diff options
Diffstat (limited to '3821')
120 files changed, 3096 insertions, 0 deletions
diff --git a/3821/CH10/EX10.1/Example10_1.sce b/3821/CH10/EX10.1/Example10_1.sce new file mode 100644 index 000000000..0ed5768dc --- /dev/null +++ b/3821/CH10/EX10.1/Example10_1.sce @@ -0,0 +1,14 @@ +/////Chapter 10 Properties Of Steam
+////Example 10.1 Page No:183
+///Find Dryness fuction of steam
+///Input data
+clc;
+clear;
+mw=15; //Water steam
+ms=185; //Dry steam
+
+///Calculation
+x=((ms)/(ms+mw))*100; //Dryness fuction of steam in %
+
+///Output
+printf('Dryness fuction of steam= %f percent \n',x);
diff --git a/3821/CH10/EX10.10/Example10_10.sce b/3821/CH10/EX10.10/Example10_10.sce new file mode 100644 index 000000000..4a884b32f --- /dev/null +++ b/3821/CH10/EX10.10/Example10_10.sce @@ -0,0 +1,45 @@ +/////////Chapter 10 Properties Of Steam
+/////Example 10.10 Page No:191
+////Find Enthalpy of superheated steam
+//Input data
+clc;
+clear;
+P1=10*10^5 //Pressure of steam in bar
+tsup1=300+273; //Temperature of steam n degree celsius
+P2=1.4*10^5; //Internal energy of steam
+x2=0.8; //Dryness fraction
+Cps=2.3;
+///from steam table properties of saturated steam (temp basis)
+///at 25 degree celsius and at 10 bar(pressure basis)
+ts1=179.9+273;
+vf=0.001127; //In m^3/Kg
+vg=0.1943; //In m^3/Kg
+hf=762.6; //In KJ/Kg
+hfg=2013.6; //In KJ/Kg
+hg1=2776.2; //In KJ/Kg
+//at 1.4 bar;
+ts=109.3; //In degree celsius
+vf1=0.001051; //In m^3/Kg
+vg1=1.2363; //In m^3/Kg
+hf1=458.4; //In KJ/Kg
+hfg1=2231.9; //In KJ/Kg
+hg=2690.3; //In KJ/Kg
+
+///Calculation
+h1=hg1+Cps*(tsup1-ts1); //Enthalpy of superheated steam in KJ/Kg
+v1=vg*(tsup1/ts1); //Volume of superheated steam in m**3/Kg
+u1=h1-((P1*v1)/10^3); //Internal energy in KJ/Kg
+h2=hf1+x2*hfg1; //Enthalpy of wet steam in KJ/Kg
+Vwet=(1-x2)*vf1+x2*vg1; //Volume of wet steam in m**3/Kg
+u2=h2-((P2*Vwet)/10^3); //Internal energy in KJ/Kg
+DeltaU=u1-u2; //Change of Internal energy in KJ/Kg
+
+
+//Output
+printf('Enthalpy of superheated steam= %f KJ/Kg \n ',h1);
+printf('Volume of superheated steam=%f m^3/kg \n ',v1);
+printf('Internal energy= %f KJ/Kg \n ',u1);
+printf('Enthalpy of wet steam= %f KJ/Kg \n ',h2);
+printf('Volume of wet steam=%f m^3/kg \n ',Vwet);
+printf('Internal energy= %f KJ/Kg \n',u2);
+printf('Change of Internal energy= %f KJ/Kg \n ',DeltaU);
diff --git a/3821/CH10/EX10.11/Example10_11.sce b/3821/CH10/EX10.11/Example10_11.sce new file mode 100644 index 000000000..044a68f56 --- /dev/null +++ b/3821/CH10/EX10.11/Example10_11.sce @@ -0,0 +1,25 @@ +/////////Chapter 10 Properties Of Steam
+//////Example 10.11 Page No:193
+/////Find Entropy of wet steam
+///Input data
+clc;
+clear;
+P=15; //Absolute pressure
+//From steam table (pressure basis at 15 bar)
+ts=198.3+273; //In degree celsius
+Sf=2.3145; //In KJ/KgK
+Sfg=4.1261; //In KJ/KgK
+Sg=6.4406; //In KJ/KgK
+tsup=300+273;
+Cps=2.3;
+x=0.8;
+
+////calculation
+S=Sf+x*Sfg; //Entropy of wet steam in KJ/Kg
+S1=Sg; //Entropy of superheated steam in KJ/Kg
+S2=Sg+Cps*(log(tsup/ts)); //Entropy of superheated steam in KJ/Kg
+
+///Output
+printf('Entropy of wet steam %f KJ/Kg \n' ,S);
+printf('Entropy of dry and saturated steam %f KJ/Kg \n ' ,S1);
+printf('Entropy of superheated steam %f KJ/Kg \n' ,S2);
diff --git a/3821/CH10/EX10.12/Example10_12.sce b/3821/CH10/EX10.12/Example10_12.sce new file mode 100644 index 000000000..90c37c39b --- /dev/null +++ b/3821/CH10/EX10.12/Example10_12.sce @@ -0,0 +1,77 @@ +/////////Chapter 10 Properties Of Steam
+/////Example 10.12 Page No:194
+///Entropy of 1.5Kg of superheated steam
+
+//Input data
+clc;
+clear;
+m=1.5; //Entropy of the steam
+P=10*10^5; //Absolute pressure in bar
+//From steam table properties of saturated steam
+///(pressure basis)at 10 bar
+ts=179.9+273; //Indegree celsius
+vf=0.001127; //In m**3/Kg
+vg=0.1943; //In m**3/Kg
+hf=762.6; ///In KJ/Kg
+hfg=2013.6; //In KJ/Kg
+hg=2776.2; //In KJ/Kg
+Sf=2.1382; //In KJ/KgK
+Sfg=4.4446; //In KJ/KgK
+Sg=6.5828; //In KJ/Kg
+Cps=2.3;
+tsup=250+273;
+
+
+///Calculation
+//(1)Enthalpy of dry and saturated steam
+
+h=hg; //Enthalpy of dry and saturated steam
+EODS=hg*m; //Enthalpy of 1.5Kg of dry and saturated steam
+v=vg; //volume of dry and saturated steam
+u=h-((P*v)/10^3); //Internal Energy
+IES=u*m; //Internal energy of the steam
+s=6.5858; //Entropy of dry and saturated steam
+EODSS=s*m; //Entropy of 1.5Kg dry and saturated steam
+x=0.75;
+//(2)Enthalpy of wet steam
+h1=hf+x*hfg; //Enthalpy of wet steam
+EWS=h1*m; //Enthalpy of1.5Kg of wet steam
+Vwet=x*vg; //Volume of steam
+u1=h1-((P*Vwet)/10^3); //Internal energy
+IES1=u1*m; //Internal energy of1.5Kg of the steam
+s1=Sf+x*Sfg; //Entropy of wet steam
+EWS1=s1*m; //Entropy of1.5Kg of wet steam
+
+///(3)Enthalpy of superheated steam
+h2=hg+Cps*(tsup-ts); //Enthalpy of superheated steam
+EOSHS=h2*m; //Enthalpy of 1.5Kg of superheated steam
+Vsup=vg*(tsup/ts); //Volume of superheated steam
+u2=h2-((P*Vsup)/10^3); //Internal energy
+IES2=u2*m; //Internal energy of 1.5Kg of the steam
+s2=Sg+Cps*(log(tsup/ts));//Entropy of superheated steam
+EOSHS1=s2*m; //Entropy of 1.5Kg of superheated steam
+
+///Output
+printf('Enthalpy of dry and saturated steam= %f KJ/Kg \n ',h);
+printf('Enthalpy of 1.5Kg of dry and saturated steam= %f KJ \n ',EODS);
+printf('volume of dry and saturated steam= %f m^3/Kg \n ',v);
+printf('Internal Energy= %f KJ/Kg \n ',u);
+printf('Internal energy of the steam= %f KJ \n ',IES);
+printf('Entropy of dry and saturated steam = %f KJ/KgK \n ',s);
+printf('Entropy of 1.5kg of dry and saturated steam= %f KJ/K \n ',EODSS);
+
+printf('Enthalpy of wet steam= %f KJ/Kg \n ',h1);
+printf('Enthalpy of1.5Kg of wet steam= %f KJ \n ',EWS);
+printf('Volume of steam= %f m^3/Kg \n',Vwet);
+printf('Internal energy= %f KJ/Kg \n ',u1);
+printf('Internal energy of1.5Kg of the steam= %f KJ \n ',IES1);
+printf('Entropy of wet steam= %f KJ/KgK \n ',s1);
+printf('Entropy of 1.5Kg of wet steam= %f KJ/K \n ',EWS1);
+
+printf('Enthalpy of superheated steam= %f KJ/Kg \n ',h2);
+printf('Enthalpy of 1.5Kg of superheated steam= %f KJ \n ',EOSHS);
+printf('Volume of superheated steam= %f m^3/Kg \n',Vsup);
+printf('Internal energy= %f \n ',u2);
+printf('Internal energy of1.5Kg of the steam= %f KJ \n ',IES2);
+printf('Entropy of superheated steam= %f KJ/KgK \n ',s2);
+printf('Entropy of 1.5Kg of superheated steam= %f KJ/K \n ',EOSHS1);
diff --git a/3821/CH10/EX10.13/Example10_13.sce b/3821/CH10/EX10.13/Example10_13.sce new file mode 100644 index 000000000..be68ec89f --- /dev/null +++ b/3821/CH10/EX10.13/Example10_13.sce @@ -0,0 +1,42 @@ +/////////Chapter 10 Properties Of Steam
+////Example 10.13 Page No:196
+////Find Volume occupied by water
+///Input data
+clc;
+clear;
+V=0.04; //Volume of vessel in m^3
+x=1;
+t=250+273; //Saturated steam temp in degree celsius
+mw=9; //Mass of liquid in Kg
+//From steam table(temp basis,at t=250)
+P=39.78*10^5; //in bar
+Vf=0.001251; //In m^3/kg
+Vg=0.05004; //In m^3/Kg
+hf=1085.7; //KJ/Kg
+hfg=2800.4; //KJ/Kg
+hg=1714.7; //KJ/Kg
+
+//Calculation
+Vw=mw*Vf; //Volume occupied by water in m^3
+Vs=V-Vw; //Volume of waterin m^3
+ms=Vs/Vg; //Volume of dry and saturated steam in Kg
+m=mw+ms; //Total mass of steam in Kg
+x=ms/(ms+mw); //Dryness fraction of steam
+Vwet=(1-x)*Vf+x*Vg; //Specific volume of steam in m^3/Kg
+h=hf+x*hfg; //Enthalpy of wet steam in KJ/Kg
+EOWS=h*m; //Enthalpy of 9.574 Kg of wet steam KJ
+u=h-((P*Vwet)/10^3); //Internal Energy in KJ/Kg
+IEOS=u*m; //Internal energy of 9.574 Kg of steam in KJ
+
+
+///Output
+printf('Volume occupied by water=%f m^3 \n ',Vw);
+printf('Volume of water=%f m^3 \n ',Vs);
+printf('Volume of dry and saturated steam=%f Kg \n',ms);
+printf('Total mass of steam= %f Kg \n ',m);
+printf('Dryness fraction of steam= %f \n',x);
+printf('Specific volume of steam=%f m^3/Kg \n ',Vwet);
+printf('Enthalpy of wet steam=%f KJ/Kg \n ',h);
+printf('Enthalpy of 9.574 Kg of wet steam=%f KJ \n ',EOWS);
+printf('Internal Energy= %f KJ/Kg \n',u);
+printf('Internal energy of 9.574 Kg of steam=%f KJ \n ',IEOS);
diff --git a/3821/CH10/EX10.14/Example10_14.sce b/3821/CH10/EX10.14/Example10_14.sce new file mode 100644 index 000000000..ceeb753ac --- /dev/null +++ b/3821/CH10/EX10.14/Example10_14.sce @@ -0,0 +1,18 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.14 Page No:197
+//// Find Degree of superheat
+///Input Data
+clc;
+clear;
+P=7; //Absolute pressure in bar
+t=200; //Absolute temperature
+ts=165; //In degree celsius from steam table
+
+//Calculation
+dos1=t-ts; //Degree of superheat in degree celcius
+
+//Output
+printf('Degree of superheat=%f degree celsius \n ',dos1);
+
+
+
diff --git a/3821/CH10/EX10.15/Example10_15.sce b/3821/CH10/EX10.15/Example10_15.sce new file mode 100644 index 000000000..eaf537449 --- /dev/null +++ b/3821/CH10/EX10.15/Example10_15.sce @@ -0,0 +1,20 @@ +/////////Chapter 10 Properties Of Steam
+///Example 15 Page No:197
+///Find Enthalpy of wet steam
+///Input data
+clc;
+clear;
+P=15; ///Absolute pressure in bar
+///From steam table (pressure basis at 15 bar)
+h=1950; //In KJ/Kg
+ts=198.3; //In degreee celsius
+hf=844.7; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+
+///Calculation
+x=((h-hf)/hfg); ///Enthalpy of wet steam
+
+///Output
+printf('Enthalpy of wet steam=%f \n ',x);
+
diff --git a/3821/CH10/EX10.16/Example10_16.sce b/3821/CH10/EX10.16/Example10_16.sce new file mode 100644 index 000000000..5e7c8163c --- /dev/null +++ b/3821/CH10/EX10.16/Example10_16.sce @@ -0,0 +1,23 @@ +/////Chapter 10 Properties Of Steam
+////Example 10.16 Page No:197
+///Find Enthalpy of superheated steam
+//Input data
+clc;
+clear;
+P=15; //Absolute pressure in bar
+//From steam table (pressure basis at 15 bar)
+h=3250; //In KJ/Kg
+ts=198.3; //In degree celsius
+hf=844.7; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+Cps=2.3;
+
+//Calculation
+tsup=(h-hg+(Cps*ts))/2.3, //Enthalpy of superheated steam in degree celsius
+dos1=tsup-ts; //Degree of superheated in degree celsius
+ //The value of ts in not used according to data in book instead of ts=198.3 author used ts=165
+
+//Output
+printf('Enthalpy of superheated steam= %f degree celcius\n ',tsup);
+printf('Degree of superheated=%f degree celcius \n ',dos1);
diff --git a/3821/CH10/EX10.17/Example10_17.sce b/3821/CH10/EX10.17/Example10_17.sce new file mode 100644 index 000000000..4aebc4922 --- /dev/null +++ b/3821/CH10/EX10.17/Example10_17.sce @@ -0,0 +1,18 @@ +///Chapter 10 Properties Of Steam
+///Example 10.17 Page No:198
+///Find Volume of steam dryness fraction
+//Input data
+clc;
+clear;
+P=7; //Absolute pressure in bar
+v=0.2; //Specific volume in m^3/Kg
+//from steam table (pressure basis at 7 bar)
+ts=165; //In degree celsius
+vf=0.001108; //In m^3/Kg
+vg=0.2727; //In m^3/Kg
+
+//Calculation
+x=v/vg; //Volume of steam dryness fraction
+
+//Output
+printf('Volume of steam dryness fraction= %f \n',x);
diff --git a/3821/CH10/EX10.18/Example10_18.sce b/3821/CH10/EX10.18/Example10_18.sce new file mode 100644 index 000000000..1a59ca032 --- /dev/null +++ b/3821/CH10/EX10.18/Example10_18.sce @@ -0,0 +1,21 @@ +////////Chapter 10 Properties Of Steam
+////Example 10.18 Page No:198
+///Find Temp of superheated steam
+///Input data
+clc;
+clear;
+P=7; //Absolute pressure in bar
+v=0.3; //Specific volume in m^3/Kg
+//From steam table (pressure basis at 7 bar)
+ts=165+273; //In degree celsius
+vf=0.001108; //In m^3/Kg
+vg=0.2727; //In m^3/Kg
+
+///Calculation
+//v=vg*tsup/ts;
+tsup=((v/vg)*ts)-273; //Temp of superheated steam in degree celsius
+DOS=tsup+273-ts; //Degree of superheated in degree celsius
+
+//Output
+printf('Temp of superheated steam=%f degree celsius \n ',tsup);
+printf('Degree of superheated=%fdegree celsius \n ',DOS);
diff --git a/3821/CH10/EX10.19/Example10_19.sce b/3821/CH10/EX10.19/Example10_19.sce new file mode 100644 index 000000000..2fb5cc318 --- /dev/null +++ b/3821/CH10/EX10.19/Example10_19.sce @@ -0,0 +1,15 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.19 Page No:198
+///Find Quality of steam
+///Input data
+clc;
+clear;
+m=2; //steam of vessel in Kg
+V=0.1598; //volume of vessel in M**3
+P=25; //Absolute pressure of vessel in bar
+
+//Calculation
+v=V/m; //Quality of steam in m**3/Kg
+
+//Output
+printf('Quality of steam %f m^3/Kg \n' ,v);
diff --git a/3821/CH10/EX10.2/Example10_2.sce b/3821/CH10/EX10.2/Example10_2.sce new file mode 100644 index 000000000..259ecfd06 --- /dev/null +++ b/3821/CH10/EX10.2/Example10_2.sce @@ -0,0 +1,11 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.2 Page No:183
+/// Find saturation pressure of the steam
+//Input data
+clc;
+clear;
+sps=150; //saturation pressure of the steam in degree celsius
+
+//Output
+P=4.76; //From steam table
+printf('saturation pressure= %f bar \n',P);
diff --git a/3821/CH10/EX10.20/Example10_20.sce b/3821/CH10/EX10.20/Example10_20.sce new file mode 100644 index 000000000..92459b8ad --- /dev/null +++ b/3821/CH10/EX10.20/Example10_20.sce @@ -0,0 +1,36 @@ +/////////Chapter 10 Properties Of Steam
+////Example 10.20 Page No:200
+/// Find Initial enthalpy of steam
+//Input data
+clc;
+clear;
+P=10*10^2; //Absolute pressure in bar
+x1=0.9; //Dryness enters
+tsup2=300+273; //Temperature in degree celsius
+//From steam table at 10 bar
+ts=179.9+273; //In degree celsius
+Vg=0.1943; //In m^3/Kg
+hf=762.6; //In KJ/Kg
+hfg=2013.6; //InK/Kg
+hg=2776.2; //In KJ/Kg
+Cps=2.3;
+
+//Calculation
+h1=hf+x1*hfg; //Initial enthalpy of steam in KJ/Kg
+V1=x1*Vg; //Initial specific volume of steam
+u1=h1-P*V1; //Initial internal energy of steam in KJ/Kg
+h2=hg+Cps*(tsup2-ts); //Final enthalpy of steam in KJ/Kg
+V2=Vg*(tsup2/ts); //Final specific volume of steam in m**3/Kg
+u2=h2-P*V2; //Final internal energy of steam in KJ/K
+deltah=h2-h1; //Heat gained by steam in KJ/Kg
+deltaU=(u2-u1); //Change in internal energy in KJ/Kg
+
+//Output
+printf('Initial enthalpy of steam=%f KJ/Kg \n',h1);
+printf('Initial specific volume of steam=%f \n ',V1);
+printf('Initial internal energy of steam=%f KJ/Kg \n',u1);
+printf('Final enthalpy of steam= %f KJ/Kg \n ',h2);
+printf('Final specific volume of steam= %f m^3/kg \n',V2);
+printf('Final internal energy of steam=%f Kj/Kg \n ',u2);
+printf('Heat gained by steam= %f KJ/Kg \n ',deltah);
+printf('Change in internal energy=%f KJ/Kg \n ',deltaU);
diff --git a/3821/CH10/EX10.21/Example10_21.sce b/3821/CH10/EX10.21/Example10_21.sce new file mode 100644 index 000000000..53df24b39 --- /dev/null +++ b/3821/CH10/EX10.21/Example10_21.sce @@ -0,0 +1,29 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.21 Page No:201
+/// Find Final enthalpy of steam
+//Input data
+clc;
+clear;
+m=4; //Steam in Kg
+P=13; //Absolute pressure in bar
+tsup1=450; //Absolute temp in degree celsius
+deltaH=2.8*10^3;
+Cps=2.3; //loses in MJ
+//from steam table at 13 bar
+ts=191.6; //In degree celsius
+Vg=0.1511; //In m^3/Kg
+hf=814.7; //In m^3/Kg
+hfg=1970.7; //In KJ/Kg
+hg=2785.4; //In KJ/Kg
+
+///Calculation
+h1=hg+Cps*(tsup1-ts); //Initial enthalpy of steam in KJ/Kg
+Deltah=deltaH/m; //Change in enthalpy/unit mass in KJ/Kg
+h2=h1-Deltah; //Final enthalpy of steam in KJ/Kg
+x2=(h2-hf)/hfg; //wet & dryness fraction
+
+//Output
+printf('Initial enthalpy of steam=%f Kj/Kg \n ',h1);
+printf('Change in enthalpy/unit mass=%f Kj/Kg \n ',Deltah);
+printf('Final enthalpy of steam= %f KJ/Kg \n',h2);
+printf('wet & dryness fraction=%f \n',x2);
diff --git a/3821/CH10/EX10.22/Example10_22.sce b/3821/CH10/EX10.22/Example10_22.sce new file mode 100644 index 000000000..340ee56ae --- /dev/null +++ b/3821/CH10/EX10.22/Example10_22.sce @@ -0,0 +1,38 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.22 Page No:201
+//Find Initial specific volume of steam
+//Input data
+clc;
+clear;
+m=2; //Steam in Kg
+x=0.7; //Initial dryness
+P=15; //Constant pressure in bar
+//V2=2V1
+//from steam table properties of
+//saturated steam(pressure basis) at 15 bar
+Ts=198.3+273; //In degree celsius
+Vg=0.1317; //In m^3/Kg
+hf=844.7; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+Cps=2.3;
+
+///Calculation
+V1=x*Vg; //Initial specific volume of steam in m**3/Kg
+V2=2*V1; //Final specific volume of steam in m**3/Kg
+Tsup=(V2/Vg)*Ts; //Steam is superheated in degree celsius
+FSS=Tsup-Ts; //Degree of superheated in degree celsius
+h1=hf+x*hfg; //Initial enthalpy of steam in KJ/Kg
+h2=hg+Cps*(Tsup-Ts); //Final enthalpy of steam in KJ/Kg
+Q=(h2-h1)*m; //Heat transferred in the process in KJ
+W1=P*(m*V2-m*V1); //Work transferred in the process in KJ
+
+//Output
+printf('Initial specific volume of steam=%f m^3/kg \n',V1);
+printf('Final specific volume of steam= %f m^3/kg \n',V2);
+printf('Steam is superheated= %f K \n ',Tsup);
+printf('Degree of superheated=%f degree celsius \n ',FSS);
+printf('Initial enthalpy of steam=%f KJ/Kg \n ',h1);
+printf('Final enthalpy of steam=%f KJ/Kg \n ',h2);
+printf('Heat transferred in the process=%f KJ \n ',Q);
+printf('Work transferred in the process= %f KJ \n',W1);
diff --git a/3821/CH10/EX10.23/Example10_23.sce b/3821/CH10/EX10.23/Example10_23.sce new file mode 100644 index 000000000..d0335c351 --- /dev/null +++ b/3821/CH10/EX10.23/Example10_23.sce @@ -0,0 +1,31 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.23 Page No:203
+///Find Constant pressure process
+//Input data
+clc;
+clear;
+ms=1000; //Steam in Kg/h
+P=16; //Absolute pressure in bar
+x2=0.9; //Steam is dry
+t1=30+273; //temperature in degree celsius
+tsup=380; //tmperature rised in degree celsius
+
+//from steam table(pressure basis at 16 bar)
+h1=125.7; //in KJ/Kg
+ts=201.4; //In degree celsius
+hf=858.5; //in kJ/Kg
+hfg=1933.2; //in kJ/Kg
+hg=2791.7; //in kJ/Kg
+Cps=2.3;
+
+//Calculation
+h2=hf+x2*hfg; //Final enthalpy of wet steam in KJ/Kg
+Q1=(ms*(h2-h1))*(10^(-3)); //Constant pressure process in KJ/h
+h3=hg+Cps*(tsup-ts); //Final enthalpy of superheated steam in KJ/g
+Q2=(ms*(h3-h2))*(10^(-3)); //Suprheated steam in KJ/h
+
+//Output
+printf('Final enthalpy of wet steam= %f KJ/Kg \n ',h2);
+printf('Constant pressure process= %f KJ/h \n',Q1);
+printf('Final enthalpy of superheated steam= %f KJ/g \n',h3);
+printf('Suprheated steam= %f KJ/h \n',Q2);
diff --git a/3821/CH10/EX10.24/Example10_24.sce b/3821/CH10/EX10.24/Example10_24.sce new file mode 100644 index 000000000..2c53ee79a --- /dev/null +++ b/3821/CH10/EX10.24/Example10_24.sce @@ -0,0 +1,28 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.24 Page No:204
+///Find Enthalpy of steam of first boiler
+clc;
+clear;
+//Input data;
+FB=15; //First boiler in bar
+SB=15; //Second boiler in bar
+tsup1=300; //Temperature of the steam in degree celsius
+tsup2=200; //Temperature of the steam in degree celsius
+//From steam table (pressure basis at 15 bar )
+ts=198.3; //In degree celsius
+hf=844.7; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/I
+Cps=2.3;
+
+//Calculation
+h1=hg+Cps*(tsup1-ts); //Enthalpy of steam of first boiler in KJ/Kg
+h3=hg+Cps*(tsup2-ts); //Enthalpy of steam in steam main in KJ/Kg
+h2=2*h3-h1; //Energy balance in KJ/Kg
+x2=(h2-hf)/hfg; //Enthalpy of wet steam
+
+//OUTPUT
+printf('Enthalpy of steam of first boiler= %f KJ/Kg\n',h1);
+printf('Enthalpy of steam in steam main=%f KJ/Kg \n ',h3);
+printf('Energy balance=%f KJ/Kg \n ',h2);
+printf('Enthalpy of wet steam= %f \n ',x2);
diff --git a/3821/CH10/EX10.25/Example10_25.sce b/3821/CH10/EX10.25/Example10_25.sce new file mode 100644 index 000000000..c78ea5cb6 --- /dev/null +++ b/3821/CH10/EX10.25/Example10_25.sce @@ -0,0 +1,45 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.25 Page No:205
+// Find Initial specific volume of steam
+clc;
+clear;
+///Input data
+V=0.35; //Capacity of vessel in m^3
+P1=10*10^2; //Absolute pressure in bar
+tsup1=250+273; //Absolute temperature in degree celsius
+P2=2.5*102; //Absolute pressure in the vessel fall in bar
+
+//From steam table (pressure basis at 10 bar)
+ts1=179.9+273; //In degree celsius
+Vg1=0.1943; //In m^3/Kg
+hf1=762.6; //In KJ/Kg
+hfg1=2013.6; //In KJ/Kg
+hg1=2776.2; //In KJ/Kg
+
+//From steam table(pressure basis at 2.5 bar)
+V2=0.2247; //In m^3/Kg
+ts2=127.4; //In degree celsius
+Vg2=0.7184; //In m^3/Kg
+hf2=535.3; //In KJ/Kg
+hfg2=2181.0; //In KJ/Kg
+hg2=2716.4; //In KJ/Kg
+Cps=2.3;
+///Calculation
+V1=Vg1*(tsup1/ts1); //Initial specific volume of steam in m^3/Kg
+m=V/V1; //Initial mass of steam in Kg
+x2=V2/Vg2; //Final condition of wet steam
+h1=hg1+Cps*(tsup1-ts1); //Initial enthalpy of steam in KJ/Kg
+u1=h1-P1*V1; //Initial internal energy of steam in KJ/Kg
+h2=hf2+x2*hfg2; //Final enthalpy of steam in KJ/Kg
+u2=h2-P2*V2; //Final internal energy of steam in KJ/Kg
+deltaU=(u2-u1)*m; //Change in internal energy in KJ
+
+//Output
+printf('Initial specific volume of steam=%f m^3/Kg \n ',V1);
+printf('Initial mass of steam=%fKg \n ',m);
+printf('Final condition of wet steam= %f \n ',x2);
+printf('Initial enthalpy of steam=%f KJ/Kg \n ',h1);
+printf('Initial internal energy of steam= %f KJ/Kg \n',u1);
+printf('Final enthalpy of steam=%f KJ/Kg \n ',h2);
+printf('Final internal energy of steam=%f KJ/Kg \n',u2);
+printf('Change in internal energy= %f KJ/Kg \n',deltaU);
diff --git a/3821/CH10/EX10.26/Example10_26.sce b/3821/CH10/EX10.26/Example10_26.sce new file mode 100644 index 000000000..02474da0f --- /dev/null +++ b/3821/CH10/EX10.26/Example10_26.sce @@ -0,0 +1,34 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.26 Page No:207
+//Find Constant volume process
+clc;
+clear;
+//Input data
+m=1.5; //Saturated steam in Kg
+x1=1;
+x2=0.6;
+P1=5*10^5; //Absolute pressure in bar
+//From steam table at pressure basis 5 bar
+hg1=2747.5; //In KJ/Kg
+Vg1=0.3747; //In m^3/Kg
+V1=0.3747; //In m^3/Kg
+V2=0.3747; //In m^3/Kg
+//From steam table at Vg2 is 2.9 bar
+P2=2.9*10^5; //Absolute pressure in bar
+t2=132.4; //In degree celsius
+hf2=556.5; //In KJ/Kg
+hfg2=2166.6; //In KJ/Kg
+
+
+
+//Calculation
+Vg2=V2/x2; //Constant volume process in m^3/Kg
+u1=hg1-((P1*Vg1)/1000); //Initial internal energy in KJ/Kg
+u2=(hf2+x2*hfg2)-((P2*V2)/1000); //Final internal energy in KJ
+deltaU=(u1-u2)*m; //Heat supplied in KJ
+
+//Output
+printf('Constant volume process=%f m^3/Kg \n ',Vg2);
+printf('Initial internal energy=%f KJ/Kg \n ',u1);
+printf('Final internal energy= %f KJ \n',u2);
+printf('Heat supplied=%f KJ \n ',deltaU);
diff --git a/3821/CH10/EX10.27/Example10_27.sce b/3821/CH10/EX10.27/Example10_27.sce new file mode 100644 index 000000000..53d28f86f --- /dev/null +++ b/3821/CH10/EX10.27/Example10_27.sce @@ -0,0 +1,34 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.27 Page No:208
+//Find Enthalpy of steam
+clc;
+clear;
+//Input data
+P1=20; //Initial steam in bar
+x1=0.95; //dryness throttled
+P2=1.2; //Absolute pressure in bar
+
+//From steam table (pressure basis at 20 bar)
+ts=212.4; //In degree celsius
+hf=908.6; //In KJ/Kg
+hfg=1888.6; //In KJ/Kg
+hg=2797.2; //In KJ/Kg
+//From steam table (pressure basis at 1.2 bar)
+//h2=h1; //In KJ/Kg
+ts2=104.8; //In degree celsius
+hf2=439.3; //In KJ/Kg
+hfg2=2244.1; //In KJ/Kg
+hg2=2683.4; //In KJ/Kg
+Cps=2.3;
+
+
+//Calculation
+h1=hf+x1*hfg; //Enthalpy of steam in KJ/Kg
+tsup2=((h1-hg2)/Cps)+ts2; //Enthalpy of wet steam in degree celsius
+DOS=tsup2-ts2; //Degree of superheat in degree celsius
+
+
+//Output
+printf('Enthalpy of steam=%f KJ/Kg \n ',h1);
+printf('Enthalpy of wet steam=%f degree celsius \n ',tsup2);
+printf('Degree of superheat=%f degree celsius \n',DOS);
diff --git a/3821/CH10/EX10.28/Example10_28.sce b/3821/CH10/EX10.28/Example10_28.sce new file mode 100644 index 000000000..d8eedce54 --- /dev/null +++ b/3821/CH10/EX10.28/Example10_28.sce @@ -0,0 +1,23 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.28 Page No:209
+///Enthalpy after throttling
+//Input data
+clc;
+clear;
+P1=12; //Throttled steam
+x1=0.96; //Dryness is brottled
+x2=1; //Constant enthalpy process
+//From steam table at12 bar
+ts=188; //In degree celsius
+hf=798.4; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hg=2782.7; //In KJ/Kg
+
+
+//Calculation
+h1=hf+x1*hfg; //Enthalpy of the steam in KJ/Kg
+h2=h1; //Enthalpy after throttling in KJ/Kg
+
+///Output
+printf('Enthalpy of the steam=%f KJ/Kg \n ',h1);
+printf('Enthalpy after throttlin= %f KJ/Kg \n',h2);
diff --git a/3821/CH10/EX10.29/Example10_29.sce b/3821/CH10/EX10.29/Example10_29.sce new file mode 100644 index 000000000..3947f6ae8 --- /dev/null +++ b/3821/CH10/EX10.29/Example10_29.sce @@ -0,0 +1,40 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.29 Page No:210
+///Find Entropy of superheated steam
+//Input data
+clc;
+clear;
+P1=15; //Initial steam in bar
+tsup1=250+273; //Temperature of steam in degree celsius
+P2=0.5; //Steam turbine in bar
+
+//From steam table at 15 bar
+ts1=198.3+273; //In degree celsius
+hg1=2789.9; //In KJ/Kg
+sf1=2.3145; //In KJ/KgK
+sfg1=4.1261; //In KJ/KgK
+sg1=6.4406; //In KJ/KgK
+//From steam table at 0.5 bar
+ts2=81.53; //In degree celsius
+sf2=1.0912; //In KJ/Kg
+sfg2=6.5035; //In KJ/Kg
+sg2=7.5947; //In KJ/Kg
+hf2=340.6;
+Cps=2.3;
+hfg2=2646;
+
+//Calculation
+S1=sg1+Cps*(log(tsup1/ts1)); //Entropy of superheated steam in KJ/KgK
+S2=S1 //Entropy after isentropic processes in KJ/KgK
+x2=(S2-sf2)/sfg2; //Enthalpy of wet steam
+h1=hg1+Cps*(tsup1-ts1); //Enthalpy of steam at 15 bar
+h2=hf2+x2*hfg2; //Enthalpy of wet steam at 0.5 bar
+WOT=h1-h2; //Work output of the turbine
+
+///OUTPUT
+printf('Entropy of superheated steam= %f KJ/KgK \n ',S1);
+printf('Entropy after isentropic processes=%f KJ/KgK \n' ,S2);
+printf('Enthalpy of wet steam= %f \n',x2);
+printf('Enthalpy of steam= %f KJ/Kg',h1);
+printf('Enthalpy of wet steam= %f KJ/Kg \n',h2);
+printf('Work output of the turbine=%f KJ/Kg \n ',WOT);
diff --git a/3821/CH10/EX10.3/Example10_3.sce b/3821/CH10/EX10.3/Example10_3.sce new file mode 100644 index 000000000..e5bde3601 --- /dev/null +++ b/3821/CH10/EX10.3/Example10_3.sce @@ -0,0 +1,19 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.3 Page No:184
+//Find Saturation temperature of bar
+///Input data
+clc;
+clear;
+P1=28; ///Absolute pressure in bar
+P2=5.5; //Absolute pressure in MPa
+P3=77; ///Absolute pressure in mm of Hg
+
+///Calcutation
+ts1=230.05; //Saturation temperature in degree celsius
+ts2=269.93; //Saturation temperature in degree celsius
+ts3=45.83; //Saturation temperature in degree celsius
+
+///Output
+printf('Saturation temperature= %f degree celsius \n ',ts1);
+printf('Saturation temperature= %f degree celsius \n ',ts2);
+printf('Saturation temperature= %f degree celsius \n',ts3);
diff --git a/3821/CH10/EX10.4/Example10_4.sce b/3821/CH10/EX10.4/Example10_4.sce new file mode 100644 index 000000000..8a01ed8f8 --- /dev/null +++ b/3821/CH10/EX10.4/Example10_4.sce @@ -0,0 +1,27 @@ +//////Chapter 10 Properties Of Steam
+////Example 10.4 Page No:185
+///
+////#Input data
+clc;
+clear;
+P=15; //Absolute pressure in bar
+//From steam table (pressure basis at 15 bar)
+ts=198.3; //In degree celsius
+hf=844.7; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+tsup=300; //In degree celsius
+x=0.8;
+Cps=2.3;
+hg=2789.9;
+
+//Calculation
+h1=hf+x*hfg; //Enthalpy of wet steam in KJ/KG
+h=hg; //Enthalpy of dry and saturated steam in KJ/KG
+h2=hg+Cps*(tsup-ts); //Enthalpy of superheated steam in KJ/KG
+
+
+//Output
+printf('Enthalpy of wet steam=%f KJ/Kg \n ",h1);
+printf('Enthalpy of dry and saturated steam=%f KJ/Kg \n ',h);
+printf('Enthalpy of superheated steam= %f KJ/Kg \n ',h2);
diff --git a/3821/CH10/EX10.5/Example10_5.sce b/3821/CH10/EX10.5/Example10_5.sce new file mode 100644 index 000000000..256444ad8 --- /dev/null +++ b/3821/CH10/EX10.5/Example10_5.sce @@ -0,0 +1,27 @@ +/////////Chapter 10 Properties Of Steam
+///Example 7.5 Page No:186
+///Find Final Enthalpy of the steam
+//Input data
+clc;
+clear;
+ti=30; //Temperature in degree celsius
+m=2; //Water in Kg
+pf=8; //Steam at 8 bar
+x=0.9; //Water to dry
+tb=30;
+///From steam table at 30 degree celsius
+hf=125.7;
+///h1=hf initial enthalpy of water
+///From steam table at 8 bar
+ts=170.4; //In degree celsius
+hf1=720.9; //In KJ/KG
+hfg=2046.6; //In KJ/KG
+hg=2767.5; //In KJ/KG
+
+///Calculation
+h=hf1+(x*hfg); //Final Enthalpy of the steam in KJ/Kg
+Qha=m*(h-hf); //Quantity of the heat in KJ/Kg ///Calculation mistake m is not multiplied by (h-hf) in book
+
+///Output
+printf('Final Enthalpy of the steam=%f KJ/Kg \n ',h);
+printf('Quantity of the heat=%f KJ/Kg \n',Qha);
diff --git a/3821/CH10/EX10.6/Example10_6.sce b/3821/CH10/EX10.6/Example10_6.sce new file mode 100644 index 000000000..ac2e1a25d --- /dev/null +++ b/3821/CH10/EX10.6/Example10_6.sce @@ -0,0 +1,27 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.6 Page No:186
+///Find Enthalpy of superheated steam
+///Input data
+clc;
+clear;
+IT=25; //Initial temperature
+m=5; //Heat required to generate steam in kg
+pf=10; //Final pressure in bar
+tsup=250; //Water temperature
+///From steam table (temp basis)at 25 degree celsius
+//and at 10 bar(pressure basis)
+hf=104.8; //In KJ/KG
+h1=104.8; //In KJ/KG
+ts=179.9; //In degree celsius
+hf1=792.6; //In KJ/KG
+hfg=2013.6; //In KJ/KG
+hg=2776.2; //In KJ/KG
+Cps=2.1;
+
+///Calculation
+h=hg+Cps*(tsup-ts); //Enthalpy of superheated steam in KJ/Kg
+H=m*(h-h1); //Quantity of heat added in KJ/Kg
+
+///Output
+printf('Enthalpy of superheated steam=%f KJ/Kg \n ",h);
+printf('Quantity of heat added= %f KJ/Kg \n ",H);
diff --git a/3821/CH10/EX10.7/Example10_7.sce b/3821/CH10/EX10.7/Example10_7.sce new file mode 100644 index 000000000..787aea388 --- /dev/null +++ b/3821/CH10/EX10.7/Example10_7.sce @@ -0,0 +1,25 @@ +/////////Chapter 10 Properties Of Steam
+////Example 10.7 Page No:188
+///Find Volume of wet steam
+///#Input data
+clc;
+clear;
+P=15; //Absolute pressure in bar
+//From steam table (pressure basis at 15 bar)
+ts=198.3+273; //In degree celsius
+vg=0.1317; //In m^3/Kg
+vf=0.001154; //In m^3/Kg
+x=0.8;
+Tsup=300+273; //Degree celsius
+
+
+//Calculation
+v=(1-x)*vf+x*vg; //Volume of wet steam in m**3/Kg
+vg=0.1317; //Dry and saturated steam in m**3/Kg
+vsup=vg*(Tsup/ts); //Volume of superheated steam m**3/Kg
+
+
+///Output
+printf('Volume of wet steam= %f m^3/Kg \n ',v);
+printf('Dry and Saturated Steam= %f m^3/Kg \n ',vg);
+printf('volume of superheated steam= %f m^3/kg \n ',vsup);
diff --git a/3821/CH10/EX10.8/Example10_8.sce b/3821/CH10/EX10.8/Example10_8.sce new file mode 100644 index 000000000..f46731430 --- /dev/null +++ b/3821/CH10/EX10.8/Example10_8.sce @@ -0,0 +1,19 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.8 Page No:188
+////Find Mass of steam
+///Input data
+clc;
+clear;
+P=25; //Absolute pressure
+ts=223.9; //Volume
+//Frome steam table (pressure basis at 25 bar)
+vf=0.001197; //In m^3/Kg
+vg=0.0799; //In m^3/Kg
+v=8; //In m^3/Kg
+
+
+///Calculation
+m=v/vg; //Mass of steam in Kg
+
+///Output
+printf('Mass of steam=%f Kg \n ',m);
diff --git a/3821/CH10/EX10.9/Example10_9.sce b/3821/CH10/EX10.9/Example10_9.sce new file mode 100644 index 000000000..f9e2de84d --- /dev/null +++ b/3821/CH10/EX10.9/Example10_9.sce @@ -0,0 +1,41 @@ +/////////Chapter 10 Properties Of Steam
+///Example 10.9 Page No:190
+////Find Volume of wet steam
+
+//Input data
+clc;
+clear;
+P=12*10^5; //Absolute pressure
+//From steam table (pressure basis at 12 bar)
+ts=188+273; //In degree celsius
+vf=0.001139; //In m**3/Kg
+vg=0.1632; //In m**3/Kg
+hf=798.4; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hg=2782.7; //In KJ/Kg
+x=0.94;
+Cps=2.3;
+tsup=350+273; //In degree celsius
+
+//Calcuation
+h=hf+x*hfg; //Enthalpy of wet steam in KJ/Kg
+v=(1-x)*vf+x*vg; //Volume of wet steam m**3/Kg
+u=h-((P*v)/10^3); //Internal Energy in KJ/Kg
+hg=2782.7; //Enthalpy of dry & saturated steam in KJ/Kg
+v1=vg; //Volume of dry & saturated steam m**3/Kg
+u1=hg-((P*vg)/10^3); //Internal Energy in KJ/Kg
+h1=hg+Cps*(tsup-ts); //Enthalpy of superheated steam in KJ/Kg
+vsup=vg*(tsup/ts); //Volume of superheated steam in m**3/Kg
+u2=h1-((P*v)/10**3); //Internal Energy in KJ/Kg
+
+
+///Output
+printf('Enthalpy of wet steam=%f KJ/Kg \n ',h);
+printf('Volume of wet steam= %f m^3/Kg \n ',v);
+printf('Internal Energy= %f KJ/Kg \n ',u);
+printf('Enthalpy of dry & saturated steam=%f KJ/Kg \n',hg);
+printf('Volume of dry & saturated steam=%f m^3/Kg \n ',v1);
+printf('Internal Energy= %f KJ/Kg \n',u1);
+printf('Enthalpy of superheated steam=%f KJ/Kg \n ',h1);
+printf('Volume of superheated steam= %f m^3/kg \n',vsup);
+printf('Internal Energy= %f KJ/Kg \n ',u2);
diff --git a/3821/CH11/EX11.1/Example11_1.sce b/3821/CH11/EX11.1/Example11_1.sce new file mode 100644 index 000000000..ab90694a0 --- /dev/null +++ b/3821/CH11/EX11.1/Example11_1.sce @@ -0,0 +1,25 @@ +///Chapter No 11 Steam Boilers
+////Example 11.1 Page No 228
+///Find Mass of evaporation
+//Input data
+clc;
+clear;
+ms=5000; //Boiler produces wet steam in Kg/h
+x=0.95; //Dryness function
+P=10; //Operating pressure in bar
+mf=5500; //Bour in the furnace in Kg
+Tw=40; //Feed water temp in degree celsius
+
+//Calculation
+//from steam table
+hfw=167.45; //In KJ/Kg
+hf=762.61; //In KJ/Kg
+hfg=2031.6; //In KJ/Kg
+hs=(hf+x*hfg); //Enthalpy of wet stream in KJ/Kg
+me=ms/mf; //Mass of evaporation
+E=((me*(hs-hfw))/(2257))*10; //Equivalent evaporation in Kg/Kg of coal
+
+//Output
+printf('Enthalpy of wet stream=%f KJ/Kg \n',hs);
+printf('Mass of evaporation=%f KJ/Kg \n',me);
+printf('Equivalent evaporation = %f Kg/Kg of coal \n',E);
diff --git a/3821/CH11/EX11.10/Example11_10.sce b/3821/CH11/EX11.10/Example11_10.sce new file mode 100644 index 000000000..96836ec28 --- /dev/null +++ b/3821/CH11/EX11.10/Example11_10.sce @@ -0,0 +1,17 @@ +///Chapter No 11 Steam Boilers
+////Example 11.10 Page No 242
+//Find Draught produce in terms of water
+//Input data
+clc;
+clear;
+ma=18; //Boileruses of per Kg of fuel in Kg/Kg
+hw=25*10^-3; //Chimney height to produce draught in mm
+Tg=315+273; //Temperature of chimney gases in degree celsius
+Ta=27+273; //Out side air temp in degree celsius
+
+//Calculation
+//Draught produce in terms of water column in m
+H=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*1000;
+
+//Output
+printf('Draught produce in terms of water column=%f m \n',H);
diff --git a/3821/CH11/EX11.11/Example11_11.sce b/3821/CH11/EX11.11/Example11_11.sce new file mode 100644 index 000000000..21b76bd64 --- /dev/null +++ b/3821/CH11/EX11.11/Example11_11.sce @@ -0,0 +1,19 @@ +///Chapter No 11 Steam Boilers
+////Example 11.11 Page No 242
+///Find Draught produce in terms of hot gas
+//Input data
+clc;
+clear;
+H=40; //High discharge in m
+ma=19; //Fuel gases per Kg of fuel burnt
+Tg=220+273; //Average temp of fuel gases in degree celsius
+Ta=25+273; //Ambient temperature in degreee celsius
+
+
+//Calculation
+hw=353*H*(1/Ta-1/Tg*((ma+1)/ma)); //Draught produce in terms of water column in mm
+H1=H*((Tg/Ta)*(ma/(ma+1))-1); //Draught produce in terms of hot gas column in m
+
+//Output
+printf('Draught produce in terms of water column=%f mm \n',hw);
+printf('Draught produce in terms of hot gas column=%f m \n',H1);
diff --git a/3821/CH11/EX11.12/Example11_12.sce b/3821/CH11/EX11.12/Example11_12.sce new file mode 100644 index 000000000..8d7c7fec7 --- /dev/null +++ b/3821/CH11/EX11.12/Example11_12.sce @@ -0,0 +1,18 @@ +///Chapter No 11 Steam Boilers
+////Example 11.12 Page No 243
+///Find Mean temperature of fuel gases
+//Input data
+clc;
+clear;
+H=27; //Chimney height in m
+hw=15; //Draught produces of water column in mm
+ma=21; //Gases formed per Kg of fuel burnt in Kg/Kg
+Ta=25+273; //Temperature of the ambient air in degree celsius
+
+
+//Calculation
+Tg=-(((ma+1)/ma)/((hw/(353*H))-(1/Ta))) //Mean temperature of fuel gases in K
+
+//Output
+printf('Mean temperature of fuel gases= %f K \n',Tg);
+
diff --git a/3821/CH11/EX11.13/Example11_13.sce b/3821/CH11/EX11.13/Example11_13.sce new file mode 100644 index 000000000..ad9cb6e29 --- /dev/null +++ b/3821/CH11/EX11.13/Example11_13.sce @@ -0,0 +1,16 @@ +///Chapter No 11 Steam Boilers
+////Example 11.13 Page No 244
+//Find Air-fuel ratio
+//Input data
+clc;
+clear;
+hw=20; //Static draught of water in mm
+H=50; //Chimney height in m
+Tg=212+273; //Temperature of the fuel degree celsius
+Ta=27+273; //Atmospheric air in degree celsius
+
+//Calculation
+ma=(-((hw/(353*H))-Ta*Tg))*10^-4 //Air-fuel ratio in Kg/Kg of fuel burnt-3
+
+//Output
+printf('Air-fuel ratio= %f Kg/Kg of fuel burnt \n',ma);
diff --git a/3821/CH11/EX11.14/Example11_14.sce b/3821/CH11/EX11.14/Example11_14.sce new file mode 100644 index 000000000..39afce846 --- /dev/null +++ b/3821/CH11/EX11.14/Example11_14.sce @@ -0,0 +1,24 @@ +///Chapter No 11 Steam Boilers
+////Example 11.14 Page No 245
+///Find Theoretical draught in millimeters of water
+//Input data
+clc;
+clear;
+H=24; //Chimney height in m
+Ta=25+273; //Ambient temperature in degree celsius
+Tg=300+273; //Temperature of fuel gases in degree celsius
+ma=20; //Combustion space of fuel burnt in Kg/Kgof fuel
+g=9.81;
+
+
+//Calculation
+hw=((353*H)*((1/Ta)-((1/Tg)*((ma+1)/ma))));//Theoretical draught in millimeters of water in mm
+H1=H*((Tg/Ta)*(ma/(ma+1))-1); //Theoretical draught produced in hot gas column in m
+H2=H1-9.975; //Draught lost in friction at the grate and passage in m
+V=round(sqrt(2*g*H2)); //Actual draught produced in hot gas column in m
+
+///Output
+printf('Theoretical draught in millimeters of water= %f mm \n',hw);
+printf('Theoretical draught produced in hot gas column=%f m \n',H1);
+printf('Draught lost in friction at the grate and passage=%f m \n',H2);
+printf('Actual draught produced in hot gas column= %f m \n ',V);
diff --git a/3821/CH11/EX11.15/Example11_15.sce b/3821/CH11/EX11.15/Example11_15.sce new file mode 100644 index 000000000..0d153e836 --- /dev/null +++ b/3821/CH11/EX11.15/Example11_15.sce @@ -0,0 +1,32 @@ +///Chapter No 11 Steam Boilers
+////Example 11.15 Page No 246
+////Find Draught lost in friction at the grate and pasage
+//Input data
+clc;
+clear;
+H=38; //Stack height in m
+d=1.8; //Stack diameter discharge in m
+ma=17; //Fuel gases per Kg of fuel burnt Kg/Kg
+Tg=277+273; //Average temperature of fuel gases in degree celsius
+Ta=27+273; //Temperature of outside air in degree celsius
+h1=0.4; //Theoretical draught is lost in friction in
+g=9.81;
+pi=3.142;
+
+//Calculation
+H1=H*(((Tg/Ta)*(ma/(ma+1))-1)); //Theoretical draught produce in hot gas column in m
+gp=0.45*27.8; //Draught lost in friction at the grate and pasage in m
+C=H1-gp; //Actual draught produce in hot gas column in m
+V=sqrt(2*9.81*C); //Velocity of the flue gases in the chimney in m/s
+rhog=((353*(ma+1))/(ma*Tg)); //Density of flue gases in Kg/m^3
+mg=round(rhog*((pi/4)*(d**(2))*V)); //Mass of gas flowing through the chimney in Kg/s
+
+
+///Output
+printf('Theoretical draught produce in hot gas column=%f m \n',H1);
+printf('Draught lost in friction at the grate and pasage=%f m \n',gp);
+printf('Actual draught produce in hot gas column=%f m \n ',C);
+printf('Velocity of the flue gases in the chimney =%f m/s \n',V);
+printf('Density of flue gases=%f Kg/m^3 \n',rhog);
+printf('Mass of gas flowing through the chimney=%f Kg/s \n',mg);
+
diff --git a/3821/CH11/EX11.16/Example11_16.sce b/3821/CH11/EX11.16/Example11_16.sce new file mode 100644 index 000000000..3c161203a --- /dev/null +++ b/3821/CH11/EX11.16/Example11_16.sce @@ -0,0 +1,27 @@ +///Chapter No 11 Steam Boilers
+////Example 11.16 Page No 247
+///Find Theoretical draught produced in water
+//Input data
+clc;
+clear;
+hw=1.9; //Drauhgt water in cm
+Tg=290+273; //Temp of flue gases in degree celsius
+Ta=20+273; //Ambient temp in degree celsius
+ma=22; //Flue gases formed in kg/Kg of coal
+d=1.8; //Fuel burnt in m
+pi=3.142;
+g=9.81;
+
+//Calculation
+H=(hw/(353*(1/Ta-1/Tg*((ma+1)/ma))))*10; //Theoretical draught produced in water column in m
+H1=round(H*(((Tg/Ta)*(ma/(ma+1))-1))); //Theoretical draught produced in hot gas column n m
+V=sqrt(2*g*H1); //Velocity of tthe flue gases in the chimney in m/s
+rhog=((353*(ma+1))/(ma*Tg)); //Density of flue gases in Kg/m^3
+mg=rhog*((pi/4)*d^2)*V; //Mass of gas flowing through the chimney in Kg/s
+
+//Output
+printf('Theoretical draught produced in water column= %f m \n ',H);
+printf('Theoretical draught produced in hot gas column= %f m \n',H1);
+printf('Velocity of tthe flue gases in the chimney= %f m \n',V);
+printf('Density of flue gases=%f Kg/m^3 \n',rhog);
+printf('Mass of gas flowing through the chimney= %f Kg/s \n',mg);
diff --git a/3821/CH11/EX11.17/Example11_17.sce b/3821/CH11/EX11.17/Example11_17.sce new file mode 100644 index 000000000..04fca964d --- /dev/null +++ b/3821/CH11/EX11.17/Example11_17.sce @@ -0,0 +1,36 @@ +///Chapter No 11 Steam Boilers
+////Example 11.17 Page No 248
+///Find Actual draught produced in hot gas
+//Input data
+clc;
+clear;
+mf1=8000; //Average coal consumption in Kg/h
+ma1=19; //Flue gases formed in Kg/Kg
+Tg1=270+273; //Average temperature of the chimney in degree celsius
+Ta1=27+273; //Ambient temperature in degree celsius
+hw1=18; //Theoretical draught produced by the chimney in mm
+h11=0.6; //Draught is lost in friction H1
+g1=9.81;
+pi1=3.142;
+
+
+//Calculation
+H2=(hw1/(353*(1/Ta1-1/Tg1*((ma1+1)/ma1)))); //Theoretical draught produced in water column in m
+H3=H2*(((Tg1/Ta1)*(ma1/(ma1+1)))-1); //Theoretical draught produced in hot gas column in m
+gp1=h11*H3; //Draught is lost in friction at the grate and passing in m
+hgc1=H3-gp1; //Actual draught produced in hot gas column in m
+V1=sqrt(2*g1*(hgc1)); //Velocity of the flue gases in the chimney in m/s
+rhog1=((353*(ma1+1))/(ma1*Tg1)); //Density of flue gases in Kg/m^3
+mg1=((mf1/3600)*ma1); //Mass of gas fowing throgh the chimney in Kg/s
+d1=sqrt(mg1/(rhog1*(pi1/4)*V1)); //Diameter of the chimney in m
+
+
+//Output
+printf('Theoretical draught produced in water column=%f m \n',H2);
+printf('Theoretical draught produced in hot gas column=%f m \n',H3);
+printf('Draught is lost in friction at the grate and passing=%f m \n',gp1);
+printf('Actual draught produced in hot gas column=%f m \n ',hgc1);
+printf('Velocity of the flue gases in the chimney=%f \n',V1);
+printf('Density of flue gases=%f Kg/m^3 \n',rhog1);
+printf('Mass of gas fowing throgh the chimney=%f Kg/s \n',mg1);
+printf('Diameter of the chimney=%f m \n',d1);
diff --git a/3821/CH11/EX11.18/Example11_18.sce b/3821/CH11/EX11.18/Example11_18.sce new file mode 100644 index 000000000..6c78ce225 --- /dev/null +++ b/3821/CH11/EX11.18/Example11_18.sce @@ -0,0 +1,27 @@ +///Chapter No 11 Steam Boilers
+////Example 11.18 Page No 251
+///Find Actual draught produced in hot gas
+//Input data
+clc;
+clear;
+H2=24; //Chimney height in m
+Ta1=25+273; //Ambient temperature in degree celsius
+Tg1=300+273; //Temp of flue gases passing through the chimney in degree celsius
+ma1=20; //Combustion space of fuel burnt in Kg/kg of fuel
+g1=9.81;
+
+//Calculation
+hw1=((353*H2)*((1/Ta1)-((1/Tg1)*((ma1+1)/ma1)))); //Theoretical draught produced in water column in m
+ //Calculation mistake in book of hw1 it is correct according to data &calculation
+H3=H2*(((Tg1/Ta1)*(ma1/(ma1+1))-1)); //Theoretical draught produced in hot gas column in m
+H4=0.5*H3; //Draught is lost in friction at the grate and passing in m
+hgc1=H3-H4; //Actual draught produced in hot gas column in m
+V1=sqrt(2*g1*H4); //Velocity of the flue gases in the chimney in m/s
+
+
+//Output
+printf('Theoretical draught produced in water column=%f m \n',hw1);
+printf('Theoretical draught produced in hot gas column= %f m \n',H3);
+printf('Draught is lost in friction at the grate and passing=%f m \n ',H4);
+printf('Actual draught produced in hot gas column= %f m \n',hgc1);
+printf('Velocity of the flue gases in the chimney= %f m/s \n',V1);
diff --git a/3821/CH11/EX11.19/Example11_19.sce b/3821/CH11/EX11.19/Example11_19.sce new file mode 100644 index 000000000..8dd5a4031 --- /dev/null +++ b/3821/CH11/EX11.19/Example11_19.sce @@ -0,0 +1,31 @@ +///Chapter No 11 Steam Boilers
+////Example 11.19 Page No 252
+///Find Velocity of the flue gases in the chimney
+//Input data
+clc;
+clear;
+H2=38; //Stack height in m
+d1=1.8; //Stack diameter in m
+ma1=18; //Flue gases per kg of the fuel burnt
+Tg1=277+273; //Average temp of the flue gases in degree celsius
+Ta1=27+273; //Temperature of outside air in degree celsius
+h11=0.4; //Theorical draught is lost in friction in %
+g1=9.81;
+pi1=3.142
+
+//Calculation
+H3=H2*(((Tg1/Ta1)*(ma1/(ma1+1))-1)); //Theoretical draught produced in hot gas column in m
+gp1=0.40*H3; //Draught is lost in friction at the grate and passing in m
+hgc1=H3-gp1; //Actual draught produced in hot gas column in m
+V1=sqrt(2*g1*hgc1); //Velocity of the flue gases in the chimney in m/s
+rhog1=((353*(ma1+1))/(ma1*Tg1)); //Density of flue gases in Kg/m^3
+mg1=rhog1*((pi1/4)*d1^2)*V1; //Mass of gas fowing throgh the chimney in Kg/s
+
+
+//Output
+printf('Theoretical draught produced in hot gas column= %f m \n',H3);
+printf('Draught is lost in friction at the grate and passing= %f m \n',gp1);
+printf('Actual draught produced in hot gas column=%f m \n',hgc1);
+printf('Velocity of the flue gases in the chimney=%f m/s \n',V1);
+printf('Density of flue gases=%f Kg/m^3 \n',rhog1);
+printf('Mass of gas fowing throgh the chimney=%f Kg/s \n',mg1);
diff --git a/3821/CH11/EX11.2/Example11_2.sce b/3821/CH11/EX11.2/Example11_2.sce new file mode 100644 index 000000000..dca842dda --- /dev/null +++ b/3821/CH11/EX11.2/Example11_2.sce @@ -0,0 +1,25 @@ +///Chapter No 11 Steam Boilers
+////Example 11.2 Page No 229
+///Find Enthalpy of wet stream
+///Input data
+clc;
+clear;
+p=14; //Boiler pressure in bar
+me=9; //Evaporates of water in Kg
+Tw=35; //Feed water entering in degree celsius
+x=0.9; //Steam stop value
+CV=35000; //Calorific value of the coal
+
+///Calculation
+//From Steam Table
+hfw=146.56; //In KJ/Kg
+hf=830.07; //In KJ/Kg
+hfg=1957.7; //In KJ/Kg
+hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
+etaboiler=((me*(hs-hfw))/CV)*100;//Boiler efficiency in %
+
+///Output
+printf('Enthalpy of wet stream=%f KJ/Kg \n',hs);
+printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
+printf('Boiler efficiency=%f percent \n',etaboiler);
diff --git a/3821/CH11/EX11.20/Example11_20.sce b/3821/CH11/EX11.20/Example11_20.sce new file mode 100644 index 000000000..e5db3fdc2 --- /dev/null +++ b/3821/CH11/EX11.20/Example11_20.sce @@ -0,0 +1,29 @@ +///Chapter No 11 Steam Boilers
+////Example 11.20 Page No 253
+////Find Density of flue gases
+//Input data
+clc;
+clear;
+hw1=19; //Draught produced water in cm
+Tg1=290+273; //Temperature of flue gases in degree celsius
+Ta1=20+273; //Ambient temperature in degree celsius
+ma1=22; //Flue gases formed per kg of fuel burnt in kg/kg of coal
+d1=1.8; //Diameter of chimney
+g1=9.81;
+pi1=3.142
+
+
+//Calculation
+H2=(hw1/((353)*((1/Ta1)-((1/Tg1)*((ma1+1)/ma1))))); //Theoretical draught produced in hot gas column in m
+H3=round(H2*(((Tg1/Ta1)*(ma1/(ma1+1))-1))); //Draught is lost in friction at the grate and passing in m
+V1=(sqrt(2*g1*H3)); //Velocity of the flue gases in the chimney in m/s
+rhog1=((353*(ma1+1))/(ma1*Tg1)); //Density of flue gases in Kg/m**3
+mg1=rhog1*((pi1/4)*d1^2)*V1; //Mass of gas fowing throgh the chimney in Kg/s
+
+
+//Output
+printf('Theoretical draught produced in hot gas column= %f m \n',H2);
+printf('Draught is lost in friction at the grate and passing=%f m \n',H3);
+printf('Velocity of the flue gases in the chimney= %f m/s \n',V1);
+printf('Density of flue gases=%f Kg/m^2 \n',rhog1);
+printf('Mass of gas fowing throgh the chimney= %f Kg/s \n',mg1);
diff --git a/3821/CH11/EX11.21/Example11_21.sce b/3821/CH11/EX11.21/Example11_21.sce new file mode 100644 index 000000000..a94cc8198 --- /dev/null +++ b/3821/CH11/EX11.21/Example11_21.sce @@ -0,0 +1,35 @@ +///Chapter No 11 Steam Boilers
+////Example 11.21 Page No 254
+///Find Mass of gas fowing throgh the chimney
+//Input data
+clc;
+clear;
+mf=8000; //Average coal consumption in m
+ma=18; //Fuel gases formed ccoal fired in m
+Tg=270+273; //Average temp of the chimney of water in degree celsius
+Ta=27+273; //Ambient temp in degree celsius
+hw=18; //Theoretical draught produced by the chimney in mm
+h1=0.6; //Draught is lost in friction in H1
+g=9.81;
+pi=3.142;
+
+
+//Calculation
+H=(hw/((353)*((1/Ta)-((1/Tg)*((ma+1)/ma))))); //Theoretical draught produced in water column in m
+H1=H*(((Tg/Ta)*(ma/(ma+1))-1)); //Theoretical draught produced in hot gas column in m
+gp=0.6*H1; //Draught is lost in friction at the grate and passing in m
+hgc=H1-gp; //Actual draught produced in hot gas column in m
+V=sqrt(2*g*hgc); //Velocity of the flue gases in the chimney in m/s
+rhog=((353*(ma+1))/(ma*Tg)); //Density of flue gases in Kg/m^3
+mg=mf/3600*(ma+1); //Mass of gas fowing throgh the chimney in Kg/s
+d=sqrt(mg/(rhog*(pi/4)*V)); //Diameter of flue gases in Kg/m^3
+
+///Output
+printf('Theoretical draught produced in water column= %f m \n ',H);
+printf('Theoretical draught produced in hot gas column= %f m \n',H1);
+printf('Draught is lost in friction at the grate and passing= %f m \n',gp);
+printf('Actual draught produced in hot gas column= %f \n',hgc);
+printf('Velocity of the flue gases in the chimney= %f m/s \n',V);
+printf('Density of flue gases= %f Kg/m^3 \n ',rhog);
+printf('Mass of gas fowing throgh the chimney= %f Kg/s \n ',mg);
+printf('Diameter of flue gases= %f Kg/m^3 \n ',d);
diff --git a/3821/CH11/EX11.22/Example11_22.sce b/3821/CH11/EX11.22/Example11_22.sce new file mode 100644 index 000000000..e77ed0512 --- /dev/null +++ b/3821/CH11/EX11.22/Example11_22.sce @@ -0,0 +1,19 @@ +///Chapter No 11 Steam Boilers
+////Example 11.22 Page No 256
+///Find Efficeincy of chimney draught
+///Input data
+clc;
+clear;
+H=45; //Chimney height in m
+Tg=370+273; //Temperature of flue gases in degree celsius
+T1=150+273; //Temperature of flue gases in degree celsius
+ma=25; //Mass of the flue gas formed in Kg/kg of a cosl fired
+Ta=35+273; //The boiler temperature in degree celsius
+Cp=1.004; //fuel gas
+
+//Calculation
+//Efficeincy of chimney draught in %
+A=(H*(((Tg/Ta)*(ma/(ma+1)))-1))/(Cp*(Tg-T1))*100;
+
+//Output
+printf('Efficeincy of chimney draught= %f percent \n',A);
diff --git a/3821/CH11/EX11.3/Example11_3.sce b/3821/CH11/EX11.3/Example11_3.sce new file mode 100644 index 000000000..6200c2f09 --- /dev/null +++ b/3821/CH11/EX11.3/Example11_3.sce @@ -0,0 +1,28 @@ +///Chapter No 11 Steam Boilers
+////Example 11.3 Page No 230
+///Find mass of evaporation
+//Input data
+clc;
+clear;
+ms=2500; //Saturated steam per bour in Kg
+x=1;
+P=15; //Boiler pressure in bar
+Tw=25; //Feed water entering in degree celsius
+mf=350; //Coal burnt in Kg/bour
+CV=32000; //Calorific value in Kj/Kg
+
+//Calculation
+//steam table
+hfw=104.77; //In KJ/Kg
+hf=844.66; //In KJ/Kg
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+hs=2789.9; //Enthalpy of dry steam in KJ/Kg
+me=ms/mf; //mass of evaporation
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg ofcoal
+etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %
+
+//Output
+printf('mass of evaporation= %f \n',me);
+printf('Equivalent evaporation= %f Kg/Kg of coal\n',E);
+printf('Boiler efficiency= %f percent \n',etaboiler);
diff --git a/3821/CH11/EX11.4/Example11_4.sce b/3821/CH11/EX11.4/Example11_4.sce new file mode 100644 index 000000000..7761a6f2b --- /dev/null +++ b/3821/CH11/EX11.4/Example11_4.sce @@ -0,0 +1,33 @@ +///Chapter No 11 Steam Boilers
+////Example 11.4 Page No 231
+///Find Enthalpy of superheated steam
+//Input data
+clc;
+clear;
+mf=500; //Boiler plant consumes of coal in Kg/h
+CV=32000; //Calorific value in Kj/Kg
+ms=3200; //plant generates in Kg/h
+P=1.2; //Absolute pressure MN/m^2
+MN=12;
+Tsup=300; //Absolute temperature in degree celsius
+Tw=35; //Feed water temperature
+Cps=2.3;
+
+//Calculation
+hfw=146.56; //In KJ/Kg
+Ts=187.96; //In Degree celsius
+hf=798.43; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hg=2782.7; //In KJ/Kg
+hs=hg+Cps*(Tsup-Ts); //Enthalpy of superheated steam in KJ/Kg
+me=ms/mf; //mass of evaporation
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg ofcoal
+etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %
+
+
+///Output
+printf('Enthalpy of superheated steam= %f KJ/Kg\n',hs);
+printf('mass of evaporation=%f \n',me);
+printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
+printf('Boiler efficiency %f percent \n ',etaboiler);
+
diff --git a/3821/CH11/EX11.5/Example11_5.sce b/3821/CH11/EX11.5/Example11_5.sce new file mode 100644 index 000000000..384e5085f --- /dev/null +++ b/3821/CH11/EX11.5/Example11_5.sce @@ -0,0 +1,30 @@ +///Chapter No 11 Steam Boilers
+////Example 11.5 Page No 232
+//Find Enthalpy of wet stream
+//Input data
+clc;
+clear;
+ms=5000; //Steam generted in Kg/h
+mf=700; //Coal burnt in Kg/h
+CV=31402; //Cv of coal in KJ/Kg
+x=0.92; //quality of steam
+P=1.2; //Boiler pressure in MPa
+Tw=45; //Feed water temperature in degree celsius
+
+
+//Calculation
+hfw=188.35; //In KJ/Kg
+hf=798.43; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hs=hf+x*hfg; //Enthalpy of wet stream in KJ/Kg
+me=ms/mf; //mass of evaporation
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
+etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency in %
+
+
+
+//Output
+printf('Enthalpy of wet stream= %f KJ/Kg \n',hs);
+printf('mass of evaporation=%f \n',me);
+printf('Equivalent evaporation=%f Kg/Kg of coal \n',E);
+printf('Boiler efficiency=%f percent \n',etaboiler);
diff --git a/3821/CH11/EX11.6/Example11_6.sce b/3821/CH11/EX11.6/Example11_6.sce new file mode 100644 index 000000000..1c52df746 --- /dev/null +++ b/3821/CH11/EX11.6/Example11_6.sce @@ -0,0 +1,32 @@ +///Chapter No 11 Steam Boilers
+////Example 11.6 Page No 233
+///Enthalpy of superheated steam
+//Input data
+clc;
+clear;
+ms=6000; //Boiler produce of steam Kg/h
+P=25; //Boiler pressure in bar
+Tsup=350; //Boiler temperature in degree celsius
+Tw=40; //Feed water temperature indegree celsius
+CV=42000; //Calorific value in Kj/Kg
+etaboiler=75/100; //Expected thermal efficiency in %
+
+
+//Calculation
+hfw=167.45; //In KJ/Kg
+Ts=223.94; //In degree celsius
+hf=961.96; //In KJ/Kg
+hfg=1839.0; //In KJ/Kg
+hg=2800.9; //In KJ/Kg
+Cps=2.3;
+hs=((hg)+(Cps)*(Tsup-Ts)); //Enthalpy of superheated steam KJ/Kg
+mf=((ms*(hs-hfw))/(CV*etaboiler)); //Boiler efficiency in %
+me=ms/mf; //Equivalent mass of evaporation
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of oil
+
+
+//Output
+printf('Enthalpy of superheated steam=%f KJ/Kg \n',hs);
+printf('Boiler efficiency=%f percent \n',mf);
+printf('Equivalent mass of evaporation=%f \n',me);
+printf('Equivalent evaporation=%fKg/Kg of oil \n' ,E);
diff --git a/3821/CH11/EX11.7/Example11_7.sce b/3821/CH11/EX11.7/Example11_7.sce new file mode 100644 index 000000000..59c089fd0 --- /dev/null +++ b/3821/CH11/EX11.7/Example11_7.sce @@ -0,0 +1,34 @@ +///Chapter No 11 Steam Boilers
+////Example 11.7 Page No 234
+///Find Boiler efficiency
+///Input data
+clc;
+clear;
+E=12; //Boiler found steam in Kg/Kg
+CV=35000; //Calorific value in KJ/Kg
+ms=15000; //Boiler produces in Kg/h
+P=20; //Boiler pressure in bar
+Tw=40; //Feed water in degree celsius
+mf=1800; //Fuel consumption
+
+
+//Calculation
+//R=me(hs-hfw)
+hfw=167.45; //In KJ/Kg
+hg=2797.2; //In KJ/Kg
+Ts=211.37; //In degree celsius
+Cps=2.3;
+R=E*2257; //Equivalent evaporation in KJ/Kg of coal
+etaboiler=(R/CV)*100; //Boiler efficiency in %
+me=ms/mf; //Equivalent mass evaporation in KJ/Kg of coal
+hs=(R/me)+hfw; //In KJ/Kg
+Tsup=((hs-hg)/Cps)+Ts; //Enthalpy of superheated steam in degree celsius
+
+
+
+//Output
+printf('Equivalent evaporation=%f KJ/Kg of coal \n',R);
+printf('Boiler efficiency=%f percent \n',etaboiler);
+printf('Equivalent mass evaporation= %f KJ/Kg of coal \n',me);
+printf('hs=%f KJ/Kg \n',hs);
+printf('Enthalpy of superheated steam=%f degree celsius \n',Tsup);
diff --git a/3821/CH11/EX11.8/Example11_8.sce b/3821/CH11/EX11.8/Example11_8.sce new file mode 100644 index 000000000..8bf400f80 --- /dev/null +++ b/3821/CH11/EX11.8/Example11_8.sce @@ -0,0 +1,38 @@ +///Chapter No 11 Steam Boilers
+////Example 11.8 Page No 236
+///Find Equivalent mass evaporation
+//Input data
+clc;
+clear;
+ms=6000; //Steam generated in Kg/h
+mf=700; //Coal burnt in Kg/h
+CV=31500; //Cv of coal in KJ/Kg
+x=0.92; //Dryness in fraction of steam
+P=12; //Boiler pressure in bar
+Tsup=259; //Temperature of steam in degree celsius
+Tw=45; //Hot well temperature in degree celsius
+
+//Calculation
+hfw=188.35; //In KJ/Kg
+Ts=187.96; //In degree celsius
+hf=798.43; //In KJ/Kg
+hfg=1984.3; //In KJ/Kg
+hg=2782.7; //In KJ/Kg
+Cps=2.3;
+me=ms/mf; //Equivalent mass evaporation
+hs=hf+x*hfg; //Enthalpy of wet steam in KJ/Kg
+E=((me*(hs-hfw))/2257); //Equivalent evaporation in Kg/Kg of coal
+hs1=(hg+Cps*(Tsup-Ts)); //Enthalpy of superheated steam in KJ/Kg
+E1=((me*(hs1-hfw))/2257); //Equivalent evaporation(with superheater) in Kg/Kg of coal
+etaboiler=((me*(hs-hfw))/CV)*100; //Boiler efficiency without superheater in %
+etaboiler1=((me*(hs1-hfw))/CV)*100;//Boiler efficiency with superheater in %
+
+
+//Output
+printf('Equivalent mass evaporation=%f \n',me);
+printf('Enthalpy of wet steam=%f KJ/Kg \n',hs);
+printf('Equivalent evaporation=%f Kg/Kg of coal\n',E);
+printf('Enthalpy of superheated steam=%f KJ/Kg \n',hs1);
+printf('Equivalent evaporation(with superheater)=%f Kg/Kg of coal\n',E1);
+printf('Boiler efficiency without superheater=%f percent \n',etaboiler);
+printf('Boiler efficiency without superheater=%f percent \n',etaboiler1);
diff --git a/3821/CH11/EX11.9/Example11_9.sce b/3821/CH11/EX11.9/Example11_9.sce new file mode 100644 index 000000000..3f1142bae --- /dev/null +++ b/3821/CH11/EX11.9/Example11_9.sce @@ -0,0 +1,34 @@ +///Chapter No 11 Steam Boilers
+////Example 11.9 Page No 237
+///Find Mass of steam consumption
+///Input data
+clc;
+clear;
+P=15; //Boiler produces steam in bar
+Tsup=250; //Boiler temperature in degree celsius
+Tw=35; //Feed water in degree celsius
+MWh=1.5; //steam supplied to the turbine
+CV=32000; //Coal of calorific value in KJ/Kg
+etaboiler=80/100; //Thermal efficiency in %
+fr=210; //Firing rate in Kg/m^2/h
+//From steam table(temp basis at 35 degree celsius)
+hfw=146.56; //In KJ/Kg
+Ts=198.29; //In degree celsius
+hfg=1945.2; //In KJ/Kg
+hg=2789.9; //In KJ/Kg
+Cps=2.3;
+
+
+//calculator
+hs=hg+Cps*(Tsup-Ts); //Enthalpy of superheated steam(with superheater) in KJ/Kg
+ms=9000/MWh; //Steam rate in Kg/MWh
+mf=((ms*(hs-hfw))/(etaboiler*CV)); //Mass of steam consumption in Kg/h
+GA=mf/fr; //Grate rate in m^2
+
+
+
+//Output
+printf('Enthalpy of superheated steam(with superheater)=%f KJ/Kg \n',hs);
+printf('Steam rate= %f Kg/h \n',ms);
+printf('Mass of steam consumption=%f Kg/h \n',mf);
+printf('Grate rate=%f m^2 \n',GA);
diff --git a/3821/CH13/EX13.1/Example13_1.sce b/3821/CH13/EX13.1/Example13_1.sce new file mode 100644 index 000000000..25456c463 --- /dev/null +++ b/3821/CH13/EX13.1/Example13_1.sce @@ -0,0 +1,16 @@ +////Chapter 13 Steam Engines
+////Example 13.1 Page No 281
+///Find Therotical mean effective pressure
+//Input data
+clc;
+clear;
+Pa=10; //Single cylinder double acting steam engine pressure in bar
+Pb=1.5; //Single cylinder double acting steam engine pressure in bar
+rc=100/35; //Cut-off of the stroke in %
+
+
+//Calculation
+Pm=((Pa/rc)*(1+log(rc))-Pb); //Therotical mean effective pressure
+
+//Output
+printf('Therotical mean effective pressure= %f bar \n',Pm);
diff --git a/3821/CH13/EX13.10/Example13_10.sce b/3821/CH13/EX13.10/Example13_10.sce new file mode 100644 index 000000000..af6b11d40 --- /dev/null +++ b/3821/CH13/EX13.10/Example13_10.sce @@ -0,0 +1,30 @@ +////Chapter 13 Steam Engines
+////Example 13.10 Page No 290
+///Find Indicated power of steam engine
+//Input data
+clc;
+clear;
+IP=343; //Steam engine develop indicated power in Kw
+N=180; //power In rpm
+P1=15; //Steam supplied i bar
+Pb=1.25; //Steam is exhausted in bar
+rc=100/25; //Cut-off take place of stroke
+K=0.78; //Diagram factor
+//x=L/D=4/3
+x=4/3; //Stroke to bore ratio
+pi=3.142;
+
+
+//Calculation
+Pm=((P1/rc)*(1+log(rc))-Pb); //Therotical mean effective pressure Pm
+Pma=Pm*K; //Actual mean effective pressure Pma
+D=(((60000*IP)/(2*(Pma*10^5)*(4/3)*N))/(pi/4))^(1/3);//Indicated power of steam engine
+A=((pi/4)*(D^2));
+L=(x)*D;
+
+
+//Output
+printf('Therotical mean effective pressure= %f bar \n',Pm);
+printf('Actual mean effective pressure=%f bar \n',Pma);
+printf('Indicated power of steam engine=%f mm \n',D);
+printf('Indicated power of steam engine= %f mm \n',L);
diff --git a/3821/CH13/EX13.11/Example13_11.sce b/3821/CH13/EX13.11/Example13_11.sce new file mode 100644 index 000000000..c899ec049 --- /dev/null +++ b/3821/CH13/EX13.11/Example13_11.sce @@ -0,0 +1,27 @@ +////Chapter 13 Steam Engines
+////Example 13.11 Page No 290
+///Find Actual mean effective pressure
+//Input data
+clc;
+clear;
+D=240*10^-3; //Steam engine bor
+L=300*10^-3; //Stroke of engine
+N=220; //Speed of engine 220 in rpm
+IP=36; //Indicated power in Kw
+Pb=1.3; //Exhaust pressure in bar
+re=2.5; //Expansion ratio
+K=0.8; //Diagram factor
+pi=3.142
+A=((pi/4)*(D^2));
+
+
+
+//Calculation
+Pma=((IP*60000)/(2*10^5*L*A*N)); //Indicated power of steam engine in bar
+Pm=Pma/K; //Actual mean effective pressure in bar
+P1=((Pm+Pb)*re)/(1+log(re)); //Theoretical mean effective pressure in bar
+
+//Output
+printf('Indicated power of steam engine= %f bar \n',Pma);
+printf('Actual mean effective pressure= %f bar \n',Pm);
+printf('theoretical mean effective pressure= %f bar \n',P1);
diff --git a/3821/CH13/EX13.12/Example13_12.sce b/3821/CH13/EX13.12/Example13_12.sce new file mode 100644 index 000000000..69aab7686 --- /dev/null +++ b/3821/CH13/EX13.12/Example13_12.sce @@ -0,0 +1,26 @@ +////Chapter 13 Steam Engines
+////Example 13.12 Page No 291
+///Find Indicated power of steam engine
+//Input data
+clc;
+clear;
+D=700*10^-3; //Steam engine diameter in mm
+L=900*10^-3; //Steam engine diameter in mm
+Ip=450; //Develop indicated power Kw
+N=90; //Speed of steam engine in rpm
+P2=12; //Pressure at cut-off in bar
+P1=12; //Pressure at cut-off in bar
+Pb=1.3; //Back pressure in bar
+K=0.76; //Diameter factor
+pi=3.142;
+A=((pi/4)*0.7^2);
+
+//Calculation
+Pma=(Ip*60000)/(2*10^5*L*A*90); //Indicated power of steam engine in bar
+Pm=Pma/K; //Theoretical mean effective pressure in bar
+//using trial and error method
+re=1/0.241; //Expansion ratio
+///Output
+printf('Indicated power of steam engine= %f bar \n',Pma);
+printf('Theoretical mean effective pressure= %f bar \n',Pm);
+printf('Expansion ratio= %f \n',re);
diff --git a/3821/CH13/EX13.13/Example13_13.sce b/3821/CH13/EX13.13/Example13_13.sce new file mode 100644 index 000000000..d31871187 --- /dev/null +++ b/3821/CH13/EX13.13/Example13_13.sce @@ -0,0 +1,19 @@ +////Chapter 13 Steam Engines
+////Example 13.13 Page No 293
+///Find Brake Power
+//Input data
+clc;
+clear;
+Db=900*10^-3; //Diameter of break drum in mm
+dr=50*10^-3; //Diameter of rope in mm
+W=105*9.81; //dead weight on the tight side of the rope in Kg
+S=7*9.81; //Spring balance of the rope in N
+N=240; //Speed of the engine in rpm
+pi=3.142;
+//Calculation
+T=(W-S)*((Db+dr)/2); //Torque Nm
+Bp=2*pi*N*T/60000; //Brake Power in Kw
+
+//Output
+printf('Torque= %f Nm \n',T);
+printf('Brake Power= %f Kw \n',Bp);
diff --git a/3821/CH13/EX13.14/Example13_14.sce b/3821/CH13/EX13.14/Example13_14.sce new file mode 100644 index 000000000..29a620c82 --- /dev/null +++ b/3821/CH13/EX13.14/Example13_14.sce @@ -0,0 +1,29 @@ +////Chapter 13 Steam Engines
+////Example 13.14 Page No 294
+///Example Mechanical efficiency
+//Input data
+clc;
+clear;
+D=300*10^-3; //steam engine bor
+L=400*10^-3; //stroke
+Db=1.5; //effective brake diameter
+W=6.2*10^3; //net load on the brake
+N=180; //speed of engine in rpm
+Pma=6.5*10^3; //mean effective pressure in bar
+pi=3.142;
+A=((pi/4)*0.3^2);
+dr=0;
+S=0;
+
+//Calculation
+Ip=((2*Pma*L*A*N)/60000)*100; //Indicated power of steam engine in Kw
+T=(W-S)*((Db+dr)/2); //Torque in Nm
+Bp=2*pi*N*T/ 60000; //Break power Kw
+eta=(Bp/Ip)*100; //Mechanical efficiency in%
+
+
+//Output
+printf('Indicated power of steam engine= %f Kw \n',Ip);
+printf('Torque=%f Nm \n',T);
+printf('Break power= %f Kw \n ',Bp);
+printf('Mechanical efficiency= %f percent \n ',eta);
diff --git a/3821/CH13/EX13.2/Example13_2.sce b/3821/CH13/EX13.2/Example13_2.sce new file mode 100644 index 000000000..972789717 --- /dev/null +++ b/3821/CH13/EX13.2/Example13_2.sce @@ -0,0 +1,17 @@ +////Chapter 13 Steam Engines
+////Example 13.2 Page No 283
+///Find Therotical mean effective pressure
+//Input data
+clc;
+clear;
+a=5/100; //Engine cylinder of the stroke valume in %
+P1=12; //Pressure of the stream
+rc=3; //Cut-off is one-third
+Pb=1.1; //Constant the back pressure in bar
+
+//Calulation
+//Therotical mean effective pressure Pm
+Pm=P1*(1/rc+((1/rc)+a)*log((1+a)/((1/rc)+a)))-Pb;
+
+//Output
+printf('Therotical mean effective pressure=%f N/m^2 \n',Pm);
diff --git a/3821/CH13/EX13.3/Example13_3.sce b/3821/CH13/EX13.3/Example13_3.sce new file mode 100644 index 000000000..3117d4673 --- /dev/null +++ b/3821/CH13/EX13.3/Example13_3.sce @@ -0,0 +1,21 @@ +////Chapter 13 Steam Engines
+////Example 13.2 Page No 285
+///Find Mean Effective pressure
+///Input data
+clc;
+clear;
+P1=14; //Steam is ssupplied in bar
+P6=6; //Pressure at the end in bar
+Pb=1.2; //Pressure at back in bar
+a=0.1;
+re=4;
+//From hyperbolic process
+b=0.4;
+
+///Calculation
+//Mean Effective pressure in N/m^2
+Pm=P1*((1/re)+((1/re)+a)*log((1+a)/((1+re)+a)))-Pb*((1+b)+(a+b)*log((a+b)/a));
+
+
+//Output
+printf('Mean Effective pressure= %f N/m^2 \n',-Pm);
diff --git a/3821/CH13/EX13.4/Example13_4.sce b/3821/CH13/EX13.4/Example13_4.sce new file mode 100644 index 000000000..045736158 --- /dev/null +++ b/3821/CH13/EX13.4/Example13_4.sce @@ -0,0 +1,22 @@ +////Chapter 13 Steam Engines
+////Example 13.2 Page No 285
+///Find Cover end mean effective pressure
+//Input data
+clc;
+clear;
+Cover=1200; //Area of the indicator diagram for cover
+Crank=1100; //Area of the indicator diagram for crank
+ID=75;
+PS=0.15;
+
+
+///Calculation
+CoverMEP=Cover/ID*PS; //Cover end mean effective pressure
+CrankMEP=Crank/ID*PS; //Crank end mean effective pressure
+AverageMEP=(CoverMEP+CrankMEP)/2; //Average end mean effective pressure
+
+
+///Output
+printf('Cover end mean effective pressure= %f bar \n',CoverMEP);
+printf('Crank end mean effective pressure= %f bar \n',CrankMEP);
+printf('Average end mean effective pressure= %f bar \n',AverageMEP);
diff --git a/3821/CH13/EX13.5/Example13_5.sce b/3821/CH13/EX13.5/Example13_5.sce new file mode 100644 index 000000000..6f594606d --- /dev/null +++ b/3821/CH13/EX13.5/Example13_5.sce @@ -0,0 +1,19 @@ +////Chapter 13 Steam Engines
+////Example 13.5 Page No 286
+///Find Mean effective pressure
+//Input data
+clc;
+clear;
+a=25; //Area of indicator diagram cm^2
+Vs=0.15; //swept volume m^2
+S=1; //Scale in cm
+cm=0.02; //pressure axis m^3
+
+
+///Calculation
+b=Vs/cm; //Base length of diagram
+Pm=a/b*S; //Mean effective pressure
+
+//Output
+printf('Base length of diagram=%f bar \n',b);
+printf('Mean effective pressure= %f bar \n',Pm);
diff --git a/3821/CH13/EX13.6/Example13_6.sce b/3821/CH13/EX13.6/Example13_6.sce new file mode 100644 index 000000000..95d3e8077 --- /dev/null +++ b/3821/CH13/EX13.6/Example13_6.sce @@ -0,0 +1,18 @@ +////Chapter 13 Steam Engines
+////Example 13.6 Page No 287
+///Find Therotical mean effective pressure
+//Input data
+clc;
+clear;
+P1=14; //Steam Engine pressure in bar
+Pb=0.15; //Back pressure in bar
+K=0.72; //Diagram factor
+rc=100/20;
+
+//Calculation
+Pm=((P1/rc)*(1+log(rc))-Pb); //Therotical mean effective pressure Pm
+Pma=Pm*K; //Actual mean effective pressure Pma
+
+//Output
+printf('Therotical mean effective pressure= %f bar \n',Pm);
+printf('Actual mean effective pressure= %f bar \n',Pma);
diff --git a/3821/CH13/EX13.7/Example13_7.sce b/3821/CH13/EX13.7/Example13_7.sce new file mode 100644 index 000000000..54be62f8c --- /dev/null +++ b/3821/CH13/EX13.7/Example13_7.sce @@ -0,0 +1,18 @@ +////Chapter 13 Steam Engines
+////Example 13.7 Page No 287
+////Find Actual mean effective pressure
+//Input data
+clc;
+clear;
+P1=9; //Reciprocating engine pressure in bar
+Pb=1.5; //Back pressure in bar
+rc=100/25; //Cut-off
+K=0.8; //Diagram factor
+
+//Calculation
+Pm=((P1/rc)*(1+log(rc))-Pb); //Therotical mean effective pressure Pm
+Pma=Pm*K; //Actual mean effective pressure Pma
+
+///Output
+printf('Therotical mean effective pressure= %f bar \n ',Pm);
+printf('Actual mean effective pressure= %f bar \n',Pma);
diff --git a/3821/CH13/EX13.8/Example13_8.sce b/3821/CH13/EX13.8/Example13_8.sce new file mode 100644 index 000000000..6f64b2048 --- /dev/null +++ b/3821/CH13/EX13.8/Example13_8.sce @@ -0,0 +1,23 @@ +////Chapter 13 Steam Engines
+////Example 13.8 Page No 288
+////Find Diagram factor
+//Input data
+clc;
+clear;
+P1=10; //Inlet pressure
+Pb=1; //Back pressure
+rc=3; //Expansion ratio
+a=12.1; //Area of indicator diagram
+b=7.5; //Length of indicator diagram
+S=3; //Pressure scale
+
+
+//Calculation
+Pm=round((P1/rc)*(1+log(rc))-Pb ); //Therotical mean effective pressure Pm
+Pma=a/b*S; //Actual mean effective pressure Pma
+K=Pma/Pm; //Diagram factor
+
+///Output
+printf('Therotical mean effective pressure= %f bar \n',Pm);
+printf('Actual mean effective pressure= %f bar \n',Pma);
+printf('Diagram factor= %f \n',K);
diff --git a/3821/CH13/EX13.9/Example13_9.sce b/3821/CH13/EX13.9/Example13_9.sce new file mode 100644 index 000000000..541b84d58 --- /dev/null +++ b/3821/CH13/EX13.9/Example13_9.sce @@ -0,0 +1,27 @@ +////Chapter 13 Steam Engines
+////Example 13.9 Page No 289
+//Input data
+clc;
+clear;
+D=200*10^-3; //Steam engine cylinder in mm
+L=300*10^-3; //Bore of steam engine cylinder in mm
+rc=100/40; //Cut-off of the sroke
+P1=7; //Admission pressure of steam in bar
+Pb=0.38; //Exhaust pressure of steam in bar
+K=0.8; //Diagram factor
+N=200; //Indicator factor of engine
+pi=3.142; //Constant value
+//Indicated power of the engine in rpm
+A1=pi*(200*10^-3)^2/4;
+
+
+//Calculation
+Pm=((P1/rc)*(1+log(rc))-Pb); //Therotical mean effective pressure Pm
+Pma=round(Pm*K); //Actual mean effective pressure Pma
+IP=(2*Pma*L*A1*N/60000)*10^5; //Indicated power of steam engine in Kw
+
+
+//Output
+printf('Therotical mean effective pressure= %f bar \n ',Pm);
+printf('Actual mean effective pressure= %f bar \n',Pma);
+printf('Indicated power of steam engine= %f Kw \n',IP);
diff --git a/3821/CH14/EX14.1/Example14_1.sce b/3821/CH14/EX14.1/Example14_1.sce new file mode 100644 index 000000000..653f1fc50 --- /dev/null +++ b/3821/CH14/EX14.1/Example14_1.sce @@ -0,0 +1,18 @@ +////Chapter No 14 Air Standard Cycles
+////Example 14.1 Page No:302
+///Find thermal efficiency of the carnot cycle eta
+///Input data
+clc;
+clear;
+Tmax=477+273; //Temperature limits for the engine 477 degree celcius
+Tmin=27+273; //Temperature limits for the engine 27 degree celcius
+wd=150; //Carnot cycle produce in KJ
+
+//Calculatkion
+eta=(1-(Tmin/Tmax)); //Thermal efficiency of the carnot cycle in %
+Qs=(wd/eta); //Added during the process in Kj
+
+
+//Output
+printf('thermal efficiency of the carnot cycle eta= %f percent \n',100*eta);
+printf('added during the process Qs= %f KJ \n',Qs);
diff --git a/3821/CH14/EX14.10/Example14_10.sce b/3821/CH14/EX14.10/Example14_10.sce new file mode 100644 index 000000000..9241bba54 --- /dev/null +++ b/3821/CH14/EX14.10/Example14_10.sce @@ -0,0 +1,26 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.10 Page No 312
+///Find Middle temperature
+//Input data
+clc;
+clear;
+T1=300; //Initial temp in K
+T3=2500; //Final temp in K
+P1=1; //Initial pressure in N/m^2
+P3=50; //Final pressure in N/m^2
+gamma1=1.4;
+Cv=0.718;
+
+//Calculation
+r=(P3*T1)/(P1*T3); //Compression ratio
+eta=(1-(1/r^(gamma1-1))); //Standard effeciency in %
+T2=T1*((P3/P1)^((gamma1-1)/gamma1)); //Middle temperature in K
+Qs=Cv*(T3-T2); //Heat supplied in KJ/Kg
+WD=eta*Qs; //Work done KJ/Kg
+
+//Output
+printf('Compression ratio= %f \n',r);
+printf('Standard effeciency= %f percent \n',eta);
+printf('Middle temperature= %f K \n',T2);
+printf('Heat supplied= %f KJ/Kg \n',Qs);
+printf('Work done= %f KJ/Kg \n',WD);
diff --git a/3821/CH14/EX14.11/Example14_11.sce b/3821/CH14/EX14.11/Example14_11.sce new file mode 100644 index 000000000..207b49291 --- /dev/null +++ b/3821/CH14/EX14.11/Example14_11.sce @@ -0,0 +1,17 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.11 Page No 316
+///Find diesel engine air standard efficiency
+//input data
+clc;
+clear;
+r=18; //compression ratio of diesel engine
+K=6; //cut-off ratio of the stroke in%
+rho=2.02;
+gamma1=1.4;
+
+///Calculation
+//diesel engine air standard efficiency
+eta=100*((1-(1/r^(gamma1-1)))*(1/gamma1*(rho^(gamma1-1)/(rho-1))));
+
+//Output
+printf('diesel engine air standard efficiency %f percent \n',eta);
diff --git a/3821/CH14/EX14.12/Example14_12.sce b/3821/CH14/EX14.12/Example14_12.sce new file mode 100644 index 000000000..001098e9a --- /dev/null +++ b/3821/CH14/EX14.12/Example14_12.sce @@ -0,0 +1,19 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.12 Page No 317
+///Find cut-off ratio
+//Input Data
+clc;
+clear;
+r=22; //compression ratio of diesel engine r=v1/v2
+r1=11; //expansion ratio r1=v4/v3
+gamma1=1.4;
+rho=1.4;
+
+//Calculation
+rho=r/r1; //cut-off ratio
+//diesel engine air standard efficiency
+eta=100*((1-(1/r^(gamma1-1)))*(1/gamma1*(rho^(gamma1-1)/(rho-1))));
+
+//Output
+printf('cut-off ratio= %f \n',rho);
+printf('diesel engine air standard efficiency= %f percent \n',eta);
diff --git a/3821/CH14/EX14.13/Example14_13.sce b/3821/CH14/EX14.13/Example14_13.sce new file mode 100644 index 000000000..9d8a92e13 --- /dev/null +++ b/3821/CH14/EX14.13/Example14_13.sce @@ -0,0 +1,21 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.13 Page No 317
+///Find Compression ratio
+//Input data
+clc;
+clear;
+Vc=10/100; //Clearance volume in %
+Vs=Vc/0.1;
+K=0.05; //Cut-off of the strok in
+gamma1=1.4;
+
+//Calculation
+r=((Vs+Vc)/(Vc)); //Compression ratio
+rho=1+K*(r-1); //Cut-off ratio
+//Effeciency in %
+eta=(1-(1/r^(gamma1-1))*((1/gamma1)*(((rho^(gamma1))-1)/(rho-1))))*100;
+
+//Output
+printf('Compression ratio= %f Vs \n',r);
+printf('Cut-off ratio= %f \n',rho);
+printf('Effeciency= %f \n',eta);
diff --git a/3821/CH14/EX14.14/Example14_14.sce b/3821/CH14/EX14.14/Example14_14.sce new file mode 100644 index 000000000..91a3dd5fc --- /dev/null +++ b/3821/CH14/EX14.14/Example14_14.sce @@ -0,0 +1,23 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.14 Page No 317
+///Find air standard efficiency
+//Input data
+clc;
+clear;
+T1=50+273; //Temperature at the beginning of the compression
+T2=700+273; //Temperature at the end of the compression
+T3=2000+273; //Temperature at the beginning of the expansion
+gamma1=1.4;
+
+//Calculation
+r=((T2/T1)^(1/(gamma1-1))); //Compression ratio
+rho=(T3/T2); //Cut-off ratio
+K=((rho-1)/(r-1)); //Also cut-off ratio
+//Air standard efficiency
+eta=(1-(1/r^(gamma1-1))*((1/gamma1)*(((rho^(gamma1))-1)/(rho-1))))*100;
+
+//Output
+printf('compression ratio= %f \n',r);
+printf('cut-off ratio= %f \n',rho);
+printf('also cut-off ratio= %f \n',K);
+printf('air standard efficiency= %f percent',eta);
diff --git a/3821/CH14/EX14.15/Example14_15.sce b/3821/CH14/EX14.15/Example14_15.sce new file mode 100644 index 000000000..436bbf20f --- /dev/null +++ b/3821/CH14/EX14.15/Example14_15.sce @@ -0,0 +1,32 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.15 Page No 317
+///Find maximum temperatureof the cycle
+//Input data
+clc;
+clear;
+P1=0.1; //Diesel cycle is supplied# with air in MPa
+T1=40+273; //Diesel cycle is supplied with temperature in degree celsius
+r=18; //Compression ratio
+Qs=1500; //Heat supplied
+v1=18;
+v2=1;
+Cp=1.005;
+gamma1=1.4;
+
+
+//Calculation
+T2=T1*((v1/v2)^(gamma1-1)); //For isentropic process the temperature is
+P2=P1*((v1/v2)^(gamma1)); //For isentropic process the pressure is
+T3=(Qs/Cp)+T2; //Maximum temperatureof the cycle
+rho=T3/T2; //Cut-off ratio
+//Air standard efficiency
+eta=(1-(1/r^(gamma1-1))*((1/gamma1)*(((rho^(gamma1))-1)/(rho-1))))*100;
+NWD=(Qs*eta)*10^-2; //Net work done
+
+//Output
+printf('for isentropic process the temperature= %f K \n',T2);
+printf('for isentropic process the pressure= %f MPa \n',P2);
+printf('maximum temperatureof the cycle= %f K \n ',T3);
+printf('cut-off ratio= %f MPa \n',rho);
+printf('air standard efficiency= %f percent \n',eta);
+printf('net work done= %f KJ/Kg \n',NWD);
diff --git a/3821/CH14/EX14.16/Example14_16.sce b/3821/CH14/EX14.16/Example14_16.sce new file mode 100644 index 000000000..835eefb04 --- /dev/null +++ b/3821/CH14/EX14.16/Example14_16.sce @@ -0,0 +1,42 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.16 Page No 317
+///Find constant pressure
+//Input data
+clc;
+clear;
+r=14; //compression ratio of standard diesel cycle
+P1=1; //compression stroke in bar
+T1=300; //temperature of air in k
+T3=2774; //temperature rises in k
+CP=1.005;
+v1=14;
+v2=1;
+gamma1=1.4;
+Qs=1921.43;
+R=0.287*10^3;
+
+
+//Calculation
+T2=T1*((v1/v2)^(gamma1-1)); //Constant pressure
+rho=T3/T2; //cut-off ratio
+eta=(1-(1/r^(gamma1-1))*((1/gamma1)*(((rho^(gamma1))-1)/(rho-1))))*100; //air standard efficiency
+HS=(CP*(T3-T2)); //heat supplied
+WD=(Qs*eta)*10^-2; //Net work done
+v1=(R*T1/P1)*10^-5; //characteristics gas equation
+v2=(v1/r ); //characteristics gas equation
+Sv=(v1-v2); //Swept volume
+Pme=(WD/Sv )*10^-2; //Mean effective pressur
+Pm=((P1*r)/((r-1)*(gamma1-1)))*((gamma1*(r^(gamma1-1)))*(rho-1)-((rho^(gamma1))-1)); // mean effective pressure
+
+
+//utput
+printf('constant pressure= %f K \n',T2);
+printf('cut-off ratio= %f \n ',rho);
+printf('air standard efficiency= %f percent \n',eta);
+printf('heat supplied= %f KJ/Kg \n',HS);
+printf('Net work done= %f KJ/Kg \n',WD);
+printf('characteristics gas equation= %f m^3/Kg \n',v1);
+printf('characteristics gas equation=%f m^3/Kg \n ',v2);
+printf('Swept volume=%f m^3/Kg \n ',Sv);
+printf('Mean effective pressure= %f bar \n',Pme);
+printf('Mean effective pressure= %f bar \n ',Pm);
diff --git a/3821/CH14/EX14.2/Example14_2.sce b/3821/CH14/EX14.2/Example14_2.sce new file mode 100644 index 000000000..dea73eede --- /dev/null +++ b/3821/CH14/EX14.2/Example14_2.sce @@ -0,0 +1,24 @@ +////Chapter No 14 Air Standard Cycles
+////Example 2 Page No:302
+///Find Engin work on carnot cycle
+//Input data
+clc;
+clear;
+QR=1.5; //tau=QS-QR
+ //T=Tmax-Tmin
+T=300; //temperature limit of the cycle in degree celsius
+
+
+//Calculation
+//QR=1.5*(QS-QR)
+QR=(1.5/2.5); //Engin work on carnot cycle
+eta=(1-QR); //Thermal effeciency
+Tmax=round((T/eta)-273.15); //Maximum temperataure
+Tmin=(Tmax-T); //Minimum temperataure
+
+
+//Output
+printf('Engin work on carnot cycle= %f QS \n',QR);
+printf('Thermal effeciency= %f percent \n',100*eta);
+printf('Maximum temperataure= %f degree celsius \n ',Tmax);
+printf('Minimum temperataure= %f degree celsius \n ',Tmin);
diff --git a/3821/CH14/EX14.3/Example14_3.sce b/3821/CH14/EX14.3/Example14_3.sce new file mode 100644 index 000000000..9290b92b6 --- /dev/null +++ b/3821/CH14/EX14.3/Example14_3.sce @@ -0,0 +1,35 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.3 Page No 303
+///Find pressure at intermediate salient points
+//Input data
+clc;
+clear;
+//Refer figure
+
+T1=300; //Carnot engine work in minimum temperature in kelvin
+T2=750; //Carnot engine work in maximum temperature kelvin
+P2=50; //pressure of carnot engine N/m^2
+P4=1; //pressure of carnot engine N/m^2
+//Considering air as the working fluid therefore
+R=0.287; //Air as the working fluid in KJ/Kg K
+Cp=1.005; //KJ/Kg K
+Cv=0.718; //KJ/Kg K
+K=1.4;
+gamma1=1.4;
+
+//Calculation
+//T2/T1=(P2/P1)**(gamma1-1)/gamma1;
+P1=P2*(T1/T2)^(gamma1/(gamma1-1)); //Pressure at intermediate salient points(1-2) in bar
+P3=P4*(T2/T1)**(gamma1/(gamma1-1)); //Pressure at intermediate salient points(3-4) in bar
+QS=R*T2*log(P2/P3 ); //Heat supplied and rejected per Kg of air in KJ/Kg
+QR=R*T1*log(P1/P4 ); //Heat supplied and rejected per Kg of air in KJ/Kg
+W=QS-QR; //Work done in KJ/Kg
+eta=(1-(T1/T2)); //Thermal of the carnot cycle
+
+//Output
+printf('pressure at intermediate salient points(1-2)= %f bar \n',P1);
+printf('pressure at intermediate salient points(3-4)= %f bar \n',P3);
+printf('heat supplied and rejected per Kg of air(2-3)= %f KJ/Kg \n',QS);
+printf('heat supplied and rejected per Kg of air(4-1)= %f KJ/Kg \n',QR);
+printf('work done= %f KJ/Kg \n',W);
+printf('thermal of the carnot cycle= %f percent \n',100*eta);
diff --git a/3821/CH14/EX14.4/Example14_4.sce b/3821/CH14/EX14.4/Example14_4.sce new file mode 100644 index 000000000..fa5686b73 --- /dev/null +++ b/3821/CH14/EX14.4/Example14_4.sce @@ -0,0 +1,31 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.4 Page No 304
+///Find Heat supplied process
+//input data
+clc;
+clear;
+T2=377+273; //Carnot cycle temperature in bar
+P2=20*10^5; //Carnot cycle pressure in bar
+V2=1;
+V1=5;
+V3=2;
+//Consider air as the working fluid therefore
+R=0.287; //In KJ/Kg K
+Cp=1.005; //In KJ/Kg K
+Cv=0.718; //In KJ/Kg K
+K=1.4;
+gamma1=1.4;
+
+//Calculation
+T1=T2*((V2/V1)^(gamma1-1)); //Minimum temp in degree celsius
+Qs=R*T2*log(V3/V2); //Heat supplied process in KJ/Kg
+QR=R*T1*log((V1/V2)*(V2/V3)*((T2/T1)^(1/(gamma1-1)))); //Heat Rejected Process in KJ/Kg
+etath=(1-(T1/T2))*100; //Thermal Effeiciency of the carnot cycle in %
+
+
+
+//Output
+printf('Minimum temp= %f degree celsius \n',T1);
+printf('Heat supplied process= %f KJ/Kg \n',Qs);
+printf('Heat Rejected Process= %f KJ/Kg \n',QR);
+printf('Thermal Effeiciency of the carnot cycle=%f percent \n',etath);
diff --git a/3821/CH14/EX14.5/Example14_5.sce b/3821/CH14/EX14.5/Example14_5.sce new file mode 100644 index 000000000..16faa5a57 --- /dev/null +++ b/3821/CH14/EX14.5/Example14_5.sce @@ -0,0 +1,20 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.5 Page No 308
+///Find compression ratio
+///Input data
+clc;
+clear;
+P1=1; //Isentropic Compression in bar
+P2=20; //Isentropic Compression in bar
+//Consider air as the working fluid therefore
+gamma1=1.4;
+
+
+//Calculation
+r=(P2/P1)**(1/gamma1); //Isentropic process
+eta=100*(1-(1/(r^(gamma1-1)))); //Otto cycle air standard effeciency in %
+
+
+//Output
+printf('compression ratio= %f \n ',r);
+printf('standard efficiency= %f percent \n',eta);
diff --git a/3821/CH14/EX14.6/Example14_6.sce b/3821/CH14/EX14.6/Example14_6.sce new file mode 100644 index 000000000..f2ee380cd --- /dev/null +++ b/3821/CH14/EX14.6/Example14_6.sce @@ -0,0 +1,17 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.6 Page No 308
+///Find standard efficiency
+//Input data
+clc;
+clear;
+T1=27+273; //Initial temp in degree celsius
+T2=450+273; //Final temp in degree celsius
+gamma1=1.4;
+
+//Calculation
+r=(T2/T1)^(1/(gamma1-1)); //Isentropic process
+eta=100*(1-(1/(r^(gamma1-1)))); //Otto cycle air standard effeciency in %
+
+//Output
+printf('compression ratio= %f \n ',r);
+printf('standard efficiency= %f percent \n',eta);
diff --git a/3821/CH14/EX14.7/Example14_7.sce b/3821/CH14/EX14.7/Example14_7.sce new file mode 100644 index 000000000..68825c667 --- /dev/null +++ b/3821/CH14/EX14.7/Example14_7.sce @@ -0,0 +1,21 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.7 Page No 309
+///Find Swept volume
+//Input data
+clc;
+clear;
+D=200*10^-3; //Otto cycle cylindrical bore in mm
+L=450*10^-3; //Otto cycle Stroke in mm
+vc=2*10^-3; //Clearance volume in mm^3
+gamma1=1.4;
+pi=3.142;
+
+//Calculation
+vs=(pi/4)*(D^2*L); //Swept volume
+r=((vs+vc)/vc); //Compression ratio
+eta=100*(1-(1/(r**(gamma1-1)))); //Standard efficiency
+
+//Output
+printf('Swept volume= %f m^3 \n',vs);
+printf('compression ratio= %f \n',r);
+printf('standard efficiency= %f percent \n',eta);
diff --git a/3821/CH14/EX14.8/Example14_8.sce b/3821/CH14/EX14.8/Example14_8.sce new file mode 100644 index 000000000..eb256c643 --- /dev/null +++ b/3821/CH14/EX14.8/Example14_8.sce @@ -0,0 +1,44 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.8 Page No 309
+///Find Max temp of cycle
+//Input data
+clc;
+clear;
+P1=0.1*10^6; //Otto cycle air
+T1=35+273; //Otto cycle temp degree celsius
+r=9; //Compression ratio
+Qs=1800; //Supplied heat in kJ/kg
+v1=9;
+v2=1;
+R=0.287*10^3;
+gamma1=1.4;
+Cv=0.718;
+
+//Calculation
+T2=(T1*((v1/v2)^(gamma1-1))); //Temperature at point 2 in K
+P2=(P1*((v1/v2)^1.4))*10^-6; //pressure at point 2 in MPa
+T3=((Qs/Cv)+(T2)); //Max temp of cycle in degree celsius
+P3=(T3/T2*P2); //Max pressure of cycle in MPa
+eta=100*(1-(1/(r^(gamma1-1))));//Otto cycle thermal efficiency in %
+WD=(Qs*eta)*10^-2; //Work done during the cycle in KJ/Kg
+v1=((R*T1)/P1); //Char gass equation in m^3/Kg
+v2=v1/r; //Char gass equation in m^3/Kg
+Sv=v1-v2; //Swept volume in m^3/Kg
+Pme=(WD/Sv)*10^-3; //Mean effective pressure in MPa
+alpha=P3/P2; //Explosion ratio
+Pm=(((P1*r)/((r-1)*(gamma1-1)))*(((r^(gamma1-1))-1)*(alpha-1)))*10^-6;//Mean effective pressure in MPa
+
+
+//Output
+printf('Temperature at point= %f K \n',T2);
+printf('pressure at point= %f MPa \n',P2);
+printf('Max temp of cycle= %f K \n',T3);
+printf('Max pressure= %f MPa \n',P3);
+printf('Otto cycle thermal efficiency= %f percent \n',eta);
+printf('Work done during the cycle= %f KJ/Kg \n',WD);
+printf('Char gass equation= %f m^3/Kg \n',v1);
+printf('Char gass equation= %f m^3/Kg \n',v2);
+printf('Swept volume= %f m^3/Kg \n',Sv);
+printf('Mean effective pressure= %f MPa \n',Pme);
+printf('Explosion ratio= %f \n',alpha);
+printf('Mean effective pressure= %f MPa \n',Pm);
diff --git a/3821/CH14/EX14.9/Example14_9.sce b/3821/CH14/EX14.9/Example14_9.sce new file mode 100644 index 000000000..709db7a0a --- /dev/null +++ b/3821/CH14/EX14.9/Example14_9.sce @@ -0,0 +1,30 @@ +////Chapter No 14 Air Standard Cycles
+////Example No 14.9 Page No 311
+///Find Work done per Kg of air
+//Input data
+clc;
+clear;
+P1=0.1; //Beginning compression in MPa
+T1=40+273; //Beginning temp in degree celsius
+eta=0.55; //Standard effeciency in %
+QR=540; //Rejected heat in KJ/Kg
+r=7.36; //Compression ratio
+gamma1=1.4;
+Cv=0.718;
+
+//Calculation
+//eta=(1-(1/(r^(gamma-1))))
+QS=(-QR/(eta-1)); //Heat supplied/unit mass in KJ/Kg
+WD=QS-QR; //Work done per Kg of air in KJ/Kg
+T2=T1*(r^(gamma1-1)); //Temp at end of compression in K
+P2=P1*((r)^gamma1); //pressure at point 2 in MPa
+T3=(QS/Cv)+T2; //max temp of the cycle in K
+P3=(T3/T2)*P2; //max pressure of the cycle in MPa
+
+//Output
+printf('Heat supplied/unit mass= %f KJ/Kg \n',QS);
+printf('Work done per Kg of air= %f KJ/Kg \n',WD);
+printf('Temp at end of compression= %f K \n ',T2);
+printf('pressure at point two= %f MPa \n',P2);
+printf('max temp of the cycle= %f K \n',T3);
+printf('max pressure of the cycle= %f MPa \n',P3);
diff --git a/3821/CH2/EX2.1/Example2_1.sce b/3821/CH2/EX2.1/Example2_1.sce new file mode 100644 index 000000000..d54bc5a48 --- /dev/null +++ b/3821/CH2/EX2.1/Example2_1.sce @@ -0,0 +1,22 @@ +///Example 1.2 Page No:20
+///Find Cross-Section Area
+///Input data
+clc;
+clear;
+L1=5; //Length of steel bar in m
+d1=25*10^-3; //Diametr of steel bar in mm
+deltaLt1=25*10^-3; ///Steel
+pt1=800;
+pi1=3.142; //Power load of steel bar in N
+
+////Calculation
+A1=(pi1/4)*((deltaLt1)^2); ///Cross-section area
+sigmat1=pt1/A1; //Stress in steel bar
+et1=deltaLt1/L1; ///Strain in steel bar
+E1=sigmat1/et1; //Young's modulus
+
+///Output
+mprintf('value of Cross-section area= %f \n',A1);
+printf('value of stress in steel bar= %f MN/m^2 \n',sigmat1);
+printf('value of strain in steel bar= %f \n',et1);
+printf('value of Youngs modulus= %f N/m^2 \n',E1);
diff --git a/3821/CH2/EX2.2/Example2_2.sce b/3821/CH2/EX2.2/Example2_2.sce new file mode 100644 index 000000000..744aa48f5 --- /dev/null +++ b/3821/CH2/EX2.2/Example2_2.sce @@ -0,0 +1,19 @@ +///Example 1.2 Page No:20
+///Find Stress in Steel bar
+///Input data
+clc;
+clear;
+L1=300*10^-3; //Length of hexagonal prismatic steel bar in mm
+A1=500*10^-6; //Area of cross section of steel bar mm**2
+Pt1=500*10^3; //Load of steel bar in KN
+E1=210*10^9; //Modulus of elasticity GN/m**2
+
+///Calculation
+sigmat1=Pt1/A1; //Stress in steel bar
+et1=sigmat1/E1; //Strain steel bar is
+deltaLt1=et1*L1; //Therefore,elongation of the steel bar is given by
+
+////Output
+printf('stress in steel bar= %f N/m^2 \n',sigmat1);
+printf('therefore,strain steel bar is given by= %f \n',et1);
+printf('therefore,elongation of the steel bar is given by= %f m',deltaLt1);
diff --git a/3821/CH2/EX2.3/Example2_3.sce b/3821/CH2/EX2.3/Example2_3.sce new file mode 100644 index 000000000..c4d4fdd4f --- /dev/null +++ b/3821/CH2/EX2.3/Example2_3.sce @@ -0,0 +1,26 @@ +///Example 1.3 Page No:21
+///Find Stress in the Steel wire
+//Input Data
+clc;
+clear;
+Pt1=600; //Tensils force in N
+d1=2*10^-3; //Diameter of steel wire in mm
+L1=15; //Length of wire in m
+E1=210*10^9; //Modulus of elasticity of the material in GN/M**2
+pi1=3.1482;
+
+
+//Calculation
+A1=(pi1/4)*(d1^2); //(1)cross section area
+sigmat1=(Pt1)/(A1); //stress in the steel wire
+et1=((sigmat1)/(E1)); //(2)Therefore, strain in steel wire is given by
+deltaLt1=et1*L1; //(3)Enlongation of the steel wire is given by
+pe=((deltaLt1/L1)*100); //(4)Percentage elongation
+
+
+/////Output
+printf('cross section area= %f m^2\n',A1);
+printf('stress in the steel wire= %f GN/m^2 \n',sigmat1);
+printf('modulus of elasticity=%f \n',et1);
+printf('strain in steel wire=%f mm \n',deltaLt1)
+printf('percentage elongation=%f percent \n',pe)
diff --git a/3821/CH2/EX2.4/Example2_4.sce b/3821/CH2/EX2.4/Example2_4.sce new file mode 100644 index 000000000..865109e13 --- /dev/null +++ b/3821/CH2/EX2.4/Example2_4.sce @@ -0,0 +1,23 @@ +////Example 1.4 Page No:22
+///Find Stress in square rod
+//Input data
+clc;
+clear;
+A1=30*30*10^-6; //Area of square rod in mm**2
+L1=5; ///Length of square rod in m
+Pc=150*10^3; //Axial comperessive load of a rod in kN
+E1=215*10^9; //Modulus of elasticity in GN/m**2
+
+
+//Calculation
+sigmac=((Pc)/(A1)); //Stress in square rod
+ec=(sigmac)/(E1); //Modulusof elasticity is E1=sigmac/ec ,therefore strain in square rod is
+deltaLc=ec*5; ///Therefore shortening of length of the rod
+
+
+///Output
+printf('stress in square rod %f N/m^2',sigmac);
+printf('\n');
+printf('strain in square rod ec= %f\n',ec);
+printf('shortening of length of the rod= %f m \n',deltaLc);
+
diff --git a/3821/CH2/EX2.5/Example2_5.sce b/3821/CH2/EX2.5/Example2_5.sce new file mode 100644 index 000000000..1d80a7f2f --- /dev/null +++ b/3821/CH2/EX2.5/Example2_5.sce @@ -0,0 +1,31 @@ +////Example 1.5 Page No:23
+////Find Stress in metallic rod
+////input data
+clc;
+clear;
+d1=50*10^-3; //Diameter of metalic rod in mm**2
+L1=220*10^-3; //Length of metalic rod in mm
+Pt1=40*10^3; //Load of metalic rod in KN
+deltaLt1=0.03*10^-3; //Elastic enlongation in mm
+ypl=160*10^3; //Yield point load in KN
+ml=250*10^3; //Maximum load in KN
+lsf=270*10^-3; //Length of specimen at fracture in mm
+pi=3.142;
+
+//calculation
+A1=(((pi)/(4))*((d1)^2)); //(1)Cross section area
+sigmat1=Pt1/A1; //Stress in metallic rod
+et1=deltaLt1/L1; //Strain n metallic rod
+E1=sigmat1/et1; //Young's modulus
+ys=ypl/A1; //(2)Yeild strength
+uts=ml/A1; //(3)Ultimate tensile strength
+Pebf1=((lsf-L1)/L1)*100; //Percentage elongation before fracture
+
+//output
+printf('cross section area = %f m^2\n',A1);
+printf('stress in metallic rod= %f N/m^2 \n',sigmat1);
+printf('strain n metallic rod= %f \n',et1);
+printf('youngs modulus= %f GN/m^2\n',E1);
+printf('yeild strength= %f MN/m^2\n',ys);
+printf('ultimate tensile strength= %f MN/m^2 \n',uts);
+printf('percentage elongation before fracture= %f percent \n ',Pebf1);
diff --git a/3821/CH2/EX2.6/Example2_6.sce b/3821/CH2/EX2.6/Example2_6.sce new file mode 100644 index 000000000..2de7a49ba --- /dev/null +++ b/3821/CH2/EX2.6/Example2_6.sce @@ -0,0 +1,25 @@ +////Example 1.6 Page No:24
+///Find Stress in square metal bar
+//Input data
+clc;
+clear;
+A1=50*50*10^-6; //Area ofsquare metal bar in mm**2
+Pc=600*10^3; //Axial compress laod in KN
+L1=200*10^-3; //Gauge length of metal bar in mm
+deltaLc=0.4*10^-3; //Contraction length of metal bar in mm
+deltaLlateral=0.05*10^-3; //Lateral length of metal bar in mm
+
+//Calculation
+sigmac=Pc/A1; //Stress in square metal bar
+ec=deltaLc/L1; //Longitudinal or linear strain in square metal bar
+E1 =sigmac/ec; //Smodule of elasticity
+elateral=deltaLlateral/L1; //Lateral strain in square metal bar
+poissonsratio=elateral/ec;
+
+
+//Output
+printf('stress in bar=%f N/m^2 \n',sigmac);
+printf('longitudinal or linear strain in square metal bar= %f \n',ec);
+printf('module of elasticity= %f N/m^2 \n',E1);
+printf('lateral strain in square metal bar=%f \n',elateral);
+printf('poissons ratio=%f \n',poissonsratio);
diff --git a/3821/CH5/EX5.1/Example5_1.sce b/3821/CH5/EX5.1/Example5_1.sce new file mode 100644 index 000000000..1290e5e80 --- /dev/null +++ b/3821/CH5/EX5.1/Example5_1.sce @@ -0,0 +1,14 @@ +//Example 5.1 Page No:81
+//Find Diameter of the rod
+//Input data
+clc;
+clear;
+MSR=3.2; //Main scale reading of cylindrical rod in cm
+NCD=7; //Number of coinciding Vernier Scale division
+Lc=0.1*10^-3; //Least count of the instrument in mm
+
+//Calculation
+DOR=MSR+(NCD*Lc); //Diameter of the rod
+
+//Output
+printf('Diameter of the rod= %f cm \n',DOR);
diff --git a/3821/CH5/EX5.2/Example5_2.sce b/3821/CH5/EX5.2/Example5_2.sce new file mode 100644 index 000000000..675a1618c --- /dev/null +++ b/3821/CH5/EX5.2/Example5_2.sce @@ -0,0 +1,18 @@ +///Example 5.2 Page No:82
+///Measured length of bar
+//Input data
+clc;
+clear;
+MSR=5.3; //Main scale reading of prismatic bar in cm
+NCD=6; //Number of coinciding Vernier Scale division
+Lc=0.1*10^-3; //Least count of the instrument in mm
+Ne=(-0.2*10^-3); //Instrument bears a nagative error in mm
+
+//Calulation
+Mlb=MSR+(NCD*Lc); //Measured length of the bar in cm
+Tlb=(Mlb-(Ne)); //True length of the bar in cm
+
+
+//Output
+printf('Measured length of the bar= %f cm \n ',Mlb);
+printf('True length of the bar= %f cm ",Tlb);
diff --git a/3821/CH5/EX5.3/Example5_3.sce b/3821/CH5/EX5.3/Example5_3.sce new file mode 100644 index 000000000..583bcd87e --- /dev/null +++ b/3821/CH5/EX5.3/Example5_3.sce @@ -0,0 +1,17 @@ +///Example 5.3 Page No:88
+///Find Height required to setup of bar
+//Input data
+clc;
+clear;
+//Import maths
+L=100; //Height of sine bar
+theta=12.8 //angle in degree minut
+//Z=sin(theta)=0.22154849
+Z=0.22154849
+
+///Calculation
+b=Z*L; //Height required to setup in mm
+
+
+///Output
+printf('Height required= %f mm \n',b);
diff --git a/3821/CH7/EX7.1/Example7_1.sce b/3821/CH7/EX7.1/Example7_1.sce new file mode 100644 index 000000000..c8d4f7c72 --- /dev/null +++ b/3821/CH7/EX7.1/Example7_1.sce @@ -0,0 +1,25 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.1 Page No:113
+////Find mass density of liquid
+//Input data
+clc;
+clear;
+V=5; //volume of the liquid in m**3
+W=45*10^3; //weight of the liquid in KN
+g=9.81; //acceleration due to gravity in m/s**2
+rhow=1000; //constant value
+
+////Calculation
+m=W/g; //mass in Kg
+rho=m/V; //Mass density in kg/m**3
+w=W/V; //Weight Density in N/m**3
+v=V/m; //Specific volume in m**3/kg
+S=rho/rhow; //Specific gravity
+
+
+//Output
+printf('mass=%f kg \n',m);
+printf('Mass density= %f kg/m^3 \n ',rho);
+printf('Weight Density= %f N/m^3\n ',w);
+printf('Specific volume=%f m^3/kg \n',v);
+printf('Specific gravity= %f \n',S);
diff --git a/3821/CH7/EX7.10/Example7_10.sce b/3821/CH7/EX7.10/Example7_10.sce new file mode 100644 index 000000000..168a0ce19 --- /dev/null +++ b/3821/CH7/EX7.10/Example7_10.sce @@ -0,0 +1,28 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.10 Page No:119
+/// Find Kinematic viscosity
+//Input data
+clc;
+clear;
+//import math
+A=120*10^-3; //Side of square plate in mm
+W=30; //Side weight in N
+du=3.75; //Uniform velocity in m/s
+theta=30; //Lubricated inclined plane making an angle in degree at horizontal
+dy=6*10^-3; //Thickness lubricating oil film in mm
+rho=800; //Lubricating oil film density in Kg/m**3
+
+
+//Calculation
+sin30=0.5;
+F=W*sin30; //Component of force in N
+tau=(F/(A**2)); //Shear stress in Ns/m**2
+mu=tau/(du/dy); //From Newton's law of Shear stress in Ns/m**2
+V=(mu/rho)*10^3; //Kinematic viscosity in m**2/s
+
+
+///Output
+printf('Component of force=%f N \n ',F);
+printf('Shear stress=%f Ns/m^2 \n ',tau);
+printf('From Newtons law of Shear stress=%f Ns/m^2 \n ',mu);
+printf('Kinematic viscosity= %f m^2/s \n ',V);
diff --git a/3821/CH7/EX7.11/Example7_11.sce b/3821/CH7/EX7.11/Example7_11.sce new file mode 100644 index 000000000..29c3e59d4 --- /dev/null +++ b/3821/CH7/EX7.11/Example7_11.sce @@ -0,0 +1,19 @@ +///Chapter No 7 Fluid Mechanics
+//Example 7.11 Page No 121
+//#Input data
+clc;
+clear;
+Z=15; //Pressure due to column in m
+S=0.85; //Oil of specific gravity
+g=9.81; //Gravity
+
+
+
+///Calculation
+rho=S*10^3; //Density of oil in kg/m**3
+P=rho*g*Z; //Pressure in N/m**2 or kPa
+
+
+///Output
+printf('Density of oil= % f kg/m^3 \n ',rho);
+printf('Pressure= %f N/m**2 \n ',P);
diff --git a/3821/CH7/EX7.12/Example7_12.sce b/3821/CH7/EX7.12/Example7_12.sce new file mode 100644 index 000000000..f2c38953b --- /dev/null +++ b/3821/CH7/EX7.12/Example7_12.sce @@ -0,0 +1,21 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.12 Page No 122
+/// Find Intensity of pressure of water
+///Input data
+clc;
+clear;
+Z1=1.5; //open tank contain water in m
+Z2=2.5; //oil of specific gravity for depth in m
+S=0.9; //oil of specific gravity
+rho1=1000; //density of water in Kg/m**3
+rho2=S*10^3; //density of oil in Kg/m**3
+g=9.81; //gravity
+
+
+
+///calculation
+P=rho1*g*Z1+rho2*g*Z2; //Intensity of pressure in kPa
+
+
+///output
+printf('intensity of pressure=%f N/m^2 \n',P);
diff --git a/3821/CH7/EX7.13/Example7_13.sce b/3821/CH7/EX7.13/Example7_13.sce new file mode 100644 index 000000000..bcc5efe74 --- /dev/null +++ b/3821/CH7/EX7.13/Example7_13.sce @@ -0,0 +1,19 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.13 Page No:124
+///Find Discharge through pipe
+///Input data
+clc;
+clear;
+D1=0.2; //Diameter of pipe section 1 in m
+D2=0.3; //Diameter of pipe section 2 in m
+V1=15; //Velocity of water in m/s
+pi=3.14;
+
+///calculation
+Q=((3.14/4)*(0.2)^2)*15; //Discharge through pipe in m**3/s
+V2=(((3.14/4)*(0.2)^2)*15)/((3.14/4)*(0.3)^2); //velocity of section2 in m/s
+
+
+///Output
+printf('Discharge through pipe= %f m^3/s \n ',Q);
+printf('velocity of section2= %f m/s \n ',V2);
diff --git a/3821/CH7/EX7.14/Example7_14.sce b/3821/CH7/EX7.14/Example7_14.sce new file mode 100644 index 000000000..78cf3c686 --- /dev/null +++ b/3821/CH7/EX7.14/Example7_14.sce @@ -0,0 +1,19 @@ +///Chapter No 7 Fluid Mechanics
+////Example 7.14 Page No:126
+////Find Total energy per unit weight
+//Input data
+clc;
+clear;
+V=13; //Velocity of water flowing throgh pipe in m/s
+P=200*10^3; //Pressure of water in Kpa
+Z=25; //Height above the datum in m
+g=9.81;
+rho=1000;
+
+
+///Calculation
+E=(P/(rho*g))+((V^2)/(2*g))+(Z); //Total energy per unit weight in m
+
+
+///Output
+printf('Total energy per unit weight= %f m \n',E);
diff --git a/3821/CH7/EX7.15/Example7_15.sce b/3821/CH7/EX7.15/Example7_15.sce new file mode 100644 index 000000000..3c997555f --- /dev/null +++ b/3821/CH7/EX7.15/Example7_15.sce @@ -0,0 +1,26 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.15 Page No:127
+/// Find Total energy per unit weight
+///Input data
+clc;
+clear;
+S=0.85; //Specific gravity of oil
+D=0.08; //Diameter of pipe in m
+P=1*10^5; //Intenity of presssure in N/m^2
+Z=15; //Total energy bead in m
+E=45; //Datum plane in m
+Mdw=1*10^3; //Mass density of water constant
+g1=9.81; //Gravity constant
+rho=S*Mdw; //Mass density of oil
+pi1=3.14;
+
+///calculation
+rho=S*Mdw; //Mass density of oil
+//E=(P/(rho*g1))+((V**2)/(2*g1))+(Z);
+V=sqrt((E-((P/(rho*g1))+Z))*(2*g1)); ///Total energy per unit weight in m/s
+Q1=(pi1/4)*D^2*V //Discharge in m^3/Kg
+
+///output
+printf('mass density of oil=%f Kg/m^3 \n',rho);
+printf('Total energy per unit weight= %f m/s \n ',V);
+printf('discharge= %f m^3/kg',Q1);
diff --git a/3821/CH7/EX7.16/Example7_16.sce b/3821/CH7/EX7.16/Example7_16.sce new file mode 100644 index 000000000..19b88eb8d --- /dev/null +++ b/3821/CH7/EX7.16/Example7_16.sce @@ -0,0 +1,30 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.16 Page No:127
+///Find continuity discharge equation
+///input data
+clc;
+clear;
+///refer figure 11
+ZA=2; //water flows section A-A in m
+DA=0.3; //datum pipe diameter at section A-A in m
+PA=550*10^3; //pressure in kPa
+VA=6; //flow velocity in m/s
+ZB=18; //water flows at section B-B in m
+DB=0.15; //datum pipe diameter at section B-B in m
+pi1=3.14; //constant
+rho=1000; //constant
+g1=9.81; //constant
+Aa=(pi1/4)*(DA)^2;
+Ab=(pi1/4)*(DB)^2;
+pi1=3.14;
+
+///calculation
+VB=((Aa*VA)/Ab); //continuity discharge equation in m/s
+//bernoulli's equation Kpa
+//(PA/rho*g)+(VA**2/2*g)+ZA=(PB/rho*g)+(VB**2/2*g)+ZB
+PB=(((PA/(rho*g1))+(VA**2/(2*g1))+ZA)-((VB**2/(2*g1))+ZB))*(rho*g1);
+
+
+///output
+printf('continuity discharge equation= %f m/s \n',VB);
+printf('bernoullis equation= %f pa \n ',PB);
diff --git a/3821/CH7/EX7.17/Example7_17.sce b/3821/CH7/EX7.17/Example7_17.sce new file mode 100644 index 000000000..4720ff717 --- /dev/null +++ b/3821/CH7/EX7.17/Example7_17.sce @@ -0,0 +1,27 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.17 Page No:128
+/// Find bernoulli's equation for discharge
+//input data
+//refer figure 12
+clc;
+clear;
+Q1=0.04; //Water flows at rate in m**2/s
+DA=0.22; //Pipe diameter at section A in m
+DB=0.12; //Pipe diameter at section B in m
+PA=400*10^3; //Intensity of pressure at setion A in kPa
+PB=150*10^3; //Intensity of pressure at setion B in kPa
+pi1=3.14; //Pi constant
+g1=9.81; //Gravity constant
+rho=1000;
+
+///Calculation
+VA=Q1/(pi1/4*(DA)^2); //contuity equation for discharge
+VB=Q1/(pi1/4*(DB)^2); //bernoulli's equation for discharge
+///Z=ZB-ZA
+Z=(PA/(rho*g1))+(VA^2/(2*g1))-(PB/(rho*g1))-(VB^2/(2*g1));
+
+
+///Output
+printf('Contuity equation for discharge= %f m63 \n ',VA);
+printf('Contuity equation for discharge= %f m^3 \n ',VB);
+printf('Bernoullis equation for discharge=%f m \n',Z);
diff --git a/3821/CH7/EX7.18/Example7_18.sce b/3821/CH7/EX7.18/Example7_18.sce new file mode 100644 index 000000000..c16fd79c9 --- /dev/null +++ b/3821/CH7/EX7.18/Example7_18.sce @@ -0,0 +1,34 @@ +///Chapter No 7 Fluid Mechanics
+///Example 18 Page No:129
+//// Find rate of water flow l/min
+//Input data
+clc;
+clear;
+L1=200; //length of pipe in m
+D11=1; //Diameter at high end in m
+D12=0.4; //Diameter at low end in m
+P1=50*10^3; //Pressure at high end in kPa
+Q1=4000; //Rate of water flow l/min
+S=1; //Slope of pipe 1 in 100
+Z2=0; //Datum line is passing through the center of the low end,therefore
+pi1=3.14;
+rho=1000;
+g1=9.81;
+
+
+///Calculation
+Q1=(4000*10^-3)/60; //rate of water flow l/min in m**3/s
+Z1=1/100*L1; //slope of pipe 1 in 100 is in m
+//Q=A1*V11=A2V2 //continuity eqation ,discharge
+V11=Q1/((pi1/4)*(D11^2));//in m^3
+V12=Q1/((pi1/4)*(D12**2));//in m^3
+//bernoulli's equation
+P2=(((((P1/(rho*g1))+(V11^2/(2*g1))+Z1)-(V12^2/(2*g1))-Z2))*(rho*g1))*10^-3;
+
+
+///Output
+printf('rate of water flow=%f m^3/s \n ',Q1);
+printf('slope of pipe=%f m \n',Z1);
+printf('continuity eqation ,discharge= %f m^3 \n ",V11);
+printf('continuity eqation ,discharge= %f m^3 \n ",V12);
+printf('bernoullis equation for discharge= %f kpa \n ',P2);
diff --git a/3821/CH7/EX7.19/Example7_19.sce b/3821/CH7/EX7.19/Example7_19.sce new file mode 100644 index 000000000..d9f39c972 --- /dev/null +++ b/3821/CH7/EX7.19/Example7_19.sce @@ -0,0 +1,31 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.19 Page No:130
+///Find pipe inclined 30 degree,therefore Z2
+///Input data
+clc;
+clear;
+L1=36; //Length of pipe in m
+D11=0.15; //Diameter at upper side in m
+D12=0.3; //Diameter at lower side in m
+sin30=0.5;
+theta=sin(30); //Pipe slope upward at angle in degree
+V11=2; //Velocity of water at smaller section in m/s
+pi1=3.14; //Pi constant
+rho=1000; //Roh constant
+g1=9.81; //Gravity constant
+
+
+///Calculation
+//datum line is passing through the center of the low end,therefore
+Z1=0;
+Z2=Z1+L1*(0.5); //pipe inclined 30 degree,therefore in m
+//Q=A1*V1=A2*V2 continuity eqation ,discharge
+V12=(pi1/4*(D11^2)*2)/(pi1/4*(D12^2));
+//Z=P1-P2 bernoulli's equation
+Z=((((-V11^2)/(2*g1))+((V12^2)/(2*g1))-Z1+Z2)*(rho*g1))*10^-3;
+
+
+///Output
+printf('pipe inclined 30 degree,therefore Z2= %f m \n',Z2);
+printf('continuity eqation discharge V2= %f m/s \n',V12);
+printf('bernoullis equation Z=%f kpa \n',Z);
diff --git a/3821/CH7/EX7.2/Example7_2.sce b/3821/CH7/EX7.2/Example7_2.sce new file mode 100644 index 000000000..12f2d372c --- /dev/null +++ b/3821/CH7/EX7.2/Example7_2.sce @@ -0,0 +1,26 @@ +///Chapter No 7 Fluid Mechanics
+////Find mass density of oil
+///Example 7.2 Page No:114
+///Input data
+clc;
+clear;
+V=3*10^-3; //3l of oil in m**3
+W=24; //Weight of oil in N
+g=9.81; //Gravity in m/s**2
+rhow=1000; //Constant value
+
+
+//Calculation
+m=W/g; //Mass in Kg
+rho=m/V; //Mass density in kg/m**3
+w=W/V; //Weight Density in N/m**3
+v=V/m; //Specific volume in m**3/kg
+S=rho/rhow; //Specific gravity
+
+//Output
+printf('mass= %f kg \n',m);
+printf('Mass density= %f kg/m^3 \n',rho);
+printf('Weight Density= %f N/m^3\n ',w);
+printf('Specific volume= %f m^3/kg \n ',v);
+printf('Specific gravity= %f \n ',S);
+
diff --git a/3821/CH7/EX7.20/Example7_20.sce b/3821/CH7/EX7.20/Example7_20.sce new file mode 100644 index 000000000..2cc30073c --- /dev/null +++ b/3821/CH7/EX7.20/Example7_20.sce @@ -0,0 +1,30 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.20 Page No:130-131
+/// Find Continuity eqation
+//Input data
+clc;
+clear;
+D11=0.25; //Diameter at inlet in m
+D12=0.175; //Diameter at outlet in m
+P1=450*10^3; //Intensity of pressure at inlet in kPa
+P2=200*10^3; //Intensity of pressure at outlet in kPa
+pi1=3.14; //pi constant
+rho=1000; //Rho constant
+g1=9.81; //Gravity constant
+//Z1=Z2;
+
+///Calculation
+///A1*V11=A2*V12 Continuity eqation in V1
+V12=((pi1/4)*(D11^2))/((pi1/4)*(D12^2));
+///Z=V12^2-V11^2 Bernoulli's equation in m/s
+Z=-(((P2/(rho*g1))-(P1/(rho*g1)))*(2*g1));
+X=Z/((V12^2)-1);
+V11=sqrt(X);
+Q1=(pi1/4)*(D11^2)*V11; //Flow rate Water in m**3/Kg
+
+
+///Output
+printf('Continuity eqation=%f V1 \n ",V12);
+printf('Bernoullis equation=%f m/s \n ",Z);
+printf('V1= %f \n',V11);
+printf('Flow rate Water= %f m^3/kg \n ',Q1);
diff --git a/3821/CH7/EX7.21/Example7_21.sce b/3821/CH7/EX7.21/Example7_21.sce new file mode 100644 index 000000000..29010c20c --- /dev/null +++ b/3821/CH7/EX7.21/Example7_21.sce @@ -0,0 +1,34 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.21 Page No:131-132
+///Find Bernoulli's equation
+///Input data
+clc;
+clear;
+L1=300; //Length of pipe in m
+D11=0.9; //Diameter at higher end in m
+D12=0.6; //Diameter at lower end in m
+S=0.85; //Specific gravity
+Q1=0.08; //Flow in l/s
+P1=40*10^3; //Pressure at higher end in kPa
+pi1=3.14; //pi constant
+rho=1000; //Rho constant
+g1=9.81; //Gravity constant
+slop=1/50; //1 in 50
+
+
+//Calculation
+//Datum line is passing through the center of the low end,therefore
+Z2=0;
+Z1=slop*L1;
+//Q=A1*V1=A2*V2 Continuity eqation
+V11=Q1/((pi1/4)*(D11^2)); //Frome continuity eqation, discharge
+V12=Q1/((pi1/4)*(D12^2)); //Frome continuity eqation, discharge
+///Bernoulli's equation
+P2=(((((P1/(rho*S*g1))+(V11^2/(2*g1))+Z1)-(V12^2/(2*g1))+Z2))*(S*rho*g1))*10^-3;
+
+
+///Output
+printf('Z1= %f m \n ',Z1);
+printf('continuity eqation discharge V11= %f m^3 \n ',V11);
+printf('continuity eqation, discharge V12= %f m^3 \n',V12);
+printf('bernoullis equation= %f Kpa \n ',P2);
diff --git a/3821/CH7/EX7.3/Example7_3.sce b/3821/CH7/EX7.3/Example7_3.sce new file mode 100644 index 000000000..0891570ce --- /dev/null +++ b/3821/CH7/EX7.3/Example7_3.sce @@ -0,0 +1,22 @@ +///Chapter No 7 Fluid Mechanics
+////find mass density of liquid
+///Example 7.3 Page No:114
+//Input data
+clc;
+clear;
+S=0.85; //Specific gravity of a liquid
+g=9.81; //Acceleration due to gravity in m/s**2(constant)
+rhow=1000; //Constant value
+
+
+///Calculation
+//Specific gravity S=roh/rohw
+rho=S*rhow; //Mass density in Kg/m**3
+w=rho*g; //Weight Density in N/m**3
+v=1/rhow; //Specific volume in m**3/kg
+
+
+///Output
+printf('Mass densit= %f kg/m^3 \n ',rho);
+printf('Weight Density=%f N/m^3 \n ',w);
+printf('Specific volume= %f m^3/kg \n ',v);
diff --git a/3821/CH7/EX7.4/Example7_4.sce b/3821/CH7/EX7.4/Example7_4.sce new file mode 100644 index 000000000..c7f334ed2 --- /dev/null +++ b/3821/CH7/EX7.4/Example7_4.sce @@ -0,0 +1,15 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.4 Page No:116
+////Find mass density of liquid
+//Input data
+clc;
+clear;
+dy=21*10^-3; //Horizontal plates in mm
+du=1.4; //Relative velocity between the plates in m/s
+mu=0.6; //Oil of viscosity 6 poise in Ns/m^2
+
+///Calculation
+tau=mu*(du/dy); //Shear in the oil in N/m^2
+
+///Output
+printf('shear in the oil= %f N/m^2 \n',tau);
diff --git a/3821/CH7/EX7.5/Example7_5.sce b/3821/CH7/EX7.5/Example7_5.sce new file mode 100644 index 000000000..3e24903b7 --- /dev/null +++ b/3821/CH7/EX7.5/Example7_5.sce @@ -0,0 +1,17 @@ +///Chapter No 7 Fluid Mechanics
+///Find viscosity of the liquid
+///Example 7.5 Page No:116
+//Input data
+clc;
+clear;
+v=4*10^-4; ///kinematic viscosity is 4 stoke inm**2/s
+S=1.2; //specific gravity
+dow=1000; ///density of water Kg/m**3
+
+
+///Calculation
+rho=S*dow;
+vol=rho*v; //viscosity of the liquid in Ns/m**2 or poise
+
+///Output
+printf('viscosity of the liquid= %f Ns/m^2 \n',vol);
diff --git a/3821/CH7/EX7.6/Example7_6.sce b/3821/CH7/EX7.6/Example7_6.sce new file mode 100644 index 000000000..e55ac7fab --- /dev/null +++ b/3821/CH7/EX7.6/Example7_6.sce @@ -0,0 +1,21 @@ +///Chapter No 7 Fluid Mechanics
+//// Find newton's law of viscosity in shear stress
+///Example 7.6 Page No:116
+///Input data
+clc;
+clear;
+S=0.9; //Specific gravity
+tau=2.4; //shear stress in N/m**2
+vg=0.125; //velocitty gradientin per s
+dow=1000; //density of water Kg/m**3
+
+
+///Calculation
+mu=tau/vg; //newton's law of viscosity in shear stress in Ns/m**2
+rho=S*dow; //Density of oil in Kg/m**3
+v=mu/rho; //Kinematic viscosity in m**2/s or stoke
+
+///Output
+printf('newtons law of viscosity in shear stress= %f Ns/m^2 \n',mu);
+printf('Density of oil= %f kg/m^3 \n ',rho);
+printf('Kinematic viscosity=%f m^2/s \n ',v);
diff --git a/3821/CH7/EX7.7/Example7_7.sce b/3821/CH7/EX7.7/Example7_7.sce new file mode 100644 index 000000000..31e63f3da --- /dev/null +++ b/3821/CH7/EX7.7/Example7_7.sce @@ -0,0 +1,29 @@ +///Chapter No 7 Fluid Mechanics
+////Example 7.7 Page No:117
+///Find Density of oil
+///Input data
+clc;
+clear;
+A=6*10^-2; //Space between two square plates in mm
+dy=8*10^-3; //Thickness of fluid in mm
+u1=0; //Lower pate is stationary
+u2=2.4; //Upper plate in m/s
+F=5; //Speed of force in N
+s=1.6; //Specific gravity of the liquid
+dow=1000; //Density of water Kg/m**3
+
+
+//(1)Calculation
+du=u2-u1; //change in velocity in m/s
+tau=F/((A)^2); //shear stress N/m**2
+mu=tau/(du/dy); //Newton's law of viscosity in Ns/m**2 or poise
+rho=s*dow; //Density of oil in kg/m**3
+v=mu/rho; ///kinematic viscosity is given by m**2/s or stoke
+
+
+///Output
+printf('change in velocity=%f m/s \n ',du);
+printf('shear stress=%f N/m^2 \n ',tau);
+printf('Newtons law of viscosity=%f Ns/m^2 \n',mu);
+printf('Density of oil=%f kg/m^3 \n ',rho);
+printf('kinematic viscosity=%f m^2/s ',v);
diff --git a/3821/CH7/EX7.8/Example7_8.sce b/3821/CH7/EX7.8/Example7_8.sce new file mode 100644 index 000000000..0214e4013 --- /dev/null +++ b/3821/CH7/EX7.8/Example7_8.sce @@ -0,0 +1,22 @@ +///Chapter No 7 Fluid Mechanics
+////Example 7.8 Page No:118
+/// Find Power required to maintain the speed of upper plate
+//Input data
+clc;
+clear;
+dy=1.5*10^-4; //Two horizontal plates are placed in m
+mu=0.12; //Space between plates Ns/m**2
+A=2.5; //Upper area is required to move in m**2
+du=0.6; //Speed rerlated to lower plate in m/s
+
+
+////(1)Calculation
+tau=mu*(du/dy); //Shear stress N/m**2
+F=tau*A; //Force in N
+P=F*du; //Power required to maintain the speed of upper plate in W
+
+
+//Output
+printf('Shear stress=%f N/m^2 \n ',tau);
+printf('Force=%f N \n ',F);
+printf('Power required to maintain the speed of upper plate=%f W \n ',P);
diff --git a/3821/CH7/EX7.9/Example7_9.sce b/3821/CH7/EX7.9/Example7_9.sce new file mode 100644 index 000000000..43e04be2e --- /dev/null +++ b/3821/CH7/EX7.9/Example7_9.sce @@ -0,0 +1,20 @@ +///Chapter No 7 Fluid Mechanics
+///Example 7.9 Page No 118
+///Find Tangential speed of shaft
+//Input data
+clc;
+clear;
+mu=0.1; //Oil of viscosity used for lubricant in poise or Ns/m**2
+D=0.15; //Clearance between the shaft of diameter in m
+dy=3*10^-4; //Clearance in m
+N=90; //Shaft rorates in rpm
+pi=3.14;
+
+
+///Calculation
+du=(pi*D*N)/60; //Tangential speed of shaft in m/s
+tau=mu*(du/dy); //The shear force in N/m**2
+
+///Output
+printf('Tangential speed of shaft=%f m/s \n ',du);
+printf('The shear force= %f N/m^2 \n ',tau);
diff --git a/3821/CH9/EX9.1/Example9_1.sce b/3821/CH9/EX9.1/Example9_1.sce new file mode 100644 index 000000000..0ed0b5c20 --- /dev/null +++ b/3821/CH9/EX9.1/Example9_1.sce @@ -0,0 +1,23 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.1 Page No:165
+/// Find Work interaction during the 4th processes
+///Input data
+clc;
+clear;
+Qab=720; //Heat transfer of 1st processes in KJ
+Qbc=-80; //Heat transfer of 2nd processes in KJ
+Qcd=40; //Heat transfer of 3rd processes in KJ
+Qda=-640; //Heat transfer of 4th processes in KJ
+Wab=-90; //Work transfer of 1st processes in KJ
+Wbc=-50; //Work transfer of 2nd processes in KJ
+Wcd=130; //Work transfer of 3rd processes in KJ
+
+
+///Calculation
+///From the 1st law of thermodynamic for close system undergoing a cycle.
+
+//Work interaction during the 4th processes
+Wda=((Qab+Qbc+Qcd+Qda)-(Wab+Wbc+Wcd));
+
+///Output
+printf('Work interaction during the 4th processes= %f KJ \n",Wda);
diff --git a/3821/CH9/EX9.2/Example9_2.sce b/3821/CH9/EX9.2/Example9_2.sce new file mode 100644 index 000000000..b2d948472 --- /dev/null +++ b/3821/CH9/EX9.2/Example9_2.sce @@ -0,0 +1,21 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.2 Page No:166
+///Find Quantity of heat transferred
+///Input data
+clc;
+clear;
+ //During compression
+W1=-9200; //Stroke work done by the piston in Nm
+Nm1=-9.2; //Nm of work done
+Q1=-50; //Heat rejected during copression in KJ
+ //During expansion
+W2=8400; //Stroke work done by the piston in Nm
+Nm2=8.4; //Nm of work done
+
+///Calculation;
+ //Quantity of heat transferred
+Q2=-((Nm1+Nm2)+Q1); //-sign for indicate heat is transferred
+
+
+///Output
+printf('Quantity of heat transferred= %f KJ \n',Q2);
diff --git a/3821/CH9/EX9.3/Example9_3.sce b/3821/CH9/EX9.3/Example9_3.sce new file mode 100644 index 000000000..220b60b4c --- /dev/null +++ b/3821/CH9/EX9.3/Example9_3.sce @@ -0,0 +1,18 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.3 Page No:166
+/// Find Magnitude and direction of the third heat interation
+///input data
+clc;
+clear;
+W1=-20; //Work interaction to the fluid in KJ
+W2=42; //Work interaction from the fluid in KJ
+Q1=85; //Heat interaction to the fluid in KJ
+Q2=85; //Heat interaction to the fluid in KJ
+Q3=-50; //Heat interaction from the fluid in KJ
+
+///Calculation
+W3=((Q1+Q2+Q3)-(W1+W2)); //Magnitude and direction of the third heat interation
+
+
+///Output
+printf('Magnitude and direction of the third heat interation=%f KJ \n',W3);
diff --git a/3821/CH9/EX9.4/Example9_4.sce b/3821/CH9/EX9.4/Example9_4.sce new file mode 100644 index 000000000..351e4568f --- /dev/null +++ b/3821/CH9/EX9.4/Example9_4.sce @@ -0,0 +1,14 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.4 Page No:168
+/// Find Work done and compression process of heat
+///Input data
+clc;
+clear;
+Q1=-2100; //Non flow process losses heat in KJ
+deltaU=420; //Gain heat
+
+///Calculation
+W=Q1-deltaU; //Work done and compression process in KJ
+
+///Output
+printf('Work done and compression process= %f KJ \n',W);
diff --git a/3821/CH9/EX9.5/Example9_5.sce b/3821/CH9/EX9.5/Example9_5.sce new file mode 100644 index 000000000..6945e2832 --- /dev/null +++ b/3821/CH9/EX9.5/Example9_5.sce @@ -0,0 +1,14 @@ +///Chapter 9 Law Of Thermodynamics
+//Example 9.5 Page No:168
+/// Find Change in interval energy
+///Input data
+clc;
+clear;
+W=-2000; //Work input of panddle wheel in KJ
+Q1=-6000; //Heat transferred to the surrounding from tank
+
+//Calculation
+deltaU=Q1-W; //Change in interval energy
+
+///Output
+printf('change in interval energy drop= %f KJ \n',deltaU);
diff --git a/3821/CH9/EX9.6/Example9_6.sce b/3821/CH9/EX9.6/Example9_6.sce new file mode 100644 index 000000000..a7dd3c363 --- /dev/null +++ b/3821/CH9/EX9.6/Example9_6.sce @@ -0,0 +1,16 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.6 Page No:169
+/// Find Heat transferred during the process
+///Input data
+clc;
+clear;
+U1=520; //internal energy in KJ/Kg
+U2=350; //internal energy in KJ/Kg
+W=-80; //work done by the air in the cylinder KJ/kg
+
+///Calculation
+deltaU=U2-U1;
+Q1=deltaU+W; //Heat transferred during the process
+
+///Output
+printf('Heat transferred during the process= %f KJ \n',Q1);
diff --git a/3821/CH9/EX9.7/Example9_7.sce b/3821/CH9/EX9.7/Example9_7.sce new file mode 100644 index 000000000..48409f141 --- /dev/null +++ b/3821/CH9/EX9.7/Example9_7.sce @@ -0,0 +1,17 @@ +///Chapter 9 Law Of Thermodynamics
+////Example 9.7 Page No:169
+///Find Steam flow rate
+///Input data
+clc;
+clear;
+W1=800; //Power of turbine shaft Kw
+W2=-5; //Work pump to feed in Kw
+Q1=2700; //Heat for steam generation KJ/Kg
+Q2=-1800; //Condenser rejected heat KJ/Kg
+
+//Calculation
+m=((W1+W2)/(Q1+Q2)); //Steam flow rate in Kg/h
+
+
+//Output
+printf('Steam flow rate= %f Kg/s \n",m);
diff --git a/3821/CH9/EX9.8/Example9_8.sce b/3821/CH9/EX9.8/Example9_8.sce new file mode 100644 index 000000000..530a739ba --- /dev/null +++ b/3821/CH9/EX9.8/Example9_8.sce @@ -0,0 +1,38 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.8 Page No:170
+/// Find DeltaUab
+///input data
+clc;
+clear;
+///Data consistent with first law pf thermodynamics
+Qabcd=-22; //In KJ
+N=150; //In Cycles/min
+Qab=17580; //In KJ/min
+Qbc=0;
+Qcd=-3660; //In KJ/min
+Wab=-8160; //In KJ/min
+Wbc=4170; //In KJ/min
+DeltaUcd=-21630; //In KJ/min
+
+
+///Calculation
+DeltaUab=Qab-Wab; //In KJ/min
+DeltaUbc=Qbc-Wbc; //In KJ/min
+Wcd=Qcd-DeltaUcd; //In KJ/min
+Qabcd1=-220*150; //In KJ/min
+Qda=((Qabcd1)-(Qab+Qbc+Qcd)); //In KJ/min
+Wda=((Qabcd1)-(Wab+Wbc+Wcd)); //In KJ/min
+DeltaUabcd=0;
+DeltaUda=((DeltaUabcd)-(DeltaUab+DeltaUbc+DeltaUcd)); //In KJ/min
+NWO=Qabcd1/60; //In KW
+
+///Output
+printf('DeltaUab= %f Kj/min \n ',DeltaUab);
+printf('DeltaUbc= %f KJ/min \n ',DeltaUbc);
+printf('Wcd=%f KJ/min \n ',Wcd);
+printf('Qabcd1= %f KJ/min \n ',Qabcd1);
+printf('Qda= %f KJ/min \n ',Qda);
+printf('Wda= %f KJ/min \n ',Wda);
+printf('DeltaUabcd= %f KJ/min \n ',DeltaUabcd);
+printf('DeltaUda= %f KJ/min \n',DeltaUda);
+printf('NWO= %f Kw \n',NWO);
diff --git a/3821/CH9/EX9.9/Example9_9.sce b/3821/CH9/EX9.9/Example9_9.sce new file mode 100644 index 000000000..d68b5a2c6 --- /dev/null +++ b/3821/CH9/EX9.9/Example9_9.sce @@ -0,0 +1,32 @@ +///Chapter 9 Law Of Thermodynamics
+///Example 9.9 Page No:171
+///Find Net heat transfer in 1st cycle
+///Input data
+clc;
+clear;
+Qab=-6500; //Heat transferred in 1st process KJ/min
+Qbc=0; //Heat transferred in 2nd process
+Qcd=-10200; //Heat transferred in 3rd process KJ/min
+Qda=32600; //Heat transferred in 4th process KJ/min
+Wab=-1050; //Heat transferred in 1st process KJ
+Wbc=-3450; //Heat transferred in 2nd process KJ
+Wcd=20400; //Heat transferred in 3rd process KJ
+Wda=0; //Heat transferred in 4th process
+
+///Calculator
+dQ=Qab+Qbc+Qcd+Qda; //Net heat transfer in 1st cycle
+dW=Wab+Wbc+Wcd+Wda; //Net work done in 1st cycle
+dW1=dW/60; //Net work done in 1st cycle
+DeltaUab=Qab-Wab; //ab process
+DeltaUbc=Qbc-Wbc; //bc processes
+DeltaUcd=Qcd-Wcd; //cd processes
+DeltaUda=Qda-Wda; //dc processes
+
+///Output
+printf('Net heat transfer in 1st cycle= %f KJ/min \n',dQ);
+printf('Net work done in 1st cycle= %f KJ/min \n',dW);
+printf('Net work done in 1st cycle=%f KW \n ',dW1);
+printf('ab process= %f KJ/min \n',DeltaUab);
+printf('bc processes= %f KJ/min \n ',DeltaUbc);
+printf('cd processes= %f KJ/min \n ',DeltaUcd);
+printf('dc processes= %f KJ/min \n',DeltaUda);
|