composition of Oxygen
+CO2=0.13;//Volumetric composition of CO2
+CO=0.003;//Volumetric composition of CO
+O2=0.06;//Volumetric composition of Oxygen
+N2=0.807;//Volumetric composition of Nitrogen
+M1=44;//Molecular mass of CO2
+M2=28;//Molecular mass of CO
+M3=32;//Molecular mass of O2
+M4=28;//Molecular mass of N2
+mw=0.378;//Mass of H2O in kg
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)));//Minimum air required in kg
+c1=CO2*M1;//Proportional mass of CO2
+c2=CO*M2;//Proportional mass of CO
+c3=O2*M3;//Proportional mass of O2
+c4=N2*M4;//Proportional mass of N2
+c=c1+c2+c3+c4;//Total proportional mass of constituents
+m1=c1/c;//Mass of CO2 per kg of flue gas in kg
+m2=c2/c;//Mass of CO per kg of flue gas in kg
+m3=c3/c;//Mass of O2 per kg of flue gas in kg
+m4=c4/c;//Mass of N2 per kg of flue gas in kg
+d1=m1*100;//Mass analysis of CO2
+d2=m2*100;//Mass analysis of CO
+d3=m3*100;//Mass analysis of O2
+d4=m4*100;//Mass analysis of N2
+mC=((3*m1)/11)+((3*m2)/7);//Mass of carbon in kg
+md=C/mC;//Mass of dry flue gas in kg
+mact=(md+mw)-(C+H+O);//Actual air supplied per kg of fuel in kg
+me=mact-m;//Mass of excess air per kg of fuel in kg
+
+//OUTPUT
+mprintf('Minimum air required to burn 1 kg of coal is %3.2f kg \n Mass of air actually supplied per kg of coal is %3.3f kg \n Amount of excess air supplied per kg of coal burnt is %3.3f kg',m,mact,me)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.6/Chapter1_Example6.sce b/1775/CH1/EX1.6/Chapter1_Example6.sce new file mode 100755 index 000000000..6b95d1115 --- /dev/null +++ b/1775/CH1/EX1.6/Chapter1_Example6.sce @@ -0,0 +1,30 @@ +//Chapter-1, Illustration 6, Page 19
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.78;//Percentage composition of Carbon
+H=0.03;//Percentage composition of Hydrogen
+O=0.03;//Percentage composition of Oxygen
+S=0.01;//Percentage composition of Sulphur
+me=0.3;//Mass of excess air supplied
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)))+(4.3*S);//Mass of air per kg of coal in kg
+mec=me*m;//Excess air supplied per kg of coal in kg
+mact=m+mec;//Actual mass of air supplied per kg of coal in kg
+mCO2=(11*C)/3;//Mass of CO2 produced per kg of coal in kg
+mHw=9*H;//Mass of H2O produced per kg of coal in kg
+mSO2=2*S;//Mass of SO2 produced per kg of coal in kg
+mO2=0.232*mec;//Mass of excess O2 produced per kg of coal in kg
+mN2=0.768*mact;//Mass of N2 produced per kg of coal in kg
+
+//OUTPUT
+mprintf('Mass of air to be supplied is %3.2f kg \n Mass of CO2 produced per kg of coal is %3.2f kg \n Mass of H2O produced per kg of coal is %3.2f kg \n Mass of SO2 produced per kg of coal is %3.2f kg \n Mass of excess O2 produced per kg of coal is %3.2f kg \n Mass of N2 produced per kg of coal is %3.2f kg \n',m,mCO2,mHw,mSO2,mO2,mN2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.7/Chapter1_Example7.sce b/1775/CH1/EX1.7/Chapter1_Example7.sce new file mode 100755 index 000000000..58929075d --- /dev/null +++ b/1775/CH1/EX1.7/Chapter1_Example7.sce @@ -0,0 +1,55 @@ +//Chapter-1, Illustration 7, Page 20
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.9;//Percentage composition of Carbon
+H=0.033;//Percentage composition of Hydrogen
+O=0.03;//Percentage composition of Oxygen
+N=0.008;//Percentage composition of Nitrogen
+S=0.009;//Percentage composition of Sulphur
+M1=44;//Molecular mass of CO2
+M2=64;//Molecular mass of SO2
+M3=32;//Molecular mass of O2
+M4=28;//Molecular mass of N2
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)))+(4.3*S);//Minimum mass of air per kg of coal in kg
+mCO2=(11*C)/3;//Mass of CO2 produced per kg of coal in kg
+mHw=9*H;//Mass of H2O produced per kg of coal in kg
+mSO2=2*S;//Mass of SO2 produced per kg of coal in kg
+mt=11.5*1.5;//Total mass of air supplied per kg of coal in kg
+me=mt-m;//Excess air supplied in kg
+mO2=0.232*me;//Mass of excess O2 produced per kg of coal in kg
+mN2=0.768*mt;//Mass of N2 produced per kg of coal in kg
+mtN2=mN2+N;//Total mass of Nitrogen in exhaust in kg
+md=mCO2+mSO2+mO2+mtN2;//Total mass of dry flue gases per kg of fuel in kg
+CO2=(mCO2/md)*100;//Percentage composition of CO2 by mass in percent
+SO2=(mSO2/md)*100;//Percentage composition of SO2 by mass in percent
+O2=(mO2/md)*100;//Percentage composition of O2 by mass in percent
+N2=(mN2/md)*100;//Percentage composition of N2 by mass in percent
+c1=CO2/M1;//Proportional volume of CO2
+c2=SO2/M2;//Proportional volume of SO2
+c3=O2/M3;//Proportional volume of O2
+c4=N2/M4;//Proportional volume of N2
+c=c1+c2+c3+c4;//Total proportional volume of constituents
+m1=c1/c;//Volume of CO2 in 1 (m^3) of flue gas
+m2=c2/c;//Volume of SO2 in 1 (m^3) of flue gas
+m3=c3/c;//Volume of O2 in 1 (m^3) of flue gas
+m4=c4/c;//Volume of N2 in 1 (m^3) of flue gas
+d1=m1*100;//Volume analysis of CO2
+d2=m2*100;//Volume analysis of SO2
+d3=m3*100;//Volume analysis of O2
+d4=m4*100;//Volume analysis of N2
+
+//OUTPUT
+mprintf('Minimum mass of air required is %3.1f kg \n Total mass of dry flue gases per kg of fuel is %3.2f kg \n Percentage composition of CO2 by volume is %3.2f percent \n Percentage composition of SO2 by volume is %3.3f percent \n Percentage composition of O2 by volume is %3.1f percent \n Percentage composition of N2 by volume is %3.2f percent',m,md,d1,d2,d3,d4)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.8/Chapter1_Example8.sce b/1775/CH1/EX1.8/Chapter1_Example8.sce new file mode 100755 index 000000000..d7be45407 --- /dev/null +++ b/1775/CH1/EX1.8/Chapter1_Example8.sce @@ -0,0 +1,49 @@ +//Chapter-1, Illustration 8, Page 21
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.88;//Percentage composition of Carbon
+H=0.036;//Percentage composition of Hydrogen
+O=0.048;//Percentage composition of oxygen
+CO2=0.109;//Volumetric composition of CO2
+CO=0.01;//Volumetric composition of CO
+O2=0.071;//Volumetric composition of Oxygen
+N2=0.81;//Volumetric composition of Nitrogen
+M1=44;//Molecular mass of CO2
+M2=28;//Molecular mass of CO
+M3=32;//Molecular mass of O2
+M4=28;//Molecular mass of N2
+
+//CALCULATIONS
+m=(11.5*C)+(34.5*(H-(O/8)));//Theoretical air required in kg
+c1=CO2*M1;//Proportional mass of CO2
+c2=CO*M2;//Proportional mass of CO
+c3=O2*M3;//Proportional mass of O2
+c4=N2*M4;//Proportional mass of N2
+c=c1+c2+c3+c4;//Total proportional mass of constituents
+m1=c1/c;//Mass of CO2 per kg of flue gas in kg
+m2=c2/c;//Mass of CO per kg of flue gas in kg
+m3=c3/c;//Mass of O2 per kg of flue gas in kg
+m4=c4/c;//Mass of N2 per kg of flue gas in kg
+mC=((3*m1)/11)+((3*m2)/7);//Mass of carbon in kg
+md=C/mC;//Mass of dry flue gas in kg
+hc=H*9;//Hydrogen combustion in kg of H2O
+mair=(md+hc)-(C+H+O);//Mass of air supplied per kg of coal in kg
+me=mair-m;//Excess air per kg of coal in kg
+mN2=m4*md;//Mass of nitrogen per kg of coal in kg
+mact=mN2/0.768;//Actual mass of air per kg of coal in kg
+pe=(me/m)*100;//Perccentage excess air in percent
+
+//OUTPUT
+mprintf('Mass of air actually supplied per kg of coal is %3.2f kg \n Percentage of excess air is %3.2f percent',mact,pe)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.9/Chapter1_Example9.sce b/1775/CH1/EX1.9/Chapter1_Example9.sce new file mode 100755 index 000000000..f35e7a7ae --- /dev/null +++ b/1775/CH1/EX1.9/Chapter1_Example9.sce @@ -0,0 +1,52 @@ +//Chapter-1, Illustration 9, Page 22
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C=0.84;//Percentage composition of Carbon
+H=0.14;//Percentage composition of Hydrogen
+O=0.02;//Percentage composition of oxygen
+CO2=8.85;//Volumetric composition of CO2
+CO=1.2;//Volumetric composition of CO
+O2=6.8;//Volumetric composition of Oxygen
+N2=83.15;//Volumetric composition of Nitrogen
+M1=44;//Molecular mass of CO2
+M2=28;//Molecular mass of CO
+M3=32;//Molecular mass of O2
+M4=28;//Molecular mass of N2
+a=8/3;//O2 required per kg C
+b=8;//O2 required per kg H2
+mair=0.23;//Mass of air
+
+//CALCULATIONS
+c=C*a;//O2 required per kg of fuel for C
+d=H*b;//O2 required per kg of fuel for H2
+tO2=c+d+O;//Theoreticcal O2 required in kg/kg of fuel
+tm=tO2/mair;//Theoretical mass of air in kg/kg of fuel
+c1=CO2*M1;//Proportional mass of CO2 by Volume
+c2=CO*M2;//Proportional mass of CO by Volume
+c3=O2*M3;//Proportional mass of O2 by Volume
+c4=N2*M4;//Proportional mass of N2 by Volume
+c=c1+c2+c3+c4;//Total proportional mass of constituents
+m1=c1/c;//Mass of CO2 per kg of flue gas in kg
+m2=c2/c;//Mass of CO per kg of flue gas in kg
+m3=c3/c;//Mass of O2 per kg of flue gas in kg
+m4=c4/c;//Mass of N2 per kg of flue gas in kg
+mC=((m1*12)/M1)+((m2*12)/M2);//Mass of carbon per kg of dry flue gas in kg
+md=C/mC;//Mass of dry flue per kg of fuel in kg
+p=(4*m2)/7;//Oxygen required to burn CO in kg
+meO2=md*(m3-p);//Mass of excess O2 per kg of fuel in kg
+me=meO2/mair;//Mass of excess air in kg/kg fuel
+mt=tm+me;//Total air required per kg fuel
+
+//OUTPUT
+mprintf('Mass of excess air supplied per kg of fuel burnt is %3.1f kg/kg of fuel \n Air-fuel ratio is %3.1f:1',me,mt)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
+
diff --git a/1775/CH2/EX2.1/Chapter2_Example1.sce b/1775/CH2/EX2.1/Chapter2_Example1.sce new file mode 100755 index 000000000..75371c726 --- /dev/null +++ b/1775/CH2/EX2.1/Chapter2_Example1.sce @@ -0,0 +1,35 @@ +//Chapter-2, Illustration 1, Page 55
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=0.1;//Pressure of air supplied in MPa
+T1=308;//Temperature of air supplied in K
+rv=8;//Compression ratio
+q1=2100;//Heat supplied in kJ/kg
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+Cv=0.718;//Specific heat at constant volume in kJ/kg-K
+R=0.287;//Universal gas constant in kJ/kg-K
+
+//CALCULATIONS
+y=Cp/Cv;//Ratio of specific heats
+n=(1-(1/(rv^(y-1))))*100;//Cycle efficiency
+v1=(R*T1)/(P1*1000);//Specific volume at point 1 in (m^3)/kg
+v2=v1/rv;//Specific volume at point 2 in (m^3)/kg
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+T3=(q1/Cv)+T2;//Temperature at point 3 in K
+P2=P1*(rv^y);//Pressure at point 2 in MPa
+P3=P2*(T3/T2);//Pressure at point 3 in MPa
+wnet=(q1*n)/100;//Net workdone in J/kg
+MEP=(wnet/(v1-v2))/1000;//Mean effective pressure in MPa
+
+//OUTPUT
+mprintf('Maximum pressure of the cycle is %3.3f MPa \n Maximum temperature of the cycle is %3.0f K \n Cycle efficiency is %3.1f percent \n Mean effective pressure is %3.3f MPa',P3,T3,n,MEP)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.10/Chapter2_Example10.sce b/1775/CH2/EX2.10/Chapter2_Example10.sce new file mode 100755 index 000000000..a3d2400af --- /dev/null +++ b/1775/CH2/EX2.10/Chapter2_Example10.sce @@ -0,0 +1,35 @@ +//Chapter-2, Illustration 10, Page 66
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=20;//Compression ratio
+P1=95;//Pressure at point 1 in kPa
+T1=293;//Temperature at point 1 in K
+T3=2200;//Temperature at point 3 in K
+y=1.4;//Ratio of specific heats
+R=287;//Universal gas constant in J/kg-K
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+P2=P1*(rv^y);//Pressure at point 2 in kPa
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+v2=(R*T2)/(P2*1000);//Specific volume at point 2 in (m^3)/kg
+v3=v2*(T3/T2);//Specific volume at point 3 in (m^3)/kg
+rc=v3/v2;//Cut-off ratio
+nth=(1-(((rc^y)-1)/((rv^(y-1))*y*(rc-1))))*100;//Thermal efficiency
+q23=Cp*(T3-T2);//Heat flow between points 2 and 3 in kJ/kg
+wnet=(nth*q23)/100;//Net workdone in kJ/kg
+MEP=wnet/(v2*(rv-1));//Mean effective pressure in kPa
+
+//OUTPUT
+mprintf('Thermal efficiency is %3.1f percent \n Mean effective pressure is %3.2f kPa',nth,MEP)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.11/Chapter2_Example11.sce b/1775/CH2/EX2.11/Chapter2_Example11.sce new file mode 100755 index 000000000..687920f54 --- /dev/null +++ b/1775/CH2/EX2.11/Chapter2_Example11.sce @@ -0,0 +1,26 @@ +//Chapter-2, Illustration 11, Page 68
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=21;//Compression ratio
+re=10.5;//Expansion ratio
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+rc=rv/re;//Cut-off ratio
+nth=(1-(((rc^y)-1)/((rv^(y-1))*y*(rc-1))))*100;//Air standard efficiency
+
+//OUTPUT
+mprintf('Cut-off ratio is %3.0f \n Air standard efficiency is %3.2f percent',rc,nth)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.12/Chapter2_Example12.sce b/1775/CH2/EX2.12/Chapter2_Example12.sce new file mode 100755 index 000000000..c764eb2ff --- /dev/null +++ b/1775/CH2/EX2.12/Chapter2_Example12.sce @@ -0,0 +1,25 @@ +//Chapter-2, Illustration 12, Page 69
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=16;//Compression ratio
+rp=1.5;//Pressure ratio
+y=1.4;//Ratio of specific heats
+cp=8;//Cut-off percentage
+
+//CALCULATIONS
+rc=2.2;//Cut-off ratio
+ntd=(1-((rp*(rc^y)-1)/((rv^(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;//Dual cycle efficiency
+
+//OUTPUT
+mprintf('Ideal efficiency of engine is %3.1f percent',ntd)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.13/Chapter2_Example13.sce b/1775/CH2/EX2.13/Chapter2_Example13.sce new file mode 100755 index 000000000..dcaf50df2 --- /dev/null +++ b/1775/CH2/EX2.13/Chapter2_Example13.sce @@ -0,0 +1,30 @@ +//Chapter-2, Illustration 13, Page 69
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=0.2;//Bore in m
+L=0.5;//Stroke in m
+c=0.06;//Cut-off percentage
+y=1.4;//Ratio of specific heats
+rv=15;//Compression ratio
+rp=1.4;//Pressure ratio
+
+//CALCULATIONS
+Vs=(3.147/4)*(d^2)*L;//Stroke volume in m^3
+DV=c*Vs;//Difference in volumes at points 4 and 3
+V3=Vs/(rv-1);//Specific volume at point 3 in m^3
+V4=V3+DV;//Specific volume at point 4 in m^3
+rc=V4/V3;//Cut-off ratio
+ntd=(1-((rp*(rc^y)-1)/((rv^(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;//Ideal efficiency
+
+//OUTPUT
+mprintf('Ideal efficiency of the engine is %3.1f percent',ntd)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.14/Chapter2_Example14.sce b/1775/CH2/EX2.14/Chapter2_Example14.sce new file mode 100755 index 000000000..3f9e798d5 --- /dev/null +++ b/1775/CH2/EX2.14/Chapter2_Example14.sce @@ -0,0 +1,47 @@ +//Chapter-2, Illustration 14, Page 70
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=0.2;//Bore in m
+L=0.3;//Stroke in m
+c=0.04;//Cut-off percentage
+y=1.4;//Ratio of specific heats
+rv=8;//Compression ratio
+P1=1;//Pressure at point 1 in bar
+P3=60;//Pressure at point 3 in bar
+T1=298;//Temperature at point 1 in K
+R=287;//Universal gas constant in J/kg
+Cv=0.718;//Speific heat at constant volume in kJ/kg-K
+Cp=1.005;//Speific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+Vs=(3.147/4)*(d^2)*L;//Stroke volume in m^3
+V2=Vs/(rv-1);//Specific volume at point 2 in m^3
+V3=V2;//Specific volume at point 3 in m^3
+V1=V2+Vs;//Specific volume at pont 1 in m^3
+V5=V1;//Specific volume at pont 5 in m^3
+P2=P1*(rv^y);//Pressure at point 2 in bar
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+T3=T2*(P3/P2);//Temperature at point 3 in K
+V4=V3+(c*(V1-V2));//Specific volume at point 4 in m^3
+T4=T3*(V4/V3);//Temperature at point 4 in K
+T5=T4*((V4/V5)^(y-1));//Temperature at point 5 in K
+q1=(Cv*(T3-T2))+(Cp*(T4-T3));//Heat added in kJ/kg
+q2=Cv*(T5-T1);//Heat rejected in kJ/kg
+nth=(1-(q2/q1))*100;//Thermal efficiency
+m=(P1*V1*(10^5))/(R*T1);//Mass of air supplied in kg
+W=m*(q1-q2);//Workdone in kJ/cycle
+
+//OUTPUT
+mprintf('Amount of heat added is %3.1f kJ/kg \n Amount of heat rejected is %3.2f kJ/kg \n Workdone per cycle is %3.2f kJ/cycle \n Thermal efficiency is %3.2f percent',q1,q2,W,nth)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.15/Chapter2_Example15.sce b/1775/CH2/EX2.15/Chapter2_Example15.sce new file mode 100755 index 000000000..381d67d2e --- /dev/null +++ b/1775/CH2/EX2.15/Chapter2_Example15.sce @@ -0,0 +1,45 @@ +//Chapter-2, Illustration 15, Page 72
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+P3=70;//Pressure at point 3 in bar
+T1=310;//Temperature at point 1 in K
+rv=10;//Compression ratio
+y=1.4;//Ratio of specific heats
+qin=2805;//Heat added in kJ/kg
+m=1;//Mass of air in kg
+R=287;//Universal gas constant in J/kg
+Cv=0.718;//Speific heat at constant volume in kJ/kg-K
+Cp=1.005;//Speific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+V1=(m*R*T1)/(P1*(10^5));//Volume at point 1 in m^3
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+P2=P1*(rv^y);//Pressure at point 2 in K
+T3=T2*(P3/P2);//Temperature at point 3 in K
+q23=Cv*(T3-T2);//Heat supplied at constant volume in kJ/kg
+q34=qin-q23;//Heat supplied at constant pressure in kJ/kg
+T4=(q34/Cp)+T3;//Temperature at point 4 in K
+V2=V1/rv;//Volume at point 2 in m^3
+V4=V2*(T4/T3);//Volume at point 4 in m^3
+V5=V1;//Volume at point 5 in m^3
+T5=T4*((V4/V5)^(y-1));//Temperature at point 5 in K
+qout=Cv*(T5-T1);//Heat rejected in kJ/kg
+nth=(1-(qout/qin))*100;//Thermal efficiency
+W=qin-qout;//Workdone in kJ/kg
+Vs=V1*(1-(1/rv));//Swept volume in (m^3)/kg
+MEP=(W/Vs)/100;//Mean effective pressure in bar
+
+//OUTPUT
+mprintf('Mean effective pressure is %3.2f bar \n Thermal efficiency is %3.2f percent',MEP,nth)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.16/Chapter2_Example16.sce b/1775/CH2/EX2.16/Chapter2_Example16.sce new file mode 100755 index 000000000..0bded73f9 --- /dev/null +++ b/1775/CH2/EX2.16/Chapter2_Example16.sce @@ -0,0 +1,33 @@ +//Chapter-2, Illustration 16, Page 74
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+T1=298;//Temperature at point 1 in K
+P2=3;//Pressure at point 2 in bar
+T3=923;//Temperature at point 3 in K
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Speific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+rp=P2/P1;//Pressure ratio
+nth=(1-(1/(rp^x)))*100;//Cycle efficiency
+T2=T1*(rp^x);//Temperature at point 2 in K
+q1=Cp*(T3-T2);//Heat supplied in kJ/kg
+Wout=(nth*q1)/100;//Work output in kJ/kg
+q2=q1-Wout;//Heat rejected in kJ/kg
+T4=T3*((1/rp)^x);//Temperature at point 4 in K
+
+//OUTPUT
+mprintf('Cycle efficiency is %3.2f percent \n Heat supplied to air is %3.1f kJ/kg \n Work available at the shaft is %3.2f kJ/kg \n Heat rejected in the cooler is %3.2f kJ/kg \n Turbine exit temperature is %3.2f K',nth,q1,Wout,q2,T4)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.17/Chapter2_Example17.sce b/1775/CH2/EX2.17/Chapter2_Example17.sce new file mode 100755 index 000000000..23da8e04e --- /dev/null +++ b/1775/CH2/EX2.17/Chapter2_Example17.sce @@ -0,0 +1,32 @@ +//Chapter-2, Illustration 17, Page 75
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=283;//Temperature at point 1 in K
+T3=1353;//Temperature at point 3 in K
+y=1.41;//Ratio of specific heats
+Cp=1.007;//Specific heat constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+rpmax=((T3/T1)^(1/x));//Maximum pressure ratio
+rpopt=sqrt(rpmax);//Optimum pressure ratio
+T2=T1*(rpopt^x);//Temperature at point 2 in K
+T4=T2;//Maximum temperature at point 4 in K
+Wmax=Cp*((T3-T4)-(T2-T1));//Maximum net specific work output in kJ/kg
+nth=(Wmax/(Cp*(T3-T2)))*100;//Thermal efficiency
+WR=nth/100;//Work ratio
+nc=((T3-T1)/T3)*100;//Carnot efficiency
+
+//OUTPUT
+mprintf('Optimum pressure ratio is %3.2f \n Maximum net specific work output %3.0f kJ/kg \n Thermal efficiency %3.0f percent \n Work ratio is %3.2f \n Carnot efficiency is %3.0f percent',rpopt,Wmax,nth,WR,nc)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.18/Chapter2_Example18.sce b/1775/CH2/EX2.18/Chapter2_Example18.sce new file mode 100755 index 000000000..679c1d90a --- /dev/null +++ b/1775/CH2/EX2.18/Chapter2_Example18.sce @@ -0,0 +1,26 @@ +//Chapter-2, Illustration 18, Page 76
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tmin=300;//Minimum temperature in K
+Tmax=1073;//Maximum temperature in K
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+Wmax=Cp*((sqrt(Tmax)-sqrt(Tmin))^2);//Maximum work output in kJ/kg
+nB=(1-sqrt(Tmin/Tmax))*100;//Brayton cycle efficiency
+nC=(1-(Tmin/Tmax))*100;//Carnot efficiency
+r=nB/nC;//Ratio of brayton cycle efficiency to carnot efficieny
+
+//OUTPUT
+mprintf('Maximum work per kg of air is %3.2f kJ/kg \n Cycle efficiency is %3.0f percent \n Ratio of brayton cycle efficiency to carnot efficieny is %3.3f',Wmax,nB,r)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.19/Chapter2_Example19.sce b/1775/CH2/EX2.19/Chapter2_Example19.sce new file mode 100755 index 000000000..7fc7b60ba --- /dev/null +++ b/1775/CH2/EX2.19/Chapter2_Example19.sce @@ -0,0 +1,38 @@ +//Chapter-2, Illustration 19, Page 77
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=291;//Temperature at point 1 in K
+P1=100;//Pressure at point 1 in kN/(m^2)
+nC=0.85;//Isentropic efficiency of compressor
+nT=0.88;//Isentropic effficiency of turbine
+rp=8;//Pressure ratio
+T3=1273;//Temperature at point 3 in K
+m=4.5;//Mass flow rate of air in kg/s
+y=1.4;//Ratio of speciifc heats
+Cp=1.006;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2s=T1*(rp^x);//Temperature at point 2s in K
+T2=T1+((T2s-T1)/nC);//Temperature at point 2 in K
+t2=T2-273;//Temperature at point 2 in oC
+T4s=T3*((1/rp)^x);//Temperature at point 4s in K
+T4=T3-((T3-T4s)*nT);//Temperature at point 4 in K
+t4=T4-273;//Temperature at point 4 in oC
+W=m*Cp*((T3-T4)-(T2-T1));//Net power output in kW
+nth=(((T3-T4)-(T2-T1))/(T3-T2))*100;//Thermal efficiency
+WR=W/(m*Cp*(T3-T4));//Work ratio
+
+//OUTPUT
+mprintf('Net power output of the turbine is %3.0f kW \n Thermal efficiency of the plant is %3.0f percent \n Work ratio is %3.3f',W,nth,WR)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.2/Chapter2_Example2.sce b/1775/CH2/EX2.2/Chapter2_Example2.sce new file mode 100755 index 000000000..f85fd4b7d --- /dev/null +++ b/1775/CH2/EX2.2/Chapter2_Example2.sce @@ -0,0 +1,30 @@ +//Chapter-2, Illustration 2, Page 57
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=80;//Bore in mm
+L=85;//Stroke in mm
+Vc=0.06;//Clearance volume in litre
+n=0.22;//Actual thermal efficiency
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+Vs=(3.147/4)*(d^2)*L;//Stroke volume in mm^3
+Vt=Vs+(Vc*(10^6));//Total volume in mm^3
+rv=Vt/(Vc*(10^6));//Compression ratio
+ni=(1-(1/(rv^(y-1))));//Ideal thermal efficiency
+nr=(n/ni)*100;//Relative efficiency
+
+//OUTPUT
+mprintf('Relative efficiency of the engine is %3.1f percent',nr)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.20/Chapter2_Example20.sce b/1775/CH2/EX2.20/Chapter2_Example20.sce new file mode 100755 index 000000000..b774d6e1b --- /dev/null +++ b/1775/CH2/EX2.20/Chapter2_Example20.sce @@ -0,0 +1,45 @@ +//Chapter-2, Illustration 20, Page 79
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=0.1;//Pressure at point 1 in MPa
+T1=303;//Temperature at point 1 in K
+T3=1173;//Temperature at point 3 in K
+rp=6;//Pressure ratio
+nC=0.8;//Compressor efficiency
+nT=nC;//Turbine efficiency
+e=0.75;//Regenerator effectiveness
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2s=T1*(rp^x);//Temperature at point 2s in K
+T4s=T3/(rp^x);//Temperature at point 4s in K
+DTa=(T2s-T1)/nC;//Difference in temperatures at point 2 and 1 in K
+DTb=(T3-T4s)*nT;//Difference in temperatures at point 3 and 4 in K
+wT=Cp*DTb;//Turbine work in kJ/kg
+wC=Cp*DTa;//Compressor work in kJ/kg
+T2=DTa+T1;//Temperature at point 2 in K
+q1=Cp*(T3-T2);//Heat supplied in kJ/kg
+nth1=((wT-wC)/q1)*100;//Cycle efficiency without regenerator
+T4=T3-DTb;//Temperature at point 4 in K
+T5=T2+(e*(T4-T2));//Temperature at point 5 in K
+q2=Cp*(T3-T5);//Heat supplied with regenerator in kJ/kg
+nth2=((wT-wC)/q2)*100;//Cycle efficiency with regenerator
+p=((nth2-nth1)/nth1)*100;//Percentage increase due to regeneration
+
+//OUTPUT
+mprintf('Percentage increase in the cycle efficiency due to regeneration is %3.2f percent',p)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.21/Chapter2_Example21.sce b/1775/CH2/EX2.21/Chapter2_Example21.sce new file mode 100755 index 000000000..44be75112 --- /dev/null +++ b/1775/CH2/EX2.21/Chapter2_Example21.sce @@ -0,0 +1,31 @@ +//Chapter-2, Illustration 21, Page 80
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in atm
+P3=5;//Pressure at point 3 in atm
+T1=288;//Temperature at point 1 in K
+T4=1143;//Temperature at point 4 in K
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+rp=P3/P1;//Pressure ratio
+x=(y-1)/y;//Ratio
+T3=T1*(rp^x);//Temperature at point 3 in K
+T5=T4-(T3-T1);//Temperature at point 5 in K
+T6=T4/(rp^x);//Temperature at point 6 in K
+C6=sqrt(2000*Cp*(T5-T6));//Velocity of air leaving the nozzle in m/s
+
+//OUTPUT
+mprintf('Velocity of air leaving the nozzle is %3.1f m/s',C6)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.22/Chapter2_Example22.sce b/1775/CH2/EX2.22/Chapter2_Example22.sce new file mode 100755 index 000000000..cc6ebb314 --- /dev/null +++ b/1775/CH2/EX2.22/Chapter2_Example22.sce @@ -0,0 +1,45 @@ +//Chapter-2, Illustration 22, Page 81
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C1=280;//Velocity of aircraft in m/s
+P1=48;//Pressure at point 1 kPa
+T1=260;//Temperature at point 1 in K
+rp=13;//Pressure ratio
+T4=1300;//Temperature at point 4 in K
+Cp=1005;//Specific heat at constant pressure in J/kg
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2=T1+((C1^2)/(2*Cp));//Temperature at point 2 in K
+P2=P1*((T2/T1)^(1/x));//Pressure at point 2 in kPa
+P3=rp*P2;//Pressure at point 3 in kPa
+P4=P3;//Pressure at point 4 in kPa
+T3=T2*(rp^x);//Temperature at point 3 in K
+T5=T4-T3+T2;//Temperature at point 5 in K
+P5=P4*((T5/T4)^(1/x));//Pressure at point 5 in kPa
+P6=P1;//Pressure at point 6 in kPa
+T6=T5*((P6/P5)^x);//Temperature at point 6 in K
+C6=sqrt(2*Cp*(T5-T6));//Velocity of air at nozzle exit in m/s
+W=(C6-C1)*C1;//Propulsive power in J/kg
+Q=Cp*(T4-T3);//Total heat transfer rate in J/kg
+nP=(W/Q)*100;//Propulsive efficiency
+
+//OUTPUT
+mprintf('Pressure at the turbine exit is %3.1f kPa \n Velocity of exhaust gases are %3.1f m/s \n Propulsive efficiency is %3.1f percent',P5,C6,nP)
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.3/Chapter2_Example3.sce b/1775/CH2/EX2.3/Chapter2_Example3.sce new file mode 100755 index 000000000..0e0188a15 --- /dev/null +++ b/1775/CH2/EX2.3/Chapter2_Example3.sce @@ -0,0 +1,27 @@ +//Chapter-2, Illustration 3, Page 57
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=0.137;//Bore in m
+L=0.13;//Stroke in m
+Vc=280*(10^-6);//Clearance volume in m^3
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+Vs=(3.147/4)*(d^2)*L;//Stroke volume in m^3
+rv=(Vc/Vs)*100;//Compression ratio
+rvf=(Vs+Vc)/Vc;//final compression ratio
+n=(1-(1/rvf^(y-1)))*100;//Cycle efficiency
+
+//OUTPUT
+mprintf('Clearance volume is %3.1f percent of swept volume \n Otto cycle efficiency is %3.2f percent',rv,n)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.4/Chapter2_Example4.sce b/1775/CH2/EX2.4/Chapter2_Example4.sce new file mode 100755 index 000000000..3002c430c --- /dev/null +++ b/1775/CH2/EX2.4/Chapter2_Example4.sce @@ -0,0 +1,37 @@ +//Chapter-2, Illustration 4, Page 58
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=9.5;//Compression ratio
+P1=100;//Air pressure in kPa
+T1=290;//Air temperature in K
+V1=600*(10^-6);//Volume of air in m^3
+T4=800;//Final temperature in K
+R=287;//Universal gas constan in J/kg.K
+Cv=0.718;//Specific heat at constant volume in kJ/kg.K
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+T3=T4*(rv^(y-1));//Temperature at the end of constant volume heat addition in K
+P2=P1*(rv^y);//Pressure at point 2 in kPa
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+P3=P2*(T3/T2);//Pressure at point 3 in kPa
+m=(P1*1000*V1)/(R*T1);//Specific mass in kg/s
+Q=m*Cv*(T3-T2);//Heat transferred in kJ
+n=(1-(1/rv^(y-1)))*100;//Thermal efficiency
+Wnet=(n*Q)/100;//Net workdone in kJ
+MEP=Wnet/(V1*(1-(1/rv)));//Mean effective pressure in kPa
+
+//OUTPUT
+mprintf('Maximum pressure of the cycle is %3.2f kPa \n Maximum temperature of the cycle is %3.1f K \n Amount of heat transferred is %3.2f kJ \n Thermal efficiency is %3.1f percent \n Mean effective pressure is %3.1f kPa',P3,T3,Q,n,MEP)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.5/Chapter2_Example5.sce b/1775/CH2/EX2.5/Chapter2_Example5.sce new file mode 100755 index 000000000..c64e5a791 --- /dev/null +++ b/1775/CH2/EX2.5/Chapter2_Example5.sce @@ -0,0 +1,35 @@ +//Chapter-2, Illustration 5, Page 60
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=8;//Compression ratio
+P1=95;//Pressure at point 1 in kPa
+T1=300;//Temperature at point 1 in K
+q23=750;//Heat transferred during constant volume heat addition process in kJ/kg
+y=1.4;//Ratio of specific heats
+Cv=0.718;//Specific heat at constant volume in kJ/kg-K
+R=287;//Universal gas constant in J/kg-K
+
+//CALCULATIONS
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+P2=P1*(rv^y);//Pressure at point 2 in kPa
+T3=(q23/Cv)+T2;//Temperature at point 3 in K
+P3=P2*(T3/T2);//Pressure at point 3 in kPa
+nth=(1-(1/(rv^(y-1))))*100;//Thermal efficiency
+Wnet=(nth*q23)/100;//Net work output in kJ/kg
+v1=(R*T1)/(P1*1000);//Speific volume at point 1 in (m^3)/kg
+MEP=Wnet/(v1*(1-(1/rv)));//Mean effective pressure in kPa
+
+//OUTPUT
+mprintf('Pressure at the end of heat addition process is %3.1f kPa \n Temperature at the end of heat addition process is %3.1f K \n Net work output is %3.2f kJ/kg \n Thermal efficiency is %3.2f percent \n Mean effective pressure is %3.0f kPa',P3,T3,Wnet,nth,MEP)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.6/Chapter2_Example6.sce b/1775/CH2/EX2.6/Chapter2_Example6.sce new file mode 100755 index 000000000..90217347d --- /dev/null +++ b/1775/CH2/EX2.6/Chapter2_Example6.sce @@ -0,0 +1,25 @@ +//Chapter-2, Illustration 6, Page 61
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=14;//Compression ratio
+c=0.06;//Cut-off percentage
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+rc=1.78;//Cut-off ratio
+nth=(1-(((rc^y)-1)/((rv^(y-1))*y*(rc-1))))*100;//Thermal efficiency
+
+//OUTPUT
+mprintf('Air standard efficiency is %3.1f percent',nth)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.7/Chapter2_Example7.sce b/1775/CH2/EX2.7/Chapter2_Example7.sce new file mode 100755 index 000000000..628ef516e --- /dev/null +++ b/1775/CH2/EX2.7/Chapter2_Example7.sce @@ -0,0 +1,32 @@ +//Chapter-2, Illustration 7, Page 62
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+rv=16;//Compression ratio
+P1=0.1;//Pressure at point 1 in MPa
+T1=288;//Temperature at point 1 in K
+T3=1753;//Temperature at point 3 in K
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+R=0.287;//Universal gas constant in kJ/kg-K
+
+//CALCULATIONS
+T2=T1*(rv^(y-1));//Temperature at point 2 in K
+rc=T3/T2;//Cut-off ratio
+q1=Cp*(T3-T2);//Heat supplied in kJ/kg
+nth=(1-(((rc^y)-1)/((rv^(y-1))*y*(rc-1))))*100;//Cycle efficiency
+wnet=(q1*nth)/100;//Net work done in kJ/kg
+v1=(R*T1)/(P1*1000);//Speific volume at point 1 in (m^3)/kg
+v2=v1/rv;//Speific volume at point 2 in (m^3)/kg
+MEP=wnet/(v1-v2);//Mean effective pressure in kPa
+
+//OUTPUT
+mprintf('Cut-off ratio is %3.2f \n Heat supplied is %3.1f kJ/kg \n Cycle efficiency is %3.1f percent \n Mean effective pressure is %3.2f kPa',rc,q1,nth,MEP)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.8/Chapter2_Example8.sce b/1775/CH2/EX2.8/Chapter2_Example8.sce new file mode 100755 index 000000000..f6316a233 --- /dev/null +++ b/1775/CH2/EX2.8/Chapter2_Example8.sce @@ -0,0 +1,34 @@ +//Chapter-2, Illustration 8, Page 64
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=0.15;//Bore in m
+L=0.25;//Stroke in m
+Vc=400*(10^-6);//Clearance volume in m^3
+V2=Vc;//Clearance volume in m^3
+c1=0.05;//Cut-off percentage 1
+c2=0.08;//Cut-off percentage 2
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+Vs=(3.147/4)*(d^2)*L;//Stroke volume in m^3
+V31=V2+(c1*Vs);//Volume at the point of cut-off in m^3
+rc1=V31/V2;//Cut-off ratio 1
+rv=(Vc+Vs)/Vc;//Compression ratio
+nth1=(1-(((rc1^y)-1)/((rv^(y-1))*y*(rc1-1))))*100;//Air standard efficiency 1
+V32=V2+(c2*Vs);//Volume at the point of cut-off in m^3
+rc2=V32/V2;//Cut-off ratio 2
+nth2=(1-(((rc2^y)-1)/((rv^(y-1))*y*(rc2-1))))*100;//Air standard efficiency 2
+pl=nth1-nth2;//Percentage loss in efficiency
+
+//OUTPUT
+mprintf('Air standard efficiency at 5 percent cut-off is %3.2f percent \n Air standard efficiency at 8 percent cut-off is %3.2f percent \n Percentage loss in efficiency is %3.2f percent',nth1,nth2,pl)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH2/EX2.9/Chapter2_Example9.sce b/1775/CH2/EX2.9/Chapter2_Example9.sce new file mode 100755 index 000000000..0e5ece4f1 --- /dev/null +++ b/1775/CH2/EX2.9/Chapter2_Example9.sce @@ -0,0 +1,30 @@ +//Chapter-2, Illustration 9, Page 65
+//Title: Gas Power Cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+e=7.5;//Expansion ratio
+c=15;//Compression ratio
+P1=98;//Pressure at point 1 in kN/(m^2)
+P4=258;//Pressure at point 4 in kN/(m^2)
+T1=317;//Temperature at point 1 in K
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+T4=T1*(P4/P1);//Temperature at point 4 in K
+T3=T4*(e^(y-1));//Temperature at point 3 in K
+t3=T3-273;//Temperature at point 3 in oC
+T2=T1*(c^(y-1));//Temperature at point 2 in K
+n=(1-((T4-T1)/(y*(T3-T2))))*100;//Thermal efficiency
+
+//OUTPUT
+mprintf('Maximum temperature attained during the cycle is %3.1f oC \n Thermal efficiency of the cycle is %3.1f percent',t3,n)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.1/Chapter3_Example1.sce b/1775/CH3/EX3.1/Chapter3_Example1.sce new file mode 100755 index 000000000..e8b7372af --- /dev/null +++ b/1775/CH3/EX3.1/Chapter3_Example1.sce @@ -0,0 +1,32 @@ +//Chapter-3, Illustration 1, Page 139
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+d=200;//diameter of cylinder in mm
+L=300;//stroke of cylinder in mm
+Vc=1.73;//Clearance volume in litres
+imep=650;//indicated mean effective pressure in kN/(m^2)
+g=6.2;//gas consumption in (m^3)/h
+CV=38.5;//Calorific value in MJ/(m^3)
+y=1.4;//Ratio of specific heats
+N=150;//No. of firing cycles per minute
+
+//CALCULATIONS
+Vs=((3.1415/4)*(d^2)*L)*(10^-6);//Stroke volume in litres
+Vt=Vs+Vc;//Total volume in litres
+rv=(Vt/Vc);//Compression ratio
+n=(1-(1/rv^(y-1)))*100;//Air standard efficiency
+IP=imep*(Vs*10^-3)*(N/60);//Indicated power in kW
+F=(g*CV*1000)/3600;//Fuel energy input in kW
+nT=(IP/F)*100;//Indicated thermal efficiency
+
+//OUTPUT
+mprintf('Air Standard Efficiency is %3.1f percent \n Indicated Power is %3.1f kW \n Indicated thermal efficiency is %3.0f percent',n,IP,nT)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.2/Chapter3_Example2.sce b/1775/CH3/EX3.2/Chapter3_Example2.sce new file mode 100755 index 000000000..5b02d32e1 --- /dev/null +++ b/1775/CH3/EX3.2/Chapter3_Example2.sce @@ -0,0 +1,27 @@ +//Chapter-3, Illustration 2, Page 140
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Vs=0.0008;//Swept volume in m^3
+Vc=0.00015;//Clearance volume in m^3
+CV=38;//Calorific value in MJ/(m^3)
+v=0.45;//volume in m^3
+IP=81.5;//Indicated power in kW
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+rv=(Vs+Vc)/Vc;//Compression ratio
+n=(1-(1/rv^(y-1)));//Air standard efficiency
+Ps=(v*CV*1000)/60;//Power supplied in kW
+nact=IP/Ps;//Actual efficiency
+nr=(nact/n)*100;//Relative efficiency
+
+//OUTPUT
+mprintf('Relative Efficiency is %3.2f percent',nr)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.3/Chapter3_Example3.sce b/1775/CH3/EX3.3/Chapter3_Example3.sce new file mode 100755 index 000000000..d38421706 --- /dev/null +++ b/1775/CH3/EX3.3/Chapter3_Example3.sce @@ -0,0 +1,32 @@ +//Chapter-3, Illustration 3, Page 141
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+n=6;//No. of cylinders
+d=0.61;//Diameter in m
+L=1.25;//Stroke in m
+N=2;//No.of revolutions per second
+m=340;//mass of fuel oil in kg
+CV=44200;//Calorific value in kJ/kg
+T=108;//Torque in kN-m
+imep=775;//Indicated mean efective pressure in kN/(m^2)
+
+//CALCULATIONS
+IP=(imep*L*3.1415*(d^2)*N)/(8);//Indicated power in kW
+TotalIP=(n*IP);//Total indicated power in kW
+BP=(2*3.1415*N*T);//Brake power in kW
+PI=(m*CV)/3600;//Power input in kW
+nB=(BP/PI)*100;//Brake thermal efficiency
+bmep=(BP*8)/(n*L*3.1415*(d^2)*2);//Brake mean effective pressure in kN/(m^2)
+nM=(BP/TotalIP)*100;//Mechanical efficiency
+bsfc=m/BP;//Brake specific fuel consumption in kg/kWh
+
+//OUTPUT
+mprintf('Total Indicated Power is %3.1f kW \n Brake Power is %3.1f kW \n Brake thermal efficiency is %3.1f percent \n Brake mean effective pressure is %3.1f kN/(m^2) \n Mechanical efficiency is %3.1f percent \n Brake specific fuel consumption is %3.3f kg/kW.h',TotalIP,BP,nB,bmep,nM,bsfc)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.4/Chapter3_Example4.sce b/1775/CH3/EX3.4/Chapter3_Example4.sce new file mode 100755 index 000000000..0e8ea93ce --- /dev/null +++ b/1775/CH3/EX3.4/Chapter3_Example4.sce @@ -0,0 +1,42 @@ +//Chapter-3, Illustration 4, Page 142
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Hm=21;//Mean height of indicator diagram in mm
+isn=27;//indicator spring number in kN/(m^2)/mm
+Vs=14;//Swept volume in litres
+N=6.6;//Speed of engine in rev/s
+Pe=77;//Effective brake load in kg
+Re=0.7;//Effective vrake radius in m
+mf=0.002;//fuel consumed in kg/s
+CV=44000;//Calorific value of fuel in kJ/kg
+mc=0.15;//cooling water circulation in kg/s
+Ti=311;//cooling water inlet temperature in K
+To=344;//cooling water outlet temperature in K
+C=4.18;//specific heat capacity of water in kJ/kg-K
+Ee=33.6;//Energy to exhaust gases in kJ/s
+g=9.81;//Acceleration due to geravity in m/(s^2)
+
+//CALCULATIONS
+imep=isn*Hm;//Indicated mean efective pressure in kN/(m^2)
+IP=(imep*Vs*N)/(2000);//Indicated Power in kW
+BP=(2*3.1415*N*g*Pe*Re)/1000;//Brake Power in kW
+nM=(BP/IP)*100;//Mechanical efficiency
+Ef=mf*CV;//Eneergy from fuel in kJ/s
+Ec=mc*C*(To-Ti);//Energy to cooling water in kJ/s
+Es=Ef-(BP+Ec+Ee);//Energy to surroundings in kJ/s
+p=(BP*100)/Ef;//Energy to BP in %
+q=(Ec*100)/Ef;//Energy to coolant in %
+r=(Ee*100)/Ef;//Energy to exhaust in %
+w=(Es*100)/Ef;//Energy to surroundings in %
+
+//OUTPUT
+mprintf('Indicated Power is %3.1f kW \n Brake Power is %3.0f kW \n Mechanical Efficiency is %3.0f percent \n \nENERGY BALANCE kJ/s Percentage \nEnergy from fuel %3.0f 100\nEnergy to BP %3.0f %3.0f\nEnergy to coolant %3.01f %3.1f\nEnergy to exhaust %3.1f %3.1f\nEnergy to surroundings, etc %3.1f %3.1f',IP,BP,nM,Ef,BP,p,Ec,q,Ee,r,Es,w)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.5/Chapter3_Example5.sce b/1775/CH3/EX3.5/Chapter3_Example5.sce new file mode 100755 index 000000000..a9ac95449 --- /dev/null +++ b/1775/CH3/EX3.5/Chapter3_Example5.sce @@ -0,0 +1,42 @@ +//Chapter-3, Illustration 5, Page 143
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+t=30;//duration of trial in minutes
+N=1750;//speed in rpm
+T=330;//brake torque in Nm
+m=9.35;//mass of fuel in kg
+CV=42300;//Calorific value in kJ/kg
+mj=483;//jacket cooling water circulation in kg
+Ti=290;//inlet temperature in K
+T0=350;//outlet temperature in K
+ma=182;//air consumption in kg
+Te=759;//exhaust temperature in K
+Ta=256;//atmospheric temperature in K
+nM=0.83;//Mechanical efficiency
+ms=1.25;//mean specific heat capacity of exhaust gas in kJ/kg-K
+Cw=4.18;//specific heat capacity of water in kJ/kg-K
+
+//CALCULATIONS
+BP=(2*3.1415*T*N)/(60*1000);//Brake power in kW
+sfc=(m*2)/BP;//specific fuel consumption in kg/kWh
+IP=BP/nM;//Indicated power in kW
+nIT=((IP*3600)/(m*CV*2))*100;//Indicated thermal efficiency
+Ef=(m*CV)/t;//Eneergy from fuel in kJ/min
+EBP=BP*60;//Energy to BP in kJ/min
+Ec=(mj*Cw*(T0-Ti))/t;//Energy to cooling water in kJ/min
+Ee=((ma+m)*ms*(Te-Ti))/30;//Energy to exhaust in kJ/min
+Es=Ef-(EBP+Ec+Ee);//Energy to surroundings in kJ/min
+
+//OUTPUT
+mprintf('Brake power is %3.1f kW \n Specific fuel consumption is %3.3f kg/kWh \n Indicated thermal efficiency is %3.1f percent \n Energy from fuel is %3.0f kJ/min \n Energy to BP is %3.0f kJ/min \n Energy to cooling water is %3.0f kJ/min \n Energy to exhaust is %3.0f kJ/min \n Energy to surroundings is %3.0f kJ/min',BP,sfc,nIT,Ef,EBP,Ec,Ee,Es)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.6/Chapter3_Example6.sce b/1775/CH3/EX3.6/Chapter3_Example6.sce new file mode 100755 index 000000000..df491caca --- /dev/null +++ b/1775/CH3/EX3.6/Chapter3_Example6.sce @@ -0,0 +1,32 @@ +//Chapter-3, Illustration 6, Page 144
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+BP0=12;//Brake Power output in kW
+BP1=40.5;//Brake Power in trial 1 in kW
+BP2=40.2;//Brake Power in trial 2 in kW
+BP3=40.1;//Brake Power in trial 3 in kW
+BP4=40.6;//Brake Power in trial 4 in kW
+BP5=40.7;//Brake Power in trial 5 in kW
+BP6=40.0;//Brake Power in trial 6 in kW
+
+//CALCULATIONS
+BPALL=BP0+BP6;//Total Brake Power in kW
+IP1=BPALL-BP1;//Indicated Power in trial 1 in kW
+IP2=BPALL-BP2;//Indicated Power in trial 2 in kW
+IP3=BPALL-BP3;//Indicated Power in trial 3 in kW
+IP4=BPALL-BP4;//Indicated Power in trial 4 in kW
+IP5=BPALL-BP5;//Indicated Power in trial 5 in kW
+IP6=BPALL-BP6;//Indicated Power in trial 6 in kW
+IPALL=IP1+IP2+IP3+IP4+IP5+IP6;//Total Indicated Power in kW
+nM=(BPALL/IPALL)*100;//Mechanical efficiency
+
+//OUTPUT
+mprintf('Indicated Power of the engine is %3.1f kW \n Mechanical efficiency of the engine is %3.1f percent',IPALL,nM)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH3/EX3.7/Chapter3_Example7.sce b/1775/CH3/EX3.7/Chapter3_Example7.sce new file mode 100755 index 000000000..78caaf11e --- /dev/null +++ b/1775/CH3/EX3.7/Chapter3_Example7.sce @@ -0,0 +1,35 @@ +//Chapter-3, Illustration 7, Page 145
+//Title: Internal Combustion Engines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+n=2;//No. of cylinders
+N=4000;//speed of engine in rpm
+nV=0.77;//Volumetric efficiency
+nM=0.75;//Mechanical efficiency
+m=10;//fuel consumed in lit/h
+g=0.73;//spcific gravity of fuel
+Raf=18;//air-fuel ratio
+Np=600;//piston speed in m/min
+imep=5;//Indicated mean efective pressure in bar
+R=281;//Universal gas constant in J/kg-K
+T=288;//Standard temperature in K
+P=1.013;//Standard pressure in bar
+
+//CALCULATIONS
+L=Np/(2*N);//Piston stroke in m
+mf=m*g;//mass of fuel in kg/h
+ma=mf*Raf;//mass of air required in kg/h
+Va=(ma*R*T)/(P*60*(10^5));//volume of air required in (m^3)/min
+D=sqrt((2*Va)/(nV*L*N*3.1415));//Diameter in m
+IP=(2*imep*100*L*3.1415*(D^2)*N)/(4*60);//Indicated Power in kW
+BP=nV*IP;//Brake Power in kW
+
+//OUTPUT
+mprintf('Piston Stroke is %3.3f m \n Bore diameter is %3.4f m \n Brake power is %3.1f kW',L,D,BP)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.1/Chapter4_Example1.sce b/1775/CH4/EX4.1/Chapter4_Example1.sce new file mode 100755 index 000000000..9b035a170 --- /dev/null +++ b/1775/CH4/EX4.1/Chapter4_Example1.sce @@ -0,0 +1,35 @@ +//Chapter-4, Illustration 1, Page 161
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=3.5;//Pressure at entry in MN/(m^2)
+T1=773;//Temperature at entry in K
+P2=0.7;//Pressure at exit in MN/(m^2)
+ma=1.3;//mass flow rate of air in kg/s
+y=1.4;//Ratio of specific heats
+R=0.287;//Universal gas constant in KJ/Kg-K
+
+//CALCULATIONS
+c=y/(y-1);//Ratio
+Pt=((2/(y+1))^c)*P1;//Throat pressure in MN/(m^2)
+v1=(R*T1)/(P1*1000);//Specific volume at entry in (m^3)/kg
+Ct=((2*c*P1*v1*(1-((Pt/P1)^(1/c))))^0.5)*1000;//Velocity at throat in m/s
+vt=v1*((P1/Pt)^(1/y));//Specific volume at throat in (m^3)/kg
+At=((ma*vt)/Ct)*(10^6);//Area of throat in (mm^2)
+C2=((2*c*P1*v1*(1-((P2/P1)^(1/c))))^0.5)*1000;//Velocity at exit in m/s
+v2=v1*((P1/P2)^(1/y));//Specific volume at exit in (m^3)/kg
+A2=((ma*v2)/C2)*(10^6);//Area of exit in (mm^2)
+M=C2/Ct;//Mach number at exit
+
+//OUTPUT
+mprintf('Throat area is %3.0f (mm^2) \n Exit area is %3.0f (mm^2) \n Mach number at exit is %3.2f',At,A2,M)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.10/Chapter4_Example10.sce b/1775/CH4/EX4.10/Chapter4_Example10.sce new file mode 100755 index 000000000..5ad442899 --- /dev/null +++ b/1775/CH4/EX4.10/Chapter4_Example10.sce @@ -0,0 +1,30 @@ +//Chapter-4, Illustration 10, Page 173
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=10.5;//Pressure at point 1 in bar
+x1=0.95;//Dryness fraction
+n=1.135;//Adiabatic gas constant
+P2=0.85;//Pressure at point 2 in bar
+vg=0.185;//Specific volume in m^3/kg
+
+
+//CALCULATIONS
+c=n/(n-1);//Ratio
+Pt=((2/(n+1))^c)*P1;//Throat pressure in MN/(m^2)
+v1=x1*vg;//Specific volume at point 1 in m^3/kg
+Ct=sqrt((2*n*P1*v1*(10^5)/(n+1)));//Velocity at throat in m/s
+vt=((P1/Pt)*(v1^n))^(1/1.135);//Specific volume at throat in m^3/kg
+m=Ct/vt;//Mass flow rate per unit throat area in kg/(m^2)
+
+//OUTPUT
+mprintf('Throat velocity is %3.2f m/s \n Mass flow rate of steam is %3.2f kg/(m^2)',Ct,m)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.11/Chapter4_Example11.sce b/1775/CH4/EX4.11/Chapter4_Example11.sce new file mode 100755 index 000000000..7ba02eb0c --- /dev/null +++ b/1775/CH4/EX4.11/Chapter4_Example11.sce @@ -0,0 +1,29 @@ +//Chapter-4, Illustration 11, Page 174
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=10;//Pressure at point 1 in bar
+T1=452.9;//Temperature at point 1 in K
+P2=4;//Pressure at point 2 in bar
+n=1.3;//Adiabatic gas constant
+Ps=0.803;//Saturation pressure at T2 in bar
+Ts=143.6;//Saturation temperature at P2 in oC
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+T2=((P2/P1)^x)*T1;//Temperature at point 2 in K
+Ds=P2/Ps;//Degree of supersaturation
+Du=Ts-(T2-273);//Degree of undercooling
+
+//OUTPUT
+mprintf('Degree of supersaturation is %3.2f \n Degree of undercooling %3.0f oC',Ds,Du)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.12/Chapter4_Example12.sce b/1775/CH4/EX4.12/Chapter4_Example12.sce new file mode 100755 index 000000000..f08eaadaf --- /dev/null +++ b/1775/CH4/EX4.12/Chapter4_Example12.sce @@ -0,0 +1,35 @@ +//Chapter-4, Illustration 12, Page 174
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=9;//Pressure at point 1 in bar
+P2=1;//Pressure at point 2 in bar
+Dt=0.0025;//Throat diameter in m
+nN=0.9;//Nozzle efficiency
+n=1.135;//Adiabatic gas consstant
+h1=2770;//Enthalpy at point 1 in kJ/kg
+ht=2670;//Throat enthlapy in kJ/kg
+h3=2400;//Enthlapy at point 2 in kJ/kg
+x2=0.96;//Dryness fraction 2
+vg2=0.361;//Specific volume in m^3/kg
+
+//CALCULATIONS
+x=n/(n-1);//Ratio
+Pt=((2/(n+1))^x)*P1;//Throat pressure in bar
+Ct=sqrt(2000*(h1-ht)*nN);//Throat velocity in m/s
+At=(3.147*2*(Dt^2))/4;//Throat area in m^2
+vt=x2*vg2;//Specific volume at throat in m^3/kg
+m=(At*Ct)/vt;//Mass flow rate of steam in kg/s
+hact=nN*(h1-h3);//Actual enthalpy drop in kJ/kg
+C2=sqrt(2000*hact);//Exit velocity of steam in m/s
+
+//OUTPUT
+mprintf('Quantity of steam used per second is %3.3f kg/s \n Exit velocity of steam is %3.2f m/s',m,C2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.13/Chapter4_Example13.sce b/1775/CH4/EX4.13/Chapter4_Example13.sce new file mode 100755 index 000000000..0266e4094 --- /dev/null +++ b/1775/CH4/EX4.13/Chapter4_Example13.sce @@ -0,0 +1,30 @@ +//Chapter-4, Illustration 13, Page 202
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C1=1000;//Steam velocity in m/s
+a1=20;//Nozzle angle in degrees
+U=400;//Mean blade speed in m/s
+m=0.75;//Mass flow rate of steam in kg/s
+b1=33;//Blade angle at inlet from the velocity triangle in degrees
+b2=b1;//Blade angle at exit from the velocity triangle in degrees
+Cx=1120;//Change in whirl velocity from the velocity triangle in m/s
+Ca=0;//Change in axial velocity from the velocity triangle in m/s
+
+//CALCULATIONS
+Fx=m*Cx;//Tangential force on blades in N
+Fy=m*Ca;//Axial thrust in N
+W=(m*Cx*U)/1000;//Diagram power in kW
+ndia=((2*U*Cx)/(C1^2))*100;//Diagram efficiency
+
+//OUTPUT
+mprintf('Blade angles are %3.0f degrees,%3.0f degrees \n Tangential force on blades is %3.0f N \n Axial thrust is %3.0f \n Diagram power is %3.0f kW \n Diagram efficiency %3.1f percent',b1,b2,Fx,Fy,W,ndia)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.14/Chapter4_Example14.sce b/1775/CH4/EX4.14/Chapter4_Example14.sce new file mode 100755 index 000000000..083a1ff51 --- /dev/null +++ b/1775/CH4/EX4.14/Chapter4_Example14.sce @@ -0,0 +1,34 @@ +//Chapter-4, Illustration 14, Page 203
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=2.5;//Mean diameter of blade ring in m
+N=3000;//Speed in rpm
+a1=20;//Nozzle angle in degrees
+r=0.4;//Ratio blade velocity to steam velocity
+Wr=0.8;//Blade friction factor
+m=10;//Steam flow in kg/s
+x=3;//Sum in blade angles in degrees
+b1=32.5;//Blade angle at inlet from the velocity triangle in degrees
+W1=626.7;//Relative velocity at inlet from the velocity triangle in m/s
+Cx=967;//Change in whirl velocity from the velocity triangle in m/s
+
+//CALCULATIONS
+U=(3.147*D*N)/60;//Blade velocity in m/s
+C1=U/r;//Steam velocity in m/s
+b2=b1-x;//Blade angle at exit in degrees
+W2=Wr*W1;//Relative velocity at outlet from the velocity triangle in m/s
+W=(m*Cx*U)/1000;//Power developed in kW
+ndia=((2*U*Cx)/(C1^2))*100;//Blade efficiency
+sc=(m*3600)/W;//Steam consumption in kg/kWh
+
+//OUTPUT
+mprintf('Power developed is %3.0f kW \n Blade efficiency is %3.1f percent \n Steam consumed is %3.2f kg/kWh',W,ndia,sc)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.15/Chapter4_Example15.sce b/1775/CH4/EX4.15/Chapter4_Example15.sce new file mode 100755 index 000000000..595d4258a --- /dev/null +++ b/1775/CH4/EX4.15/Chapter4_Example15.sce @@ -0,0 +1,31 @@ +//Chapter-4, Illustration 15, Page 204
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+m=3;//Mass flow rate of steam in kg/s
+C1=425;//Steam velocity in m/s
+r=0.4;//Ratio of blade speed to jet speed
+W=170;//Stage output in kW
+IL=15;//Internal losses in kW
+a1=16;//Nozzle angle in degrees
+b2=17;//Blade angle at exit in degrees
+W1=265;//Relative velocity at inlet from the velocity triangle in m/s
+W2=130;//Relative velocity at outlet from the velocity triangle in m/s
+
+//CALCULATIONS
+U=C1*r;//Blade speed in m/s
+P=(W+IL)*1000;//Total power developed in W
+Cx=P/(m*W);//Change in whirl velocity in m/s
+ndia=((2*U*Cx)/(C1^2))*100;//Blading efficiency
+Wr=W2/W1;//Blade velocity co-efficient
+
+//OUTPUT
+mprintf('Blading efficiency is %3.1f percent \n Blade velocity co-efficient is %3.2f',ndia,Wr)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.16/Chapter4_Example16.sce b/1775/CH4/EX4.16/Chapter4_Example16.sce new file mode 100755 index 000000000..c855331b3 --- /dev/null +++ b/1775/CH4/EX4.16/Chapter4_Example16.sce @@ -0,0 +1,29 @@ +//Chapter-4, Illustration 16, Page 205
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C1=375;//Steam velocity in m/s
+a1=20;//Nozzle angle
+U=165;//Blade speed in m/s
+m=1;//Mass flow rate of steam in kg/s
+Wr=0.85;//Blade friction factor
+Ca1=130;//Axial velocity at inlet from the velocity triangle in m/s
+Ca2=Ca1;//Axial velocity at outlet in m/s
+W1=230;//Relative velocity at inlet from the velocity triangle in m/s
+Cx=320;//Change in whirl velocity from the velocity triangle in m/s
+
+//CALCULATIONS
+b2=41;//Blade angle at exit from the velocity triangle in degrees
+b1=34;//Blade angle at exit from the velocity triangle in degrees
+W=(m*Cx*U)/1000;//Power developed by turbine in kW
+
+//OUTPUT
+mprintf('Blade angles assumed are %3.0f degrees,%3.0f degrees \n Power developed by turbine is %3.1f kW',b1,b2,W)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.17/Chapter4_Example17.sce b/1775/CH4/EX4.17/Chapter4_Example17.sce new file mode 100755 index 000000000..f095bd555 --- /dev/null +++ b/1775/CH4/EX4.17/Chapter4_Example17.sce @@ -0,0 +1,29 @@ +//Chapter-4, Illustration 17, Page 206
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+m=2;//Mass flow rate of steam in kg/s
+W=130;//Turbine power in kW
+U=175;//Blade velocity in m/s
+C1=400;//Steam velocity in m/s
+Wr=0.9;//Blade friction factor
+W1=240;//Realtive velocity at inlet from the velocity triangle in m/s
+
+//CALCULATIONS
+Cx1=(W*1000)/(m*U);//Whirl velocity at inlet in m/s
+W2=Wr*W1;//Realtive velocity at outlet from the velocity triangle in m/s
+a1=19;//Nozzle angle from the velocity triangle in degrees
+b1=33;//Blade angle at inlet from the velocity triangle in degrees
+b2=36;//Blade angle at outlet from the velocity triangle in degrees
+
+//OUTPUT
+mprintf('Nozzle angle is %3.0f degrees \n Blade angles are %3.0f degrees,%3.0f degrees',a1,b1,b2)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.18/Chapter4_Example18.sce b/1775/CH4/EX4.18/Chapter4_Example18.sce new file mode 100755 index 000000000..8414b2e2f --- /dev/null +++ b/1775/CH4/EX4.18/Chapter4_Example18.sce @@ -0,0 +1,38 @@ +//Chapter-4, Illustration 18, Page 207
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+U=150;//Blade speed in m/s
+m=3;//Mass flow rate of steam in kg/s
+P=10.5;//Pressure in bar
+r=0.21;//Ratio blade velocity to steam velocity
+a1=16;//Nozzle angle in first stage in degrees
+b2=20;//Blade angle at exit in first stage in degrees
+a3=24;//Nozzle angle in second stage in degrees
+b4=32;//Blade angle at exit in second stage in degrees
+Wr=0.79;//Blade friction factor for first stage
+Wr2=0.88;//Blade friction factor for second stage
+Cr=0.83;//Blade velocity coefficient
+W1=570;//Relative velocity at inlet from the velocity triangle for first stage in m/s
+C2=375;//Velocity in m/s
+W3=185;//Relative velocity at inlet from the velocity triangle for second stage in m/s
+
+//CALCULATIONS
+C1=U/r;//Steam speed at exit in m/s
+W2=Wr*W1;//Relative velocity at outlet for first stage in m/s
+C3=Cr*C2;//Steam velocity at inlet for second stage in m/s
+W4=Wr2*W3;//Relative velocity at exit for second stage in m/s
+DW1=W1+W2;//Change in relative velocity for first stage in m/s
+DW2=275;//Change in relative velocity from the velocity triangle for second stage in m/s
+ndia=((2*U*(DW1+DW2))/(C1^2))*100;//Diagram efficiency
+
+//OUTPUT
+mprintf('Diagram efficiency is %3.1f percent',ndia)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.19/Chapter4_Example19.sce b/1775/CH4/EX4.19/Chapter4_Example19.sce new file mode 100755 index 000000000..66395ea87 --- /dev/null +++ b/1775/CH4/EX4.19/Chapter4_Example19.sce @@ -0,0 +1,46 @@ +//Chapter-4, Illustration 19, Page 208
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+b1=30;//Blade angle at inlet in first stage in degrees
+b2=30;//Blade angle at exit in first stage in degrees
+b3=30;//Blade angle at inlet in second stage in degrees
+b4=30;//Blade angle at exit in second stage in degrees
+t1=240;//Temperature at entry in oC
+P1=11.5;//Pressure at entry in bar
+P2=5;//Pressure in wheel chamber in bar
+vl=10;//Loss in velocity in percent
+h=155;//Enthalpy at P2 in kJ/kg
+W4=17.3;//Relative velocity at exit from the velocity triangle for second stage in m/s
+a4=90;//Nozzle angle in second stage in degrees
+C3=33;//Steam velocity at inlet from the velocity triangle for second stage in m/s
+W2=49;//Relative velocity at outlet from the velocity triangle for first stage in m/s
+x=15;//Length of AB assumed for drawing velocity triangle in mm
+y=67;//Length of BC from the velocity triangle in mm
+
+//CALCULATIONS
+C1=sqrt(2000*h);//Velocity of steam in m/s
+W3=W4/0.9;//Relative velocity at inlet for second stage in m/s
+C2=C3/0.9;//Velocity in m/s
+W1=W2/0.9;//Relative velocity at inlet for first stage in m/s
+C1n=C1/y;//Velocity of steam in m/s
+U=x*C1n;//Blade speed in m/s
+a3=17;//Nozzle angle in second stage from the velocity triangle in degrees
+a2=43;//Nozzle angle from the velocity triangle in degrees
+DW1=731.5;//Change in relative velocity from the velocity triangle for first stage in m/s
+DW2=257.5;//Change in relative velocity from the velocity triangle for second stage in m/s
+ndia=((2*U*(DW1+DW2))/(C1^2))*100;//Diagram efficiency
+
+//OUTPUT
+mprintf('Blade speed is %3.1f m/s \n Blade tip angles of the fixed blade are %3.0f degrees and %3.0f degrees \n Diagram efficiency is %3.1f percent',U,a3,a2,ndia)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.2/Chapter4_Example2.sce b/1775/CH4/EX4.2/Chapter4_Example2.sce new file mode 100755 index 000000000..5ab73b906 --- /dev/null +++ b/1775/CH4/EX4.2/Chapter4_Example2.sce @@ -0,0 +1,32 @@ +//Chapter-4, Illustration 2, Page 163
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=273;//Temperature at section 1 in K
+P1=140;//Pressure at section 1 in KN/(m^2)
+v1=900;//Velocity at section 1 in m/s
+v2=300;//Velocity at section 2 in m/s
+Cp=1.006;//Specific heat at constant pressure in kJ/kg-K
+Cv=0.717;//Specific heat at constant volume in kJ/kg-K
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+c=y/(y-1);//Ratio
+R=Cp-Cv;//Universal gas constant in KJ/Kg-K
+T2=T1-(((v2)^2-(v1)^2)/(2000*c*R));//Temperature at section 2 in K
+DT=T2-T1;//Increase in temperature in K
+P2=P1*((T2/T1)^c);//Pressure at section 2 in KN/(m^2)
+DP=(P2-P1)/1000;//Increase in pressure in MN/(m^2)
+IE=Cv*(T2-T1);//Increase in internal energy in kJ/kg
+
+//OUTPUT
+mprintf('Increase in temperature is %3.0f K \n Increase in pressure is %3.2f MN/(m^2) \n Increase in internal energy is %3.0f kJ/kg',DT,DP,IE)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.20/Chapter4_Example20.sce b/1775/CH4/EX4.20/Chapter4_Example20.sce new file mode 100755 index 000000000..5b045db43 --- /dev/null +++ b/1775/CH4/EX4.20/Chapter4_Example20.sce @@ -0,0 +1,32 @@ +//Chapter-4, Illustration 20, Page 210
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+C1=600;//Steam velocity in m/s
+b1=30;//Blade angle at inlet in first stage in degrees
+b2=30;//Blade angle at exit in first stage in degrees
+b3=30;//Blade angle at inlet in second stage in degrees
+b4=30;//Blade angle at exit in second stage in degrees
+a4=90;//Nozzle angle in second stage in degrees
+m=3;//Mass of steam in kg/s
+x=15;//Length for drawing velocity triangle in mm
+y=56;//Length of BC from the velocity triangle in mm
+
+//CALCUALTIONS
+C1n=C1/y;//Velocity of steam in m/s
+U=x*C1n;//Blade speed in m/s
+l=103;//Length from velocity triangle in mm
+P=(m*l*C1n*U)/1000;//Power developed in kW
+
+//OUTPUT
+mprintf('Blade speed is %3.1f m/s \n Power developed by the turbine is %3.2f kW',U,P)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.21/Chapter4_Example21.sce b/1775/CH4/EX4.21/Chapter4_Example21.sce new file mode 100755 index 000000000..0f864083f --- /dev/null +++ b/1775/CH4/EX4.21/Chapter4_Example21.sce @@ -0,0 +1,40 @@ +//Chapter-4, Illustration 21, Page 211
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+N=400;//Speed in rpm
+m=8.33;//Mass of steam in kg/s
+P=1.6;//Pressure of steam in bar
+x=0.9;//Dryness fraction
+W=10;//Stage power in kW
+r=0.75;//Ratio of axial flow velocity to blade velocity
+a1=20;//Nozzle angle at inlet in degrees
+a2=35;//Nozzle angle at exit in degrees
+b1=a2;//Blade tip angle at exit in degrees
+b2=a1;//Blade tip angle at inlet in degrees
+a=25;//Length of AB from velocity triangle in mm
+vg=1.091;//Specific volume of steam from steam tables in (m^3)/kg
+
+//CALCULATIONS
+Cx=73.5;//Change in whirl velocity from the velocity triangle by measurement in mm
+y=Cx/a;//Ratio of change in whirl velocity to blade speed
+U=sqrt((W*1000)/(m*y));//Blade speed in m/s
+D=((U*60)/(3.147*N))*1000;//Mean diameter of drum in mm
+v=m*x*vg;//Volume flow rate of steam in (m^3)/s
+
+//OUTPUT
+mprintf('Mean diameter of drum is %3.0f mm \n Volume of steam flowing per second is %3.2f m^3/s',D,v)
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.22/Chapter4_Example22.sce b/1775/CH4/EX4.22/Chapter4_Example22.sce new file mode 100755 index 000000000..1273d8a87 --- /dev/null +++ b/1775/CH4/EX4.22/Chapter4_Example22.sce @@ -0,0 +1,40 @@ +//Chapter-4, Illustration 22, Page 212
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+N=300;//Speed in rpm
+m=4.28;//Mass of steam in kg/s
+P=1.9;//Pressure of steam in bar
+x=0.93;//Dryness fraction
+W=3.5;//Stage power in kW
+r=0.72;//Ratio of axial flow velocity to blade velocity
+a1=20;//Nozzle angle at inlet in degrees
+b2=a1;//Blade tip angle at inlet in degrees
+l=0.08;//Tip leakage steam
+vg=0.929;//Specific volume of steam from steam tables in (m^3)/kg
+
+//CALCULATIONS
+mact=m-(m*l);//Actual mass of steam in kg/s
+a=(3.147*N)/60;//Ratio of blade velocity to mean dia
+b=r*a;//Ratio of axial velocity to mean dia
+c=46;//Ratio of change in whirl velocity to mean dia
+D=sqrt((W*1000)/(mact*c*a));//Mean dia in m
+Ca=b*D;//Axial velocity in m/s
+h=((mact*x*vg)/(3.147*D*Ca))*1000;//Blade height in mm
+D1=D-(h/1000);//Drum dia in m
+
+//OUTPUT
+mprintf('Drum diameter is %3.3f m \n Blade height is %3.0f mm',D1,h)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.23/Chapter4_Example23.sce b/1775/CH4/EX4.23/Chapter4_Example23.sce new file mode 100755 index 000000000..bdf5111c7 --- /dev/null +++ b/1775/CH4/EX4.23/Chapter4_Example23.sce @@ -0,0 +1,40 @@ +//Chapter-4, Illustration 23, Page 214
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P0=800;//Steam pressure in kPa
+P2=100;//Pressure at point 2 in kPa
+T0=973;//Steam temperature in K
+a1=73;//Nozzle angle in degrees
+ns=0.9;//Steam efficiency
+m=35;//Mass flow rate in kg/s
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+y=1.4;//Ratio of specific heats
+
+//CALCULATIONS
+b1=atand(tand(a1)/2);//Blade angle at inlet in degrees
+b2=b1;//Blade angle at exit in degrees
+p=2/tand(a1);//Flow coefficient
+s=p*(tand(b1)+tand(b2));//Blade loading coefficient
+Dh=ns*Cp*T0*(1-((P2/P0)^((y-1)/y)));//Difference in enthalpies in kJ/kg
+W=(m*Dh)/1000;//Power developed in MW
+
+//OUTPUT
+mprintf('Rotor blade angles are %3.2f degrees and %3.2f degrees \n Flow coefficient is %3.3f \n Blade loading coefficient is %3.0f \n Power developed is %3.1f MW',b1,b2,p,s,W)
+
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.24/Chapter4_Example24.sce b/1775/CH4/EX4.24/Chapter4_Example24.sce new file mode 100755 index 000000000..4d9b95590 --- /dev/null +++ b/1775/CH4/EX4.24/Chapter4_Example24.sce @@ -0,0 +1,55 @@ +//Chapter-4, Illustration 24, Page 215
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P0=100;//Steam pressure in bar
+T0=773;//Steam temperature in K
+a1=70;//Nozzle angle in degrees
+ns=0.78;//Steam efficiency
+m=100;//Mass flow rate of steam in kg/s
+D=1;//Turbine diameter in m
+N=3000;//Turbine speed in rpm
+h0=3370;//Steam enthalpy from Moiller chart in kJ/kg
+v2=0.041;//Specific volume at P2 from steam tables in (m^3)/kg
+v4=0.05;//Specific volume at P4 from steam tables in (m^3)/kg
+
+//CALCULATIONS
+U=(3.147*D*N)/60;//Blade speed in m/s
+C1=(2*U)/sind(a1);//Steam speed in m/s
+b1=atand(tand(a1)/2);//Blade angle at inlet for first stage in degrees
+b2=b1;//Blade angle at exit for first stage in degrees
+b3=b1;//Blade angle at inlet for second stage in degrees
+b4=b2;//Blade angle at exit for second stage in degrees
+Wt=(4*m*(U^2))/(10^6);//Total workdone in MW
+Dh=(2*(U^2))/1000;//Difference in enthalpies in kJ/kg
+Dhs=Dh/ns;//Difference in enthalpies in kJ/kg
+h2=h0-Dh;//Enthalpy at point 2 in kJ/kg
+h2s=h0-Dhs;//Enthalpy at point 2s in kJ/kg
+Dh2=(2*(U^2))/1000;//Difference in enthalpies in kJ/kg
+Dh2s=Dh2/ns;//Difference in enthalpies in kJ/kg
+h4=h2-Dh2;//Enthalpy at point 4 in kJ/kg
+h4s=h2-Dh2s;//Enthalpy at point 4s in kJ/kg
+Ca=C1*cosd(a1);//Axial velocity in m/s
+hI=(m*v2)/(3.147*D*Ca);//Blade height at first stage in m/s
+hII=(m*v4)/(3.147*D*Ca);//Blade height at second stage in m/s
+
+//OUTPUT
+mprintf('Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees \n Rotor blade angles for second stage are %3.2f degrees and %3.2f degrees \n Power developed is %3.2f MW \n Final state of steam at first stage is %3.2f kJ/kg \n Final state of steam at second stage is %3.2f kJ/kg \n Blade height at first stage is %3.4f m \n Blade height at second stage is %3.4f m',b1,b2,b3,b4,Wt,h2s,h4s,hI,hII)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.25/Chapter4_Example25.sce b/1775/CH4/EX4.25/Chapter4_Example25.sce new file mode 100755 index 000000000..aa1aeea8d --- /dev/null +++ b/1775/CH4/EX4.25/Chapter4_Example25.sce @@ -0,0 +1,50 @@ +//Chapter-4, Illustration 25, Page 218
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P0=100;//Steam pressure in bar
+T0=773;//Steam temperature in K
+a1=70;//Nozzle angle in degrees
+ns=0.78;//Steam efficiency
+m=100;//Mass flow rate of steam in kg/s
+D=1;//Turbine diameter in m
+N=3000;//Turbine speed in rpm
+h0=3370;//Steam enthalpy from Moiller chart in kJ/kg
+P4=27;//Pressure at point 4 in bar
+T4=638;//Temperature at point 4 in K
+v4=0.105;//Specific volume at P4 from mollier chart in (m^3)/kg
+ns=0.65;//Stages efficiency
+
+//CALCULATIONS
+U=(3.147*D*N)/60;//Blade speed in m/s
+C1=(4*U)/sind(a1);//Steam speed in m/s
+Ca=C1*cosd(a1);//Axial velocity in m/s
+b1=atand((3*U)/Ca);//Blade angle at inlet for first stage in degrees
+b2=b1;//Blade angle at exit for first stage in degrees
+b4=atand(U/Ca);//Blade angle at exit for second stage in degrees
+b3=b4;//Blade angle at inlet for second stage in degrees
+WI=m*6*(U^2);//Power developed in first stage in MW
+WII=m*2*(U^2);//Power developed in second stage in MW
+W=(WI+WII)/(10^6);//Total power developed in MW
+Dh=(W*1000)/100;//Difference in enthalpies in kJ/kg
+Dhs=(W*1000)/(ns*100);//Difference in enthalpies in kJ/kg
+h4=h0-Dh;//Enthalpy at point 4 in kJ/kg
+h4s=h0-Dhs;//Enthalpy at point 4s in kJ/kg
+h=(m*v4)/(3.147*D*Ca);//Rotor blade height in m
+
+//OUTPUT
+mprintf('Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees \n Rotor blade angles for second stage are %3.2f degrees and %3.2f degrees \n Power developed is %3.2f MW \n Final state of steam at first stage is %3.1f kJ/kg \n Final state of steam at second stage is %3.2f kJ/kg \n Rotor blade height is %3.4f m',b1,b2,b3,b4,W,h4,h4s,h)
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.26/Chapter4_Example26.sce b/1775/CH4/EX4.26/Chapter4_Example26.sce new file mode 100755 index 000000000..c1cc4c14a --- /dev/null +++ b/1775/CH4/EX4.26/Chapter4_Example26.sce @@ -0,0 +1,30 @@ +//Chapter-4, Illustration 26, Page 221
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+a1=30;//Nozzle angle in degrees
+Ca=180;//Axial velocity in m/s
+U=280;//Rotor blade speed in m/s
+R=0.5;//Degree of reaction
+
+//CALCULATIONS
+a1n=90-a1;//Nozzle angle measured from axial direction in degrees
+Cx1=Ca*tand(a1n);//Whirl velocity in m/s
+b1=atand((Cx1-U)/Ca);//Blade angle at inlet in degrees
+b2=a1n;//Blade angle at exit in degrees
+
+//OUTPUT
+mprintf('Blade angle at inlet is %3.0f degrees \n Blade angle at exit is %3.0f degrees',b1,b2)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.27/Chapter4_Example27.sce b/1775/CH4/EX4.27/Chapter4_Example27.sce new file mode 100755 index 000000000..74e60abcf --- /dev/null +++ b/1775/CH4/EX4.27/Chapter4_Example27.sce @@ -0,0 +1,35 @@ +//Chapter-4, Illustration 27, Page 222
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P0=800;//Steam pressure in kPa
+T0=900;//Steam temperature in K
+a1=70;//Nozzle angle in degrees
+ns=0.85;//Steam efficiency
+m=75;//Mass flow rate of steam in kg/s
+R=0.5;//Degree of reaction
+U=160;//Blade speed in m/s
+
+//CALCULATIONS
+C1=U/sind(a1);//Steam speed in m/s
+Ca=C1*cosd(a1);//Axial velocity in m/s
+b1=0;//Blade angle at inlet from velocity triangle in degrees
+b2=a1;//Blade angle at exit in degrees
+a2=b1;//Nozzle angle in degrees
+W=(m*(U^2))/(10^6);//Power developed in MW
+Dhs=(W*1000)/(ns*m);//Isentropic enthalpy drop in kJ/kg
+
+//OUTPUT
+mprintf('Rotor blade angles are %3.0f degrees and %3.0f degrees \n Power developed is %3.2f MW \n Isentropic enthalpy drop is %3.2f kJ/kg',b1,b2,W,Dhs)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.3/Chapter4_Example3.sce b/1775/CH4/EX4.3/Chapter4_Example3.sce new file mode 100755 index 000000000..9a0731536 --- /dev/null +++ b/1775/CH4/EX4.3/Chapter4_Example3.sce @@ -0,0 +1,34 @@ +//Chapter-4, Illustration 3, Page 163
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=2;//Pressure at entry in MN/(m^2)
+T1=598;//Temperature at entry in K
+P2=0.36;//Pressure at exit in MN/(m^2)
+m=7.5;//mass flow rate of steam in kg/s
+n=1.3;//Adiabatic gas constant
+v1=0.132;//Volume at entry in (m^3)/kg from steam table
+Ts=412.9;//Saturation temperature in K
+
+//CALCULATIONS
+c=n/(n-1);//Ratio
+Pt=((2/(n+1))^c)*P1;//Throat pressure in MN/(m^2)
+Ct=((2*c*P1*v1*(1-((Pt/P1)^(1/c))))^0.5)*1000;//Velocity at throat in m/s
+vt=v1*((P1/Pt)^(1/n));//Specific volume at throat in (m^3)/kg
+At=((m*vt)/Ct)*(10^6);//Area of throat in (mm^2)
+C2=((2*c*P1*v1*(1-((P2/P1)^(1/c))))^0.5)*1000;//Velocity at exit in m/s
+v2=v1*((P1/P2)^(1/n));//Specific volume at exit in (m^3)/kg
+A2=((m*v2)/C2)*(10^6);//Area of exit in (mm^2)
+T2=T1*((P2/P1)^(1/c));//Temperature at exit in K
+D=Ts-T2;//Degree of undercooling at exit in K
+
+//OUTPUT
+mprintf('Throat area is %3.0f (mm^2) \n Exit area is %3.0f (mm^2) \n Degree of undercooling at exit is %3.1f K',At,A2,D)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.4/Chapter4_Example4.sce b/1775/CH4/EX4.4/Chapter4_Example4.sce new file mode 100755 index 000000000..a0ca9dec8 --- /dev/null +++ b/1775/CH4/EX4.4/Chapter4_Example4.sce @@ -0,0 +1,32 @@ +//Chapter-4, Illustration 4, Page 165
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=2.2;//Pressure at entry in MN/(m^2)
+T1=533;//Temperature at entry in K
+P2=0.4;//Pressure at exit in MN/(m^2)
+m=11;//mass flow rate of steam in kg/s
+n=0.85;//Efficiency of expansion
+h1=2940;//Enthalpy at entrance in kJ/kg from Moiller chart
+ht=2790;//Enthalpy at throat in kJ/kg from Moiller chart
+h2s=2590;//Enthalpy below exit level in kJ/kg from Moiller chart
+vt=0.16;//Throat volume in (m^3)/kg
+v2=0.44;//Volume at exit in (m^3)/kg
+
+//CALCULATIONS
+Ct=(2000*(h1-ht))^0.5;//Throat velocity in m/s
+h2=ht-(0.85*(ht-h2s));//Enthalpy at exit in kJ/kg
+C2=(2000*(h1-h2))^0.5;//Exit velocity in m/s
+At=((m*vt)/Ct)*(10^6);//Area of throat in (mm^2)
+A2=((m*v2)/C2)*(10^6);//Area of exit in (mm^2)
+
+//OUTPUT
+mprintf('Throat velocity is %3.0f m/s \n Exit velocity is %3.0f m/s \n Throat area is %3.0f (mm^2) \n Exit area is %3.0f (mm^2) \n',Ct,C2,At,A2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.5/Chapter4_Example5.sce b/1775/CH4/EX4.5/Chapter4_Example5.sce new file mode 100755 index 000000000..c79974e5e --- /dev/null +++ b/1775/CH4/EX4.5/Chapter4_Example5.sce @@ -0,0 +1,45 @@ +//Chapter-4, Illustration 5, Page 166
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=35;//Pressure at entry in bar
+T1=573;//Temperature at entry in K
+P2=8;//Pressure at exit in bar
+Ts=443.4;//Saturation temperature in K
+Ps=3.1;//Saturation pressure in bar
+m=5.2;//mass flow rate of steam in kg/s
+n=1.3;//Adiabatic gas constant
+v1=0.06842;//Specific volume at entry in (m^3)/kg from steam table
+v3=0.2292;//Specific volume at exit in (m^3)/kg from steam table
+h1=2979;//Enthalpy in kJ/kg from Moiller chart
+h3=2673.3;//Enthalpy in kJ/kg from Moiller chart
+
+//CALCULATIONS
+c=n/(n-1);//Ratio
+C2=((2*c*P1*(10^5)*v1*(1-((P2/P1)^(1/c))))^0.5);//Velocity at exit in m/s
+v2=v1*((P1/P2)^(1/n));//Specific volume at exit in (m^3)/kg
+A2=((m*v2)/C2)*(10^4);//Area of exit in (cm^2)
+a=((A2/18)^0.5)*10;//Length in mm
+b=3*a;//Breadth in mm
+T2=T1*((P2/P1)^(1/c));//Temperature at exit in K
+D=Ts-T2;//Degree of undercooling in K
+Ds=P2/Ps;//Degree of supersaturation
+hI=h1-h3;//Isentropic enthalpy drop in kJ/kg
+ha=(C2^2)/2000;//Actual enthalpy drop in kJ/kg
+QL=hI-ha;//Loss in available heat in kJ/kg
+DS=QL/Ts;//Increase in entropy in kJ/kg-K
+C3=(2000*(h1-h3))^0.5;//Exit velocity from nozzle
+mf=((A2*C3*(10^-4))/v3);//Mass flow rate in kg/s
+Rm=m/mf;//Ratio of mass rate
+
+//OUTPUT
+mprintf('Cross section of nozzle is %3.1f mm * %3.1f mm \n Degree of undercooling is %3.1f K and Degree of supersaturation is %3.2f \n Loss in available heat drop due to irreversibility is %3.2f kJ/kg \n Increase in entropy is %3.5f kJ/kg-K \n Ratio of mass flow rate with metastable expansion to the thermal expansion is %3.3f',b,a,D,Ds,QL,DS,Rm)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.6/Chapter4_Example6.sce b/1775/CH4/EX4.6/Chapter4_Example6.sce new file mode 100755 index 000000000..1182c53be --- /dev/null +++ b/1775/CH4/EX4.6/Chapter4_Example6.sce @@ -0,0 +1,32 @@ +//Chapter-4, Illustration 6, Page 169
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+m=14;//Mass flow rate of steam in kg/s
+P1=3;//Pressure of Steam in MN/(m^2)
+T1=300;//Steam temperature in oC
+h1=2990;//Enthalpy at point 1 in kJ/kg
+h2s=2630;//Enthalpy at point 2s in kJ/kg
+ht=2850;//Enthalpy at point t in kJ/kg
+n=1.3;//Adiabatic gas constant
+C2=800;//Exit velocity in m/s
+v2=0.4;//Specific volume at exit in (m^3)/kg
+
+//CALCULATIONS
+x=n/(n-1);//Ratio
+Pt=((2/(n+1))^x)*P1;//Temperature at point t in MN/(m^2)
+h2=h1-((C2^2)/2000);//Exit enthalpy in kJ/kg
+nN=((h1-h2)/(h1-h2s))*100;//Nozzle efficiency
+A2=((m*v2)/C2)*(10^6);//Exit area in (mm^2)
+Ct=sqrt(2000*(h1-ht));//Throat velocity in m/s
+
+//OUTPUT
+mprintf('Nozzle efficiency is %3.1f percent \n Exit area is %3.0f (mm^2) \n Throat velocity is %3.0f m/s',nN,A2,Ct)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.7/Chapter4_Example7.sce b/1775/CH4/EX4.7/Chapter4_Example7.sce new file mode 100755 index 000000000..22c03f53c --- /dev/null +++ b/1775/CH4/EX4.7/Chapter4_Example7.sce @@ -0,0 +1,40 @@ +//Chapter-4, Illustration 7, Page 170
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=10;//Pressure at point 1 in bar
+P2=0.5;//Pressure at point 2 in bar
+h1=3050;//Enthalpy at point 1 in kJ/kg
+h2s=2480;//Enthalpy at point 2s in kJ/kg
+ht=2910;//Enthalpy at throat in kJ/kg
+n=1.3;//Adiabatic gas constant
+r=0.1;//Total available heat drop
+v1=0.258;//Specific volume at point 1 in (m^3)/kg
+h2f=340.6;//Enthalpy for exit pressure from steam tables in kJ/kg
+hfg=2305.4;//Enthalpy for exit pressure from steam tables in kJ/kg
+m=0.5;//Mass flow rate in kg/s
+
+//CALCULATIONS
+x=n/(n-1);//Ratio
+Pt=((2/(n+1))^x)*P1;//Temperature at throat in bar
+h2=h2s+(r*(h1-h2s));//Enthalpy at point 2 in kJ/kg
+vt=((P1/Pt)^(1/n))*v1;//Specific volume at throat in (m^3)/kg
+v2=((P1/P2)^(1/n))*v1;//Specific volume at point 2 in (m^3)/kg
+Ct=sqrt(2000*(h1-ht));//Throat velocity in m/s
+At=((m*vt)/Ct)*(10^6);//Throat area in (mm^2)
+C2=sqrt(2000*(h1-h2));//Exit velocity in m/s
+A2=((m*v2)/C2)*(10^6);//Exit area in (mm^2)
+x2=((h2-h2f)/hfg)*100;//Steam quality at exit
+
+//OUTPUT
+mprintf('Throat area is %3.0f (mm^2) \n Exit area is %3.0f (mm^2) \n Steam quality at exit is %3.0f percent',At,A2,x2)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.8/Chapter4_Example8.sce b/1775/CH4/EX4.8/Chapter4_Example8.sce new file mode 100755 index 000000000..d57cbf248 --- /dev/null +++ b/1775/CH4/EX4.8/Chapter4_Example8.sce @@ -0,0 +1,33 @@ +//Chapter-4, Illustration 8, Page 171
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=3.5;//Dry saturated steam in bar
+P2=1.1;//Exit pressure in bar
+At=4.4;//Throat area in cm^2
+h1=2731.6;//Enthalpy at P1 in kJ/kg
+v1=0.52397;//Specific volume at P1 in m^3/kg
+n=1.135;//Adiabatic gas constant
+ht=2640;//Enthalpy at Pt in kJ/kg
+vt=0.85;//Specific volume at throat in m^3/kg
+h2=2520;//Enthalpy at P2 in kJ/kg
+v2=1.45;//Specific volume at P2 in m^3/kg
+
+//CALCULATIONS
+x=n/(n-1);//Ratio
+Pt=((2/(n+1))^x)*P1;//Throat pressure in bar
+Ct=sqrt(2000*(h1-ht));//Throat velocity in m/s
+mmax=((At*Ct*(10^-4))/vt)*60;//Maximum discharge in kg/min
+C2=sqrt(2000*(h1-h2));//Exit velocity in m/s
+A2=((mmax*v2)/(C2*60))*(10^6);//Exit area in mm^2
+
+//OUTPUT
+mprintf('Maximum discharge is %3.3f kg/min \n Exit area is %3.2f mm^2',mmax,A2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH4/EX4.9/Chapter4_Example9.sce b/1775/CH4/EX4.9/Chapter4_Example9.sce new file mode 100755 index 000000000..6cf39451e --- /dev/null +++ b/1775/CH4/EX4.9/Chapter4_Example9.sce @@ -0,0 +1,30 @@ +//Chapter-4, Illustration 9, Page 172
+//Title: Steam Nozzles and Steam Turbines
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=10;//Pressure at point 1 in bar
+T1=200;//Temperature at point 1 in oC
+P2=5;//Pressure at point 2 in bar
+n=1.3;//Adiabatic gas constant
+h1=2830;//Enthalpy at P1 in kJ/kg
+ht=2710;//Enthalpy at point Pt in kJ/kg
+vt=0.35;//Specific volume at Pt in m^3/kg
+m=3;//Nozzle flow in kg/s
+
+//CALCULATIONS
+x=n/(n-1);//Ratio
+Pt=((2/(n+1))^x)*P1;//Throat pressure in bar
+Ct=sqrt(2000*(h1-ht));//Throat velocity in m/s
+At=(m*vt)/Ct;//Throat area in m^2
+
+//OUTPUT
+mprintf('Since throat pressure is greater than exit pressure,nozzle used is convergent-divergent nozzle \n Minimum area of nozzle required is %3.5f m^2',At)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.1/Chapter5_Example1.sce b/1775/CH5/EX5.1/Chapter5_Example1.sce new file mode 100755 index 000000000..70e5a7281 --- /dev/null +++ b/1775/CH5/EX5.1/Chapter5_Example1.sce @@ -0,0 +1,32 @@ +//Chapter-5, Illustration 1, Page 250
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=0.2;//Cylinder diameter in m
+L=0.3;//Cylinder Stroke in m
+P1=1;//Pressure at entry in bar
+T1=300;//Temperature at entry in K
+P2=8;//Pressure at exit in bar
+n=1.25;//Adiabatic gas constant
+N=100;//Speed in rpm
+R=287;//Universal gas constant in J/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+V1=(3.147*L*(D^2))/4;//Volume of cylinder in m^3/cycle
+W=(P1*(10^5)*V1*(((P2/P1)^x)-1))/x;//Work done in J/cycle
+Pc=(W*100)/(60*1000);//Indicated power of compressor in kW
+m=(P1*(10^5)*V1)/(R*T1);//Mass of air delivered in kg/cycle
+md=m*N;//Mass delivered per minute in kg
+T2=T1*((P2/P1)^x);//Temperature of air delivered in K
+
+//OUTPUT
+mprintf('Indicated power of compressor is %3.2f kW \n Mass of air delivered by compressor per minute is %3.2f kg \n Temperature of air delivered is %3.1fK',Pc,md,T2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.10/Chapter5_Example10.sce b/1775/CH5/EX5.10/Chapter5_Example10.sce new file mode 100755 index 000000000..b05b4a4b7 --- /dev/null +++ b/1775/CH5/EX5.10/Chapter5_Example10.sce @@ -0,0 +1,50 @@ +//Chapter-5, Illustration 10, Page 259
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=0.98;//Pressure at point 1 in bar
+P4=P1;//Pressure at point 4 in bar
+P2=7;//Pressure at point 2 in bar
+P3=P2;//Pressure at point 3 in bar
+n=1.3;//Adiabatic gas constant
+Ta=300;//Air temperature in K
+Pa=1.013;//Air pressure in bar
+T1=313;//Temperature at point 1 in K
+c=0.04;//Ratio of clearance volume to swept volume
+Va=15;//Volume of air delivered in m^3
+R=0.287;//Universal gas constant in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+r=(P2/P1)^(1/n);//Ratio of volumes
+a=r*c;//Ratio of volume at point 4 to swept volume
+DV=1+c-a;//Difference in volumes
+V=(P1*DV*Ta)/(T1*Pa);//Volume of air delivered per cycle
+nv=V*100;//Volumetric efficiency
+DV1=(Pa*Va*T1)/(Ta*P1);//Difference in volumes
+T2=T1*((P2/P1)^x);//Temperature at point 2 in K
+ma=(Pa*100*Va)/(R*Ta);//Mass of air delivered in kg/min
+IP=(ma*R*(T2-T1))/(x*60);//Indicated power in kW
+Piso=(ma*R*T1*log(P2/P1))/60;//Isothermal indicated power in kW
+niso=(Piso/IP)*100;//Isothermal efficiency
+
+//OUTPUT
+mprintf('Volumetric efficiency is %3.1f percent \n Indicated power is %3.2f kW \n Isothermal efficiency is %3.0f percent',nv,IP,niso)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.11/Chapter5_Example11.sce b/1775/CH5/EX5.11/Chapter5_Example11.sce new file mode 100755 index 000000000..ecad8c08b --- /dev/null +++ b/1775/CH5/EX5.11/Chapter5_Example11.sce @@ -0,0 +1,30 @@ +//Chapter-5, Illustration 11, Page 261
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+V1=7*(10^-3);//Volume of air in (m^3)/s
+P1=1.013;//Pressure of air in bar
+T1=288;//Air temperature in K
+P2=14;//Pressure at point 2 in bar
+n=1.3;//Adiabatic gas constant
+nm=0.82;//Mechanical efficiency
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+W=(P1*100*V1*(((P2/P1)^x)-1))/x;//Work done by compressor in kW
+P=W/nm;//Power requred to drive compressor in kW
+
+//OUTPUT
+mprintf('Power requred to drive compressor is %3.2f kW',P)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.12/Chapter5_Example12.sce b/1775/CH5/EX5.12/Chapter5_Example12.sce new file mode 100755 index 000000000..0b7075ac7 --- /dev/null +++ b/1775/CH5/EX5.12/Chapter5_Example12.sce @@ -0,0 +1,35 @@ +//Chapter-5, Illustration 12, Page 261
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+L=0.15;//Stroke in mm
+D=0.15;//Bore in mm
+N=8;//Speed in rps
+P1=100;//Pressure at point 1 in kN/(m^2)
+P2=550;//Pressure at point 2 in kN/(m^2)
+n=1.32;//Adiabatic gas constant
+C=0.06;//Ratio of clearance volume to swept volume
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+nv=(1+C-(C*((P2/P1)^(1/n))))*100;//Volumetric efficiency
+DV=(3.147*(D^2)*L)/4;//Difference in volumes at points 1 and 3
+DV1=(nv*DV)/100;//Difference in volumes at points 1 and 4
+V2=DV1*((P1/P2)^(1/n))*N;//Volume of air delivered per second
+W=(P1*DV1*(((P2/P1)^x)-1))*N/x;//Power of compressor in kW
+
+//OUTPUT
+mprintf('Theoretical volume efficiency is %3.1f percent \n Volume of air delivered is %3.5f (m^3)/s \n Power of compressor is %3.3f kW',nv,V2,W)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.13/Chapter5_Example13.sce b/1775/CH5/EX5.13/Chapter5_Example13.sce new file mode 100755 index 000000000..dbd8bb632 --- /dev/null +++ b/1775/CH5/EX5.13/Chapter5_Example13.sce @@ -0,0 +1,45 @@ +//Chapter-5, Illustration 13, Page 262
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+V=16;//Volume of air compresssed in m^3
+P1=1;//Pressure at point 1 in bar
+P3=10.5;//Pressure at point 3 in bar
+T1=294;//Temperature at point 1 in K
+Tc=25;//Temperature of cooling water in oC
+n=1.35;//Adiabatics gas constant
+R=0.287;//Universal gas constant in kJ/kg-K
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+Cw=4.187;//Specific heat of water in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+P2=sqrt(P1*P3);//Pressure at point 2 in bar
+W1=(2*P1*100*V*(((P2/P1)^x)-1))/(x*60);//Indicated power of compressor from P1 to P2 in kW
+W2=(P1*100*V*(((P3/P1)^x)-1))/(x*60);//Indicated power of compressor from P1 to P3 in kW
+T4=T1*((P2/P1)^x);//Maximum temperature for two stage compression in K
+T2=T1*((P3/P1)^x);//Maximum temperature for single stage compression in K
+m=(P1*100*V)/(R*T1);//Mass of air compressed in kg/min
+Q=m*Cp*(T4-T1);//Heat rejected by air in kJ/min
+mc=Q/(Cw*Tc);//Mass of cooling water in kg/min
+
+//OUTPUT
+mprintf('Minimum indicated power required for 2 stage compression is %3.1f kW \n Power required for single stage compression is 18 percent more than that for two stage compression with perfect intercooling \n Maximum temperature for two stage compression is %3.1f K \n Maximum temperature for single stage compression is %3.1f K \n Heat rejected by air is %3.1f kJ/min \n Mass of cooling water required is %3.1f kg/min',W1,T4,T2,Q,mc)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.14/Chapter5_Example14.sce b/1775/CH5/EX5.14/Chapter5_Example14.sce new file mode 100755 index 000000000..2b5ee5632 --- /dev/null +++ b/1775/CH5/EX5.14/Chapter5_Example14.sce @@ -0,0 +1,32 @@ +//Chapter-5, Illustration 14, Page 264
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+V=0.2;//Air flow rate in (m^3)/s
+P1=0.1;//Intake pressure in MN/(m^2)
+P3=0.7;//Final pressure in MN/(m^2)
+T1=289;//Intake temperature in K
+n=1.25;//Adiabatic gas constant
+N=10;//Compressor speed in rps
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+P2=sqrt(P1*P3);//Intermediate pressure in MN/(m^2)
+V1=(V/N)*1000;//Total volume of LP cylinder in litres
+V2=((P1*V1)/P2);//Total volume of HP cylinder in litres
+W=((2*P1*V*(((P2/P1)^x)-1))/x)*1000;//Cycle power in kW
+
+//OUTPUT
+mprintf('Intermediate pressure is %3.3f MN/(m^2) \n Total volume of LP cylinder is %3.0f litres \n Total volume of HP cylinder is %3.1f litres \n Cycle power is %3.0f kW',P2,V1,V2,W)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.15/Chapter5_Example15.sce b/1775/CH5/EX5.15/Chapter5_Example15.sce new file mode 100755 index 000000000..3804a82dc --- /dev/null +++ b/1775/CH5/EX5.15/Chapter5_Example15.sce @@ -0,0 +1,35 @@ +//Chapter-5, Illustration 15, Page 265
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+T1=290;//Temperature at point 1 in K
+P3=60;//Pressure at point 3 in bar
+P2=8;//Pressure at point 2 in bar
+T2=310;//Temperature at point 2 in K
+L=0.2;//Stroke in m
+D=0.15;//Bore in m
+n=1.35;//Adiabatic gas constant
+N=200;//Speed in rpm
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+V1=(3.147*(D^2)*L)/4;//Volume at point 1 in m^3
+V2=(P1*V1*T2)/(T1*P2);//Volume of air entering LP cylinder in m^3
+W=((P1*(10^5)*V1*(((P2/P1)^x)-1))/x)+((P2*(10^5)*V2*(((P3/P2)^x)-1))/x);//Workdone by compressor per cycle in J
+P=(W*N)/(60*1000);//Power of compressor in kW
+
+//OUTPUT
+mprintf('Power of compressor is %3.2f kW',P)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.16/Chapter5_Example16.sce b/1775/CH5/EX5.16/Chapter5_Example16.sce new file mode 100755 index 000000000..b93ec834c --- /dev/null +++ b/1775/CH5/EX5.16/Chapter5_Example16.sce @@ -0,0 +1,50 @@ +//Chapter-5, Illustration 16, Page 265
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+N=220;//Speed of compressor in rpm
+P1=1;//Pressure entering LP cylinder in bar
+T1=300;//Temperature at point 1 in K
+Dlp=0.36;//Bore of LP cylinder in m
+Llp=0.4;//Stroke of LP cylinder in m
+Lhp=0.4;//Stoke of HP cylinder in m
+C=0.04;//Ratio of clearance volumes of both cylinders
+P2=4;//Pressure leaving LP cylinder in bar
+P5=3.8;//Pressure entering HP cylinder in bar
+T3=300;//Temperature entering HP cylinder in K
+P6=15.2;//Dicharge pressure in bar
+n=1.3;//Adiabatic gas constant
+Cp=1.0035;//Specific heat at constant pressure in kJ/kg-K
+R=0.287;//Universal gas constant in kJ/kg-K
+T5=T1;//Temperature at point 5 in K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+Vslp=(3.147*(Dlp^2)*Llp*N*2)/4;//Swept volume of LP cylinder in m^3/min
+nv=1+C-(C*((P2/P1)^(1/n)));//Volumetric efficiency
+V1=nv*Vslp;//Volume of air drawn at point 1 in (m^3)/min
+m=(P1*100*V1)/(R*T1);//Mass of air in kg/min
+T2=T1*((P2/P1)^x);//Temperature at point 2 in K
+QR=m*Cp*(T2-T5);//Heat rejected in kJ/min
+V5=(m*R*T5)/(P5*100);//Volume of air drawn in HP cylinder M^3/min
+Plp=P2/P1;//Pressure ratio of LP cylinder
+Php=P6/P5;//Pressure ratio of HP cylinder
+Vshp=V5/nv;//Swept volume of HP cylinder in m^3/min
+Dhp=sqrt((Vshp*4)/(3.147*Lhp*N*2));//Bore of HP cylinder in m
+P=(m*R*(T2-T1))/(x*60);//Power required for HP cylinder in kW
+
+//OUTPUT
+mprintf('Heat rejected in intercooler is %3.1f kJ/min \n Diameter of HP cylinder is %3.4f m \n Power required for HP cylinder is %3.0f kW',QR,Dhp,P)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.17/Chapter5_Example17.sce b/1775/CH5/EX5.17/Chapter5_Example17.sce new file mode 100755 index 000000000..b6f2d5632 --- /dev/null +++ b/1775/CH5/EX5.17/Chapter5_Example17.sce @@ -0,0 +1,26 @@ +//Chapter-5, Illustration 17, Page 267
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+P3=30;//Pressure at point 3 in bar
+T1=300;//Temperature at point 1 in K
+n=1.3;//Adiabatics gas constant
+
+//CALCULATIONS
+P2=sqrt(P1*P3);//Intermediate pressure in bar
+rD=sqrt(P2/P1);//Ratio of cylinder diameters
+
+//OUTPUT
+mprintf('Ratio of cylinder diameters is %3.2f',rD)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.18/Chapter5_Example18.sce b/1775/CH5/EX5.18/Chapter5_Example18.sce new file mode 100755 index 000000000..d2781dbe8 --- /dev/null +++ b/1775/CH5/EX5.18/Chapter5_Example18.sce @@ -0,0 +1,41 @@ +//Chapter-5, Illustration 18, Page 268
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1.013;//Pressure at point 1 in bar
+T1=288;//Temperaturea at point 1 in K
+v1=8.4;//free air delivered by compressor in m^3
+P4=70;//Pressure at point 4 in bar
+n=1.2;//Adiabatic gas constant
+Cp=1.0035;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+P2=P1*((P4/P1)^(1/3));//LP cylinder delivery pressure in bar
+P3=P2*((P4/P1)^(1/3));//IP cylinder delivery pressure in bar
+r=P2/P1;//Ratio of cylinder volumes
+r1=P3/P2;//Ratio of cylinder volumes
+r2=r*r1;//Ratio of cylinder volumes
+V3=1;//Volume at point 3 in m^3
+T4=T1*((P2/P1)^x);//Three stage outlet temperature in K
+QR=Cp*(T4-T1);//Heat rejected in intercooler in kJ/kg of air
+W=((3*P1*100*v1*(((P4/P1)^(x/3))-1))/(x*60));//Total indiacted power in kW
+
+//OUTPUT
+mprintf('LP cylinder delivery pressure is %3.3f bar \n IP cylinder delivery pressure is %3.2f bar \n Ratio of cylinder volumes is %3.2f:%3.1f:%3.0f \n Temperature at end of each stage is %3.2f K \n Heat rejected in each intercooler is %3.1f kJ/kg of air \n Total indicated power is %3.2f kW',P2,P3,r2,r1,V3,T4,QR,W)
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.19/Chapter5_Example19.sce b/1775/CH5/EX5.19/Chapter5_Example19.sce new file mode 100755 index 000000000..f5e8d39b3 --- /dev/null +++ b/1775/CH5/EX5.19/Chapter5_Example19.sce @@ -0,0 +1,45 @@ +//Chapter-5, Illustration 19, Page 269
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=0.45;//Bore in m
+L=0.3;//Stroke in m
+C=0.05;//Ratio of clearance volume to swept volume
+P1=1;//Pressure at point 1 inn bar
+T1=291;//Temperature at point 1 in K
+P4=15;//Pressure at point 4 in bar
+n=1.3;//Adiabatic gas constant
+R=0.29;//Universal gas constant in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+k=(P4/P1)^(1/3);//Pressure ratio
+P2=k*P1;//Pressure at point 2 in bar
+P3=k*P2;//Pressure at point 1 in bar
+Vslp=(3.147*(D^2)*L)/4;//Swept volume of LP cylinder
+V7=C*Vslp;//Volume at point 7 in m^3
+V1=Vslp+V7;//Volume at point 1 in m^3
+V8=V7*(k^(1/n));//Volume at point 8 in m^3
+EVs=(V1-V8)*1000;//Effective swept volume in litres
+T4=T1*(k^x);//Temperature at point 4 in K
+t4=T4-273;//Delivery temperature in oC
+DV=((P1*T4*(V1-V8))/(P4*T1))*1000;//Delivery volume per stroke in litres
+W=(3*R*T1*((k^x)-1))/x;//Workdone per kg of air in kJ
+
+//OUTPUT
+mprintf('Intermediate pressures are %3.3f bar and %3.3f bar \n Effective swept volume of LP cylinder is %3.2f litres \n Temperature of air delivered per stroke is %3.1f oC \n Volume of air delivered per stroke is %3.2f litres \n Work done per kg of air is %3.1f kJ',P2,P3,EVs,t4,DV,W)
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.2/Chapter5_Example2.sce b/1775/CH5/EX5.2/Chapter5_Example2.sce new file mode 100755 index 000000000..6412643fe --- /dev/null +++ b/1775/CH5/EX5.2/Chapter5_Example2.sce @@ -0,0 +1,31 @@ +//Chapter-5, Illustration 2, Page 251
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+IP=37;//Indicated power in kW
+P1=0.98;//Pressure at entry in bar
+T1=288;//Temperature at entry in K
+P2=5.8;//Pressure at exit in bar
+n=1.2;//Adiabatic gas constant
+N=100;//Speed in rpm
+Ps=151.5;//Piston speed in m/min
+a=2;//For double acting compressor
+
+//CALCULATIONS
+L=Ps/(2*N);//Stroke length in m
+x=(n-1)/n;//Ratio
+r=(3.147*L)/4;//Ratio of volume to bore
+D=sqrt((IP*1000*60*x)/(N*a*r*P1*(10^5)*(((P2/P1)^x)-1)));//Cylinder diameter in m
+
+//OUTPUT
+mprintf('Stroke length of cylinder is %3.4f m \n Cylinder diameter is %3.4f m',L,D)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.20/Chapter5_Example20.sce b/1775/CH5/EX5.20/Chapter5_Example20.sce new file mode 100755 index 000000000..30eebce8b --- /dev/null +++ b/1775/CH5/EX5.20/Chapter5_Example20.sce @@ -0,0 +1,30 @@ +//Chapter-5, Illustration 20, Page 271
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+Pns=100;//Maximum pressure in bar
+p=4;//Pressure ratio
+
+//CALCULATIONS
+Ns=log(Pns)/log(p);//Number of stages
+y=ceil(Ns);//Rounding off to next higher integer
+ps=(Pns/P1)^(1/y);//Exact stage pressure ratio
+P2=ps*P1;//Pressure at point 2 in bar
+P3=ps*P2;//Pressure at point 3 in bar
+P4=ps*P3;//Pressure at point 4 in bar
+
+//OUTPUT
+mprintf('Number of stages are %3.0f \n Exact stage pressure ratio is %3.3f \n Intermediate pressures are %3.3f bar,%3.2f bar,%3.2f bar',y,ps,P2,P3,P4)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.3/Chapter5_Example3.sce b/1775/CH5/EX5.3/Chapter5_Example3.sce new file mode 100755 index 000000000..1b2174c9c --- /dev/null +++ b/1775/CH5/EX5.3/Chapter5_Example3.sce @@ -0,0 +1,31 @@ +//Chapter-5, Illustration 3, Page 251
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+IP=11;//Indicated power in kW
+P1=1;//Pressure at entry in bar
+P2=7;//Pressure at exit in bar
+n=1.2;//Adiabatic gas constant
+Ps=150;//Piston speed in m/s
+a=2;//For double acting compressor
+r=1.5;//Storke to bore ratio
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+y=3.147/(4*(r^2));//Ratio of volume to the cube of stroke
+z=(P1*(10^2)*y*(((P2/P1)^x)-1))/x;//Ratio of workdone to the cube of stroke
+L=(sqrt(IP/(z*Ps)))*1000;//Stroke in mm
+D=(L/r);//Bore in mm
+
+//OUTPUT
+mprintf('Stroke length of cylinder is %3.0f mm \n Bore diameter of cylinder is %3.0f mm',L,D)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.4/Chapter5_Example4.sce b/1775/CH5/EX5.4/Chapter5_Example4.sce new file mode 100755 index 000000000..298382980 --- /dev/null +++ b/1775/CH5/EX5.4/Chapter5_Example4.sce @@ -0,0 +1,35 @@ +//Chapter-5, Illustration 4, Page 252
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+x=0.05;//Ratio of clearance volume to swept volume
+P1=1;//Pressure at point 1 in bar
+T1=310;//Temperature at point 1 in K
+n=1.2;//Adiabatic gas constant
+P2=7;//Pressure at point 2 in bar
+Pa=1.01325;//Atmospheric pressure in bar
+Ta=288;//Atmospheric temperature in K
+
+//CALCULATIONS
+V1=1+x;//Ratio of volume of air sucked to stroke volume
+V4=((P2/P1)^(1/n))/20;//Ratio of volume delivered to stroke volume
+DV=V1-V4;//Difference in volumes
+nv1=DV*100;//Volumetric efficiency
+V=(P1*DV*Ta)/(T1*Pa);//Ratio of volumes referred to atmospheric conditions
+nv2=V*100;//Volumetric efficiency referred to atmospheric conditions
+W=(n*0.287*T1*((P2/P1)^((n-1)/n)-1))/(n-1);//Work required in kJ/kg
+
+//OUTPUT
+mprintf('Volumetric efficiency is %3.1f percent \n Volumetric efficiency referred to atmospheric conditions is %3.1f percent \n Work required is %3.1f kJ/kg',nv1,nv2,W)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.5/Chapter5_Example5.sce b/1775/CH5/EX5.5/Chapter5_Example5.sce new file mode 100755 index 000000000..dcea89b21 --- /dev/null +++ b/1775/CH5/EX5.5/Chapter5_Example5.sce @@ -0,0 +1,32 @@ +//Chapter-5, Illustration 5, Page 253
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=0.2;//Bore in m
+L=0.3;//Stroke in m
+lc=0.015;//Linear clearance in m
+P1=1;//Pressure at point 1 in bar
+P2=7;//Pressure at point 2 in bar
+n=1.25;//Adiabatic gas constant
+
+//CALCULATIONS
+V3=(3.147*(D^2)*lc)/4;//Clearance volume in m^3
+Vs=(3.147*(D^2)*L)/4;//Stoke volume in m^3
+C=V3/Vs;//Clearance ratio
+nv=(1+C-(C*((P2/P1)^(1/n))))*100;//Volumetric efficiency
+DV=(nv*Vs)/100;//Volume of air taken in (m^3)/stroke
+
+//OUTPUT
+mprintf('Theoretical volume of air taken in per stroke is %3.6f (m^3)/stroke',DV)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.6/Chapter5_Example6.sce b/1775/CH5/EX5.6/Chapter5_Example6.sce new file mode 100755 index 000000000..6bde597d9 --- /dev/null +++ b/1775/CH5/EX5.6/Chapter5_Example6.sce @@ -0,0 +1,36 @@ +//Chapter-5, Illustration 6, Page 254
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=0.2;//Bore in m
+L=0.3;//Stroke in m
+r=0.05;//Ratio of clearance volume to stroke volume
+P1=1;//Pressure at point 1 in bar
+T1=293;//Temperature at point 1 in K
+P2=5.5;//Pressure at point 2 in bar
+n=1.3;//Adiabatic gas constant
+N=500;//Speed of compressor in rpm
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+Vs=(3.147*L*(D^2))/4;//Stroke volume in m^3
+Vc=r*Vs;//Clearance volume in m^3
+V1=Vc+Vs;//Volume at point 1 in m^3
+V4=Vc*((P2/P1)^(1/n));//Volume at point 4 in m^3
+EVs=V1-V4;//Effective swept volume in m^3
+W=(P1*(10^5)*EVs*(((P2/P1)^x)-1))/x;//Work done in J/cycle
+MEP=(W/Vs)/(10^5);//Mean effective pressure in bar
+P=(W*N)/(60*1000);//Power required in kW
+
+//OUTPUT
+mprintf('Mean effective pressure is %3.2f bar \n Power required is %3.2f kW',MEP,P)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.7/Chapter5_Example7.sce b/1775/CH5/EX5.7/Chapter5_Example7.sce new file mode 100755 index 000000000..acd4347c6 --- /dev/null +++ b/1775/CH5/EX5.7/Chapter5_Example7.sce @@ -0,0 +1,50 @@ +//Chapter-5, Illustration 7, Page 255
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+D=0.2;//Bore in m
+L=0.3;//Stroke in m
+r=0.05;//Ratio of clearance volume to stroke volume
+P1=97;//Pressure at entry in kN/(m^2)
+P4=P1;//Pressure at point 4 in kN/(m^2)
+T1=293;//Temperature at point 1 in K
+P2=550;//Compression Pressure in kN/(m^2)
+P3=P2;//Pressure at point 3 in kN/(m^2)
+n=1.3;//Adiabatic gas constant
+N=500;//Speed of compressor in rpm
+Pa=101.325;//Air pressure in kN/(m^2)
+Ta=288;//Air temperature in K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+DV=(3.147*L*(D^2))/4;//Difference in volumes in m^3
+V3=r*DV;//Clearance volume in m^3
+V1=V3+DV;//Volume at point 1 in m^3
+V4=V3*((P3/P4)^(1/n));//Volume at point 4 in m^3
+Vs=V1-V4;//Effective swept volume in m^3
+EVs=Vs*N;//Effective swept volume per min
+Va=(P1*EVs*Ta)/(Pa*T1);//Free air delivered in (m^3)/min
+nV=((V1-V4)/(V1-V3))*100;//Volumetric effciency
+T2=T1*((P2/P1)^x);//Air delivery temperature in K
+t2=T2-273;//Air delivery temperature in oC
+W=(n*P1*(V1-V4)*(((P2/P1)^x)-1))*N/((n-1)*60);//Cycle power in kW
+Wiso=P1*V1*(log(P2/P1));//Isothermal workdone
+P=(n*P1*V1*(((P2/P1)^x)-1))/(n-1);//Cycle power neglecting clearance
+niso=(Wiso/P)*100;//Isothermal efficiency
+
+//OUTPUT
+mprintf('Free air delivered is %3.3f (m^3)/min \n Volumetric efficiency is %3.0f percent \n Air delivery temperature is %3.1f oC \n Cycle power is %3.0f kW \n Isothermal efficiency is %3.1f percent',Va,nV,t2,W,niso)
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.8/Chapter5_Example8.sce b/1775/CH5/EX5.8/Chapter5_Example8.sce new file mode 100755 index 000000000..bd8f6502b --- /dev/null +++ b/1775/CH5/EX5.8/Chapter5_Example8.sce @@ -0,0 +1,35 @@ +//Chapter-5, Illustration 8, Page 257
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Ve=30;//Volume of air entering compressor per hour in m^3
+P1=1;//Presure of air entering compressor in bar
+N=450;//Speed in rpm
+P2=6.5;//Pressure at point 2 in bar
+nm=0.8;//Mechanical efficiency
+nv=0.75;//Volumetric efficiency
+niso=0.76;//Isothermal efficiency
+
+//CALCULATIONS
+Vs=Ve/(nv*3600);//Swept volume per sec in (m^3)/s
+V=(Vs*60)/N;//Swept volume per cycle in m^3
+V1=(Ve*60)/(3600*N);//Volume at point 1 in m^3
+Wiso=P1*100*V1*log(P2/P1);//Isothermal workdone per cycle
+Wact=Wiso/niso;//Actual workdone per cycle on air
+MEP=(Wact/V)/100;//Mean effective pressure in bar
+IP=(Wact*N)/60;//Indicated power in kW
+BP=IP/nm;//Brake power in kW
+
+//OUTPUT
+mprintf('Mean effective pressure is %3.3f bar \n Brake power is %3.2f kW',MEP,BP)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH5/EX5.9/Chapter5_Example9.sce b/1775/CH5/EX5.9/Chapter5_Example9.sce new file mode 100755 index 000000000..6f81e1ef8 --- /dev/null +++ b/1775/CH5/EX5.9/Chapter5_Example9.sce @@ -0,0 +1,36 @@ +//Chapter-5, Illustration 9, Page 258
+//Title: Air Compressors
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Va=15;//Volume of air in (m^3)/min
+Pa=1.01325;//Pressure of air in bar
+Ta=302;//Air temperature in K
+P1=0.985;//Pressure at point 1 in bar
+T1=313;//Temperature at point 1 in K
+r=0.04;//Ratio of clearance volume to swept volume
+y=1.3;//Ratio of stroke to bore diameter
+N=300;//Speed in rpm
+n=1.3;//Adiabatic gas constant
+P2=7.5;//Pressure at point 2 in bar
+
+//CALCULATIONS
+x=((P2/P1)^(1/n))-1;//Ratio of volume at point 4 to clearance volume
+a=x*r;//Ratio of volume at point 4 to swept volume
+nv=1-a;//Volumetric efficiency
+V1=(Pa*Va*T1)/(Ta*P1);//Volume at point 1 in (m^3)/min
+Vs=V1/(nv*N*2);//Swept volume in m^3
+D=((Vs*4)/(3.147*y))^(1/3);//Bore in m
+L=y*D;//Stroke in m
+
+//OUTPUT
+mprintf('Cylinder bore in %3.3f m \n Cylinder stroke %3.3f m',D,L)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.1/Chapter6_Example1.sce b/1775/CH6/EX6.1/Chapter6_Example1.sce new file mode 100755 index 000000000..0e5a30cee --- /dev/null +++ b/1775/CH6/EX6.1/Chapter6_Example1.sce @@ -0,0 +1,20 @@ +//Chapter-6, Illustration 1, Page 308
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+COP=8.5;//Co-efficient of performance
+T1=300;//Room temperature in K
+T2=267;//Refrigeration temperature in K
+
+//CALCULATIONS
+COPmax=T2/(T1-T2);//Maximum COP possible
+
+//OUTPUT
+mprintf('Maximum COP possible is %3.2f \n Since the COP claimed by the inventor is more than the maximum possible COP his claim is not correct',COPmax)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.10/Chapter6_Example10.sce b/1775/CH6/EX6.10/Chapter6_Example10.sce new file mode 100755 index 000000000..ab81e67ea --- /dev/null +++ b/1775/CH6/EX6.10/Chapter6_Example10.sce @@ -0,0 +1,20 @@ +//Chapter-6, Illustration 10, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tg=470;//Heating temperature in K
+T0=290;//Cooling temperature in K
+TL=270;//Refrigeration temperature in K
+
+//CALCULATIONS
+COP=((Tg-T0)/Tg)*(TL/(T0-TL));//Ideal COP of absorption refrigeration system
+
+//OUTPUT
+mprintf('Ideal COP of absorption refrigeration system is %3.2f',COP)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.11/Chapter6_Example11.sce b/1775/CH6/EX6.11/Chapter6_Example11.sce new file mode 100755 index 000000000..96dbc4ae9 --- /dev/null +++ b/1775/CH6/EX6.11/Chapter6_Example11.sce @@ -0,0 +1,35 @@ +//Chapter-6, Illustration 11, Page 317
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=-18;//Temperature at point 1 in oC
+T3=27;//Temperature at point 3 in oC
+rp=4;//Pressure ratio
+m=0.045;//mass flow rate in kg/s
+y=1.4;//Ratio of specific heats
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2=(rp^x)*(273+T1);//Temperature at point 2 in K
+Tmax=T2-273;//Maximum temperature in oC
+T4=((1/rp)^x)*(273+T3);//Temperature at point 4 in K
+Tmin=T4-273;//Minimum temperature in oC
+qL=Cp*(T1-Tmin);//Heat rejected
+Wcin=Cp*(Tmax-T1);//Compressor work
+Wtout=Cp*(T3-Tmin);//Turbine work
+Wnet=Wcin-Wtout;//Net work done
+COP=qL/Wnet;//Co-efficient of performance
+Qref=m*qL;//Rate of refrigeration in kW
+
+//OUTPUT
+mprintf('Maximum temperature in the cycle is %3.0foC \n Minimum temperature in the cycle is %3.0foC \n COP is %3.2f \n Rate of refrigeration is %3.2f kW',Tmax,Tmin,COP,Qref)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.12/Chapter6_Example12.sce b/1775/CH6/EX6.12/Chapter6_Example12.sce new file mode 100755 index 000000000..077290251 --- /dev/null +++ b/1775/CH6/EX6.12/Chapter6_Example12.sce @@ -0,0 +1,31 @@ +//Chapter-6, Illustration 12, Page 318
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1;//Pressure at point 1 in bar
+T1=268;//Temperature at point 1 in K
+P2=5;//Pressure at point 2 in bar
+T3=288;//Temperature at point 3 in K
+n=1.3;//Adiabatic gas constant
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+
+//CALCULATIONS
+x=(n-1)/n;//Ratio
+T2=((P2/P1)^x)*T1;//Temperature at point 2 in K
+T4=((P1/P2)^x)*T3;//Temperature at point 4 in K
+W=Cp*(T3-T4);//Work developed per kg of air in kJ/kg
+Re=Cp*(T1-T4);//Refrigerating effect per kg of air in kJ/kg
+Wnet=Cp*((T2-T1)-(T3-T4));//Net work output in kJ/kg
+COP=Re/Wnet;//Co-efficient of performance
+
+//OUTPUT
+mprintf('Work developed per kg of air is %3.3f kJ/kg \n Refrigerating effect per kg of air is %3.3f kJ/kg \n COP of the cycle is %3.2f',W,Re,COP)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.13/Chapter6_Example13.sce b/1775/CH6/EX6.13/Chapter6_Example13.sce new file mode 100755 index 000000000..15cf6908e --- /dev/null +++ b/1775/CH6/EX6.13/Chapter6_Example13.sce @@ -0,0 +1,36 @@ +//Chapter-6, Illustration 13, Page 319
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=277;//Temperature at point 1 in K
+T3=328;//Temperature at point 3 in K
+P1=0.1;//Pressure at point 1 in MPa
+P2=0.3;//Pressure at point 2 in MPa
+nc=0.72;//Isentropic efficiency of compressor
+nt=0.78;//Isentropic efficiency of turbine
+y=1.4;//Adiabatic gas constant
+Cp=1.005;//Specific heat at constant pressure in kJ/kg-K
+m=3;//Cooling load in tonnes
+
+//CALCULATIONS
+x=(y-1)/y;//Ratio
+T2s=T1*((P2/P1)^x);//Temperature at point 2s in K
+T2=((T2s-T1)/nc)+T1;//Temerature at point 2 in K
+T4s=T3*((P1/P2)^x);//Temperature at point 4s in K
+T4=T3-((T3-T4s)*nt);//Temperature at point 4 in K
+Re=Cp*(T1-T4);//Refrigerating effect in kJ/kg
+Wnet=Cp*((T2-T1)-(T3-T4));//Net work output in kJ/kg
+COP=Re/Wnet;//Co-efficient of performance
+P=(m*3.52)/COP;//Driving power required in kW
+ma=(m*3.52)/Re;//Mass flow rate of air in kg/s
+
+//OUTPUT
+mprintf('COP of refrigerator is %3.2f \n Driving power required is %3.0f kW \n Mass flow rate of air is %3.3f kg/s',COP,P,ma)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.14/Chapter6_Example14.sce b/1775/CH6/EX6.14/Chapter6_Example14.sce new file mode 100755 index 000000000..f2fa9f579 --- /dev/null +++ b/1775/CH6/EX6.14/Chapter6_Example14.sce @@ -0,0 +1,35 @@ +//Chapter-6, Illustration 14, Page 321
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=2.5;//Pressure at point 1 in bar
+P3=9;//Pressure at point 3 in bar
+COPr=0.65;//Ratio of actual COP to the theoretical COP
+m=5;//Refrigerant flow in kg/min
+T1=309;//Temperature at point 1 in K
+T2s=300;//Temperature at point 2s in K
+h1=570.3;//Enthalpy at P1 from the given tables in kJ/kg
+h4=456.4;//Enthalpy at P3 from the given tables in kJ/kg
+h2g=585.3;//Enthalpy at P3 from the given tables in kJ/kg
+s2=4.76;//Entropy at P1 from the given tables in kJ/kg-K
+s2g=4.74;//Entropy at P3 from the given tables in kJ/kg-K
+Cp=0.67;//Specific heat at P3 in kJ/kg-K
+
+//CALCULATIONS
+T2=(2.718^((s2-s2g)/Cp))*T2s;//Temperature at point 2 in K
+h2=h2g+(Cp*(T2-T2s));//Enthalpy at point 2 in kJ/kg
+COPR=(h1-h4)/(h2-h1);//Refrigerant COP
+COPact=COPr*COPR;//Actual COP
+qL=COPact*(h2-h1);//Heat rejected in kJ/kg
+QL=((m*qL*60)/3600)/3.516;//Cooling produced per kg of refrigerant in tonnes of refrigeration
+
+//OUTPUT
+mprintf('Theoretical COP is %3.2f \n Net cooling produced per hour is %3.2f TR',COPR,QL)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.15/Chapter6_Example15.sce b/1775/CH6/EX6.15/Chapter6_Example15.sce new file mode 100755 index 000000000..761db224d --- /dev/null +++ b/1775/CH6/EX6.15/Chapter6_Example15.sce @@ -0,0 +1,31 @@ +//Chapter-6, Illustration 15, Page 322
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=298;//Temperature at point 2 in K
+T1=268;//Temperature at point 1 in K
+hf1=-7.54;//Liquid Enthalpy at T1 in kJ/kg
+x1=0.6;//Quality factor 1
+hfg1=245.3;//Latent heat at T1 in kJ/kg
+sf1=0.251;//Liquid Entropy at T1 in kJ/kg-K
+s1=0.507;//Entropy at point 1 in kJ/kg-K
+hfg2=121.4;//Latent heat at T2 in kJ/kg
+hf2=81.3;//Liquid Enthalpy at T2 in kJ/kg
+h4=hf2;//Enthalpy at point 4 in kJ/kg
+
+//CALCULATIONS
+h1=hf1+(x1*hfg1);//Enthalpy at point 1 in kJ/kg
+x2=((s1-sf1)*T2)/hfg2;//Quality factor 2
+h2=hf2+(x2*hfg2);//Enthalpy at point 2 in kJ/kg
+COP=(h1-h4)/(h2-h1);//COP of the machine
+
+//OUTPUT
+mprintf('COP of the machine is %3.2f',COP)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.16/Chapter6_Example16.sce b/1775/CH6/EX6.16/Chapter6_Example16.sce new file mode 100755 index 000000000..57c2150e4 --- /dev/null +++ b/1775/CH6/EX6.16/Chapter6_Example16.sce @@ -0,0 +1,31 @@ +//Chapter-6, Illustration 16, Page 323
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=25;//Pressure at point 1 in bar
+P2=60;//Pressure at point 2 in bar
+h2=208.1;//Vapour enthalpy at P2 in kJ/kg
+h3=61.9;//Liquid enthalpy at P2 in kJ/kg
+h4=h3;//Liquid enthalpy at P2 in kJ/kg
+s2=0.703;//Vapour entropy at P2 in kJ/kg-K
+sf1=-0.075;//Liquid entropy at P1 in kJ/kg-K
+sfg1=0.971;//Entropy in kJ/kg-K
+hf1=-18.4;//Liquid Enthalpy at P1 in kJ/kg
+hfg1=252.9;//Latent heat at P1 in kJ/kg
+m=5;//Refrigerant flow in kg/min
+
+//CALCULATIONS
+x1=(s2-sf1)/sfg1;//Quality factor 1
+h1=hf1+(x1*hfg1);//Enthalpy at point 1 in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+QL=(m*(h1-h4))/60;//Capacity of the refrigerator in kW
+
+//OUTPUT
+mprintf('COP of refrigerator is %3.2f \n Capacity of refrigerator is %3.2f kW',COP,QL)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.17/Chapter6_Example17.sce b/1775/CH6/EX6.17/Chapter6_Example17.sce new file mode 100755 index 000000000..1699e40ee --- /dev/null +++ b/1775/CH6/EX6.17/Chapter6_Example17.sce @@ -0,0 +1,45 @@ +//Chapter-6, Illustration 17, Page 324
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=271;//Temperature at point 1 in K
+T=265;//Temperature at point 1' in K
+Ta=303;//Temperature at point 2' in K
+Cpv=0.733;//Specific heat of vapour in kJ/kg
+Cpl=1.235;//Specific heat of liquid in kJ/kg
+h=184.07;//Liquid enthalpy at T in kJ/kg
+s=0.7;//Entropy at point 1' in kJ/kg-K
+sa=0.685;//Vapour entropy at Ta in kJ/kg-K
+ha=199.62;//Enthalpy at point 2' in kJ/kg
+hfb=64.59;//Liquid enthalpy at Ta in kJ/kg
+DT3=5;//Temperature difference in oC
+Q=2532;//Refrigeration capacity in kJ/min
+
+//CALCULATIONS
+s2=s+(Cpv*((log(T1/T))/(log(2.718))));//Entropy at point 1 in kJ/kg-K
+h1=h+(Cpv*(T1-T));//Enthalpy at point 1 in kJ/kg-K
+T2=(2.718^((s2-sa)/Cpv))*Ta;//Temperature at point 2 in K
+h2=ha+(Cpv*(T2-Ta));//Enthalpy at point 2 in kJ/kg
+h4=hfb-(Cpl*DT3);//Enthalpy at point 4 in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+m=Q/(h1-h4);//Mass flow rate of refrigerant in kJ/min
+P=(m*(h2-h1))/(60*12);//Power required in kW/TR
+
+//OUTPUT
+mprintf('COP is %3.2f \n Theoretical power required per tonne of refrigeration is %3.3f kW/TR',COP,P)
+
+
+
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.2/Chapter6_Example2.sce b/1775/CH6/EX6.2/Chapter6_Example2.sce new file mode 100755 index 000000000..014616dfc --- /dev/null +++ b/1775/CH6/EX6.2/Chapter6_Example2.sce @@ -0,0 +1,30 @@ +//Chapter-6, Illustration 2, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+TL=268;//Low temperature in K
+TH=293;//High temperature in K
+t=24;//time in hrs
+C=2100;//Capacity of refrigerator in kJ/s
+Tw=10;//Water temperature in oC
+L=335;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+COP=TL/(TH-TL);//Co-efficient of performance
+Pmin=C/COP;//Minimum power required in kW
+Qr=(4.187*(Tw-0))+L;//Heat removed from water in kJ/kg
+m=C/Qr;//mass of ice formed in kg/s
+W=(m*t*3600)/1000;//Weight of ice formed in tons
+
+//OUTPUT
+mprintf('Minimum power required is %3.2f kW \n Weight of ice formed in 24 hours is %3.2f tons',Pmin,W)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.3/Chapter6_Example3.sce b/1775/CH6/EX6.3/Chapter6_Example3.sce new file mode 100755 index 000000000..ee4bcdf7b --- /dev/null +++ b/1775/CH6/EX6.3/Chapter6_Example3.sce @@ -0,0 +1,25 @@ +//Chapter-6, Illustration 3, Page 309
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+TL=-10;//Temperature of brine in oC
+TH=20;//Temperature of water in oC
+L=335;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+Qr=(4.187*(TH-0))+L;//Heat removed from water in kJ/kg
+COP=(TL+273)/(TH-TL);//Co-efficient of performance
+mi=(COP*3600)/Qr;//mass of ice formed per kWh in kg
+
+//OUTPUT
+mprintf('Mass of ice formed per kWh is %3.1f kg',mi)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.4/Chapter6_Example4.sce b/1775/CH6/EX6.4/Chapter6_Example4.sce new file mode 100755 index 000000000..645c9248a --- /dev/null +++ b/1775/CH6/EX6.4/Chapter6_Example4.sce @@ -0,0 +1,30 @@ +//Chapter-6, Illustration 4, Page 310
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P1=1.2;//Pressure at point 1 in bar
+P2=7;//Pressure at point 2 in bar
+m=0.05;//mass flow rate of refrigerant in kg/s
+h1=340.1;//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+s1=1.57135;//Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
+s2=1.57135;//Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
+h2=372;//Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
+h3=226.575;//Enthalpy at point 3 from refrigerant-12 tables in kJ/kg
+h4=226.575;//Enthalpy at point 4 from refrigerant-12 tables in kJ/kg
+
+//CALCULATIONS
+Q2=m*(h1-h4);//Rate of heat removed from the refrigerated space in kW
+W=m*(h2-h1);//Power input to the compressor in kW
+Q1=m*(h2-h3);//Rate of heat rejection to the environment in kW
+COP=Q2/W;//Co-efficient of performance
+
+//OUTPUT
+mprintf('Rate of heat removed from the refrigerated space is %3.2f kW \n Power input to the compressor is %3.3f kW \n Rate of heat rejection to the environment is %3.2f kW \n Co-efficient of performance is %3.2f',Q2,W,Q1,COP)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.5/Chapter6_Example5.sce b/1775/CH6/EX6.5/Chapter6_Example5.sce new file mode 100755 index 000000000..abfbbad8e --- /dev/null +++ b/1775/CH6/EX6.5/Chapter6_Example5.sce @@ -0,0 +1,30 @@ +//Chapter-6, Illustration 5, Page 311
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=40;//Temperature at point 2 in oC
+T1=-10;//Temperature at point 1 in oC
+h2=367.155;//Enthalpy at point 2 from refrigerant-12 tables in kJ/kg
+s2=1.54057;//Entropy at point 2 from refrigerant-12 tables in kJ/kg-K
+s1=1.54057;//Entropy at point 1 from refrigerant-12 tables in kJ/kg-K
+sg=1.56004;//Entropy from refrigerant-12 tables in kJ/kg-K
+sf=0.96601;//Entropy from refrigerant-12 tables in kJ/kg-K
+hf=190.822;//Enthalpy from refrigerant-12 tables in kJ/kg-K
+hfg=156.319;//Enthalpy from refrigerant-12 tables in kJ/kg-K
+h3=238.533;//Enthalpy at point 3 from refrigerant-12 tables in kJ/kg-K
+h4=h3;//Enthalpy at point 4 from refrigerant-12 tables in kJ/kg-K
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+
+//OUTPUT
+mprintf('COP of the system is %3.2f',COP)
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.6/Chapter6_Example6.sce b/1775/CH6/EX6.6/Chapter6_Example6.sce new file mode 100755 index 000000000..03638b1bb --- /dev/null +++ b/1775/CH6/EX6.6/Chapter6_Example6.sce @@ -0,0 +1,41 @@ +//Chapter-6, Illustration 6, Page 311
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tc=35;//Temperature of condenser in oC
+Te=-15;//Temperature of evaporator in oC
+m=10;//Mass of ice per day in tons
+Tw=30;//Temperature of water in oC
+Ti=-5;//Temperature of ice in oC
+nv=0.65;//Volumetric efficiency
+N=1200;//Speed in rpm
+x=1.2;//Stroke to bore ratio
+na=0.85;//Adiabatic efficiency
+nm=0.95;//Mechanical efficiency
+S=4.187;//Specific heat of water in kJ/kg
+L=335;//Latent heat of ice in kJ/kg
+h1=1667.24;//Enthalpy at Te from Ammonia chart in kJ/kg
+h2=1925;//Enthalpy at Te from Ammonia chart in kJ/kg
+h4=586.41;//Enthalpy at Tc from Ammonia chart in kJ/kg
+v1=0.508;//Specific humidity at Te from Ammonia chart in (m^3)/kg
+
+//CALCULATIONS
+Qr=(((m*1000)/24)*((S*(Tw-0))+L+(1.94*(0-Ti))))/3600;//Refrigerating capacity in kW
+mr=Qr/(h1-h4);//Refrigerant mass flow rate in kg/s
+T2=112;//Discharge temperature in oC
+D=((mr*v1*4*60)/(nv*3.14*x*N))^(1/3);//Cylinder diameter in m
+L=x*D;//Stroke length in m
+W=(mr*(h2-h1))/(na*nm);//Compressor motor power in kW
+COPth=(h1-h4)/(h2-h1);//Theoretical COP
+COPact=Qr/W;//Actual COP
+
+//OUTPUT
+mprintf('Refrigerating capacity of plant is %3.2f kW \n Refrigerant mass flow rate is %3.4f kg/s \n Discharge temperature is %3.0f oC \n Cylinder diameter is %3.3f m \n Stroke length is %3.3f m \n Compressor motor power is %3.2f kW \n Theoretical COP is %3.2f \n Actual COP is %3.2f',Qr,mr,T2,D,L,W,COPth,COPact)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.7/Chapter6_Example7.sce b/1775/CH6/EX6.7/Chapter6_Example7.sce new file mode 100755 index 000000000..561ecb6bc --- /dev/null +++ b/1775/CH6/EX6.7/Chapter6_Example7.sce @@ -0,0 +1,40 @@ +//Chapter-6, Illustration 7, Page 313
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T1=-5;//Temperature at point 1 in oC
+T2=30;//Temperature at point 2 in oC
+m=13500;//mass of ice per day in kg
+Tw=20;//Temperature of water in oC
+COP=0.6;//Co-efficient of performance
+h2=1709.33;//Enthalpy at point 2 in kJ/kg
+s2=6.16259;//Entropy at point 2 in kJ/kg-K
+s1=6.16259;//Entropy at point 1 in kJ/kg-K
+sf=1.8182;//Entropy in kJ/kg-K
+sg=6.58542;//Entropy in kJ/kg-K
+hf=400.98;//Enthalpy in kJ/kg
+hfg=1278.35;//Enthalpy in kJ/kg
+h4=562.75;//Enthalpy at point 4 in kJ/kg
+S=4.187;//Specific heat of water in kJ/kg
+L=336;//Latent heat of ice in kJ/kg
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COPi=(h1-h4)/(h2-h1);//Ideal COP
+COPact=COP*COPi;//Actual COP
+Qr=((m*S*(Tw-0))+(m*L))/(24*3600);//Total amount of heat removed in kJ/s
+mr=Qr/(h1-h4);//Circulation rate of ammonia in kg/s
+W=mr*(h2-h1);//Power required in kW
+
+//OUTPUT
+mprintf('Circulation rate of ammonia is %3.3f kg/s \n Power required is %3.3f kW \n COP is %3.3f',mr,W,COPact)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.8/Chapter6_Example8.sce b/1775/CH6/EX6.8/Chapter6_Example8.sce new file mode 100755 index 000000000..d65de8ae1 --- /dev/null +++ b/1775/CH6/EX6.8/Chapter6_Example8.sce @@ -0,0 +1,40 @@ +//Chapter-6, Illustration 8, Page 314
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+Tc=20;//Temperature of condenser in oC
+Te=-25;//Temperature of evaporator in oC
+m=15;//Mass of ice per day in tons
+Ts=5;//Subcooled temperature in oC
+Tsh=10;//Superheated temperature in oC
+n=6;//No. of cylinders
+N=950;//Speed of compressor in rpm
+x=1;//Stroke to bore ratio
+h1=402;//Enthalpy at point 1 from R-22 tables in kJ/kg
+h2=442;//Enthalpy at point 2 from R-22 tables in kJ/kg
+h3=216;//Enthalpy at point 3 from R-22 tables in kJ/kg
+h4=216;//Enthalpy at point 4 from R-22 tables in kJ/kg
+v1=2.258;//Specific volume at point 1 in (m^3)/min
+
+//CALCULATIONS
+Re=h1-h4;//Refrigerating effect in kJ/kg
+mr=(m*14000)/(Re*60);//Mass flow of refrigerant in kg/min
+Pth=(mr*(h2-h1))/60;//Theoretical power in kW
+COP=(h1-h4)/(h2-h1);//Co-efficient of performance
+Dth=v1/n;//Theoretical displacement per cylinder
+D=(((Dth*4)/(3.147*N))^(1/3))*1000;//Theoretical bore of compressor in mm
+L=D;//Theoretical stroke of compressor in mm
+
+//OUTPUT
+mprintf('Refrigerating effect is %3.0f kJ/kg \n Mass flow of refrigerant per minute is %3.2f kg/min \n Theoretical input power is %3.2f kW \n COP is %3.2f \n Theoretical bore of compressor is %3.2f mm \n Theoretical stroke of compressor is %3.2f mm',Re,mr,Pth,COP,D,L)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH6/EX6.9/Chapter6_Example9.sce b/1775/CH6/EX6.9/Chapter6_Example9.sce new file mode 100755 index 000000000..2e12a43a1 --- /dev/null +++ b/1775/CH6/EX6.9/Chapter6_Example9.sce @@ -0,0 +1,35 @@ +//Chapter-6, Illustration 9, Page 316
+//Title: Refrigeration cycles
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+T2=40;//Temperature at point 2 in oC
+T1=-5;//Temperature at point 1 in oC
+h2=367.155;//Enthalpy at point 2 from F-12 tables in kJ/kg
+sg=1.55717;//Entropy from F-12 tables in kJ/kg-K
+s1=1.54057;//Entropy at point 1 from F-12 tables in kJ/kg-K
+sf=0.98311;//Entropy from F-12 tables in kJ/kg-K
+hf=195.394;//Enthalpy from F-12 tables in kJ/kg
+hfg=153.934;//Enthalpy from F-12 tables in kJ/kg
+h4=238.533;//Enthalpy at point 4 from F-12 tables in kJ/kg
+h4s=218;//Enthalpy at point 4 with subcooling from F-12 tables in kJ/kg
+
+//CALCULATIONS
+x1=(s1-sf)/(sg-sf);//Quality factor
+h1=hf+(x1*hfg);//Enthalpy at point 1 from refrigerant-12 tables in kJ/kg
+COPns=(h1-h4)/(h2-h1);//Co-efficient of performance with no subcooling
+COPs=(h1-h4s)/(h2-h1);//Co-efficient of performance with subcooling
+
+//OUTPUT
+mprintf('COP with no subcooling is %3.3f \n COP with subcooling is %3.3f',COPns,COPs)
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH7/EX7.1/Chapter7_Example1.sce b/1775/CH7/EX7.1/Chapter7_Example1.sce new file mode 100755 index 000000000..600e4435a --- /dev/null +++ b/1775/CH7/EX7.1/Chapter7_Example1.sce @@ -0,0 +1,30 @@ +//Chapter-7, Illustration 1, Page 345
+//Title: Air Conditioning
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+DBTo=10;//Out door Dry bulb temperature in oC
+WBTo=8;//Out door Wet bulb temperature in oC
+DBTi=20;//In door Dry bulb temperature in oC
+RH=0.6;//Re-Heat factor
+a=0.3;//amount of air circulated in (m^3)/min/person
+S=50;//Seating capacity of office
+BPF=0.32;//ByPass factor
+ha=25;//Enthalpy at point a from Psychrometric chart shown in Page 346 in kJ/kg
+hb=42.5;//Enthalpy at point b from Psychrometric chart shown in Page 346 in kJ/kg
+hc=42.5;//Enthalpy at point c from Psychrometric chart shown in Page 346 in kJ/kg
+Wa=0.006;//Specific humidity at point a from Psychrometric chart shown in Page 346 in kg/kg dry air
+Wc=0.009;//Specific humidity at point c from Psychrometric chart shown in Page 346 in kg/kg dry air
+Tb=27;//Temperature at point b in oC
+na=0.81;//Specific Volume from Psychrometric chart shown in page 346 in (m^3)/kg
+
+//CALCULATIONS
+ma=(a*S)/(na*60);//mass of air circulated per second in kg/s
+Hc=ma*(hb-ha);//Heating capacity of coil in kW
+Ts=(Tb-(BPF*DBTo))/(1-BPF);//Heating coil surface temperature in oC
+C=(ma*3600)*(Wc-Wa);//Capacity of humidifier in kg/hr
+
+//OUTPUT
+mprintf('Heating capacity of coil is %3.2f kW \n Surface temperature of coil is %3.0f oC \n Capacity of humidifier is %3.2f kg/hr',Hc,Ts,C)
diff --git a/1775/CH7/EX7.2/Chapter7_Example2.sce b/1775/CH7/EX7.2/Chapter7_Example2.sce new file mode 100755 index 000000000..2bb671d49 --- /dev/null +++ b/1775/CH7/EX7.2/Chapter7_Example2.sce @@ -0,0 +1,31 @@ +//Chapter-7, Illustration 2, Page 346
+//Title: Air Conditioning
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+S=60;//No. of staff
+DBTo=30;//Out door Dry bulb temperature in oC
+RHo=0.7;//Re-Heat factor at out-door
+a=0.4;//amount of air circulated in (m^3)/min/person
+DBTi=20;//In door Dry bulb temperature in oC
+RHi=0.6;//Re-Heat factor at indoor
+Td=25;//Heating coil surface temperature in oC
+ha=82.5;//Enthalpy at point a from Psychrometric chart shown in Page 347 in kJ/kg
+hb=34.5;//Enthalpy at point b from Psychrometric chart shown in Page 347 in kJ/kg
+hc=42.5;//Enthalpy at point c from Psychrometric chart shown in Page 347 in kJ/kg
+Wa=0.020;//Specific humidity at point a from Psychrometric chart shown in Page 347 in kg/kg dry air
+Wb=0.009;//Specific humidity at point b from Psychrometric chart shown in Page 347 in kg/kg dry air
+Tb=12;//Temperature at point b in oC
+na=0.89;//Specific Volume from Psychrometric chart shown in page 346 in (m^3)/kg
+
+//CALCULATIONS
+ma=(a*S)/(na*60);//mass of air circulated per second in kg/s
+Hc=(ma*(ha-hb))/3.5;//Heating capacity of cooling coil in tonnes
+Hh=ma*(hc-hb);//Heating capacity of heating coil in kW
+W=(ma*3600)*(Wa-Wb);//Amount of water vapour removed per hour in kg/hr
+BPF=(Td-DBTi)/(Td-Tb);//By-Pass factor
+
+//OUTPUT
+mprintf('Capacity of cooling coil is %3.2f tonnes \n Capacity of heating coil is %3.1f kW \n Amount of water vapour removed per hour is %3.2f kg/hr \n Bypass factor is %3.3f',Hc,Hh,W,BPF)
diff --git a/1775/CH7/EX7.3/Chapter7_Example3.sce b/1775/CH7/EX7.3/Chapter7_Example3.sce new file mode 100755 index 000000000..c5437a38f --- /dev/null +++ b/1775/CH7/EX7.3/Chapter7_Example3.sce @@ -0,0 +1,41 @@ +//Chapter-7, Illustration 3, Page 347
+//Title: Air Conditioning
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+RSH=10;//Room sensible heat in kW
+RLH=10;//Room latent heat in kW
+td1=25;//Inside temperature in oC
+RH1=0.5;//Inside Re-Heat factor
+h1=50.4;//Enthalpy at point 1 in kJ/kg
+td2=35;//Out door Dry bulb temperature in oC
+tw2=28;//Out door Wet bulb temperature in oC
+CR=4;//Cooling coil ratio
+BPF=0.1;//Cooling coil bypass factor
+tADP=10;//Apparatus dew point temperature in oC
+RH3=0.55;//Re-Heat factor at point 3
+h3=58.2;//Enthalpy at point 3 in kJ/kg
+RH4=0.95;//Re-Heat factor at point 4
+h4=32.2;//Enthalpy at point 4 in kJ/kg
+RH5=0.81;//Re-Heat factor at point 5
+h5=36.8;//Enthalpy at point 5 in kJ/kg
+RH6=0.54;//Re-Heat factor at point 6
+h6=43.1;//Enthalpy at point 5 in kJ/kg
+td6=22;//Temperature at point 6 in oC
+
+//CALCULATIONS
+td3=((td2-td1)/5)+td1;//Temperature at point 3 from Psychrometric chart shown in Page 348 in oC
+td4=(BPF*(td3-tADP))+tADP;//Temperature at point 4 from Psychrometric chart shown in Page 348 in oC
+td5=td4+((td1-td4)/5);//Temperature at point 5 from Psychrometric chart shown in Page 348 in oC
+RSHF=RSH/(RSH+RLH);//Room Sensible Heat Factor
+QR=h1-h6;//Total heat removed in kJ/kg
+S=(RSH+RLH)/QR;//Supply air quantity in kg/s
+R=(S*(h6-h5))/3.5;//Refrigeration load due to reheat in ton
+D=(S*4)/5;//Dehumidified air quantity in kg/s
+T=(D*(h3-h4))/3.5;//Total refrigerating capacity in ton
+Q=(D/5)/1.2;//Quantity of fresh air supplied in (m^3)/s
+
+//OUTPUT
+mprintf('Supply air condition to the room is %3.2f kg/s \n Refrigeration load due to reheat is %3.2f ton \n Total refrigerating capacity is %3.2f ton \n Quantity of fresh air supplied is %3.3f (m^3)/s',S,R,T,Q)
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