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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /1808 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '1808')
172 files changed, 4812 insertions, 0 deletions
diff --git a/1808/CH1/EX1.1/Chapter1_Example1.sce b/1808/CH1/EX1.1/Chapter1_Example1.sce new file mode 100644 index 000000000..dc4b5c50f --- /dev/null +++ b/1808/CH1/EX1.1/Chapter1_Example1.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+Tmax=200;//Maximum Brake Torque in Nm
+N=3600;//Speed range in rpm
+Pmax=900;//Maximum engine torque in kPa
+n=2;// For Four stroke engine
+Mps=15*60;//mean piston speed in m/min
+
+//CALCULATIONS
+Vs=((2*3.14*Tmax*n)/(1000*Pmax));//Swept volume in m^3
+d=((Vs/3.14)^(1/3))*1000;//Bore diameter
+Nmax=(Mps*1000/(2*d));//Maximum crank speed in rpm
+Bpm=800*(Vs/2)*(Nmax/(60));//Maximum break power in kW
+
+//OUTPUT
+printf('(i)swept volume is %3.5f m^3 \n (ii)bore diameter is %3.i mm \n (iii)maximum break power is %3.2f kW ',Vs,d,Bpm)
+
+
diff --git a/1808/CH1/EX1.10/Chapter1_Example10.sce b/1808/CH1/EX1.10/Chapter1_Example10.sce new file mode 100644 index 000000000..833ebad00 --- /dev/null +++ b/1808/CH1/EX1.10/Chapter1_Example10.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+IP=50;//indicated power in kW
+pmi=7;//mean effective pressure in bar
+L=0.10;//stroke in m
+d=0.08;//bore in m
+nc=4;//number of cylinders
+n=2;//for 4 cylinders
+N=3800;//speed in rpm
+
+//CALCULATIONS
+n1=(IP*4*n*60)/(pmi*100*L*3.14*d^2*nc);//Average misfire in rpm
+n2=N/2;//Theoretical number of explosions/min
+na=N/2;//actual no.of explosion/min
+n11=n2-na;//Average number of misfires
+IP1=pmi*100*L*3.14*((0.08^2)/4)*N*nc/(n*60);//Indicated power based on actual speed
+
+//OUTPUT
+printf('(i)Average misfire is %3.d rpm \n (ii)Indicated power based on actual speed is %3.3f kW',n1,IP1)
+
+
diff --git a/1808/CH1/EX1.11/Chapter1_Example11.sce b/1808/CH1/EX1.11/Chapter1_Example11.sce new file mode 100644 index 000000000..c72cc6dee --- /dev/null +++ b/1808/CH1/EX1.11/Chapter1_Example11.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+BP=50;//Brake power in kW
+nm=80;//mechanical efficiency in percentage
+pmi=6;//mean effective pressure in bar
+N=100;//no.of explosions/min
+nc=1;//number of cylinders
+n=1;//for single cylinder
+
+//CALCULATIONS
+IP=(BP*100/nm);//Indicated power in kW
+x=(IP*60)/(pmi*100*N*nc);//dimension
+d=(x*4/(3.14*1.5))^(1/3);//bore in m
+L=1.5*d;//stroke in m
+//OUTPUT
+printf('(i)Dimensions of cylinder L %3.5f m \n d %3.5f m ',L,d)
diff --git a/1808/CH1/EX1.12/Chapter1_Example12.sce b/1808/CH1/EX1.12/Chapter1_Example12.sce new file mode 100644 index 000000000..6610b43f0 --- /dev/null +++ b/1808/CH1/EX1.12/Chapter1_Example12.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+Vc=0.009;//clearance volume in m^3
+d=0.3;//bore in m
+L=0.5;//stroke in m
+g=1.4;//constant
+cv=20000;//calorific value in kJ/m^3
+pmi=800;//mean effective pressure in bar
+N=120;//explosions per minute
+nc=1;//number of cylinders
+n=1;//for single cylinder
+mf=30;//mass fuel consumption
+
+
+//CALCULATIONS
+Vs=(3.14*d^2*L/4);//swept volume in m^3
+Rc=((Vs+Vc)/Vc);//compression ratio
+IP=(pmi*L*(3.14*(d^2)/4)*N*nc)/(60*n);//Indicated power in kW
+nit=(IP/(mf*cv/3600))*100;//Indicated thermal efficiency in percentage
+no=(1-(1/(Rc^(g-1))))*100;//Air standard efficiency in percentage
+nr=(nit/no)*100;//relative efficiency in percentage
+
+
+//OUTPUT
+printf('(i)Compression ratio is %3.3f \n (ii)Indicated thermal efficiency is %3.2f percentage \n (iii)Air standard efficiency is %3.2f percentage \n (iv)Relative efficiency is %3.2f percentage',Rc,nit,no,nr)
+
diff --git a/1808/CH1/EX1.13/Chapter1_Example13.sce b/1808/CH1/EX1.13/Chapter1_Example13.sce new file mode 100644 index 000000000..4ee46d1b1 --- /dev/null +++ b/1808/CH1/EX1.13/Chapter1_Example13.sce @@ -0,0 +1,42 @@ +clc
+clear
+//INPUT DATA
+N=450;//speed in rpm
+cv=44000;//calorific value in kJ/kg
+T=450;//torque required
+pmi=3;//mean effective pressure in bar
+L=0.27;//stroke in m
+d=0.22;//bore in m
+pa=1.014;//atm.pressure
+nc=1;//number of cylinders
+n=1;//for single cylinder
+mf=5.4;//mass flow rate in kg/hr
+ra=1.20584;//density of fuel in kg/kW.hr
+ma=167.4;//mass of air in kg/hr
+Ra=0.287;// gas constant kJ/kgk
+Te=300;//temperature in k
+mw=440;//mass of water in kg/hr
+cpw=4.187;//specific pressure
+dTc=36.1;//Rise in temperature in degree C
+Ta=20;//temperature in K
+d1=6.23;//bore
+me=172.8;//exhaust gas mass in kg/hr
+cpe=1.005;//atmospheric pressure
+
+//CALCULATIONS
+BP=(2*3.14*(N/1000)*(T/60)*(1.5/2));//Brake power in kW
+IP=(pmi*100*L*(3.14*(d^2)/4)*N*nc)/(60*n);//Indicated power in kW
+nit=(IP/((mf/3600)*cv))*100;//Indicated thermal efficiency in percentage
+bsfc=(mf/BP);//Brake specific fuel consuption in kg/kWh
+Va=(ma*Ra*(Ta+273))/(pa*100*60);//volume folw rate of air in m^3/min
+Vs=((3.14*(d^2)/4)*L*N*(nc/n));;//swept volume in m^3/min
+nv=(Va/Vs)*100;//Volumetric efficiency in percentage
+Qs=(mf*cv/3600);//Heat supplied in kW
+Qw=(mw*cpw*(dTc))/3600;//Heat loss to cooling water in kW
+Qe=(me*cpe*(Te-Ta))/3600;//Heat loss to exhaust gases in kW
+c1=(Qe/Qs)*100;// % heat lost to exhaust gases
+Qu=(Qs-(BP+Qw+Qe+d1));//Enthalpy of unaccount in kW
+e1=(Qu/Qs)*100;//unaccounted heat in percentage
+
+//OUTPUT
+printf('(i)Brake power is %3.3f kW \n (ii)Indicated power is %3.3f kW \n (iii)Indicated thermal efficiency is %3.3f percentage \n (iv)Brake specific fuel consumption is %3.4f kg/kW.hr \n (v)Volumetric efficiency is %3.1f percentage \n(vi)Draw up heat balance sheet \n (I)Heat supplied is %3.i kW \n(II)Heat utilised in the system is %3.2f perentage',BP,IP,nit,bsfc,nv,Qs,c1)
diff --git a/1808/CH1/EX1.14/Chapter1_Example14.sce b/1808/CH1/EX1.14/Chapter1_Example14.sce new file mode 100644 index 000000000..94cc8f2c9 --- /dev/null +++ b/1808/CH1/EX1.14/Chapter1_Example14.sce @@ -0,0 +1,34 @@ +clc
+clear
+//INPUT DATA
+a=450;//Area of indicator diagram mm^2
+S=9.806;//Spring number
+l=50*1.2;//Length of diagram
+d=0.15;//bore in m
+L=0.25;//stroke in m
+N=400;//engine speed in rpm
+nc=1;//number of cylinders
+n=2;//for single cylinder
+mf=3;//mass flow rate in kg/h
+cv=44200;//calorific value
+dTc=42;//rise of temperature for cooling water in Degree C
+cpw=4.18;//specific pressure
+mw=4;//mass of water
+T=225;//Brake torque in Nm
+
+//CALCULATIONS
+pmi=a*S/l;//mean effective pressure in N/cm^2
+IP=((pmi/10)*L*(3.14*(d^2)/4)*N*nc)/(60*n);//Indicated power in kW
+BP=(2*3.14*N*T)/60000;//brake power in kW
+nm=(BP/IP);//Meahanical efficiency in percentage
+nbt=(BP*3600/(mf*cv))*100;//Brake thermal efficiency in percentage
+bsfc=mf/BP;//Brake specific fuel consumption in kg/kWh
+Qs=mf*cv/3600;//Heat supplied in kW
+a11=(BP/Qs)*100;//% of heat equivalent to BP
+Qw=(mw*cpw*(dTc))/60;//Heat lost to cooling water in kW
+b11=(Qw/Qs)*100;//% of heat lost to cooling water
+Qe=(Qs-(BP+Qw));//heat utilised in the system
+c11=(Qe/Qs)*100;//% of heat lost to exhaust gases and radiation
+
+//OUTPUT
+printf('(i)Mechanical efficiency is %3.2f percentage \n(ii)Brake thermal efficiency is %3.2f percentage \n (iii)Brake specific fuel consumption is %3.3f kg/kW.hr \n(iv)\n(I)heat supplied is %3.3f kW \n (II)Heat utilised in the system %3.2f \n percentage',nm,nbt,bsfc,Qs,Qe)
diff --git a/1808/CH1/EX1.15/Chapter1_Example15.sce b/1808/CH1/EX1.15/Chapter1_Example15.sce new file mode 100644 index 000000000..c4c232781 --- /dev/null +++ b/1808/CH1/EX1.15/Chapter1_Example15.sce @@ -0,0 +1,36 @@ +clc
+clear
+//INPUT DATA
+pmi=3;//Mean effective pressure in bar
+L=0.27;//Stroke in m
+N=450;//spedd in rpm
+nc=1;//nuber of cylinders
+n=1;//for single cylinder
+mf=5.4;//mass flow rate in kg/h
+cv=42000;//calorific value
+d=0.22;//bore in m
+T=355;//Temperature to exhaust gases in Degree C
+mw=440;//mass of water in kg/h
+cpw=4.18;//specific pressure of water
+cpe=1.02;//specific pressure of air
+dTc=36;//Rise in temperature in Degree C
+me=172.8;//total mass flow in kg/s
+Ta=20;//room temperature in Degree C
+Tb=460;//Brake load in N
+
+//CALCULATIONS
+IP=(pmi*100*L*(3.14*(d^2)/4)*N*nc)/(60*n);//Indicated power in kW
+nit=(IP*3600/(mf*cv))*100;//Indicated thermal efficiency in percentage
+Qs=mf*cv/3600;//Heat supplied in kJ/s
+BP=(2*3.14*(N/60)*(Tb/1000)*(1.5/2));//Brake power in kW
+a11=(BP/Qs)*100;//% of heat equivalent to BP
+Qw=(mw*cpw*(dTc))/3600;//Heat loss to cooling water in kJ/s
+b11=(Qw/Qs)*100;//% of heat lost to cooling water
+Qe=(me*cpe*(T-Ta))/3600;//Heat loss to exhaust gases in kW
+c11=(Qe/Qs)*100;//% of heat lost to exhaust gases
+Qu=(Qs-(BP+Qw+Qe));//Enthalpy of unaccount in kJ/s
+d1=(Qu/Qs)*100;//unaccounted heat in percentage
+
+//OUTPUT
+printf('(i)Indicated thermal efficiency is %3.2f percentage \n (ii) \n (I)Heat supplied %3.i kJ/s \n (II)Heat utilised in the system is %3.2f',IP,Qs,Qe)
+
diff --git a/1808/CH1/EX1.16/Chapter1_Example16.sce b/1808/CH1/EX1.16/Chapter1_Example16.sce new file mode 100644 index 000000000..ce2c7588f --- /dev/null +++ b/1808/CH1/EX1.16/Chapter1_Example16.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+pmi=6;//Mean effective pressure in bar
+L=0.45;//Stroke in m
+d=0.3;//Rope diameter in m
+N=12000;//Total revolutions made
+nc=1;//number of cylinders
+n=2;//for four cylinders
+D=1.8;//Brake drum diameter in m
+x=0.02136;//difference of W and S
+cpw=4.18;//specific pressure of water
+cpe=1.005;//specific pressure of air
+cv=45000;//calorific value
+two=60;//outlet water temperature
+twi=15;//inlet water temperature
+te=300;//exhaust gas temperature in Degree C
+ta=20;//room temperature in Degree C
+mf=7.6;//mass flow rate in kg/h
+mw=550;//water flo rate in kg/h
+me=367.6;//total flow rate in kg/h
+
+
+//CALCULATIONS
+IP=(pmi*102*L*(3.14*(d^2)/4)*N*nc)/(60*60*n);//Indicated power in kW
+BP=((x)*3.14*(D+d)*N)/60;//Brake power in kW
+nit=(IP/(mf*cv/3600))*100;//Indicetad thermal efficiency in percentage
+nm=(BP/IP)*100;//mechanical efficiency in percentage
+Qs=mf*cv/60;//heat supplied in kJ/min
+a11=(BP/Qs)*100;//% of heat equivalent to BP
+Qw=(mw*cpw*(two-twi))/60;//Heat loss to cooling water in kW
+b11=(Qw/Qs)*100;//% of heat lost to cooling water
+Qe=(me*cpe*(te-ta))/60;//Heat loss to exhaust gases in kW
+c11=(Qe/Qs)*100;//% of heat lost to exhaust gases
+Qu=(Qs-(BP*60+Qw+Qe));//Enthalpy of unaccount in kW
+d11=(Qu/Qs)*100;//unaccounted heat in percentage
+
+//OUTPUT
+printf('(i)Indicated power is %3.2f kW \n brake power is %3.2f kW \n (ii)Indicated thermal efficiency is %3.2f percentage \n (iii)Mechanical efficiency is %3.2f percentage \n (iv)HEAT BALANCE SHEET \n (I)Heat supplied %3.i kJ/min \n (II)Heat utilised in the system is %3.2f kW',IP,BP,nit,nm,Qs,Qu)
diff --git a/1808/CH1/EX1.17/Chapter1_Example17.sce b/1808/CH1/EX1.17/Chapter1_Example17.sce new file mode 100644 index 000000000..dd4b3beb0 --- /dev/null +++ b/1808/CH1/EX1.17/Chapter1_Example17.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+BP4=16.25;//Total Brake power
+BP1c=11.55;//Brake power of 1st cylinder
+BP2c=11.65;//Brake power of 2nd cylinder
+BP3c=11.70;//Brake power of 3rd cylinder
+BP4c=11.50;//Brake power of 4th cylinder
+mf=0.08;//mass flow rate in kg/s
+cv=42500;//calorific value
+d=9;//bore
+L=9;//stroke
+Vc=65;//clearance volume in cm^3
+g=1.4;//inert gas constnat
+
+
+//CALCULATIONS
+IP1=BP4-BP1c;//Indicated power of 1st cylinder
+IP2=BP4-BP2c;//Indicated power of and cylinder
+IP3=BP4-BP3c;//Indicated power of 3rd cylinder
+IP4=BP4-BP4c;//Indicated power of 4th cylinder
+IP=IP1+IP2+IP3+IP4;//Total indicated power in kW
+nbt=(BP4*100/(mf*cv))*100;//Brake thrmal efficiency in percentage
+nit=(IP*100/(mf*cv))*100;//Indicated thermal efficiency in percentage
+Vs=(3.14*(d^2)*L/4);//swept volume in cm^3
+Rc=(Vs+Vc)/Vc;//Compression ratio
+no=(1-(1/Rc^(g-1)));//Air standard efficiency in percentage
+nr=(nit/no);//Relative efficiency in percentage
+
+//OUTPUT
+printf('(i)Indicated power is %3.2f kW \n (ii)indicated thermalefficiency %3.2f percentage \n brake efficiency is %3.2f percentage \n (iii)realtive efficiency is %3.2f percentage',IP,nit,nbt,nr)
diff --git a/1808/CH1/EX1.18/Chapter1_Example18.sce b/1808/CH1/EX1.18/Chapter1_Example18.sce new file mode 100644 index 000000000..d5ebc4e83 --- /dev/null +++ b/1808/CH1/EX1.18/Chapter1_Example18.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+W=160;//load on dynamometer in N
+N=3000;//speed of engine in rpm
+Dy=20420;//Dynamometer constnat
+L=0.09;//stroke in m
+d=0.06;//bore in m
+nc=4;//number of cylinders
+n=2;//for four cylinders
+mf=4.95;//fuel consumption in kg/h
+cv=42500;//calorific value
+BP1c=16.5;//Brake power of 1st cylinder
+BP2c=16;//Brake power of 2nd cylinder
+BP3c=15.6;//Brake power of 3rd cylinder
+BP4c=17.6;//Brake power of 4th cylinder
+
+//CALCULATIONS
+BP4=W*N/Dy;//Brake power in kW
+pmb=(BP4*60*n)/(L*(3.14*(d^2)/4)*N*nc*4);//Brake ean effective pressure in kN/m^2
+nbt=(BP4*3600/(mf*cv))*100;//Brake thermal efficiency in percentage
+IP1=BP4-BP1c;//Indicated power of 1st cylinder
+IP2=BP4-BP2c;//Indicated power of 2nd cylinder
+IP3=BP4-BP3c;//Indicated power of 3rd cylinder
+IP4=BP4-BP4c;//Indicated power of 4th cylinder
+IP=IP1+IP2+IP3+IP4;//Total indicated power in kW
+nm=(BP4/IP)*100;//Mechanical efficiency in percentage
+bsfc=mf/BP4;//Brake specific fuel consumption in kg/kWh
+
+//OUTPUT
+printf('(i)Brake power is %3.3f kW \n (ii)Brake mean effective pressure is %3.3f kN/m^2 \n (iii)brake thermal efficiency is %3.2f percentage \n (iv)mechanical efficiency is %3.3f percentage \n (v)Brake specific fuel consumption %3.3f kg/kW.hr ',BP4,pmb,nbt,nm,bsfc)
+
diff --git a/1808/CH1/EX1.19/Chapter1_Example19.sce b/1808/CH1/EX1.19/Chapter1_Example19.sce new file mode 100644 index 000000000..a4919ad59 --- /dev/null +++ b/1808/CH1/EX1.19/Chapter1_Example19.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+Sp=8;//mean piston speed in m/s
+P=101.325;//atmospheric pressure in kPa
+Ra=0.287;//Gas constnat
+Ta=303;//assumed temperature in K
+x=0.5;//fuelair ratio
+nv=0.92;//volumetric efficiency in percentage
+L=0.136;//stroke in m
+d=0.125;//borre diameter in m
+nc=6;//number of cylinders
+
+//CALCULATIONS
+N=(Sp/(2*L));//Speed in rps
+Roa=(P/(Ra*Ta));//Density of air in kg/m^3
+Vs=(3.14*d^2*L*nc/4);//swept volume
+ma=(nv*Roa*Vs*N)/2;//mass flow rate of air in kg/s
+mf=ma*x;//mass flow rate of fuel
+mf1=(2*mf*10^2)/(N*nc);//mass of fuel injected per cylinder per cycle in g/cylinder/cycle
+ //OUTPUT
+ printf('(i)Mass flow rate of air is %3.4f kg/s \n (ii)mass of fuel injected per cylinder per cycle is %3.4f g/cylinder/cycle',ma,mf1)
diff --git a/1808/CH1/EX1.2/Chapter1_Example2.sce b/1808/CH1/EX1.2/Chapter1_Example2.sce new file mode 100644 index 000000000..ddc732663 --- /dev/null +++ b/1808/CH1/EX1.2/Chapter1_Example2.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+N=4000;//Speed in rpm
+T=150;//Torque developed in Nm
+n=2;//For Four stroke engine
+L=0.1;//Stroke in m
+D=0.07;//Diameter in m
+nc=6;//number of cylinders
+mf=20;//fuel consumption in kg/h
+cv=44000;//calorific value in kJ/kg
+
+//CALCULATIONS
+BP=(2*3.14*N*T/(60*1000));//Brake power in kW
+bmep=((BP*n)/(L*(3.14*0.07^2/4)*(N/60)*nc));//Bmep in kN/m^2
+nbt=(BP/((mf/3600)*cv))*100;//Brake thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)The Brake power is %3.2f kW \n (ii)bmep is %3.2f kN/m^2 \n (iii)Brake thermal efficiency is %3.1f percentage ',BP,bmep,nbt)
diff --git a/1808/CH1/EX1.20/Chapter1_Example20.sce b/1808/CH1/EX1.20/Chapter1_Example20.sce new file mode 100644 index 000000000..6768bf400 --- /dev/null +++ b/1808/CH1/EX1.20/Chapter1_Example20.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+BP=200;//Brake power in kW
+Vs=10*10^-3;//swept volue
+n=2;//for four cylinders
+nc=1;//number of cylinders
+N=2100;//speed in rpm
+L=0.136;//stroke in m
+d=0.125;//bore in m
+
+//CALCULATIONS
+Sp=2*L*N/60;//Mean piston speed in m/s
+bmep=(BP*n*60)/(Vs*N);//Brake mean effective pressure in kPa
+P=BP/(3.14*d^2*6/4);//Specific power in kN/m^2
+
+//OUTPUT
+printf('Mean piston speed is %3.2f m/s \n Specific power is %3.1f kN/m^2',Sp,P)
diff --git a/1808/CH1/EX1.21/Chapter1_Example21.sce b/1808/CH1/EX1.21/Chapter1_Example21.sce new file mode 100644 index 000000000..591e222cd --- /dev/null +++ b/1808/CH1/EX1.21/Chapter1_Example21.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+P=101.325;//Atmospheric pressure in kPa
+Ra=0.287;//gas constant
+Ta=303;//atm.temperature in K
+L=0.092;//stroke in m
+Sp1=10;//mean piston speed
+ma=60;//air flow in g/s
+Vs=2.2*10^-3;//capacity
+
+//CALCULATIONS
+Roa=P/(Ra*Ta);//Density of air in kg/m^3
+N=Sp1/(2*L);//speed in rpm
+nv=(2*ma)/(Roa*Vs*N*1000);//volumetric efficiency in percentage
+
+//OUTPUT
+printf('Volumetric efficiency is %3.2f percentage ',nv)
diff --git a/1808/CH1/EX1.22/Chapter1_Example22.sce b/1808/CH1/EX1.22/Chapter1_Example22.sce new file mode 100644 index 000000000..ce7a6c545 --- /dev/null +++ b/1808/CH1/EX1.22/Chapter1_Example22.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+BP=50;//Brake power in kW
+bmepp=700;//mean effective pressure pickup van
+bmept=850;//mean effective pressure typical
+nc=4;//4-stroke cylinder
+n=2;//for 4 cyliners
+Sp=8;//mean piston speed
+N=3000;//speed in rpm
+L=0.107;//stroke in m
+
+//CALCULATIONS
+Nm=Sp/(2*L);//Design speed in rps
+db=(BP*n/(bmepp*4*Nm*3.14/4))^(1/3);//bore diameter in m^3
+Vt=(3.14*(db^3)*nc/4);//Capacity of the engine in litres
+Tm=(bmept*Vt)/(2*3.14*n);//Maximum torque in kNm
+
+//OUTPUT
+printf('(i)Design speed %3.2f rps \n (ii)capacity of the engine %3.5f m^3 \n (iii)Maximum torque is %3.2f kNm',Nm,Vt,Tm)
+
diff --git a/1808/CH1/EX1.23/Chapter1_Example23.sce b/1808/CH1/EX1.23/Chapter1_Example23.sce new file mode 100644 index 000000000..bcb75693c --- /dev/null +++ b/1808/CH1/EX1.23/Chapter1_Example23.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+Vs=1.5*10^-3;//capacity of cylinder in m^3
+N=3000;//speed in rpm
+BP=48;//break power in kW
+nv=0.92;//volumetric efficiency in percentage
+P=101.325;//atmospheric pressure in kPa
+Ra=0.287;//gas constant
+Ta=303;//atm.temperature in K
+x=21;//airfuel ratio
+
+
+//CALCULATIONS
+Roa=(P/(Ra*Ta));//Density of air in kg/m^3
+ma=(nv*Roa*Vs*N/(2*60));//mass of air in kg/s
+mf=ma/x;//mass of fuel in kg/s
+bsfc=mf*3600/BP;//Brake specific fuel consumption in kg/kWh
+me=ma+mf;;//mass rate of exhaust flow in kg/s
+bpo=(BP/Vs)/1000;//Brake output per displacement in kW/litres
+
+//OUTPUT
+printf('(i)Rate of air flowinto engine %3.5f kg/s \n (ii)Brake specific fuel consumption is %3.3f kg/kWh \n (iii)mass rate of exhaust flow is %3.5f kg/s \n (iv)Brake output per displacement is %3.i kW/litres',ma,bsfc,me,bpo)
diff --git a/1808/CH1/EX1.3/Chapter1_Example3.sce b/1808/CH1/EX1.3/Chapter1_Example3.sce new file mode 100644 index 000000000..895eebc2b --- /dev/null +++ b/1808/CH1/EX1.3/Chapter1_Example3.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+pmb=15;//brake ean pressure in bar
+L=200;//Stroke in cm
+d=0.8;//bore daimeter in cm
+N=100;//speed in rpm
+nc=6;//number of cylinders
+bsfc=0.4;//brake specific fuel consumption
+cv=42000;//calorific value in kJ/kg
+
+//CALCULATIONS
+BP=(pmb*L*(3.14*0.8^2/4)*N*nc)/(60);//Brake power in kW
+mf=bsfc*BP;//Total fuel consumption in kg/h
+nbt=(BP/((mf/3600)*cv))*100;//Brake thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)The Brake power is %3.2f kW \n (ii)Total fuel consumption is is %3.2f kg/hr \n (iii)Brake thermal efficiency is %3.2f percentage ',BP,mf,nbt)
diff --git a/1808/CH1/EX1.4/Chapter1_Example4.sce b/1808/CH1/EX1.4/Chapter1_Example4.sce new file mode 100644 index 000000000..99132bb8e --- /dev/null +++ b/1808/CH1/EX1.4/Chapter1_Example4.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+BP=17;//Brake power in kW
+mf=6;//Mass flow rate in kg/h
+cv=44200;//calorific value in kJ/kg
+L=0.1;//Stroke in m
+d=0.06;//bore in m
+Rc=8;//copression ratio
+n=2;//for four cylinders
+nc=4;//number of cylinders
+N=50;//speed in rps
+
+//CALCULATIONS
+nbt=(BP/((mf/3600)*cv))*100;//Brake thermal efficiency in percentage
+vs=(3.14*d^2*L)/4;//swept volume in m^3
+vc=vs/7;//Clearance volume in m^3
+pmb=((BP*n)/(L*(3.14*d^2/4)*N*nc));//brake ean pressure in kPa
+no=(1-(1/(Rc^(1.4-1))))*100;//Air standard efficiency in percentage
+
+//OUTPUT
+printf('(i)Brake thermal efficiency is %3.2f percentage \n (ii)clearance volume is %3.9f m^3 \n (iii)Brake mean effective pressure is %3.2f kPa \n (iv)air standard efficiency is %3.2f percentage',nbt,vc,pmb,no)
diff --git a/1808/CH1/EX1.5/Chapter1_Example5.sce b/1808/CH1/EX1.5/Chapter1_Example5.sce new file mode 100644 index 000000000..cd372e372 --- /dev/null +++ b/1808/CH1/EX1.5/Chapter1_Example5.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+N=1500;//speed in rpm
+T=150;//Torque developed in Nm
+mf=6;//fuel consumption in kg/h
+cv=42000;//calorific value in kJ/kg
+L=0.15;//Stroke in m
+d=0.12;//bore in m
+n=2;//for four cylinders
+pa=1;//Atmospheric pressure in bar
+Ta=293;//Room temperature in K
+Ra=0.287;//Gas constant
+hw=0.06;//Head of orifice in m
+rw=1000;//density of water in kg/m^3
+
+//CALCULATIONS
+BP=2*3.14*(N/60)*(T/1000);//Brake power in kW
+nbt=(BP/((mf/3600)*cv))*100;//Brake thermal efficiency in percentage
+Pmb=(BP*60*n/(L*(3.14*d^2/4)*N));//brake ean pressure in kPa
+ra=(pa/(Ra*Ta))*100;//density of air in kg/m^3
+ha=((hw*rw)/ra);//Air inhaled in m
+Va=0.62*(3.14*0.03^2/4)*(2*9.81*ha)^(1/2);//Air inhaled in m^3/s
+Vs=((3.14*d^2/4)*L*N/(n*60));//Swept volume in m^3/s
+nv=(Va/Vs)*100;//Volumetric efficiency in percentage
+
+//OUTPUT
+printf('(i)Brake thermal efficiency is %3.2f percentage \n (ii)Brake mean effective pressure is %3.2f kPa \n (iv)Volumetric efficiency is %3.d percentage',nbt,Pmb,nv)
diff --git a/1808/CH1/EX1.6/Chapter1_Example6.sce b/1808/CH1/EX1.6/Chapter1_Example6.sce new file mode 100644 index 000000000..cf9477856 --- /dev/null +++ b/1808/CH1/EX1.6/Chapter1_Example6.sce @@ -0,0 +1,15 @@ +clc
+clear
+//INPUT DATA
+p1=780;//pressure of gas in mm
+p2=760;//pressure of gas in mm
+v1=15;//volume of gas in m^3
+T1=288;//Temperature in k
+T2=273;//Temperature in k
+
+
+//CALCULATIONS
+v2=(p1*v1*(T2/T1))/p2;//volume of gas in m^3
+
+//OUTPUT
+printf('Gas consumption at NTP is %3.3f m^3 ',v2)
diff --git a/1808/CH1/EX1.7/Chapter1_Example7.sce b/1808/CH1/EX1.7/Chapter1_Example7.sce new file mode 100644 index 000000000..fed73a45e --- /dev/null +++ b/1808/CH1/EX1.7/Chapter1_Example7.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+BP=60;//Brake power in kW
+nm=0.8;//mechanical efficiency in percentage
+d=0.15;//bore in m
+L=0.15;//stroke in m
+n=4;//for 6 cylinders
+Ps=510;//piston speed in m/min
+pmi=5;//mean effective pressure in bar
+
+
+//CALCULATIONS
+IP=(BP/nm);//indicated power in kW
+A=(3.14*d^2/4);//area
+ne=(IP*60)/(pmi*100*L*A*n);//No.of explosions
+N=(Ps/(2*L));//speed of the engine in rpm
+
+//OUTPUT
+printf('(i)No.of explosions are %3.2f /min \n (ii)speed of the engine is %3.d rpm ',ne,N)
diff --git a/1808/CH1/EX1.8/Chapter1_Example8.sce b/1808/CH1/EX1.8/Chapter1_Example8.sce new file mode 100644 index 000000000..c870273c4 --- /dev/null +++ b/1808/CH1/EX1.8/Chapter1_Example8.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+no=0.5;//Air standard efficiency in percenatge
+nr=0.7;//relative efficiency in percenatge
+nm=80;//mechanical efficiency in percenatge
+cv=45000;//calorific value in kJ/kg
+BP=75;//Brake power in kW
+Ra=0.287;//Gas constant
+Ta=300;//suction temperature in K
+pa=1*100;//pressure in kN/m^2
+Vs=0.10352;//swept volume in m^3/s
+N=2500;//speed in rpm
+nc=1;//number of cylinders
+n=2;//for four cylinders
+
+//CALCULATIONS
+Rc=(1/(1-no))^(1/(1.4-1));//copression ratio
+nit=(no*nr)*100;//Indicated thermal efficiency in percentage
+IP=(BP/nm)*100;//indicated power in kW
+mf=((IP*100)/(nit*cv));//mass fuel consumption in kg/s
+bsfc=(mf*3600)/BP;//Brake specific fuel consumption in kg/kWh
+nbt=(BP/(mf*cv))*100;//Brake thermal efficiency in percentage
+ma=16*mf;//mass of air in kg/s
+va=(ma*Ra*Ta)/(pa);//actual volume of air consumption
+vs=va/no;//swept volume in m^3/s
+d=(4*60*n*vs/(3.14*1.5*N))^(1/3);//bore in m
+L=1.5*d;//stroke in m
+
+//OUTPUT
+printf('(i)Compression ratio is %3.3f \n(ii)Indicated thermal efficiency is %3.2f percentage \n (iii)Brake specific fuel consumption is %3.4f kg/kWh \n(iv)Brake thermal efficiency is %3.d percentage \n(v)bore %3.2f m \n stroke is %3.2f m',Rc,nit,bsfc,nbt,d,L)
+
diff --git a/1808/CH1/EX1.9/Chapter1_Example9.sce b/1808/CH1/EX1.9/Chapter1_Example9.sce new file mode 100644 index 000000000..c45f315fe --- /dev/null +++ b/1808/CH1/EX1.9/Chapter1_Example9.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+pmi=7;//mean effective pressure in bar
+L=0.45;//stroke in m
+nc=1;//number of cylinders
+N=80;//speed in rpm
+n=1;//for 2 stroke
+nm=80;//Mechanical efficiency in percentage
+
+//CALCULATIONS
+IP=(pmi*100*L*(3.14*(0.3^2)/4)*N*nc)/(60*n);//indicated power in kW
+BP=(nm*IP)/100;//Brake power in kW
+FP=IP-BP;//Frictional power in kW
+
+//OUTPUT
+printf('(i)Indicated power is %3.2f kW \n (ii)brake power is %3.2f kW \n(iii)frictional power is %3.2f kW ',IP,BP,FP)
diff --git a/1808/CH2/EX2.1/Chapter2_Example1.sce b/1808/CH2/EX2.1/Chapter2_Example1.sce new file mode 100644 index 000000000..3e7bbb96e --- /dev/null +++ b/1808/CH2/EX2.1/Chapter2_Example1.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+//GASES IN THE ORDER CO2,CO,O2,N2
+p=[16.1,0.9,7.7,75.3];//percentage of gas
+m=[44,28,32,28];//molecular weight of gas
+
+//CALCULATIONS
+x1=p(1)/m(1);//individual moles per 100 kg of CO2 mixture
+x2=p(2)/m(2);//individual moles per 100 kg of CO mixture
+x3=p(3)/m(3);//individual moles per 100 kg of O2 mixture
+x4=p(4)/m(4);//individual moles per 100 kg of N2 mixture
+x=x1+x2+x3+x4;//Total moles per 100 kg mixture
+v1=(x1/x)*100;//percentage of gases on volume basis
+v2=(x2/x)*100;//percentage of gases on volume basis
+v3=(x3/x)*100;//percentage of gases on volume basis
+v4=(x4/x)*100;//percentage of gases on volume basis
+v=v1+v2+v3+v4;//total percentage of gases on volume basis
+
+//OUTPUT
+printf('(i)percentage of gases on volume basis is %3.3f ',v)
+
+
diff --git a/1808/CH2/EX2.10/Chapter2_Example10.sce b/1808/CH2/EX2.10/Chapter2_Example10.sce new file mode 100644 index 000000000..77b1fe9c0 --- /dev/null +++ b/1808/CH2/EX2.10/Chapter2_Example10.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//0.8062 CH4+0.0541 C2H6 +0.0187C3H8+0.0160C4H10+0.1050N2+a (O2+3.76 N2)=b (0.078 CO2+0.002 CO +0.07 O2 +0.85N2)+ c H2O ;//Combustion equation for 1 kmol of fuel mixture
+//b*(0.078+0.002)=0.8062+2*(0.0541)+3*(0.0160);//by carbon balance
+c=1.93;//Carbon balance
+a=2.892;//Oxygen balance
+
+//CALCULATIONS
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 12.93 (0.078 CO2 )+0.002(0+0.07 O2 +0.85 N2)+1.93 H2O ;//Balanced chemical equation
+xm=a*4.76/1;//Air fuel ratio on molar basis
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 1.0345 CO2+1.93 H2O+7.625 N2 ;//Balanced chemical equation
+xth=2*4.76;//Theoretical air fuel ratio
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)The air fuel ratio on molar basis %3.2f kmol of air/kmol of fuel \n (b)Percentage of theoretical air %3.1f percentage ',xm,nth)
diff --git a/1808/CH2/EX2.11/Chapter2_Example11.sce b/1808/CH2/EX2.11/Chapter2_Example11.sce new file mode 100644 index 000000000..c36e7935a --- /dev/null +++ b/1808/CH2/EX2.11/Chapter2_Example11.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+t1=25;//Temperature
+p1=1;//atmospheric pressure
+//H2+(1/2)O2=H20
+//Qcv+Reactants=Products
+
+//CALCULTIONS
+Qcv=-285838;//Enthalpy in kJ/kmol from the table
+
+//OUTPUT
+printf('(i)Enthalpy of formation of H2O is %3.1f kJ/kmol',Qcv)
+
+
+
diff --git a/1808/CH2/EX2.12/Chapter2_Example12.sce b/1808/CH2/EX2.12/Chapter2_Example12.sce new file mode 100644 index 000000000..0747edef8 --- /dev/null +++ b/1808/CH2/EX2.12/Chapter2_Example12.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//H2+(1/2)O2=H20
+//Qcv+Reactants=Products
+
+//CALCULTIONS
+hf=-241827;//From the tables of enthalpy
+dh=9515;//enthalpy change
+Qcv=hf+dh;//Enthalpy of H2O on kmol basis
+
+//OUTPUT
+printf('(i)Enthalpy of H2O on kmol basis %3.1f kJ/kmol',Qcv)
+
+
+
diff --git a/1808/CH2/EX2.13/Chapter2_Example13.sce b/1808/CH2/EX2.13/Chapter2_Example13.sce new file mode 100644 index 000000000..cdfe7f236 --- /dev/null +++ b/1808/CH2/EX2.13/Chapter2_Example13.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+p=50;//Power output in kW
+m=0.05;//mass flow rate in kg/s
+//C8H18 +2*12.5 O2 +2*12.5*3.76 N2= 8 CO2 +9 H2O +12.5 O2 +99N2 ;//Rate of heat transfer from the engine in kJ/kmol
+//Qcv+x1=x2+Wcv ;//rate of heat transfer
+x1=-249952;//inlet heat transfer in kJ/kmol
+x2=-2415445.5;//exit heat transfer in kJ/kmol
+
+//CALCULATIONS
+Wcv=(p/m)*114.28;//work done in J/kmol of fuel
+Qcv=x2+Wcv-x1;//Heat transfer rate from the engine in kJ/kmol
+
+//OUTPUT
+printf('(i)Heat transfer rate from the engine is %3.1f kJ/kmol',Qcv)
+
+
+
+
+
diff --git a/1808/CH2/EX2.14/Chapter2_Example14.sce b/1808/CH2/EX2.14/Chapter2_Example14.sce new file mode 100644 index 000000000..f52371727 --- /dev/null +++ b/1808/CH2/EX2.14/Chapter2_Example14.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//CH4+2O2=CO2+H2O ;//Combustion equation
+//Q=Up-Ur ;//Energy balance for the closed system
+hfco2=-393520;//enthalpy of CO2 From the table
+dhco2=28041;//change in enthalpy in KJ/kmol
+hfh2o=-241820;//enthalpy of H2O From the table
+dhh2o=21924;//change in enthalpy in KJ/kmol
+hfch4=-74850;//enthalpy of CH4 From the table
+t1=298;//initial temperature in K
+t2=900;//final temperature in K
+p1=1;//Pressure in atm
+R=8.314;//gas constant
+
+//CALCULATIONS
+Q=(hfco2+dhco2)+2*(hfh2o+dhh2o)-(hfch4)+3*R*(t1-t2);//Amount of heat transfer in kJ/kmol
+p2=p1*(t2/t1);//Final pressure in atmosphere
+
+//OUTPUT
+printf('(i)Amount of heat transfer is %3.2f kJ/kmol \n (ii)Final pressure is %3.2f atmosphere',Q,p2 )
+
+
+
diff --git a/1808/CH2/EX2.15/Chapter2_Example15.sce b/1808/CH2/EX2.15/Chapter2_Example15.sce new file mode 100644 index 000000000..72aa0995a --- /dev/null +++ b/1808/CH2/EX2.15/Chapter2_Example15.sce @@ -0,0 +1,43 @@ +clc
+clear
+//INPUT DATA
+//CH4 + 2O2 + 7.52N2=CO2 + 2H2O + 7.52N2 ;//Combustion equation with liquid water in the products
+t1=278;//atmospheric temperature
+t2=1000;//products temperature
+p1=1;//atmospheric pressure
+hfco2=-393520;//Acc.to tables with liquid water in the products enthalpy of CO2
+hfh2o=-285830;//Acc.to tables with liquid water in the products enthalpy of H2O
+hfch4=-74850;//Acc.to tables with liquid water in the products enthalpy of CH4
+
+hfco21=-393520;///Acc.to tables with water vapour in the products enthalpy of CO2
+hfh2o1=-241820;//Acc.to tables with water vapour in the products enthalpy of H2O
+hfch41=-74850;//Acc.to tables with water vapour in the products enthalpy of CH4
+
+h21co2=33368;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of CO2
+h21h2o=25978;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of H2O
+h21n2=21468;//Acc.to tables at 1000 K ,1 atm with water vapour in the products enthalpy of N2
+
+//CALCULATIONS
+hrp=1*hfco2+2*hfh2o-hfch4;//enthalpy of reactants and products in kJ/kmol
+hrpCH4=hrp/16.04;//Enthalpy of combustion of gaseous methane with liquid water in the products in kJ/kg
+
+hrp1=1*hfco21+2*hfh2o1-hfch41;//enthalpy of reactants and products in kJ/kmol
+hrpCH41=hrp1/16.04;//Enthalpy of combustion of gaseous methane with water vapour in the products in kJ/kg
+
+hrp2=1*(hfco21)+(h21co2)+2*(h21h2o)+2*(hfh2o)+7.52*(h21n2)-1*(hfch4);//enthalpy of reactants and products in kJ/kmol
+hrpCH42=hrp2/16.04;//Enthalpy of combustion of gaseous methane at 1000 K ,1atm with water vapour in the products in kJ/kg
+
+dhco2=(42769-9364);//From tables both reactants and products enthalpy
+dhh2o=(35882-9904);//From tables both reactants and products enthalpy
+dho2=(31389-8682);//From tables both reactants and products enthalpy
+dhch4=38189;//From tables both reactants and products enthalpy
+
+hrp3=1*(hfco2+dhco2)+2*(hfh2o1+h21h2o)-(hfch41+dhch4)-2*(dho2);//enthalpy of reactants and products in kJ/kmol
+hrpCH43=hrp3/16.04;//Enthalpy of combustion of gaseous methane at 1000 K ,1atm with water vapour and liqid water in the products in kJ/kg
+
+//OUTPUT
+printf('(i)Enthalpy of combustion of gaseous methane with liquid water in the \n products %3.2f kJ/kg of fuel\n(ii)Enthalpy of combustion of gaseous methane with water \n vapour in the products %3.2f kJ/kg of fuel\n ',hrpCH4,hrpCH41)
+printf('(iii)Enthalpy of combustion of gaseous methane at 1000 K ,1atm \n with water vapour in the products is %3.3f kJ/kg of fuel\n(iv)Enthalpy of combustion of gaseous methane at 1000 K ,1atm \n with water vapour and liqid water is the products is %3.2f kJ/kg of fuel',hrpCH42,hrpCH43)
+
+
+
diff --git a/1808/CH2/EX2.16/Chapter2_Example16.sce b/1808/CH2/EX2.16/Chapter2_Example16.sce new file mode 100644 index 000000000..91385906d --- /dev/null +++ b/1808/CH2/EX2.16/Chapter2_Example16.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+//C2H2+3O2=2CO2+2H2O ;//Chemical equation
+t1=298;//initial temperature in K
+t2=800;//Final temperature in K
+R=0.287;//gas constant in kJ/kgK
+dhCO2=22815;//From tables enthalpy of CO2 kJ/kmol
+dhH2O=17991;//From tables enthalpy of H2O kJ/kmol
+hfCO2=-393520;//From tables enthalpy of CO2 kJ/kmol
+hfH2O=-241827;//From tables enthalpy of H2O kJ/kmol
+hfC2H4=52283;//From tables enthalpy of C2H4 kJ/kmol
+
+//CALCULATIONS
+Q=2*(hfCO2+dhCO2)+2*(hfH2O+dhH2O)-1*(hfC2H4)-4*R*(t2-t1);//amount of heat transfer from the reactants
+
+//OUTPUT
+printf('(i)The amount of heat transfer from the reactants is %3.1f kJ/kmol of fuel ',Q)
+
diff --git a/1808/CH2/EX2.17/Chapter2_Example17.sce b/1808/CH2/EX2.17/Chapter2_Example17.sce new file mode 100644 index 000000000..ce9eb8720 --- /dev/null +++ b/1808/CH2/EX2.17/Chapter2_Example17.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+hc=4;//cylinders
+mf=2;//mass flow rate in g/s
+vd=2.8;//capacity in litres
+N=1500;//speed in rpm
+ta=303;//temperature in K
+Pa=101.325;//atmospheric pressure in kPa
+R=0.287;//gas constant in kJ/kgK
+xs =15.14;//air fuel ratio
+//C8H18+12.5(O2+3.773N2)=8CO2+9H2O+47.16 N2;//Chemical equation
+
+//CALCULATIONS
+ma1=xs*mf;//mass of air in g/s
+ma=ma1/50;//mass of air in g/cylinder/cycle
+ro=Pa/(R*ta);//density in kg/m^3
+nv=ma/(ro*vd/4);//Volumetric efficiency
+
+//OUTPUT
+printf('Volumetric effiiciency is %3.4f ',nv)
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.18/Chapter2_Example18.sce b/1808/CH2/EX2.18/Chapter2_Example18.sce new file mode 100644 index 000000000..2ee095647 --- /dev/null +++ b/1808/CH2/EX2.18/Chapter2_Example18.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//C4H10+6.5(O2+3.773N2) = 4CO2+5H2O+24.5245N2 ;//Chemicl equation
+x=0.9;//equivalent ratio
+
+//CALCULATIONS
+xs=(12*4+1*10)/(31.0245*28.962);//air fuel ratio
+xa=x*xs;//exhaust gas composition
+
+//OUTPUT
+printf('Exhaust gas compositin is %3.4f \n',xa)
+printf('C4H10+1.11*6.5( O2+3.773 N2) = 4 CO2+5 H2O+0.7150 O2+27.22 N2' )
+
+
+
diff --git a/1808/CH2/EX2.19/Chapter2_Example19.sce b/1808/CH2/EX2.19/Chapter2_Example19.sce new file mode 100644 index 000000000..553b942f3 --- /dev/null +++ b/1808/CH2/EX2.19/Chapter2_Example19.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+//CH4 + 2 O2 = CO2+2 H2O ;//Chemical equation
+hfco2=-393.52;//enthalpy of CO2 in MJ/kmol of CH4
+hfh2o=-241.83;//enthalpy of H2O in MJ/kmol of CH4
+hfch4=-77.87;//enthalpy of CH4 in MJ/kmol of CH4
+hfo2=0;//enthalpy of O2 in MJ/kmol of CH4
+
+//CALCULATIONS
+
+Qp=hfco2+2*hfh2o-(hfch4+2*hfo2);//Lower heating value in MJ/kmol
+Qp1=-Qp/(21*1-1*4);//Lower heating value in MJ/kg
+dU=Qp1;//lower heating values
+
+//OUTPUT
+printf('np=nr \n')
+printf('Lower heating values are \n Qp %3.1f MJ/kg of CH4 \n du %3.1f MJ/kg of CH4 \n Qp1=dU \n ',Qp1,dU)
+
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.2/Chapter2_Example2.sce b/1808/CH2/EX2.2/Chapter2_Example2.sce new file mode 100644 index 000000000..74b8d1ef9 --- /dev/null +++ b/1808/CH2/EX2.2/Chapter2_Example2.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//GASES IN THE ORDER CO2,CO,O2,N2
+p=[10,8,1.5,80.5];//percentage of gas
+m=[44,32,28,28];//molecular weight of gas
+
+//CALCULATIONS
+x1=m(1)*(p(1)/100);//individual moles per 100 kg of CO2 mixture
+x2=m(2)*(p(2)/100);//individual moles per 100 kg of CO mixture
+x3=m(3)*(p(3)/100);//individual moles per 100 kg of O2 mixture
+x4=m(4)*(p(4)/100);//individual moles per 100 kg of N2 mixture
+x=x1+x2+x3+x4;//Total moles per 100 kg mixture
+v1=(x1/x)*100;//percentage of gases on volume basis
+v2=(x2/x);//percentage of gases on volume basis
+
+
+
+//OUTPUT
+printf('(a)percentage of each gas by mass is %3.2f \n (b)mass of oxygen per kg of dry flue gases is %3.3f kg',v1,v2)
+
+
+
+
diff --git a/1808/CH2/EX2.20/Chapter2_Example20.sce b/1808/CH2/EX2.20/Chapter2_Example20.sce new file mode 100644 index 000000000..1abfcd59f --- /dev/null +++ b/1808/CH2/EX2.20/Chapter2_Example20.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+mf=0.4;//flue flow rate in g/s
+ma=5.6;//air flow rate in g/s
+hbc=1.8;//gasoline with H/C ratio
+//a CH1.87+b O2+c N2=13 CO2 +2.8 CO +0.933 H2+ d H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+c=83.267;//Composition of N2
+b=22.069;//Composition of O2
+a=15.8;//Composition of CH1.87
+d=13.84;//Composition of H2O
+
+//CALCULATIONS
+xa=mf/ma;//air fuel ratio
+//CH1.87+1.4675(O2+3.773N2)=CO2+0.935 H2O+ 5.536 N2 ;//CHEMICAL COMPOSITION
+xs=(12*1+1*hbc)/(202+86);//air fuel ratio
+x=(xa/xs);//Equivlent ratio from the fuel and air flow rate
+
+//15.8 CH1.87+22.069 O2+83.267 N2=13 CO2 +2.8 CO +0.937 H2+ 13.84 H2O+ 83.267 N2 ;//EXHAUST GAS COMPOSITION
+//CH1.87+1.397 O2+5.27 N2=0.823 CO2 +0.177 CO +0.059 H2+ 0.876 H2O+ 5.27 N2 ;//CALCULATED EXHAUST GAS COMPOSITION
+xa1=(d)/(1.397*32+5.27*28);//air fuel ratio
+x1=xa1/xs;//Equivalent ratio of calculated exhaust gas composition
+
+//OUTPUT
+printf('(i)The Equivlent ratio from the fuel and air flow rate is %3.3f \n (ii)TheEquivalent ratio of calculated exhaust gas composition is %3.3f',x,x1)
+
+
+
+
+
+
diff --git a/1808/CH2/EX2.21/Chapter2_Example21.sce b/1808/CH2/EX2.21/Chapter2_Example21.sce new file mode 100644 index 000000000..8cb5a22c3 --- /dev/null +++ b/1808/CH2/EX2.21/Chapter2_Example21.sce @@ -0,0 +1,13 @@ +clc
+clear
+//INPUT DATA
+//H2+ 0.5(O2+3.773 N2)= H2O+1.887 N2 ;//CHEMICAL EQUATION
+//22.3 H2 +18.606 O2 +70.2 N2 = 22.3 H2O+ 7.44 O2+70.2 N2 ;//EXHAUST GASES CHEMICAL EQUATION
+
+//CALCULATIONS
+xs=2*1/(0.5*32+0.5*3.773*28);//air fuel ratio From the combustion equation
+xa=(1*2)/(0.8343*32+3.148*28);//air fuel ratio From the combustion equation
+x=xa/xs;//Equivalent ratio
+
+//OUTPUT
+printf('Equivalent ratio is %3.3f',x)
diff --git a/1808/CH2/EX2.22/Chapter2_Example22.sce b/1808/CH2/EX2.22/Chapter2_Example22.sce new file mode 100644 index 000000000..dd86f0d71 --- /dev/null +++ b/1808/CH2/EX2.22/Chapter2_Example22.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//C8H18+12.5(O2+3.773N2)=8 CO2 +9 H2O +47.16 N2 ;//FUEL COMPOSITION
+n=60.66;//number of moles of air
+
+//CALCULATIONS
+n1=8+9+47.16;//number of moles of air and product
+xs= 15.14/1;//air fuel ratio
+xs1=1/xs;//fuel air ratio
+Mr=(1/n)*(114.15+59.66*28.96);//Molecular weights of reactants
+Mp=(1/n1)*(8*44+9*18+47.16*28);//Molecular weights of products
+
+//OUTPUT
+printf('(i)number of moles of air and product %3.2f \n (ii)(A/F)s %3.2f \n (F/A)s %3.2f \n (iii)Molecular weights of reactants %3.2f \n Molecular weights of products %3.2f',n1,xs,xs1,Mr,Mp)
+
+
diff --git a/1808/CH2/EX2.23/Chapter2_Example23.sce b/1808/CH2/EX2.23/Chapter2_Example23.sce new file mode 100644 index 000000000..bbb074647 --- /dev/null +++ b/1808/CH2/EX2.23/Chapter2_Example23.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+//CH4+2O2=CO2+2H2O ;//STOICHIOMETRIC REACTION
+//CASE I
+//H2O in the products is liquid
+//CASE II
+//H2O in the products is gas
+Hr=-74.87;//enthalpy of reactants
+Hp1=-964.2;//enthalpy of products
+Hp2=-876.18;//enthalpy of products
+R=8.314*10^-3;//gas constant
+t=298;//initial temperature in K
+
+//CLCULATIONS
+dH1=Hp1-Hr;//Enthalpy increase in MJ/kmol
+dH2=Hp2-Hr
+dU1=dH2-((1-3)*R*t);//internal energy in MJ/kmol
+dU=Hp2;//internal energy in MJ/kmol
+
+//OUTPUT
+printf('Enthalpy increase is %3.2f MJ/kmol of CH4 \n internal energy increase is %3.2f MJ/kmol of CH4 \n',dH2,dU1)
+printf('H2O in the products and internal energy increase are same \n')
+printf('np=nr')
diff --git a/1808/CH2/EX2.24/Chapter2_Example24.sce b/1808/CH2/EX2.24/Chapter2_Example24.sce new file mode 100644 index 000000000..e93581ce1 --- /dev/null +++ b/1808/CH2/EX2.24/Chapter2_Example24.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+//CH2+(3/2) (O2+3.773 N2)= CO2+H2O+5.66N2 ;//STOICHIOMETRIC EQUATION
+dU=-43.2;//Internal energy in MJ/kg
+
+//CALCULATIONS
+dH=dU+(7.66-7.16)*8.3143*10^-3*298/14;//ENTHALPY CHANGE
+Hp=-((1*-393.52)+(-241.8))/(221.4);//enthalpy of products per kg of mixture
+Hr=Hp-((-43.1*14)/(221.4));//Enthalpy of reactants per kg of mixture
+
+//OUTPUT
+printf('enthalpy of products per kg of mixture %3.2f MJ/kg \n enthalpy of reactants per kg of mixture %3.2f MJ/kg',Hp,Hr)
+
+
+
diff --git a/1808/CH2/EX2.25/Chapter2_Example25.sce b/1808/CH2/EX2.25/Chapter2_Example25.sce new file mode 100644 index 000000000..1ed69e3fe --- /dev/null +++ b/1808/CH2/EX2.25/Chapter2_Example25.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//0.506 H2 +0.1 CO+0.26 CH4 +0.04 C4H8 +0.004 O2 +0.03 CO2 +0.06N2 +0.21*8 02+0.79 8 N2= a CO2 +b H20 +c O2 +d N2 ;//COMBUSTION EQUATION
+a=0.55;//CARBON BALANCE
+b=1.186;//HYDROGEN BALANCE
+c=0.621;//OXYGEN BALANCE
+d=6.38;//NITROGEN BALANCE
+
+//CALCULATIONS
+n=a+c+d;//Total moles of dry products
+nCO2=(a/n)*100;//Analysis of products by volume
+nO2=(c/n)*100;//Analysis of products by volume
+nN2=(d/n)*100;//Analysis of products by volume
+
+//OUTPUT
+printf('Analysis of products by volume \n(i)CO2 %3.2f \n (ii)O2 %3.2f \n (iii)N2 %3.2f ',nCO2,nO2,nN2)
diff --git a/1808/CH2/EX2.26/Chapter2_Example26.sce b/1808/CH2/EX2.26/Chapter2_Example26.sce new file mode 100644 index 000000000..e481ac915 --- /dev/null +++ b/1808/CH2/EX2.26/Chapter2_Example26.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+a=7;//Composition
+b=9;//Composition
+//C7H9O0.2813 ;//GIVEN FUEL
+//C7H9O0.2813N0.107 + m O2 + m (79/21) N2 = x CO2 + y H2O + z N2 ;//COMBUSTION EQUATION
+
+//CALCULATIONS
+x=7;//By balancing method composition of CO2
+y=4.5;//By balancing method composition of H2O
+m=9.11;//By balancing method composition of O2
+z=35.4;//By balancing method composition of N2
+
+//C7H9O0.2813N0.107 + 9.11 O2 + 9.11*(79/21) N2 = 7 CO2 + 4.5 H2O + 35.4 N2 ;//COMBUSTION EQUATION CALCULATED
+xs=m*32+m*(79/21)*28/(100);//air fuel ratio
+
+//(ii)Percentage composition of dry flue gases increased by 20% excess air
+//C7H9O0.2813N0.107 + (1.2*9.11) O2 + (1.2*9.11)*(79/21) N2 = 7 CO2 + 4.5 H2O + 1.2*35.4 N2 + (0.2*9.11) O2 ;//COMBUSTION EQUATION CALCULATED
+n=7+4.5+0.2*9.11+1.2*35.4;//Total number of moles of dry flue gases by volume
+nCO2=(7/n)*100;//Percentage composition of dry flue gases by volume
+nO2=(1.822/n)*100;//Percentage composition of dry flue gases by volume
+nN2=(42.48/n)*100;//Percentage composition of dry flue gases by volume
+
+//OUTPUT
+printf('Percentage composition of dry flue gases by volume \n(i)CO2 %3.2f \n (ii)O2 %3.2f \n (iii)N2 %3.2f ',nCO2,nO2,nN2)
+
diff --git a/1808/CH2/EX2.3/Chapter2_Example3.sce b/1808/CH2/EX2.3/Chapter2_Example3.sce new file mode 100644 index 000000000..1d53e20a5 --- /dev/null +++ b/1808/CH2/EX2.3/Chapter2_Example3.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+//C8H18 + XO2 = a Co2 + b H2O// Stoichiometric equation for combustion of octane
+
+//CALCULATIONS
+
+//Carbon balance
+a=8//8C=aC
+
+//Hydrogen balance
+b=9//18H=2*b
+
+//Oxygen balance
+X=(16+9)/2//2XO=2aO+9
+
+//Combustion equation
+Y=3.76*X//moles of nitrogen in the air for Y moles of Oxygen
+N=(X+Y)/1//moles of air to one mole of fuel
+M=(N)*(29/114)//mass of air required for 1 Kg of Fuel
+
+//OUTPUT
+printf("(i)Theoretical air fuel ratio for combustion of octane is %3.2f kg of air/kg",M)
diff --git a/1808/CH2/EX2.4/Chapter2_Example4.sce b/1808/CH2/EX2.4/Chapter2_Example4.sce new file mode 100644 index 000000000..cad029031 --- /dev/null +++ b/1808/CH2/EX2.4/Chapter2_Example4.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+//C8H18 + 12.5 O2 + (12.5*3.76) N2 = 8 Co2 + 9 H2O + 47 N2// Stoichiometric equation for combustion of octane with 100 percent of air
+//C8H18 + 12.5 O2 + (12.5*3.76) N2 = 8 Co2 + 9 H2O + 47 N2// Stoichiometric equation for combustion of octane with 200 percent of air
+a=8;//Carbon balance
+b=9;//Hydrogen balance
+d=94;//Nitrogen balance
+c=12.5;//Oxygen balance
+
+//CALCULATIONS
+x=a+b+c+d;//Total moles of products
+x1=100*a/x;//Molal analysis of CO2
+x2=100*b/x;//Molal analysis of H20
+x3=100*c/x;//Molal analysis of O2
+x4=100*d/x;//Molal analysis of N2
+
+//OUTPUT
+printf('(i)Molal analysis of CO2 is %3.2f percentage \n (ii)Molal analysis of H2O is %3.2f percentage \n (iii)Molal analysis of O2 is %3.2f percentage \n (iv)Molal analysis of N2 is %3.2f percentage',x1,x2,x3,x4)
+
diff --git a/1808/CH2/EX2.5/Chapter2_Example5.sce b/1808/CH2/EX2.5/Chapter2_Example5.sce new file mode 100644 index 000000000..9cdb45e4b --- /dev/null +++ b/1808/CH2/EX2.5/Chapter2_Example5.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+//a.CH4 +b O2 + c N2=10 CO2 + 0.53 CO+2.37 O2 +d H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+c=87.1;//Nitrogen balance
+b=23.16;//(c/b)=3.76
+d=21.60;//d=2a Hydrogen balance
+a=10.54;//Carbon balance
+//10.54.CH4 +23.16 O2 + 87.1 N2=10 CO2 + 0.53 CO+2.37 O2 +21.06 H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+//CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 // Stoichiometric equation for combustion of Methane with 100 percent of air
+
+//CALCULATIONS
+N=(2.2+8.27)/1;//air fuel ratio on mole basis
+M=N*29/(12+4);//air fuel ratio on mass basis
+Nt=(2+7.52)*(29/16);//theoritical air fuel ratio
+nt=(M/Nt)*100;//Percentage theoritical air
+
+//OUTPUT
+printf('(a) CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 \n')
+printf('(b)The air fuel ratio is %3.2f kg of air/kg of fuel \n (c)Percentage theoritical air is %f ',M,nt)
+
diff --git a/1808/CH2/EX2.6/Chapter2_Example6.sce b/1808/CH2/EX2.6/Chapter2_Example6.sce new file mode 100644 index 000000000..cb1da6afd --- /dev/null +++ b/1808/CH2/EX2.6/Chapter2_Example6.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+//CaHb + d O2 + c N2 = 8 Co2 + 0.9 CO +8.8 O2+e H2O + 82.3 N2// Stoichiometric equation for combustion of dry products with 100 percent of air
+c=82.3;//Nitrogen balance
+d=21.9;//(c/d)=3.76
+e=9.3;//Oxygen balance
+a=8.9;//Carbon balance
+b=18.6;//Hydrogen balance
+mf=125.4;//Mass of fuel
+Ma=29;//mass of air
+
+//C8.9H18.6 + 21.9 O2 + 82.3 N2 = 8 Co2 + 0.9 CO +8.8 O2+9.3 H2O + 82.3 N2// Stoichiometric equation for combustion of dry products with 100 percent of air
+
+//CALCULATIONS
+xm=((c+d)*Ma)/mf;//Air fuel ratio on mass basis
+xc=(a*12/(mf))*100;//Carbon composition on mass basis
+xh=(b*1/(mf))*100;//Hydrogen composition on mass basis
+
+//C8.9H18.6 +13.5O2 +(13.5*3.76)N2 = 8.9CO2 +9.3H2O +50.8N2//Theoretical combustion equation on mass basis
+xth=(13.5+50.8)*Ma/(mf);//Air fuel ratio of theoretical air on mass basis
+nth=(xm/xth)*100;//Percentage of theoretical air om mass basis
+
+//OUTPUT
+printf('(a)Air fuel ratio is %3.1f kg of air/kg of fuel \n (b)Composition of fuel on mass basis is %3.1f percentage \n (c)Percentage of theoretical air om mass basis %3.i percentage',xm,xh,nth)
+
+
+
+
diff --git a/1808/CH2/EX2.7/Chapter2_Example7.sce b/1808/CH2/EX2.7/Chapter2_Example7.sce new file mode 100644 index 000000000..1db5e6f6e --- /dev/null +++ b/1808/CH2/EX2.7/Chapter2_Example7.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+//aC4H10+ b O2+c N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+d H2O ;//Combustion equation
+c=82.9;//Nitrogen balance
+b=22.04;//(c/b)=3.76
+a=2.22;//Carbon balance
+d=11.1;//Hydrogen balance
+Ma=29;//mass of air
+
+//2.22C4H10+ 22.04 O2+82.9 N2= 7.8 CO2+1.1 CO +8.2 O2+82.9 N2+11.1 H2O ;//Combustion equation
+
+//CALCULATIONS
+//C4H10+ 9.92 O2+37.37 N2= 3.51 CO2+0.495 CO +3.69 O2+37.37 N2+5 H2O ;//Combustion equation dividing by 2.22 yields one mole of fuel
+xm=((9.92+37.37)*Ma)/(12*4+10);//air fuel ratio on mass basis
+
+//C4H10+ 6.5 O2+(6.5*3.76) N2= 4 CO2++24.44 N2+5 H2O ;//Theoretical Combustion equation
+xth=((6.5+24.44)*Ma)/(12*4+10);//Theoretical air fuel ratio
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)Percentage of theoretical air %3.2f percentage',nth)
+
+
+
+
diff --git a/1808/CH2/EX2.8/Chapter2_Example8.sce b/1808/CH2/EX2.8/Chapter2_Example8.sce new file mode 100644 index 000000000..bf0fddf60 --- /dev/null +++ b/1808/CH2/EX2.8/Chapter2_Example8.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+a=40;//percentage of H2 fuel
+b=10;//percentage of CO fuel
+c=3;//percentage of N2 fuel
+d=40;//percentage of CH4 fuel
+e=5;//percentage of CO2 fuel
+f=2;//percentage of O2 fuel
+N=4.76;//Amount of nitrogen require for complete combustion
+
+//CALCULATIONS
+x=a+b+c+d+e;//Total volumetric analysis of fuel
+//H2+ (1/2)O2= H2O
+X1=a/2;//Moles of oxygen required
+//CO+(1/2)O2=CO2
+X2=b/2;//Moles of oxygen required
+//CH4+2O2=CO2+2H2O
+X3=2*d;//Moles of oxygen required
+X4=-f;//Moles of oxygen required
+Y=X1+X2+X3+X4;//Moles of oxygen required for 100 moles of gas
+Z=Y*N;//Moles of air required for 100 moles of gas
+Z1=Z/x;//Theoritical air required for 1 mole of gas
+
+//OUTPUT
+printf('Theoritical air required for 1 mole of gas %3.1f mole of air',Z1)
+
+
+
+
+
diff --git a/1808/CH2/EX2.9/Chapter2_Example9.sce b/1808/CH2/EX2.9/Chapter2_Example9.sce new file mode 100644 index 000000000..2f82f7c96 --- /dev/null +++ b/1808/CH2/EX2.9/Chapter2_Example9.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+a=74;//Mass of constituent C
+b=4.3;//Mass of constituent H2
+c=2.7;//Mass of constituent S
+d=1.5;//Mass of constituent N2
+e=5.5;//Mass of constituent H2O
+f=5;//Mass of constituent O2
+g=7;//Mass of constituent ash
+
+a1=6.166;//Moles of constituent C
+b1=1.075;//Moles of constituent H2
+c1=0.084;//Moles of constituent S
+d1=0.053;//Moles of constituent N2
+e1=0.3055;//Moles of constituent H2O
+f1=0.156;//Moles of constituent O2
+g1=0;//Moles of constituent ash
+X1=26.955;//Moles of products N2
+
+
+//CALCULATIONS
+//C+O2=CO2
+x1=a1;//Moles of CO2 required
+//H2+(1/2)O2=H20
+x2=b1/2;//Moles of H2 required
+//S+O2=SO2
+x3=c1;//Moles of O2 required
+x4=d1;//Moles of O2 required
+x5=e1;//Moles of O2 required
+x5=f1;//Moles of O2 required
+x6=g1;//Moles of O2 required
+X=x1+x2+x3+x4+x5+x6;//total moles of products
+Y=a1+(b1+e1)+(2*X1)+(X)+c1;//Total moles of products required
+
+//OUTPUT
+printf('For 100 percentage excess air used,Total moles of products required is %3.3f',Y)
+
+
+
diff --git a/1808/CH3/EX3.1/Chapter3_Exampl1.sce b/1808/CH3/EX3.1/Chapter3_Exampl1.sce new file mode 100644 index 000000000..6f2a282b3 --- /dev/null +++ b/1808/CH3/EX3.1/Chapter3_Exampl1.sce @@ -0,0 +1,11 @@ +clc
+clear
+//INPUT DATA
+Tl=30+273;//engine temperature in K
+Th=500+273;//maximum temperature in K
+
+//CALCULATIONS
+nc=((Th-Tl)/Th)*100;//Efficiency of carnot cycle in percentage
+
+//OUTPUT
+printf('Efficiency of carnot cycle is %3.2f percentage ',nc)
diff --git a/1808/CH3/EX3.10/Chapter3_Exampl10.sce b/1808/CH3/EX3.10/Chapter3_Exampl10.sce new file mode 100644 index 000000000..49ff9828f --- /dev/null +++ b/1808/CH3/EX3.10/Chapter3_Exampl10.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+p1=1;//pressure in atm
+t3=1700;//temperature in K
+Rc=7;//compression ratio
+R=0.287;//gas constant
+cv=0.7234;//calorific value
+
+//CALCULATIONS
+t2=t1*(Rc^(1.4-1));//temperature in K
+p2=p1*(Rc^(1.4));//pressure in atm
+p3=p2*t3/(t2);//pressure in atm
+t4=t3/((Rc)^(1.4-1));//temperature in K
+p4=p3/(Rc^(1.4));//pressure in atm
+no=1-((1/Rc)^(1.4-1))*100;//Thermal efficiency in percentage
+Rwo=1-((t1/t3)*((Rc)^(1.4-1)));//work ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+wn=(cv*(t3-t2))-((cv*(t4-t1)));//net work
+pm=(wn/(v1*(1-(1/Rc))));//mean effective pressure in Bar
+
+//OUTPUT
+printf('(a)pressure at state point 2 is %3.2f atm \n temperature at point 2 is %3.2f K \n pressure at state point 3 is %3.2f atm \n temperature at point 4 is %3.2f K \n pressure at state point 3 is %3.2f atm \n (b)Thermal efficiency is %3.2f percentage \n (c)work ratio is %3.5f \n (d)mean effective pressure is %3.2f Bar',p2,t2,p3,t4,p4,no,Rwo,pm)
+
diff --git a/1808/CH3/EX3.11/Chapter3_Exampl11.sce b/1808/CH3/EX3.11/Chapter3_Exampl11.sce new file mode 100644 index 000000000..09e1637ef --- /dev/null +++ b/1808/CH3/EX3.11/Chapter3_Exampl11.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+d=0.26;//bore of the engine in m
+L=0.38;//stroke of the engine in m
+vc=0.0025;//clearence volume in m^3
+p1=1;//pressure in bar
+t1=313;//temperature in K
+p3=25;//pressure in bar
+v12=9.07;//volume in m^3
+
+//CALCULATIONS
+vs=(3.14*d^2*L)/4;//swept volume in m^3
+Rc=((vs+vc)/vc);//compression ratio
+no=(1-((1/Rc)^(1.4-1)))*100;//Air standard efficiency of the cycle
+p2=p1*(v12^1.4);//pressure in bar
+Rp=p3/p2;//compression pressure
+pm=(p1*Rc*((Rc^(1.4-1)-1)*(Rp-1)))/((1.4-1)*(Rc-1));//mean effective pressure in bar
+
+//OUTPUT
+printf('(a)The air standard efficiency of the cycle is %3.1f percentage \n (b)The mean effective pressure is %3.4f bar',no,pm)
+
diff --git a/1808/CH3/EX3.13/Chapter3_Exampl13.sce b/1808/CH3/EX3.13/Chapter3_Exampl13.sce new file mode 100644 index 000000000..d7b43d7b2 --- /dev/null +++ b/1808/CH3/EX3.13/Chapter3_Exampl13.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+v1=0.5;//volume in m^3
+p1=1;//pressure in bar
+t1=303;//temperature in K
+p2=12;//pressure in bar
+Qs=250;//heat is added in kJ
+wc=200;//working cycles in cycles/min
+v2=0.085;//volume in m^3
+m=1;//mass of air
+cv=0.7243;//calorific value
+
+//CALCULATIONS
+Rc=(p2/p1)^(1/1.4);//compression ratio
+t2=t1*((Rc)^(1.4-1));//temperature in K
+pc=((v2/(v1-v2))*100);//percentage clearance
+t3=(Qs/(m*cv))+t2;//temperature in K
+t4=((1/Rc)^(1.4-1))*t3;//temperature in K
+Qr=m*cv*(t4-t1);//heat rejected in kJ/kg
+no=((Qs-Qr)/Qs)*100;//thermal efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//mean effective pressure
+p=((Qs-Qr)*wc)/60;//power developed in kJ/s
+
+//OUTPUT
+printf('(a)percentage clearance is %3.2f percentage \n (b)the thermal efficiency is %3.2f percentage \n (c)mean effective pressure is %3.2f \n (d)power developed is %3.2f kJ/s',pc,no,pm,p)
+
diff --git a/1808/CH3/EX3.15/Chapter3_Exampl15.sce b/1808/CH3/EX3.15/Chapter3_Exampl15.sce new file mode 100644 index 000000000..b08bde254 --- /dev/null +++ b/1808/CH3/EX3.15/Chapter3_Exampl15.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1300;//temperature in K
+cp=5.22;//specific pressure
+cv=3.13;//specific volume
+g=1.688;//for helium as working medium
+
+//CALCULATIONS
+Rc=((t3/t1)^(1/(2*(1.4-1))));//compression ratio
+no1=(1-((1/Rc)^(1.4-1)))*100;//efficiency of air
+Rcn=((t3/t1)^(1/(2*(g-1))));//compression ratio
+no2=(1-((1/Rcn)^(g-1)))*100;//efficiency of helium
+
+//OUTPUT
+printf('(a)air as working medium efficiency is %3.2f percentage \n (b)Helium as working medium efficiency is %3.2f percentage \n Hence the change in efficiency is zero',no1,no2)
+
diff --git a/1808/CH3/EX3.16/Chapter3_Exampl16.sce b/1808/CH3/EX3.16/Chapter3_Exampl16.sce new file mode 100644 index 000000000..c0ed714e7 --- /dev/null +++ b/1808/CH3/EX3.16/Chapter3_Exampl16.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1500;//temperature in K
+cv=0.7243;//calorific value
+m=0.4;//air flow rate in kg/min
+
+//CALCULATIONS
+t2=sqrt(t1*t3);//temperature in K
+Wnmax=cv*((t3-t2)-(t2-t1));//maximum workdone in kJ/kg
+Pnmax=m*Wnmax/60;//maximum power developed in kJ/s
+
+//OUTPUT
+printf('(a)Maximum power developed is %3.3f kJ/s ',Pnmax)
+
diff --git a/1808/CH3/EX3.17/Chapter3_Exampl17.sce b/1808/CH3/EX3.17/Chapter3_Exampl17.sce new file mode 100644 index 000000000..18a0d0eaf --- /dev/null +++ b/1808/CH3/EX3.17/Chapter3_Exampl17.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+Rc=18;//compression ratio
+t1=300;//temperature in K
+t3=1700;//temperature in K
+p1=101.325;//Pressure in kN/m^2
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume in kJ/kgK
+R=0.287;//gas constant in kJ/kgK
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//temperature at point 2 in K
+p2=p1*Rc;//Pressure at point 2 in atm
+r=t3/t2;//cut off ratio
+t4=t3*((r/Rc)^(g-1));//temperature in at point 4 in K
+p4=1*(t4/t1);//Pressure at point 4 in atm
+w12=cv*(t2-t1);//workdone 1-2 statein kJ/kg
+w23=R*(t3-t2);//workdoneat 2-3 statein kJ/kg
+w34=cv*(t3-t4);//workdone at 3-4 state in kJ/kg
+wn=w23+w34-w12;//workdone by the system
+Qs=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=wn*100/Qs;//Thermal efficiency in percentage
+Rw=wn/(w23+w34);//work ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+pm=wn/(v1*(1-(1/Rc)))/100;//Mean effective pressure in bar
+
+//OUTPUT
+printf('(a)pressure in each cycle is %3.3f atm \n (b)Specific work output is %3.2f kJ/kg \n (c)the Thermal efficiency is %3.3f percentage \n (d)The work ratio is %3.4f \n (e)Mean effective pressure is %3.3f bar',p4,wn,nd,Rw,pm)
+
diff --git a/1808/CH3/EX3.18/Chapter3_Exampl18.sce b/1808/CH3/EX3.18/Chapter3_Exampl18.sce new file mode 100644 index 000000000..e91c3483e --- /dev/null +++ b/1808/CH3/EX3.18/Chapter3_Exampl18.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+p1=101.325;//Pressure in kN/m^2
+t1=303;//Temperature in K
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume in kJ/kgK
+R=0.287;//gas constant in kJ/kgK
+r=2.5;//cut off ratio
+v2=0.1;//clearance volume in m^3/kg
+
+
+//CALCULATIONS
+v1=R*t1/p1;//volume at state 1 in m^3/kg
+p2=p1*((v1/v2)^(g));//pressure at state 2 in kN/m^2
+t2=t1*((v1/v2));//temperature at state 2 in K
+v3=r*v2;//volume at state 3 in m^3/kg
+t3=t2*(v3/v2);//Temperature at state 3 in K
+t4=t3*((v3/v1)^(g-1));//Temperature at state 4 in K
+p4=p2*((v3/v1)^g);//Pressure at state 4 in kN/m^2
+c=(v2/(v1-v2))*100;//percentage clearance
+Qs=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=((Qs-Qr)/Qs)*100;//Thermal efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//Mean effective pressure in kN/m^2
+
+//OUTPUT
+printf('(a)pressure at state 2 is %3.2f kN/m^2 \n temperature at state 2 is %3.2f K \n volume at state 3 is %3.2f m^3/kg\n Temperature at state 3 is %3.2f K \n Temperature at state 4 is %3.2f K \n Pressure at state 4 is %3.2f kN/m^2 \n (b)percentage clearance is %3.2f percentage \n(c)Thermal efficiency is %3.2f percentage \n (d)Mean effective pressure is %3.2f kN/m^2',p2,t2,v3,t3,t4,p4,c,nd,pm)
+
diff --git a/1808/CH3/EX3.19/Chapter3_Exampl19.sce b/1808/CH3/EX3.19/Chapter3_Exampl19.sce new file mode 100644 index 000000000..ea0fdfa85 --- /dev/null +++ b/1808/CH3/EX3.19/Chapter3_Exampl19.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+d=0.12;////bore of the engine in m
+l=0.13;//stroke of the engine in m
+p1=101.325;//pressure in atm
+t1=298;//temperature in K
+t3=1773;//temperature in K
+n=2000;//speed in rpm
+g=1.4;//constant
+cp=1.005;//specific pressure in kJ/kgK
+cv=0.718;//specific volume inkJ/kgK
+R=0.287;//gas constant inkJ/kgK
+
+//CALCULATIONS
+Rc=1.1/0.1;//compression ratio
+v1=R*t1/p1;//specific volume in m^3/kg
+v2=v1/Rc;//specific volume in m^3/kg
+t2=t1*((v1/v2)^(g-1));//temperature in K
+p2=p1*(v1/v2)^(g);//pressure in kN/m^2
+v3=v2*(t3/t2);//specific volume in m^3/kg
+t4=t3*((v3/v1)^(g-1));//temperature in K
+p4=p2*(v3/v1)^g;//pressure in kN/m^2
+Q3=cp*(t3-t2);//heat added in kJ/kg
+Qr=cv*(t4-t1);//heat rejected in kJ/kg
+nd=((Q3-Qr)/Q3)*100;//Thermal efficiency in percentage
+V1=1.1*(3.14*d^2*l)/4;//volume in m^3
+m=4*(n/2)*(V1/(v1*240));//flow rate in kg/s
+P=(Q3-Qr)*m;//Power of the engine in kg/s
+
+//OUTPUT
+printf('(a)compression ratio is %3.i \n (b)pressure and temperature at the end of compression is %3.2f kN/m^2 \n (c)Thermal efficiency is %3.2f percentage \n (d)Power of the engine is %2.3f kg/s ',Rc,p2,nd,P)
+
+
+
diff --git a/1808/CH3/EX3.2/Chapter3_Exampl2.sce b/1808/CH3/EX3.2/Chapter3_Exampl2.sce new file mode 100644 index 000000000..8b0234ade --- /dev/null +++ b/1808/CH3/EX3.2/Chapter3_Exampl2.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+Tl=150;//engine temprature in Degree C
+Th=1100;//engine temprature in Degree C
+Qs=4000;//Heat is added in kJ/min
+
+//CALCULATIONS
+nc=((Th-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+wd=nc*Qs/100;//workdone in kJ/min
+P=wd/(60);//power developed in kJ/s
+Qr=Qs-wd;//Quality of heat rejected in kJ/min
+ds=(Qs-wd)/(Tl+273);//Change in entropy during heat rejection in kJ/min
+
+//OUTPUT
+printf('(a)power developed in the engine is %3.2f kJ/s \n (b)Quality of heat rejected is %3.2f kJ/min \n (c)Change in entropy during heat rejection is %3.2f kJ/min',P,Qr,ds)
+
diff --git a/1808/CH3/EX3.20/Chapter3_Exampl20.sce b/1808/CH3/EX3.20/Chapter3_Exampl20.sce new file mode 100644 index 000000000..d480c85f7 --- /dev/null +++ b/1808/CH3/EX3.20/Chapter3_Exampl20.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+Rc=15;//compression ratio
+r1=1.84;//cutoff ratio
+r2=1.98;//cutoff ratio
+g=1.4;//constant
+p1=101.325;//Pressure in kN/m^2
+Rc3=17;//compression ratio
+r3=1.84;//cutoff ratio
+Rc4=18;//compression ratio
+r4=1.88;//cutoff ratio
+
+//CALCULATIONS
+nd1=(1-(((1/(Rc^(g-1))))*(((r1^g)-1)/((r1-1)*g))))*100;//Air standard efficiency in precentage
+pm1=(p1/((Rc-1)*(g-1)))*(((Rc^g)*g*(r1-1))-(Rc*((r1^g)-1)));//Mean effective pressure in kN/m^2
+nd2=1-(((1/(Rc^(g-1))))*(((r2^g)-1)/((r2-1)*g)));//change in efficiency in precentage
+ndd1=nd1-nd2;//change in efficiency in precentage
+pm2=(p1/((Rc-1)*(g-1)))*(((Rc^g)*g*(r2-1))-(Rc*((r2^g)-1)));//Mean effective pressure in kN/m^2
+pmii=((pm2-pm1)/pm1)*100;//Increase in mep in percentage
+nd3=(1-(((1/(Rc3^(g-1))))*(((r3^g)-1)/((r3-1)*g))))*100;//increased efficiency in precentage
+ni=nd3-nd1;//increased efficiency in percentage
+pm3=(p1/((Rc3-1)*(g-1)))*(((Rc3^g)*g*(r3-1))-(Rc*((r3^g)-1)));//Increase in Mean effective pressure in kN/m^2
+pmi=((pm3-pm1)/(2*pm1))*100;//Increase in Mean effective pressure in percentage
+K=((r4-1)/(Rc4-1))*100;//change in cutoff of stroke
+
+//OUTPUT
+printf('(a)Air standard efficiency is %3.2f precentage \n (b)Mean effective pressure is %3.2f kN/m^2 \n (II)\n (a1)Percentage change in efficiency is %3.2f percentage \n (b1)Increase in Mean effective pressure is %3.2f percentage \n (III) \n (a3)increased efficiency is %3.3f percentage \n (b3)Increase in Mean effective pressure is %3.2f kN/m^2 \n (IV)change in cutoff of stroke %3.1f percentage',nd1,pm1,ndd1,pmii,ni,pmi,K)
+
diff --git a/1808/CH3/EX3.21/Chapter3_Exampl21.sce b/1808/CH3/EX3.21/Chapter3_Exampl21.sce new file mode 100644 index 000000000..0768747a3 --- /dev/null +++ b/1808/CH3/EX3.21/Chapter3_Exampl21.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+l=30;//Stroke in cm
+d=17;//Bore in cm
+vc=440*10^6;//Clearance volume in m^3
+r=0.05;//cutoff ratio
+g=1.4;//constant
+
+//CALCULATIONS
+vs=((3.14*(d^2)*l)*10^6)/4;//swept volume in m^3
+v1=vs+vc;//Total volume in m^3
+v3=vc+(r*(v1-vc));//volume at point of cutoff
+ro=v3/vc;//cutoff ratio
+Rc=(vs+vc)/vc;//compression ratio
+nd=(1-(((1/(Rc^(g-1))))*(((ro^g)-1)/((ro-1)*g))))*100;//Air standard efficiency in precentage
+
+//OUTPUT
+printf('Air standard efficiency is %3.2f precentage',nd)
diff --git a/1808/CH3/EX3.22/Chapter3_Exampl22.sce b/1808/CH3/EX3.22/Chapter3_Exampl22.sce new file mode 100644 index 000000000..a1967f94c --- /dev/null +++ b/1808/CH3/EX3.22/Chapter3_Exampl22.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+Rc=17;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+ro=2.28;//cutoff ratio
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+v1=0.06375;//specific volume in m^3/s
+v2=0.00375;//specific volume in m^3/s
+t4=960.62;//temperature in K
+
+//CALCULATIONS
+p3=p1*(Rc^(g));//maximum pressure in kN/m^2
+t2=t1*(Rc^(g-1));//temperature in K
+t3=t2*(ro);//maximum temperature in K
+nd=(1-(((1/(Rc^(g-1))))*(((ro^g)-1)/((ro-1)*g))))*100;//Air standard efficiency in precentage
+Qs=cp*(t3-t2);//heat supplied in kJ/kg
+wn=nd*Qs/100;//workdone in KJ/kg
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+P=wn*m;//power developed in kJ/s
+wt=(((g*(ro-1)*(Rc^(g-1)))-((ro^g)-1))/((g*(ro-1)*(Rc^(g-1)))-((ro^g)-(Rc^(g-1)))));//workdone in kJ/kg
+Rw=wn/wt;//work ratio
+w12=cv*(t1-t2);//workdone in 1-2 process
+w23=R*t2*(ro-1);//workdone in 2-3 process
+w34=cv*(t3-t4);//orkdone in 3-4 process
+pw=w23+w34;//positive work
+Rw=(wn/pw);//Work ratio
+
+//OUTPUT
+printf('(a)The maximum pressure %3.2f kN/m^2 \n temperature is %3.2f K \n (b)The thermal efficiency is %3.2f percentage \n(c)The power developed is %3.4f kg/s \n (d)work ratio is %3.4f ',p3,t3,nd,P,Rw)
+
+
+
diff --git a/1808/CH3/EX3.23/Chapter3_Exampl23.sce b/1808/CH3/EX3.23/Chapter3_Exampl23.sce new file mode 100644 index 000000000..13274156a --- /dev/null +++ b/1808/CH3/EX3.23/Chapter3_Exampl23.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+l=20;//Stroke in cm
+d=15;//Bore in cm
+N=400;//speed in rpm
+Rc=22;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+n1=1.3;//no of cycles
+n2=1.35;//no of cycles
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constnat
+v1=0.003698;//specific volume in m^3/s
+vs=0.00353;//specific volume in m^3/s
+ro=2.68;//cutoff ratio
+
+//CALCULATIONS
+p2=p1*(Rc^(n1));//pressure in kN/m^2
+t2=t1*(Rc^(n1-1));//temperature in K
+t3=t2*(ro);//temperature in K
+p4=p2*((ro/Rc)^n2);//maximum pressure in kN/m^2
+t4=t3*(1/((Rc/ro)^(n2-1)));//maximum temperature in K
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+wn=R*((t3-t2)+((t3-t4)/(n2-1))-((t2-t1)/(n1-1)));//work done in kJ/kg
+pm=wn*m/(vs);//mean effective pressure in kN/m^2
+Qs=cp*(t3-t2);//heat supplied in kJ/kg
+nd=(wn/Qs)*100;//thermal efficiency in percentage
+P=wn*m*N/60;//POWER DEVELOPED
+
+//OUTPUT
+printf('(a)The temperature and pressure at all corner points are \n pressure at point 1 is %3.2f kN/m^2 \n temperature at point 2 is %3.2f K \n temperature at point 3 is %3.2f K \n maximum pressure is %3.2f kN/m^2 \n temperature at point 4 is %3.2f K \n(b)The mean effective pressure is %3.3f kN/m^2 \n (c)Thermal efficiency is %3.2f percentage \n (d)Power developed is %3.2f kJ/s ',p2,t2,t3,p4,t4,pm,nd,P)
+
diff --git a/1808/CH3/EX3.24/Chapter3_Exampl24.sce b/1808/CH3/EX3.24/Chapter3_Exampl24.sce new file mode 100644 index 000000000..7d9fb3669 --- /dev/null +++ b/1808/CH3/EX3.24/Chapter3_Exampl24.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+Rc=10;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+Qr=350;//heat rejected in kJ/kg
+Qs=450;//heat supplied in kJ/kg
+QR=452.92;//heat supplied in kJ/kg
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+
+//CALCULATIONS
+Q=Qr+Qs;//total heat in kJ/kg
+p2=p1*(Rc^g);//pressure in kN/m^2
+t2=t1*(Rc^(g-1));//temperature in K
+t3=(Qs/cv)+t2;//temperature in K
+p4=p2*(t3/t2);//Maximum pressure in kN/m^2
+t4=(Qr/cp)+t3;//Maximum temperature in K
+wn=Q-QR;//workdone in kJ/kg
+v43=t4/t3;//volume ratio
+v12=t1/t3;//volume ratio
+v45=v43/Rc;//volume ratio
+t5=t4*(v45^(g-1));//temperature in K
+nd=(wn/Q)*100;//thermal efficiency in percentage
+v1=R*t1/p1;//specific volume in m^3/kg
+v2=v1/10;//specific volume in m^3/kg
+pm=wn/(v1-v2);//mean effective pressure in kN/m^2
+w34=R*(t4-t3);//workdone in 3-4 process
+w45=R*(t4-t5);//workdone in 4-5 process
+Rw=(wn/(2*(w34+w45)));//work ratio
+
+//OUTPUT
+printf('(a)The maximum pressure is %3.2f kN/m^2 \n maxium temperature is %3.2f K \n (b)thermal efficiency is %3.2f percentage \n (c)mean effective pressure is %3.2f kN/m^2 \n (d)work ratio is %3.3f ',p4,t4,nd,pm,Rw)
+
+
+
diff --git a/1808/CH3/EX3.25/Chapter3_Exampl25.sce b/1808/CH3/EX3.25/Chapter3_Exampl25.sce new file mode 100644 index 000000000..ba73a018d --- /dev/null +++ b/1808/CH3/EX3.25/Chapter3_Exampl25.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+Rc=20;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=300;//temperature in K
+p32=2;//pressure ratio of heating process
+v43=1.5;//volume ratio of heating process
+g=1.4;//constant
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//Temperature in K
+p2=p1*(Rc^g);//pressure in kN/m^2
+p3=p32*p2;//pressure in kN/m^2
+t3=p32*t2;//Temperature in K
+t4=v43*t3;//Temperature in K
+p5=p3/((Rc/v43)^g);//pressure in kN/m^2
+t5=t4/((Rc/v43)^(g-1));//Temperature in K
+nd=(((cv*(t3-t2))+(cp*(t4-t3))-(cv*(t5-t1)))/((cv*(t3-t2)+(cp*(t4-t3)))))*100;//The thermal efficiency in percentage
+x=((cv*(t3-t2)+(cp*(t4-t3))));//workdone
+y=(cv*(t5-t1));//workdone
+v1=R*t1/p1;//specific volume in m^3/kg
+pm=(x-y)/(v1*(1-(1/Rc)));//mean effective pressure in kN/m^2
+
+//OUTPUT
+printf('(a)The thermal efficiency is %3.2f percentage \n (b)The mean effective pressure is %3.1f kN/m^2',nd,pm)
+
+
+
diff --git a/1808/CH3/EX3.26/Chapter3_Exampl26.sce b/1808/CH3/EX3.26/Chapter3_Exampl26.sce new file mode 100644 index 000000000..c2fda1602 --- /dev/null +++ b/1808/CH3/EX3.26/Chapter3_Exampl26.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+p1=120;//Pressure in kN/m^2
+t1=303;//temperature in K
+v1=0.0708;//specific volume in m^3/s
+v2=0.004165;//specific volume in m^3/s
+t3=1423;//temperature in K
+t4=1873;//temperature in K
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+
+//CALCULATIONS
+ro=t4/t3;//cutoff ratio
+Rc=v1/v2;//Compression ratio
+t2=t1*(Rc^(g-1));//temperature in K
+v45=(ro/Rc);//specific volume in m^3/s
+t5=t4*((v45)^(g-1));//temperature in K
+Qs=cv*(t3-t2)+cp*(t4-t3);//heat added in kJ/kg
+Qr=cv*(t5-t1);//heat rejected in kJ/kg
+nd=((Qs-Qr)/Qs)*100;//thermal efficiency in percentage
+
+//OUTPUT
+printf('(a)cutoff ratio %3.3f \n (b)Compression ratio is %3.1f \n (c)Heat added is %3.2f kJ/kg \n heat rejected is %3.2f kJ/kg \n (d)The thermal efficiency in %3.2f percentage',ro,Rc,Qs,Qr,nd)
+
+
+
diff --git a/1808/CH3/EX3.27/Chapter3_Exampl27.sce b/1808/CH3/EX3.27/Chapter3_Exampl27.sce new file mode 100644 index 000000000..cc18c4b24 --- /dev/null +++ b/1808/CH3/EX3.27/Chapter3_Exampl27.sce @@ -0,0 +1,38 @@ +clc
+clear
+//INPUT DATA
+l=22;//Stroke in cm
+d=15;//Bore in cm
+Rc=10;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+g=1.4;//constnat
+cp=1.005;//specific pressure
+cv=0.718;//specific volumespecific volume
+R=0.287;//gas constant
+n=1.3;//no of flows
+v1=0.00433;//volume in m^3
+
+//CALCULATIONS
+t2=t1*(Rc^(g-1));//Temperature in K
+p2=p1*(Rc^g);//pressure in kN/m^2
+ro=Rc/6;//cutoff ratio
+t3=1133.5;//temperature in K
+t4=(Rc/6)*t3;//temperature in K
+wd=3.43;//workdone per cycle in kN/m
+p3=p2*t3/t2;//pressure in kN/m^2
+p5=p3*(1/6)^n;//pressure in kN/m^2
+pm=((p3*(ro-1))+((p3*ro-p5*Rc)-(p2-p1*Rc))*(1/(n-1)))/(Rc-1);//mean effective pressure in kN/m^2
+pm1=pm/100;//mean effective pressure in bar
+vs=3.14*d^2*l/4;//stroke volume in m^3
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+Qs=m*((cv*(t3-t2))+cp*(t4-t3));//heat supplied in kJ/cycle
+nd=(wd/Qs)*100;//thermal efficiency in percentage
+p=wd*400/60;//power of the engine in kJ/s
+Rw=((p3*(ro-1))+(1/(n-1))*((p3*ro-p5*Rc)-(p2-p1*Rc)))/((p3*(ro-1))+(1/(n-1))*(p3*ro-p5*Rc));//work ratio
+
+//OUTPUT
+printf('(a)The temperature and pressure are \n p2 %3.2f kN/m^2 \n p3 %3.2f kN/m^2 \n p5 %3.2f kN/m^2 \n t2 %3.2f K \n t3 %3.2f K \n t4 %3.2f K \n (b)mean effective pressure is %3.2f bar \n (c)thermal efficiency is %3.2f percentage \n (d)power of the engine is %3.2f kJ/s \n (e)The work ratio is %3.1f ',p2,p3,p5,t2,t3,t4,pm1,nd,p,Rw)
+
+
+
diff --git a/1808/CH3/EX3.28/Chapter3_Exampl28.sce b/1808/CH3/EX3.28/Chapter3_Exampl28.sce new file mode 100644 index 000000000..353e6574c --- /dev/null +++ b/1808/CH3/EX3.28/Chapter3_Exampl28.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+vs=0.01;//Swept volume in m^3
+Rc=18;//compression ratio
+p1=101.325;//Pressure in kN/m^2
+t1=303;//temperature in K
+p3=80*10^2;//pressure in kN/m^2
+g=1.4;//constant
+v2=0.000588;//volume in m^3
+v43=0.0006;//difference in pressure
+
+//CALCULATIONS
+p2=p1*(Rc^g);//pressure in kN/m^2
+a=p3/p2;//Pressure ratio
+v4=v2+v43;//volume in m^3
+ro=v4/v2;//cutoff ratio
+v1=v2+0.01;//volume of cylinder in m^3
+m=p1*v1/(R*t1);//mass of air contained in cylinder in kg
+t2=t1*(Rc^(g-1));//temperature in K
+t3=t2*(p3/p2);//temperature in K
+t4=t3*(v4/v2);//temperature in K
+Qs=(cv*(t3-t2)+cp*(t4-t3))*0.01234;//heat added in kJ
+t5=t4/((v1/v4)^(g-1));//temperature in K
+Qr=cv*(t5-t1)*0.01234;//Heat rejected in kJ
+wn=(Qs-Qr);//workdone per cycle
+nd=(wn/Qs)*100;//Thermal efficiency in percentage
+pm=(wn/vs);//mean effective pressure in kN/m^2
+p5=p1*(t5/t1);//pressure in kN/m^2
+wp=p3*(v4-v2)+((p3*v4-p5*v1)/(g-1));//positive work done
+Rw=wn/wp;//work ratio
+
+//OUTPUT
+printf('(a)Pressure ratio is %3.2f \n (b)cutoff ratio is %3.4f \n (c)mass of air contained in cylinder is %3.5f kg \n (d)Heat added is %3.2f kJ \n (e)Heat rejected is %3.3f kJ \n (f)workdone per cycle is %3.2f kJ\n (g)Thermal efficiency is %3.2f percentage \n (h)Mean effective pressure is %3.1f kN/m^2 \n (i)The work ratio is %3.4f ',a,ro,m,Qs,Qr,wn,nd,pm,Rw)
+
+
+
diff --git a/1808/CH3/EX3.29/Chapter3_Exampl29.sce b/1808/CH3/EX3.29/Chapter3_Exampl29.sce new file mode 100644 index 000000000..a89bcb8c3 --- /dev/null +++ b/1808/CH3/EX3.29/Chapter3_Exampl29.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=303;//temperature in Degree C
+g=1.4;//constant
+t2=700;//temperature in Degree C
+v1=0.05;//volume in m^3
+Rc=10;//compression ratio
+nr=0.9;//regenerator efficiency in percentage
+t=30
+//CALCULATIONS
+m=p1*v1/(R*t1);//mass flow rate
+
+wn=m*R*log(Rc*(t2-t));//Net workdone in kJ
+ns=(1-((t+273)/(t2+273)))*100;//Thermal efficiency with 100% refrigerator in percentage
+Qs=m*cv*(t2-t)+(m*R*(273+t2)*log(Rc));//heat added in kJ
+Qr=m*cv*(t2-t)+(m*R*(273+t)*log(Rc));//heat added in kJ
+nso=(1-(Qr/Qs))*100;//Thermal efficiency without refrigerator in percentage
+nsa=(((R*(t2-t)*log(Rc)))/((R*(273+t2)*log(Rc))+((1-nr)*cv*(t2-t))))*100;//Thermal efficiency with 90% refrigerator in percentage
+
+//OUTPUT
+printf('(i)net workdone is %3.2f kJ \n (ii)Thermal efficiency with 100 percentage efficiency is %3.2f percentage \n (iii)Thermal efficiency without regenerator is %3.2f percentage \n (iv)Thermal efficiency with 90percentage efficiency is %3.2f percentage \n',wn,ns,nso,nsa )
diff --git a/1808/CH3/EX3.3/Chapter3_Exampl3.sce b/1808/CH3/EX3.3/Chapter3_Exampl3.sce new file mode 100644 index 000000000..7e7c863b4 --- /dev/null +++ b/1808/CH3/EX3.3/Chapter3_Exampl3.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+Tl=300;//engine temprature in Degree C
+Th=1500;//engine temprature in Degree C
+Fc=0.45;//Fuel consumption in kg/hr
+cv=40000//kJ/kg
+wd=4;//workdone in kW
+
+//CALCULATIONS
+nc=((Th-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+Qs=Fc*cv;//Heat is added in kJ/min
+na=(wd/(Qs))*(3600*100);//efficiency developed by scientist in percentage
+
+//OUTPUT
+printf('(a)Efficiency of carnot cycle is %3.2f percentage \n (b)efficiency developed by scientist is %3.i percentage',nc,na)
diff --git a/1808/CH3/EX3.30/Chapter3_Exampl30.sce b/1808/CH3/EX3.30/Chapter3_Exampl30.sce new file mode 100644 index 000000000..eb1e72a8f --- /dev/null +++ b/1808/CH3/EX3.30/Chapter3_Exampl30.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=30;//temperature in Degree C
+t2=800;//temperature in Degree C
+g=1.4;//constant
+Rc=5;//compression ratio
+Qs=900;//heat supplied in kJ/kg
+nr=0.75;//regenerator efficiency in percentage
+
+
+//CALCULATIONS
+Qs1=R*(t2+273)*log(Rc)+(1-nr)*cv*(t2-t1);//heat supplied in kJ/kg
+m=Qs/Qs1;//mass flow rate in kg/min
+wn=(m/60)*R*log(Rc)*(t2-t1);//net work done in kW
+ns=(wn/(Qs/60))*100;//Thermal efficiency in percentage
+vs=((m)*R*(t1+273)*(1-(1/Rc)))/(p1*60);//swept volume in m^3/s
+pm=wn/vs;//mean effective pressure in kN/m^2
+P=wn*1;//Power developed by the engine in kW
+
+//OUTPUT
+printf('(i)The net work done is %3.2f kW \n (ii)Thermal efficiency is %3.2f percentage \n (iii)mean effective pressure is %3.2f kN/m^2 \n (iv)Power developed by the engine is %3.2f kW ',wn,ns,pm,P)
+
diff --git a/1808/CH3/EX3.31/Chapter3_Exampl31.sce b/1808/CH3/EX3.31/Chapter3_Exampl31.sce new file mode 100644 index 000000000..e00d2b8f2 --- /dev/null +++ b/1808/CH3/EX3.31/Chapter3_Exampl31.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in Degree C
+t2=700;//temperature in Degree C
+p1=1;//pressure in bar
+p3=12;//pressure in bar
+R=0.287;//gas constant
+
+//CALCULATIONS
+ns=(1-(t1/t2))*100;//Thermal efficiency in percentage
+Rc=((p3/p1)*(t1/t2));//compression ratio
+wn=R*log(Rc)*(t2-t1);//net work done in kJ/kg
+vs=(R*t1*(1-(1/Rc)))/(p1);//swept volume in m^3/kg
+pm=wn/vs;//mean effective pressure in bar
+
+//OUTPUT
+printf('(i)Thermal efficiency is %3.2f percentage \n (ii)The mean effective pressure is %3.2f bar',ns,pm)
+
+
diff --git a/1808/CH3/EX3.32/Chapter3_Exampl32.sce b/1808/CH3/EX3.32/Chapter3_Exampl32.sce new file mode 100644 index 000000000..b18fd13b8 --- /dev/null +++ b/1808/CH3/EX3.32/Chapter3_Exampl32.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//pressure in kPa
+t1=300;//temperature in K
+t3=1500;//temperature in K
+g=1.4;//constant
+v1=6;//volume in m^3/s
+Rp=10;//compression ratio
+
+//CALCULATIONS
+wn=cp*(1-((1/(Rp^((g-1)/g)))))*(t3-t1*(Rp^((g-1)/g)));//Net work done in kJ/kg
+t2=t1*(Rp^((g-1)/g));//temperature in K
+ng=(wn/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+t4=t3/(Rp^((g-1)/g));//temperature in K
+Rw=((cp*(t2-t1))/(cp*(t3-t4)))*100;//back work ratio
+m=p1*v1/(R*t1);//mass flow rate in kg/s
+P=m*wn;//Power developed in kW
+
+//OUTPUT
+printf('(i))Thermal efficiency is %3.2f percentage \n(ii)The back work ratio is %3.1f percentage \n(iii)Power developed is %3.1f kW',ng,Rw,P)
+
+
diff --git a/1808/CH3/EX3.33/Chapter3_Exampl33.sce b/1808/CH3/EX3.33/Chapter3_Exampl33.sce new file mode 100644 index 000000000..32039fff5 --- /dev/null +++ b/1808/CH3/EX3.33/Chapter3_Exampl33.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gs constant
+p1=100;//pressure in kPa
+t1=300;//temperature in K
+t3=1300;//temperature in K
+g=1.4;//index of expansion
+
+
+//CALCULATIONS
+Rp=(t3/t1)^(g/(g-1));//pressure ratio
+Rpo=sqrt(Rp);//Pressure ratio which will give maximum net work output
+t2=t1*(Rpo^((g-1)/g));//temperature in K
+t4=t3/(Rpo^((g-1)/g));//temperature in K
+wn=cp*t3*((1-(sqrt(t1/t3)))^2);//maximum net work output in kJ/kg
+ng=(1-sqrt(t1/t3))*100;//Thermal efficiency in percentage
+Rw=(1-sqrt(t1/t3));//work ratio at maximum work output
+nc=(1-(t1/t3))*100;//Carnot efficiency for the same temperature limits in percentage
+
+//OUTPUT
+printf('(i)The pressure ratio which will give maximum net work output is %3.2f \n (ii)maximum net work output is %3.2f kJ/kg \n (iii)Thermal efficiency at maximum output is %3.2f percentage \n (iv)work ratio at maximum work output is %3.4f percentage \n (v)Carnot efficiency for the same temperature limits is %3.2f percentage',Rpo,wn,ng,Rw,nc)
+
+
+
diff --git a/1808/CH3/EX3.34/Chapter3_Exampl34.sce b/1808/CH3/EX3.34/Chapter3_Exampl34.sce new file mode 100644 index 000000000..8d219cfdd --- /dev/null +++ b/1808/CH3/EX3.34/Chapter3_Exampl34.sce @@ -0,0 +1,15 @@ +clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1300;//temperature in K
+g=1.4;//constant
+
+//CALCULATIONS
+Rpm=(t3/t1)^(g/(g-1));//Solution pressure ratio
+ng=(1-(t1/t3))*100;//thermal efficiency corresponds to maximum pressure ratio
+
+//OUTPUT
+printf('(i)Solution pressure ratio is %3.2f \n (ii)net workdone corresponds to maximum pressure ratio is zero \n (iii)thermal efficiency corresponds to maximum pressure ratio is %3.2f percntage \n (iv)work ratio is zero',Rpm,ng)
+
+
diff --git a/1808/CH3/EX3.35/Chapter3_Exampl35.sce b/1808/CH3/EX3.35/Chapter3_Exampl35.sce new file mode 100644 index 000000000..da803cfd5 --- /dev/null +++ b/1808/CH3/EX3.35/Chapter3_Exampl35.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+nt=0.8;//Thermal efficiency in percentage
+nc=0.8;//compressor efficiency in percentage
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+t1=300;//temperature in K
+t3=1500;//temperature in K
+Rp=10;//pressure ratio
+
+//CALCULATIONS
+t2=t1*((Rp)^((g-1)/g));//temperature in K
+t21=t1+((t2-t1)/(nc));//temperature in K
+t4=t3/((Rp)^((g-1)/g));//temperature in K
+t41=t3-(nt*(t3-t4));//temperature in K
+wna=cp*((t3-t41)-(t21-t1));//net work done in kJ/kg
+ng=wna/(cp*(t3-t21))*100;//Thermal efficiency in percentage
+Rw=wna/(cp*(t3-t41));//work ratio
+
+//OUTPUT
+printf('(i)Net work done is %3.4f kJ/kg \n (ii)Thermal efficiency is %3.2f percentage \n (iii)Work ratio is %3.4f ',wna,ng,Rw)
+
+
+
diff --git a/1808/CH3/EX3.36/Chapter3_Exampl36.sce b/1808/CH3/EX3.36/Chapter3_Exampl36.sce new file mode 100644 index 000000000..d5b9f2ae0 --- /dev/null +++ b/1808/CH3/EX3.36/Chapter3_Exampl36.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=0.718;//specific volume
+R=0.287;//gas constant
+p1=100;//Pressure in kPa
+t1=300;//temperature in K
+g=1.35//constant
+t3=1000;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+nt=0.9;//Thermal efficiency in percentage
+
+
+//CALCULATIONS
+Rp=(t3/t1)^(g/(g-1));//maximum pressure ratio
+Rpo=sqrt(Rp*nc*nt);//Pressure ratio for maximum work
+t2=t1*(Rpo)^((g-1)/g);//temperature in K
+t21=t1+((t2-t1)/nc);//temperature in K
+t4=t3/(Rpo^((g-1)/g));//temperature in K
+t41=t3-(nt*(t3-t4));//temperature in K
+nbt=(((t3-t41)-(t21-t1))/(t3-t21))*100;//Thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)Pressure ratio for maximum work is %3.2f \n (ii)Thermal efficiency is %3.2f percentage ',Rpo,nbt)
+
+
+
+
diff --git a/1808/CH3/EX3.37/Chapter3_Exampl37.sce b/1808/CH3/EX3.37/Chapter3_Exampl37.sce new file mode 100644 index 000000000..fad7410fa --- /dev/null +++ b/1808/CH3/EX3.37/Chapter3_Exampl37.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=42000;//specific volume
+R=0.287;//gas constant
+g=1.4;//constant
+t1=300;//temperature in K
+t3=1000;//temperature in K
+Rp=5;//Pressure ratio
+ma=42.93;//mass of air
+mf=0.5;//mass of fuel
+nc=0.8;//compressor efficiency
+nt=0.85;//turbine efficiency
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wna=-ma*cp*(t21-t1)+(ma+mf)*(t3-t41);//Actual power available in kJ
+ng=(wna/(mf*cv))*100;//Actual thermal efficiency in percentage
+
+//OUTPUT
+printf('(a)Actual power available is %3.2f kJ \n (b)Actual thermal efficiency is %3.2f percentage',wna,ng)
+
diff --git a/1808/CH3/EX3.38/Chapter3_Exampl38.sce b/1808/CH3/EX3.38/Chapter3_Exampl38.sce new file mode 100644 index 000000000..1ca7bcf58 --- /dev/null +++ b/1808/CH3/EX3.38/Chapter3_Exampl38.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+cv=42000;//calorific value
+R=0.287;//gas constant
+g=1.35;//constant
+t1=293;//temperature in K
+t3=973;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+Rp=5;//pressure ratio
+nt=0.8;//turbine efficiency in percentage
+p=1000;//power in mW
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wna=cp*((t3-t41)-(t2-t1));//net work done in kJ/kg
+m=p/wna;//Quantity of air circulation
+Qsa=cp*(t3-t21)/(nc);//Heat supplied in kJ/kg
+nba=(wna/Qsa)*100;//Thermal efficiency in percentage
+
+//OUTPUT
+printf('(i)Quantity of air circulation is %3.3f kg/s \n(ii)Heat supplied is %3.2f kJ/kg \n (iii)Thermal efficiency is %3.2f percentage',m,Qsa,nba)
+
+
+
+
+
diff --git a/1808/CH3/EX3.39/Chapter3_Exampl39.sce b/1808/CH3/EX3.39/Chapter3_Exampl39.sce new file mode 100644 index 000000000..45dbea28f --- /dev/null +++ b/1808/CH3/EX3.39/Chapter3_Exampl39.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+t1=300;//Initial temperature in K
+t21=523;//intermmediate temperature in K
+t3=1073;//final temperature in K
+t41=723;//turbine outlet temperature in K
+p1=1;//pressure in bar
+p2=6;//final pressure in bar
+Rp=6;//pressure ratio
+g=1.4;//constant
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+
+nc=((t2-t1)/(t21-t1))*100;//compressor efficiency in percentage
+nt=((t3-t41)/(t3-t4))*100;//Turbine efficiency in percentage
+
+ngt=(1-(1/Rp)^((g-1)/g))*100;//Thermal efficiency in percentage
+ngt1=((((nt/100)*t3*(ngt/100))-((t1/(nc/100))*((Rp^((g-1)/g))-1)))/(t1*((t3/t1)-((Rp^((g-1)/g))))))*100;//Thermal efficiency in percentage
+
+Rw=((cp*((t3-t4)-(t2-t1)))/(cp*(t3-t4)));//Work ratio
+Rw1=((cp*((t3-t41)-(t21-t1)))/(cp*(t3-t41)));//Work ratio
+
+Rpo=sqrt((t3/t1)^(g/(g-1)));//pressure ratio for maximum output
+Rpo1=sqrt(((t3/t1)^(g/(g-1)))*(nc/100)*(nt/100));//pressure ratio for maximum output
+
+Rpm=(t3/t1)^(g/(g-1));//pressure ratio for maximum efficiency
+Rpm1=(t3/t1)^(g/(g-1))*(1/((nc/100)*(nt/100)));//pressure ratio for maximum efficiency
+
+//OUTPUT
+printf('(A)The compressor efficiency is %3.3f percentage \n turbine efficiency is %3.2f percentage \n',nc,nt)
+
+printf('(B)IDEAL CYCLE \n (i)thermal efficiency is %3.2f percentage \n(ii)Work ratio is %3.4f \n (iii)Pressure ratio for maximum output is %3.2f \n (iv)pressure ratio for maximum efficiency is %3.2f \n ',ngt,Rw,Rpo,Rpm)
+
+printf('(B)ACTUAL CYCLE \n (i)thermal efficiency is %3.2f percentage \n(ii)Work ratio is %3.4f \n (iii)Pressure ratio for maximum output is %3.2f \n (iv)pressure ratio for maximum efficiency is %3.1f ',ngt1,Rw1,Rpo1,Rpm1)
+
diff --git a/1808/CH3/EX3.4/Chapter3_Exampl4.sce b/1808/CH3/EX3.4/Chapter3_Exampl4.sce new file mode 100644 index 000000000..9308e465a --- /dev/null +++ b/1808/CH3/EX3.4/Chapter3_Exampl4.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+n=1/5;//Fraction of heat input converted into power
+ts=100;//Reduced sink temperature in Degree C
+
+//CALCULATIONS
+//4Th-5Tl=0
+//3Th-5Tl=-500
+A=[4 -5
+ 3 -5]//Coefficient matrix
+B=[0
+ -500]//Constant matrix
+X=inv(A)*B//Variable matrix
+
+//Output
+printf('(a)The temperature of the source is %3.f K \n (b)The temperature of sink is %3.f K',X(1),X(2))
diff --git a/1808/CH3/EX3.40/Chapter3_Exampl40.sce b/1808/CH3/EX3.40/Chapter3_Exampl40.sce new file mode 100644 index 000000000..87c3661b3 --- /dev/null +++ b/1808/CH3/EX3.40/Chapter3_Exampl40.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+R=0.287;//gas constant
+g=1.4;//constant
+t1=303;//temperature in K
+t3=1773;//temperature in K
+t5=1123;//temperature in K
+nc=0.85;//compressor efficiency in percentage
+Rp=6;//pressure ratio
+nt=0.8;//turbine efficiency in percentage
+
+//CALCULATIONS
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+t21=t1+((t2-t1)/nc);//Temperature in K
+t41=t3-((t3-t4)*nt);//Temperature in K
+wc=cp*(t21-t1);//compressor work in kJ/kg
+wt=cp*(t3-t41);//turbine work in kJ/kg
+nb=((wt-wc)/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+wn=wt-wc;//net work in kJ/kg
+Qs=cp*(t3-t5);//Heat supplied in kJ/kg
+ns=((wt-wc)/Qs)*100;//Thermal efficiency in percentage
+e=((t5-t21)/(t41-t21))*100;//Effectiveness of the regenerator
+
+//OUTPUT
+printf('(a)compressor work is %3.2f kJ/kg \n turbine work is %3.2f kJ/kg \n (b)Thermal efficiency is %3.3f percentage \n (c)Thermal efficiency with regenerator is %3.2f percentage \n (d)Effectiveness of the regenerator is %3.1f percentage ',wc,wt,nb,ns,e)
+
+
diff --git a/1808/CH3/EX3.41/Chapter3_Exampl41.sce b/1808/CH3/EX3.41/Chapter3_Exampl41.sce new file mode 100644 index 000000000..60fd69fb0 --- /dev/null +++ b/1808/CH3/EX3.41/Chapter3_Exampl41.sce @@ -0,0 +1,78 @@ +clc
+clear
+//INPUT DATA
+cp=1.005;//specific pressure
+R=0.287;//gas constant
+g=1.4;//constant
+t1=303;//temperature in K
+t3=1073;//temperature in K in case I
+t5=1123;//temperature in K
+Rp=4;//pressure ratio
+p1=1;//atmospheric pressure in bar
+p2=4;//exit pressure in bar
+
+
+//CALCULATIONS
+//case 1
+t2=t1*(Rp^((g-1)/g));//Temperature in K
+t4=t3/(Rp^((g-1)/g));//Temperature in K
+Qs=cp*(t3-t2);//Heat supplied in kJ/kg
+wc=cp*(t2-t1);//compressor work in kJ/kg
+wt=cp*(t3-t4);//turbine work in kJ/kg
+ng=((wt-wc)/(cp*(t3-t2)))*100;//Thermal efficiency in percentage
+
+//case 2
+//a regenerator of effectiveness 0.6 is added
+t51=0.6*(t4-t2)+t2;//temperature in K
+nbr=(((wt-wc)/(cp*(t3-t51))))*100;//Thermal efficiency eith regenerator in percentage
+//case 3
+pi=(p1*p2)^(1/2);//intermediate pressure
+t21=t1*(pi)^((g-1)/g);//temperature in K
+t41=t1*(pi)^((g-1)/g);//temperature in K
+t61=t3/((Rp)^((g-1)/g));//temperature in K
+t7=0.6*(t61-t2)+t21;//temperature in K
+Qs1=cp*(t3-t7);//heat added in kJ/kg
+wt1=cp*(t3-t61);//turbine work in kJ/kg
+wc1=cp*((t41-t1)+(t41-t1));//compressor work in kJ/kg
+nt=((wt1-wc1)/Qs1)*100;//Thermal efficiency in percentage
+//case 4
+t22=t1*(Rp)^((g-1)/g);//temperature in K
+t42=t3/(pi)^((g-1)/g);//temperature in K
+t62=t3/((pi)^((g-1)/g));//temperature in K
+t72=t22+(0.6*(t62-t22));//temperature in K
+wc2=cp*(t22-t1);//compressor work in kJ/kg
+wt2=cp*((t3-t42)+(t3-t62));//turbine work in kJ/kg
+Qs2=cp*((t3-t72)+(t3-t42));//heat added in kJ/kg
+ns=((wt2-wc2)/Qs2)*100;//Thermal efficiency in percentage
+//case 5
+t23=t1*(pi)^((g-1)/g);//temperature in K
+t43=t1*(pi)^((g-1)/g);//temperature in K
+t73=t3/(pi)^((g-1)/g);//temperature in K
+t93=t3/(pi)^((g-1)/g);//temperature in K
+t53=0.6*(t93-t43)+t43;//temperature in K
+Qs3=cp*((t3-t53)+(t3-t73));//heat added in kJ/kg
+wt3=cp*((t3-t93)+(t3-t73));//turbine work in kJ/kg
+wc3=cp*((t23-t1)+(t43-t1));//compressor work in kJ/kg
+ns1=((wt3-wc3)/Qs3)*100;//Thermal efficiency in percentage
+
+
+//OUTPUT
+printf('CASE I \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt,ng)
+printf('CASE II \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt,nbr)
+printf('CASE III \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc,wt1,nt)
+printf('CASE IV \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc2,wt2,ns)
+printf('CASE V \n (i)Compressor work %3.2f kJ/kg \n (ii)Turbine work %3.2f kJ/kg \n (iii)Thermal efficiency %3.1f percentage \n ',wc3,wt3,ns1)
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH3/EX3.5/Chapter3_Exampl5.sce b/1808/CH3/EX3.5/Chapter3_Exampl5.sce new file mode 100644 index 000000000..6d44da775 --- /dev/null +++ b/1808/CH3/EX3.5/Chapter3_Exampl5.sce @@ -0,0 +1,18 @@ +clc
+clear
+//INPUT DATA
+Tl=20;//engine temprature in Degree C
+Th=500;//engine temprature in Degree C
+g=1.4;//gas constant
+v13=25;//expansion ratio
+
+//CALCULATIONS
+v14=((Th+273)/(Tl+273))^(1/(g-1));//Isentropic volume expansion
+v43=v13/v14;//Overall expansion ratio
+
+//OUTPUT
+printf('(a)Isentropic volume expansion ratio (4-1)is %3.2f \n (b)Isentropic volume expansion ratio(4-3) is %3.2f',v14,v43)
+
+
+
+
diff --git a/1808/CH3/EX3.6/Chapter3_Exampl6.sce b/1808/CH3/EX3.6/Chapter3_Exampl6.sce new file mode 100644 index 000000000..9937b5475 --- /dev/null +++ b/1808/CH3/EX3.6/Chapter3_Exampl6.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+Th=600;//engine temprature in Degree C
+p3=5;//Pressure of air in bar
+v3=4;//volume of air in m^3
+v43=3;//Isentropic volume expansion
+v23=6;//Isentropic compression ratio
+
+//CALCULATIONS
+p4=p3*(1/(v43));//pressure of carnot cycle
+p1=p4*(1/(v23))^(1.4);//pressure of carnot cycle
+t1=(Th+273)*(1/(v23))^(0.4);//Temperature of carnot cycle
+p2=p1*(v43);//pressure of carnot cycle
+Qs=p3*10^2*v3*log(v43);//heat supplied to the cycle
+Qr=p3*v3*(t1/Th)*log(v43);//Heat rejected by the system
+nc=(((Th+273)-t1)/(273+Th))*100;//Thermal efficiency in percentage
+w=Qs-Qr;//work done in kJ
+pm=w/(17*2*v3*100);//Mean effective pressure in bar
+
+//OUTPUT
+printf('(a)The pressure of carnot cycle is %3.3f bar \n temperature of carnot cycle is %3.4f K \n (b)Heat supplied to the cycle %3.2f kJ \n (c)Thermal efficiency is %3.2f percentage \n (d)Mean effective pressure is %3.4f bar ',p1,t1,Qs,nc,pm)
+
+
diff --git a/1808/CH3/EX3.8/Chapter3_Exampl8.sce b/1808/CH3/EX3.8/Chapter3_Exampl8.sce new file mode 100644 index 000000000..f3cfb30c4 --- /dev/null +++ b/1808/CH3/EX3.8/Chapter3_Exampl8.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+p3=20;//Pressure of air in bar
+v3=0.2;//volume of air in m^3
+Th=500;//engine temprature in Degree C
+v23=7;//Isentropic compression ratio
+v43=2;//Isentropic volume expansion
+v3=0.2;//volume in m^3
+
+//CALCULATIONS
+Tl=(Th+273)/((v23)^(1.4-1));//minimum temperature in K
+p2=p3/((v23)^(1.4));//pressure in bar
+p4=p3*10^2*(1/(v43));//isentropic expansion pressure
+p1=((1/(v23))^1.4)*p4;//isentropic compression
+s43=(p3*10^2*v3/(Th+273))*log(v43);//Change in entropy in kJ/K
+nc=(((Th+273)-Tl)/(Th+273))*100;//Efficiency of carnot cycle in percentage
+v1=v43*7*v3;//volume in m^3
+vs=v1-v3;//swept volume in m^3
+wd=(p3*10^2*v3-p1*v1)*log(2);//workdone in kJ/cycle
+P=wd/2.6;//Mean effective pressure in kN/m^2
+p=wd*(200/60);//power of engine in kW
+
+//OUTPUT
+printf('(a)The minimum temperature in the cycle is %3.2f K \n (b)Change in entropy during isothermal expansion is %3.4f kJ/K \n (c)Thermal efficiency of the cycle is %3.2f percentage \n (d)The mean effective pressure is %3.2f kN/m^2 \n (e)Power of the engine is %3.2f kW',Tl,s43,nc,P,p)
+
+
+
diff --git a/1808/CH3/EX3.9/Chapter3_Exampl9.sce b/1808/CH3/EX3.9/Chapter3_Exampl9.sce new file mode 100644 index 000000000..eeea2a66a --- /dev/null +++ b/1808/CH3/EX3.9/Chapter3_Exampl9.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+p1=101;//Pressure in kPa
+t1=293;//temperature in K
+v12=7;//compression ratio
+Qs=1000;//heat added in kJ
+Ra=0.287;//gas constant
+m=1;//mass of air in kg/min
+cv=0.7243;//calorific value
+
+//CALCULATIONS
+v1=(Ra*t1)/p1;//specific volume in m^3/kg
+v2=v1/(v12);//specific volume in m^3/kg
+p2=p1*((v12)^(1.4));//pressure of cycle
+t2=t1*((v12)^(1.4-1));//temperature in K
+t3=(Qs/(m*cv))+t2;//temperature in K
+p3=Ra*t3/v2;//pressure in Kn/m^2
+t4=t3*((1/v12)^(1.4-1));//temperature in K
+Qr=m*cv*(t4-t1);//heat rejected in kJ
+no=((Qs-Qr)/Qs)*100;//otto cycle efficiency in percentage
+pm=((Qs-Qr)/(v1-v2));//mean effective pressure in kN/m^2
+P=m*(Qs-Qr);//power developed in kJ/min
+//CASE B
+nc=((t3-t1)/t3)*100;//Carnot cycle efficiency in percentage
+
+//OUTPUT
+printf('(a)Specific volume of cycle is %3.2f m^3/kg \n pressure in the cycle is %3.2f kW/m^2 \n temperature in the cycle is %3.2f K \n Specific volume of cycle at point 3 is %3.3f m^3/kg \n pressure in the cycle at point 3 is %3.2f kW/m^2 \n temperature in the cycle at point 4 is %3.2f K \n(b)the efficiency of the otto cycle is %3.2f percentage \n (c)mean effective pressure is %3.2f kN/m^2 \n (d)power developed is %3.2f kJ/min \n CASE B \n Carnot cycle efficiency is %3.2f percentage \n Carnot cylce efficiency is high compared to Otto cycle efficiency',v1,p2,t2,v2,p3,t4,no,pm,P,nc)
+
+
diff --git a/1808/CH4/EX4.1/Chapter4_Example1.sce b/1808/CH4/EX4.1/Chapter4_Example1.sce new file mode 100644 index 000000000..58013114e --- /dev/null +++ b/1808/CH4/EX4.1/Chapter4_Example1.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+p=(100*10^3);//Rate of heat source in kW
+P1=40;//Boiler pressure in bar
+P2=0.1;//Condenser pressure in bar
+S1=6.0685;//Entropy in kJ/kg.K
+S3=0.649;//Entropy in kJ/kg.K
+S5=8.15;//Entropy in kJ/kg.K
+h1=2800.5;//Enthalpy in kJ/kg
+h2=1920.67;//Enthalpy in kJ/kg
+h3=191.8;//Enthalpy in kJ/kg
+h5=2584.7;//Enthalpy in kJ/kg
+v3=0.001001;//Specific volume in m^3/kg
+
+
+//CALCULATIONS
+x2=(S1-S3)/(S5-S3);//quality of steam
+h2=h3+(x2*(h5-h3));//Enthalpy in kJ/kg
+Wp=v3*(P1-P2);//Pump work in kJ/kg
+h4=h3+Wp;//Enthalpy in kJ/kg
+n=(((h1-h2)-(h4-h3))/(h1-h4))*100;//Ideal cycle efficiency
+rw=((h1-h2)-(h4-h3))/(h1-h2);//Work ratio
+m=p/(h1-h4);//Mass flow rate in kg/s
+P=m*((h1-h2)-(h4-h3));//Output power in kW
+ssc=(m*3600)/P;//Specific flow rate of steam in kg/kW.hr
+
+//OUTPUT
+printf('(i) The cycle efficiency is %3.2f percent \n(ii) The work ratio is %f \n(iii)The required mass flow rate is %3.2f kg/s \n(iv) The power output is %3.1f kW \n(v) The specific flow rate of steam is %3.2f kg/kW.hr',n,rw,m,P,ssc)
diff --git a/1808/CH4/EX4.10/Chapter4_Example10.sce b/1808/CH4/EX4.10/Chapter4_Example10.sce new file mode 100644 index 000000000..03018163c --- /dev/null +++ b/1808/CH4/EX4.10/Chapter4_Example10.sce @@ -0,0 +1,43 @@ +clc
+clear
+//INPUT DATA
+p1=100;//pressure in bar
+p2=10;//pressure in bar
+p3=0.1;//pressure in bar
+T1=500;//Temperature of turbine in Degree C
+T2=450;//Temperature of turbine in Degree C
+h1=3240.9;//Enthalpy in kJ/kg
+h4=3370.7;//Enthalpy in kJ/kg
+h3=2776.2;//Enthalpy in kJ/kg
+h10=762.6;//Enthalpy in kJ/kg
+h6=191.8;//Enthalpy in kJ/kg
+h9=2584.7;//Enthalpy in kJ/kg
+S1=6.419;//Entropy in kJ/kg.K
+S4=7.618;//Entropy in kJ/kg.K
+S3=6.5828;//Entropy in kJ/kg.K
+S10=2.1382;//Entropy in kJ/kg.K
+S6=0.649;//Entropy in kJ/kg.K
+S9=8.15;//Entropy in kJ/kg.K
+nt=0.8;//Turbine efficiency in percentage
+v6=0.001001;//Specific volume in m^3/kg
+P=100000;//power output in kW
+
+
+//CALCULATIONS
+x2=((S1-S10)/(S3-S10));//quality of steam
+h2=h10+(x2*(h3-h10));//Enthalpy in kJ/kg
+h21=h1-(nt*(h1-h2));//Enthalpy in kJ/kg
+x5=((S4-S6)/(S9-S6));//quality of steam
+h5=h6+(x5*(h9-h6));//Enthalpy in kJ/kg
+h51=h4-(nt*(h4-h5));//Enthalpy in kJ/kg
+Wt=(h1-h21)+(h4-h51);//Turbine work in kJ/kg
+h7=h6+(v6*(p1-p3)*100);//Enthalpy in kJ/kg
+Wp=(h7-h6);//Pump work in kJ/kg
+Qs=(h1-h7)+(h4-h21);//heat supplied in kJ/kg
+nRi=((Wt-Wp)/Qs)*100;//Cycle efficiency
+m=P/(Wt-Wp);//mass flow rate in kg/s
+Qr=(h51-h6)*m;//rate of heat transfer from condenser in kW
+
+//OUTPUT
+printf('(i)Thermal efficiency is %3.2f percent \n (ii)Mass flow rate is %3.2f kg/s \n (iii)Rate of heat transfer from the condenser %3.2f kW',nRi,m,Qr)
+
diff --git a/1808/CH4/EX4.11/Chapter4_Example11.sce b/1808/CH4/EX4.11/Chapter4_Example11.sce new file mode 100644 index 000000000..7df8944bf --- /dev/null +++ b/1808/CH4/EX4.11/Chapter4_Example11.sce @@ -0,0 +1,39 @@ +clc
+clear
+//INPUT DATA
+p1=90;//pressure in bar
+p2=9;//pressure in bar
+p3=0.1;//pressure in bar
+T=450;//Temperature in Degree C
+h1=2956.6;//Enthalpy in kJ/kg
+S1=6.036;//Entropy in kJ/kg.K
+h9=2772.1;//Enthalpy in kJ/kg
+h6=742.6;//Enthalpy in kJ/kg
+S9=6.6192;//Entropy in kJ/kg.K
+S6=2.0941;//Entropy in kJ/kg.K
+V6=0.001121;//Specific volume in m^3/kg
+h10=2584.7;//Enthalpy in kJ/kg
+h4=191.8;//Enthalpy in kJ/kg
+S10=8.15;//Entropy in kJ/kg.K
+S4=0.649;//Entropy in kJ/kg.K
+V4=0.001001;//Specific volume in m^3/kg
+P=120000;//power output in kW
+
+//CALCULATIONS
+x2=((S1-S6)/(S9-S6));//quality of steam
+x3=((S1-S4)/(S10-S4));//quality of steam
+h2=h6+(x2*(h9-h6));//Enthalpy in kJ/kg
+h3=h4+(x3*(h10-h4));//Enthalpy in kJ/kg
+h5=h4+(V4*(p1-p3))*100;//Enthalpy in kJ/kg
+Wp1=h5-h4;//Pump work in kJ/kg
+h7=h6+(V6*(p1-p2))*100;//Enthalpy in kJ/kg
+Wp2=h7-h6;//Pump work in kJ/kg
+m1=((h6-h5)/(h2-h5));//Mass flow rate in kJ/s
+Wt=(h1-h2)+((1-m1)*(h2-h3));//Turbine work in kJ/kg
+Wp=(h7-h6)+((1-m1)*(h5-h4));//Pump work in kJ/kg
+Qs=(h1-h7);//heat supplied in kJ/kg
+nR=((Wt-Wp)/Qs)*100;//Rankine efficiency in percentage
+m=P/(Wt-Wp);//mass flow rate in kJ/s
+
+//OUTPUT
+printf('(i) The Thermal efficiency is %3.3f percent \n (ii) Mass flow rate of steam entering to the turbine is %3.2f kg/s ',nR,m)
diff --git a/1808/CH4/EX4.12/Chapter4_Example12.sce b/1808/CH4/EX4.12/Chapter4_Example12.sce new file mode 100644 index 000000000..2364593bc --- /dev/null +++ b/1808/CH4/EX4.12/Chapter4_Example12.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+p1=5;//pressure in bar
+p2=0.1;//pressure in bar
+m=5;//mass flow rate in kJ/s
+h4=191.8;//Enthalpy in kJ/kg
+h10=2584.7;//Enthalpy in kJ/kg
+S4=0.649;//Entropy in kJ/kg.K
+S10=8.15;//Entropy in kJ/kg.K
+V4=0.001001;//Specific volume in m^3/kg
+h6=640.1;//Enthalpy in kJ/kg
+h9=2747.5;//Enthalpy in kJ/kg
+S6=1.8604;//Entropy in kJ/kg.K
+S9=6.8192;//Entropy in kJ/kg.K
+x2=0.9;//Quality of steam
+Qs=70000;//heat added in boiler in kW
+
+
+//CALCULATIONS
+h2=h6+(x2*(h9-h6));//Enthalpy in kJ/kg
+h5=h4+(V4*(p1-p2));//Enthalpy in kJ/kg
+Wp1=h5-h4;//Pump work in kJ/kg
+mf=((m*(h2-h5))/(h6-h5));//mass flow rate in kJ/s
+h1=((Qs/mf)+h6);//Enthalpy in kJ/kg
+S2=S6+(x2*(S9-S6));//Entropy in kJ/kg.K
+x3=((S2-S4)/(S10-S4));//quality of steam
+h3=h4+(x3*(h10-h4));//Enthalpy in kJ/kg
+Wt=(mf*(h1-h2))+(mf-m)*(h2-h3);//Turbine work in kJ/kg
+nR=((Wt-Wp1)/Qs)*100;//thermal efficiency in percentage
+Wn=Wt-Wp1;//work in kJ/s
+ssc=(mf*3600)/Wn;//specific steam consumption in kg/kW.hr
+R=Wn/Wt;//Work ratio
+
+//OUTPUT
+printf('(i) The Mass flow rate of steam is %3.1f kg/s \n (ii) Thermal efficiency of rankine cycle is %3.1f percentage \n (iii) Specific steam consumption is %3.2f kg/kWhr \n (iv) Work ratio is approximately equal to %f',mf,nR,ssc,R)
+
diff --git a/1808/CH4/EX4.13/Chapter4_Example13.sce b/1808/CH4/EX4.13/Chapter4_Example13.sce new file mode 100644 index 000000000..f0f11b35f --- /dev/null +++ b/1808/CH4/EX4.13/Chapter4_Example13.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+P1=70;//Boiler pressure in Bar
+P2=0.1;//condenser pressure in Bar
+P3=10;//bled pressure in Bar
+T=400;//Boiler temperature in Degree C
+h6=1267.4;//Enthalpy in kJ/kg
+h1=3158.1;//Enthalpy in kJ/kg
+S1=6.448;//Entropy in kJ/kg.K
+h7=762.2;//Enthalpy in kJ/kg
+h9=2776.2;//Enthalpy in kJ/kg
+S7=2.1382;//Entropy in kJ/kg.K
+S9=6.5828;//Entropy in kJ/kg.K
+h4=191.8;//Enthalpy in kJ/kg
+h10=2584.7;//Enthalpy in kJ/kg
+S4=0.649;//Entropy in kJ/kg.K
+S10=8.15;//Entropy in kJ/kg.K
+V4=0.001;//Specific volume in m^3/kg
+
+
+//CALCULATIONS
+x2=((S1-S7)/(S9-S7));//quality of steam
+h2=h7+(x2*(h9-h7));//Enthalpy in kJ/kg
+x3=((S1-S4)/(S10-S4));//quality of steam
+h3=h4+(x3*(h10-h4));//Enthalpy in kJ/kg
+h5=h4+(V4*(P1-P2));//Enthalpy in kJ/kg
+Wp=h5-h4;//Pump work in kJ/kg
+m1=((h6-h5)/(h2-h7));//mass flow rate
+Wt=(h1-h2)+(1-m1)*(h2-h3);//Turbine work in kJ/kg
+Qs=(h1-h6);//Heat supplied in kJ/kg
+nR=((Wt-Wp)/Qs)*100;//Rankine efficiency in percentage
+
+//OUTPUT
+printf(' Thr Rankine cycle efficiency is %3.2f percentage ',nR)
diff --git a/1808/CH4/EX4.2/Chapter4_Example2.sce b/1808/CH4/EX4.2/Chapter4_Example2.sce new file mode 100644 index 000000000..501e91ad8 --- /dev/null +++ b/1808/CH4/EX4.2/Chapter4_Example2.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+p=(100*10^3);//Rate of heat source in kW
+P1=40;//Boiler pressure in bar
+P2=0.1;//Condenser pressure in bar
+n=0.8;//Adiabatic efficiency
+S1=6.0685;//Entropy in kJ/kg.K
+S3=0.649;//Entropy in kJ/kg.K
+S5=8.15;//Entropy in kJ/kg.K
+h1=2800.5;//Enthalpy in kJ/kg
+h2=1920.67;//Enthalpy in kJ/kg
+h3=191.8;//Enthalpy in kJ/kg
+h5=2584.7;//Enthalpy in kJ/kg
+v3=0.001001;//Specific volume in m^3/kg
+
+
+//CALCULATIONS
+Wt1=(h1-h2);//Ideal turbine work in kJ/kg
+WtA=Wt1*n;//Actual turbine work in kJ/kg
+Wp=v3*(P1-P2);//Pump work in kJ/kg
+h4=h3+Wp;//Enthalpy in kJ/kg
+Qs=(h1-h4);//heat supplied in kJ/kg
+h2x=h1-WtA;//Enthalpy in kJ/kg
+nRA=((WtA-Wp)/Qs)*100;//Cycle efficiency
+m=p/Qs;//Mass flow rate in kJ/s
+P=m*(WtA-Wp);//power output in kW
+ssc=m*3600/P;//Specific steam consumption in kg/kW.hr
+
+//OUTPUT
+printf('(i) The cycle efficiency is %3.2f percent \n(ii) The power output is %3.1f kW \n(iii) The specific flow rate of steam is %3.2f kg/kW.hr',nRA,P,ssc)
diff --git a/1808/CH4/EX4.3/Chapter4_Example3.sce b/1808/CH4/EX4.3/Chapter4_Example3.sce new file mode 100644 index 000000000..066fc8764 --- /dev/null +++ b/1808/CH4/EX4.3/Chapter4_Example3.sce @@ -0,0 +1,36 @@ +clc
+clear
+//INPUT DATA
+pb=100;//Saturated vapour pressure in bar
+pc=0.1;//Saturated liquid pressure in bar
+two=35;//Cooling water exit temperature in degree C
+twi=20;//Cooling water entry temperature in degree C
+S1=5.6198;//Entropy in kJ/kg.K
+S3=0.649;//Entropy in kJ/kg.K
+S5=8.15;//Entropy in kJ/kg.K
+h1=2727.7;//Enthalpy in kJ/kg
+h3=191.8;//Enthalpy in kJ/kg
+h5=2584.7;//Enthalpy in kJ/kg
+V3=0.001;//Specific volume in m^3/kg
+Cpw=41.8;//specific heat of water in kJ/kgk
+
+
+//CALCULATIONS
+x2=(S1-S3)/(S5-S3);//quality of steam
+S1=S3+x2*(S5-S3);//Entropy in kJ/kg.K
+h2=h3+x2*(h5-h3);//Enthalpy in kJ/kg
+Wp=V3*(pb-pc);//Pump work in kJ/kg
+h4=h3+Wp;//Enthalpy in kJ/kg
+Wt=h1-h2;//Turbine work in kJ/kg
+Wn=Wt-Wp;//Net work in kJ/kg
+nR=(Wn/(h1-h4))*100;//Thermal efficiency
+m=((pb*1000*3600)/Wn)/10^5;//Mass flow rate of steam in kg/hr *10^5
+mx=((pb*1000)/Wn);//Mass flow rate of steam in kg/s
+QS1=mx*(h1-h4);//Rate of heat transferred into fluid in kJ/kg
+QR1=mx*(h2-h3);//rate of heat transfer from condenser in kJ/s
+mw1=(((h2-h3)*m)/((two-twi)*Cpw));//Mass flow rate of water in kg/hr *10^6
+Rw=((h1-h2)-Wp)/(h1-h2);//Work ratio
+
+//OUTPUT
+printf('(i) The Thermal efficiency is %3.2f percent \n(ii)The mass flow rate of steam is %3.2f * 10^5 kJ/hr \n(iii) The rate of heat transfer into working fluid is %3.1f kJ/s \n(iv)The rate of heat transfer from condenser is %3.2f kJ/s\n(v)mass flow rate of water in condenser is %3.1f *10^6 kg/hr \n(vi) The work ratio is %f ',nR,m,QS1,QR1,mw1,Rw)
+
diff --git a/1808/CH4/EX4.4/Chapter4_Example4.sce b/1808/CH4/EX4.4/Chapter4_Example4.sce new file mode 100644 index 000000000..4a9662ed3 --- /dev/null +++ b/1808/CH4/EX4.4/Chapter4_Example4.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+pb=100;//Saturated vapour in bar
+pc=0.1;//Saturated liquid in bar
+two=35;//Cooling water exit temperature in degree C
+twi=20;//Cooling water entry temperature in degree C
+S1=5.6198;//Entropy in kJ/kg.K
+S3=0.649;//Entropy in kJ/kg.K
+S5=8.15;//Entropy in kJ/kg.K
+h1=2727.7;//Entropy in kJ/kg
+h2=1778.3;//Entropy in kJ/kg
+h3=191.8;//Enthalpy in kJ/kg
+h4=201.79;//Enthalpy in kJ/kg
+h5=2584.7;//Enthalpy in kJ/kg
+x2=0.63;//Quality of steam
+V3=0.001;//Specific volume in m^3/kg
+Cpw=4.18;//specific heat of water in kJ/kgk
+nt=0.8;//Turbine efficiency in percentage
+np=0.9;//Pump efficiency in percentage
+
+
+//CALCULATIONS
+h21=h1-nt*(h1-h2);//Entropy in kJ/kg
+h41=((h4-h3)/np)+h3;//Entropy in kJ/kg
+nRA=((h1-h21)-(h41-h3))/(h1-h4)*100;//Actual thermal efficiency
+m=pb*1000/((h1-h21)-(h41-h3));//Mass flow rate of steam
+mx=(m*3600);//Mass flow rate in kg/hr
+QS1=m*(h1-h41);//Rate of heat transfer into working medium in MW
+QR1=m*(h21-h3);//Rate of heat transfer from the condenser in MW
+mw1=(mx*(h21-h3))/((Cpw)*(two-twi))/10^7;//mass flow rate of water in the condenser in kg/s
+RwA=((h1-h21)-(h41-h3))/(h1-h21);//work ratio
+
+ //OUTPUT
+printf('(i) The Actual Thermal efficiency is %3.2f percent \n(ii)The mass flow rate of steam is %3.2f kJ/s \n(iii) The rate of heat transfer into working medium is %3.1f kJ/s \n(iv)The rate of heat transfer from condenser is %3.2f kJ/s\n(v)mass flow rate of water in condenser is %3.3f *10^7 kg/s \n(vi) The work ratio is %f ',nRA,m,QS1,QR1,mw1,RwA)
diff --git a/1808/CH4/EX4.5/Chapter4_Example5.sce b/1808/CH4/EX4.5/Chapter4_Example5.sce new file mode 100644 index 000000000..a6dafde32 --- /dev/null +++ b/1808/CH4/EX4.5/Chapter4_Example5.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+T1=500;//temperature in degree C
+pb=70;//Saturated vapour in bar
+pc=0.2;//Saturated liquid in bar
+v1=30;//Specific volume in m^3/kg
+v2=90;//Specific volume in m^3/kg
+ms=130000;//mass flow rate in kg/hr
+h1=3410.3;//Enthalpy in kJ/kg
+S1=6.798;//Entropy in kJ/kg.K
+h3=251.2;//Enthalpy in kJ/kg
+h5=2609.9;//Enthalpy in kJ/kg
+v3=0.1;//Specific volume in m^3/kg
+S3=0.8321;//Entropy in kJ/kg.K
+S5=7.9094;//Entropy in kJ/kg.K
+
+
+//CALCULATIONS
+x2=(S1-S3)/(S5-S3);//quality of steam
+h2=(h3+(x2*(h5-h3)));//Enthalpy in kJ/kg
+Wt=(h1-h2)+((v1^2-v2^2)/(2*1000));//Tyrbine work in kJ/kg
+h4=h3+(v3*(pb-pc));//enthalpy in kJ/kg
+nRi=((Wt-(h4-h3))/(h1-h4))*100;//Ideal thermal efficiency in percentage
+P=ms*((Wt-(h4-h3))/3600)/1000;//Power developed in MW
+
+
+//OUTPUT
+printf('(i) The Ideal Thermal efficiency is %3.1f percent \n(ii) The power developed is %3.1f MW ',nRi,P)
diff --git a/1808/CH4/EX4.6/Chapter4_Example6.sce b/1808/CH4/EX4.6/Chapter4_Example6.sce new file mode 100644 index 000000000..2847a33b7 --- /dev/null +++ b/1808/CH4/EX4.6/Chapter4_Example6.sce @@ -0,0 +1,38 @@ +clc
+clear
+//INPUT DATA
+pb=15;//Saturated vapour in bar
+pc=0.1;//Saturated liquid in bar
+pcr=0.05;//Saturated liquid in bar
+x2=0.95;//Quality of steam
+m=50;//Steam flow rate
+Tmax=350;//temperature in degree C
+Tmin=45.8;//temperature in degree C
+h1=3147.5;//Enthalpy in kJ/kg
+S1=7.102//Entropy in kJ/kg.K
+h41=191.8;//Enthalpy in kJ/kg
+v41=0.001001;//Specific volume in m^3/kg
+h7=2584.7;//Enthalpy in kJ/kg
+h4=137.8;//Enthalpy in kJ/kg
+v4=0.001005//Specific volume in m^3/kg
+S6=8.395;//Entropy in kJ/kg.K
+S4=0.476;//Entropy in kJ/kg.K
+h6=2561.5;//Enthalpy in kJ/kg
+
+//CALCULATIONS
+h2=h41+x2*(h7-h41);//Enthalpy in kJ/kg
+x3=((S1-S4)/(S6-S4));//quality of steam
+h3=h4+x3*(h6-h4);//Enthalpy in kJ/kg
+h51=h41+v41*(pb-pc);//Enthalpy in kJ/kg
+h5=h4+v4*(pb-pcr);//Enthalpy in kJ/kg
+nRi=(((h1-h2)-(h51-h41))/(h1-h51))*100;//Ideal rankine efficiency
+P=(m*((h1-h2)-(h51-h41)));//Power in kW
+ssc=((m*3600)/P);//Specific steam consumption in kg/kW.hr
+nC=((Tmax-Tmin)/(Tmax+273))*100;//carnot efficiency in percentage
+nRi1=(((h1-h3)-(h5-h4))/(h1-h5));//Change in rankine efficiency
+P1=(m*((h1-h3)-(h5-h4)));//power in kW
+ssc1=((m*3600)/P1);//Specific steam consumption in kg/kW.hr
+
+
+//OUTPUT
+printf('(i) The Ideal Rankine efficiency is %3.1f percent \n(ii) The specific steam consumption is %3.3f kg/kwh \n(iii)The carnot efficiency for temp limits is %3.1f percent\n(iv)change in rankine efficiency is %3.2f kg/kW.hr',nRi,ssc,nC,ssc1)
diff --git a/1808/CH4/EX4.7/Chapter4_Example7.sce b/1808/CH4/EX4.7/Chapter4_Example7.sce new file mode 100644 index 000000000..05967e0af --- /dev/null +++ b/1808/CH4/EX4.7/Chapter4_Example7.sce @@ -0,0 +1,34 @@ +clc
+clear
+//INPUT DATA
+pb=25;//Saturated vapour in bar
+pc=0.2;//Saturated liquid in bar
+T111=300;//Temperature in degree C
+h1=2800.9;//Enthalpy in kJ/kg
+hb=962;//Enthalpy in kJ/kg
+h5=2609.9;//Enthalpy in kJ/kg
+h3=251.5;//Enthalpy in kJ/kg
+S5=7.9094;//Entropy in kJ/kg.K
+S3=0.8321;//Entropy in kJ/kg.K
+Sb=2.5543;//Entropy in kJ/kg.K
+S1=6.2536;//Entropy in kJ/kg.K
+x1=0.8;////Quality of steam
+h111=3008.9;//Enthalpy in kJ/kg
+S111=6.644;////Entropy in kJ/kg.K
+
+
+//CALCULATIONS
+h11=(hb+x1*(h1-hb));//Enthalpy in kJ/kg
+S11=(Sb+x1*(S1-Sb));//Enthalpy in kJ/kg
+x21=((S11-S3)/(S5-S3));//quality of steam
+h21=(h3+(x21*(h5-h3)));//Enthalpy in kJ/kg
+nRi=(((h11-h21)/(h11-h3))*100);//Rankine cycle efficiency in percentage
+x2=((S1-S3)/(S5-S3));//quality of steam
+h2=h3+x2*(h5-h3);//Enthalpy in kJ/kg
+nRi2=(((h1-h2)/(h1-h3))*100);//Rankine cycle efficiency in percentage
+x211=((S111-S3)/(S5-S3));//quality of steam
+h211=(h3+(x211*(h5-h3)));//Enthalpy in kJ/kg
+nRi1=(((h111-h211)/(h111-h3))*100);//Rankine cycle efficiency in percentage
+
+//OUTPUT
+printf('(i) The Rankine cycle efficiency when steam is dry at turbine inlet is %3.2f percent \n(ii) The Rankine cycle efficiency when steam is saturated is %3.2f percentage \n(iii)The Rankine cycle efficiency when steam is superheated is %3.2f percent ',nRi,nRi2,nRi1)
diff --git a/1808/CH4/EX4.8/Chapter4_Example8.sce b/1808/CH4/EX4.8/Chapter4_Example8.sce new file mode 100644 index 000000000..2a1ae5fac --- /dev/null +++ b/1808/CH4/EX4.8/Chapter4_Example8.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+p1=40;//Boiler pressure in bar
+p2=4;//lp turbine pressure in bar
+p4=0.1;//condenser pressure in bar
+h1=2960.7;//Enthalpy in kJ/kg
+S1=6.362;//Entropy in kJ/kg.K
+h4=3066.8;//Enthalpy in kJ/kg
+S4=7.566;//Entropy in kJ/kg.K
+S3=6.8943;//Entropy in kJ/kg.K
+S10=1.7764;//Entropy in kJ/kg.K
+h3=2737.6;//Enthalpy in kJ/kg
+h10=604.7;//Enthalpy in kJ/kg
+h6=191.8;//Enthalpy in kJ/kg
+h9=2584.7;//Enthalpy in kJ/kg
+S6=0.649;//Entropy in kJ/kg.K
+S9=8.15;//Entropy in kJ/kg.K
+V6=0.001001;//Specific volume in m^3/kg
+
+
+//CALCULATIONS
+x2=((S1-S10)/(S3-S10));//quality of steam
+h2=(h10+(x2*(h3-h10)));//Enthalpy in kJ/kg
+x5=((S4-S6)/(S9-S6));//quality of steam
+h5=(h6+(x5*(h9-h6)));//Enthalpy in kJ/kg
+Wt=((h1-h2)+(h4-h5));//turbine work in kJ/kg
+h7=(h6+(V6*(p1-p4*100)));//Enthalpy in kJ/kg
+Wp=(h7-h6);//Pump work in kJ/kg
+Qs=((h1-h7)+(h4-h2));//heat supplied in kJ/kg
+nRr=((Wt-Wp)/Qs)*100;//Rankine cycle efficiency in percentage
+
+
+//OUTPUT
+printf('(i) The Rankine efficiency is %3.2f percent ',nRr)
diff --git a/1808/CH4/EX4.9/Chapter4_Example9.sce b/1808/CH4/EX4.9/Chapter4_Example9.sce new file mode 100644 index 000000000..334c60417 --- /dev/null +++ b/1808/CH4/EX4.9/Chapter4_Example9.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+p1=100;//Boiler pressure in Bar
+p2=20;//low pressure turbine pressure in Bar
+p3=0.1;//condenser pressure in Bar
+T=500;//Temperature inlet to turbine in Degree C
+h1=3373.7;//Enthalpy in kJ/kg
+h2=2797.2;//Enthalpy in kJ/kg
+h5=191.8;//Enthalpy in kJ/kg
+h9=2584.7;//Enthalpy in kJ/kg
+S5=0.649;//Entropy in kJ/kg.K
+S9=8.15;//Entropy in kJ/kg.K
+Wn=1500;//Net work done in kJ/kg
+nRi=401;//Rankine efficiency in percentage
+
+
+//CALCULATIONS
+Qs=(Wn/nRi)*1000;//heat supplied in kJ/kg
+h3=Qs-(h1-h5-h2);//Enthalpy in kJ/kg
+t3=450+(((h3-3357.5)*(T-450))/(3467.6-3357.5));//Temperature in degree C
+
+//OUTPUT
+printf('Temperature of steam leaving is %3.2f degree C',t3)
diff --git a/1808/CH5/EX5.1/Chapter5_Exampl1.sce b/1808/CH5/EX5.1/Chapter5_Exampl1.sce new file mode 100644 index 000000000..7e505ea82 --- /dev/null +++ b/1808/CH5/EX5.1/Chapter5_Exampl1.sce @@ -0,0 +1,43 @@ +clc
+clear
+//INPUT DATA
+g=1.4;//for isentropic compression
+n=1.3;//for polytropic compression
+p1=1;//pressure in bar
+v1=0.05;//piston displacement in m^3
+R=0.287;//gas constant
+Rp=6;//compression ratio at constant pressure
+t1=293;//temperature in K
+
+//CALCULATIONS
+//Isentropic copression
+m=(p1*10^5*v1)/(1000*R*t1);//mass of air handled in kg
+t21=t1*(Rp^((g-1)/g));//Temperature at the end of compression
+ws=p1*10^5*v1/1000;//workdone by air during suction
+wc=m*R*(t21-t1)/(g-1);//workdone by air during compression
+wd=m*R*t21;//workdone by air during delivery
+wn=wc+wd-ws;//net work done on air during cycle in kJ
+
+//Polytropic compression
+t2=t1*(Rp^((n-1)/n));//Temperature at the end of compression
+ws1=p1*10^2*v1;//workdone by air during suction
+wc1=m*R*(t2-t1)/(n-1);//workdone by air during compression
+Qc1=((g-n)/(g-1))*wc1;//Heat transferred to the cylinder walls
+wd1=m*R*t2;//workdone by air during delivery
+wn1=wc1+wd1-ws1;//net work done on air during cycle in kJ
+
+//Isothermal compression
+ws2=p1*10^2*v1;//workdone by air during suction
+wc2=p1*10^2*v1*log(Rp);//workdone by air during compression
+wd2=p1*10^2*v1;//workdone by air during delivery
+wn2=wc2+wd2-ws2;//net work done during cycle
+
+//OUTPUT
+printf('(i)isentropic compression \n (a)Temperature at the end of compression is %3.2f K \n (b)Workdone by air during suction is %3.1f kNm \n (c)workdone during compression is %3.3f kJ \n heat transfer to the cylinder walls is zero \n (d)workdone on air during delivery %3.2f kJ \n (e)Net workdone on air during cycle is %3.4f kJ \n',t21,ws,wc,wd,wn)
+
+printf('(i)Polytropic compression \n (a)Temperature at the end of compression is %3.2f K \n (b)Workdone by air during suction is %3.1f kNm \n (c)workdone during compression is %3.3f kJ \n heat transfer to the cylinder walls is %3.4f kJ \n (d)workdone on air during delivery %3.2f kJ \n (e)Net workdone on air during cycle is %3.4f kJ \n',t2,ws1,wc1,Qc1,wd1,wn1)
+
+printf('(i)isothermal compression \n (a)Temperature at the end of compression is 293K \n (b)Workdone by air during suction is %3.1f kNm \n (c)workdone during compression is %3.3f kJ \n heat transfer to the cylinder walls is equal to workdone during compression \n (d)workdone on air during delivery %3.2f kJ \n (e)Net workdone on air during cycle is %3.4f kJ \n',ws2,wc2,wd2,wn2)
+
+
+
diff --git a/1808/CH5/EX5.10/Chapter5_Exampl10.sce b/1808/CH5/EX5.10/Chapter5_Exampl10.sce new file mode 100644 index 000000000..d7ff200cb --- /dev/null +++ b/1808/CH5/EX5.10/Chapter5_Exampl10.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+c=0.04;//clearance volume
+p1=0.98;//pressure in bar
+p2=7;//pressure in bar
+n=1.3;//constant for cylinder
+pa=1.013;//ambient pressure in bar
+x=1.3;//stroke to bore ratio
+va=0.25;//volume in m^3/sec
+ta=300;//ambient temperature in K
+t1=313;//temperature in K
+
+//CALCULATIONS
+nv=(1+c-c*((p2/p1)^(1/n)))*100;//volumetric efficiency in percentage
+v14=(pa-va)*t1/(p1*ta);//volume in m^3/sec
+vs=v14/nv;//swept volume in m^3/sec
+l=(0.03141*4/(3.14*9))^(1/3);//stroke length in m
+d=3*l;//bore length in m
+ip=(n/(n-1))*p1*10^2*(v14)*(((p2/p1)^((n-1)/n))-1);//indicated power in kW
+
+//OUTPUT
+printf('(i)Volumetric efficiency is %3.2f percentage \n (ii)Cylinder dimensions \n l= %3.4f m \n d= %3.3f m \n (iii)Indicated power %3.3f kW',nv,l,d,ip)
+
+
diff --git a/1808/CH5/EX5.11/Chapter5_Exampl11.sce b/1808/CH5/EX5.11/Chapter5_Exampl11.sce new file mode 100644 index 000000000..6990a0f42 --- /dev/null +++ b/1808/CH5/EX5.11/Chapter5_Exampl11.sce @@ -0,0 +1,14 @@ +clc
+clear
+//INPUT DATA
+nv=0.8;//volumetric efficiency in percentage
+vc=3;//clearence volume in litre
+p2=8;//air compressor pressure in bar
+p1=0.98;//air compressor pressure in bar
+
+//CALCULATIONS
+vs=12.085/(1-nv);//stroke volume in m^3
+d=((vs/1000)*4/3.14)^(1/3);//cylinder length in m
+
+//OUTPUT
+printf('(i)The stroke volume is %3.5f litre \n (ii)cylinder dimensions (l=d) is %3.4f m ',vs,d)
diff --git a/1808/CH5/EX5.12/Chapter5_Exampl12.sce b/1808/CH5/EX5.12/Chapter5_Exampl12.sce new file mode 100644 index 000000000..2d07c718a --- /dev/null +++ b/1808/CH5/EX5.12/Chapter5_Exampl12.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+pd=8;//delivery pressure in bar
+p1=1;//pressure in bar
+n=1.3;//for single compression
+m=2;//mass flow rate
+R=0.287;//gas constant
+t1=293;//temperature in K
+N=2;//number of stages
+t51=303;//temperature in K
+
+//CALCULATIONS
+wd1=(n/(n-1))*(m/60)*R*t1*(((pd/p1)^((n-1)/n))-1);//work done in single stage compression
+wd2=N*(n/(n-1))*(m/60)*R*t1*(((pd/p1)^((n-1)/(N*n)))-1);//work done in two stage compression
+wd3=(n/(n-1))*(m/60)*R*(((2*(t1*t51)^(1/2))*((pd/p1)^((n-1)/(n*N))))-(t1+t51));//work done in two stage compression with imperfect inter cooling
+wd4=(m/60)*R*t1*log(pd/p1);//single stage compression in kW
+p2=((wd1-wd2)/wd1)*100;//Percentage saving in 1st and 2nd stage
+p3=((wd1-wd3)/wd1)*100;//Percentage saving in 1st and 3rd stage
+p4=((wd1-wd4)/wd1)*100;//Percentage saving in 1st and 4th stage
+
+//OUTPUT
+printf('(i)work done in single stage compression is %3.3f kW \n (ii)work done in two stage compression is %3.4f kW \n (iii)work done in two stage compression with imperfect inter cooling is %3.4f kW \n (iv)single stage compression workdone is %3.4f kW \n ',wd1,wd2,wd3,wd4 )
+
+printf('Percentage saving in 1st and 2nd stage %3.3f percentage \n Percentage saving in 1st and 3rd stage %3.3f percentage \n Percentage saving in 1st and 4th stage %3.3f percentage \n',p2,p3,p4 )
diff --git a/1808/CH5/EX5.13/Chapter5_Exampl13.sce b/1808/CH5/EX5.13/Chapter5_Exampl13.sce new file mode 100644 index 000000000..0fd56f0d1 --- /dev/null +++ b/1808/CH5/EX5.13/Chapter5_Exampl13.sce @@ -0,0 +1,20 @@ +clc
+clear
+//INPUT DATA
+n=1.2;//constant for multistage compressor
+c=4;//clearance
+p4=20;//pressure in bar
+p1=1;//pressure in bar
+v1=15;//volume of free air in m^3/min
+
+//CALCULATIONS
+N=2.16;//(4^N=20) //No.of stages
+C=(p4/p1)^(1/3);//Exact stage pressure ratio
+p2=C*p1;//Intermediate pressure in bar
+p3=C*p2;//Intermediate pressure in bar
+p4=C*p3;//Intermediate pressure in bar
+P=(3*(n/(n-1))*p1*10^5*(v1/60)*(((p4/p1)^((n-1)/(3*n)))-1))/1000;//Power required to compress
+
+//OUTPUT
+printf('(a)No.of stages is %3.2f \n (b)Intermediate pressures %3.2f bar \n pressure p3 %3.2f bar \n pressure p4 %3.2f bar \n (c)Power required to compress is %3.i kW',N,p2,p3,p4,P)
+
diff --git a/1808/CH5/EX5.14/Chapter5_Exampl14.sce b/1808/CH5/EX5.14/Chapter5_Exampl14.sce new file mode 100644 index 000000000..31f582bce --- /dev/null +++ b/1808/CH5/EX5.14/Chapter5_Exampl14.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+n=1.3;//index of compression
+p1=1;//pressure in bar
+va=2;//volume of air in m^3
+N=2;//No.of stages
+p3=50;//delivery pressure in bar
+R=0.287;//gas constant
+t1=303;//temperature in K
+t31=314;//temperature in K
+vcs=0.05;//ratio of clearance volume to stroke volume
+
+//CALCULATIONS
+ip1=(n/(n-1))*p1*10^2*(va/60)*N*(((p3/p1)^((n-1)/(n*N)))-1);//IP for perfect cooling in kW
+m=(p1**10^2*va/(R*t1));//mass flow rate in kg/min
+ip2=(n/(n-1))*(m/60)*R*(t1/3)*(2*sqrt(t1*t31)*(((p3/p1)^((n-1)/(n*N))))-(t1+t31));//IP for imperfect intercooling
+p2=sqrt(p1*p3);//pressure in bar
+nv1=1-vcs*(((p2/p1)^(1/n))-1);//volumetric efficiency in percentage
+vs1=va/nv1;//stroke volume in m^3/min
+d1=(vs1*4/(3.14*N*100))^(1/3);//Dimensions of the cylinder
+d2=d1*(p1/p2);//Dimensions of the cylinder
+v13=(p2/p1);//volume ratio
+v1=1.05*vs1;//volume in m^3
+v2=v1/((p2/p1)^(1/n));//volume in m
+t2=(p2/((p2/p1)^(1/n)))*t1;//temperature in K
+v31=v2*t31/t2;//volume in m
+v131=v1/v31;//volume ratio
+
+//OUTPUT
+printf('(a)IP for Perfect cooling %3.3f kW \n (b) IP for Imperfect intercooling is %3.2f kW \n (a1)perfect intercooling \n cylinder volume ratio is %3.2f \n (b1)Imperfect intercooling \n cylinder volume ratio is %3.3f \n ',ip1,ip2,v13,v131)
+
diff --git a/1808/CH5/EX5.15/Chapter5_Exampl15.sce b/1808/CH5/EX5.15/Chapter5_Exampl15.sce new file mode 100644 index 000000000..109b26392 --- /dev/null +++ b/1808/CH5/EX5.15/Chapter5_Exampl15.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+pa=1;//Ambient pressure in bar
+p1=0.98;//pressure in bar
+p2=4;//pressure in bar
+p3=15;//pressure in bar
+ta=293;//Ambient temperature in K
+t1=303;//temperature in K
+t5=303;//temperature in K
+n=1.3;//for two stage compressor
+c=0.05;//clearance volume
+R=0.287;//gas constant
+
+//CALCULATIONS
+nvs=1+c-c*((p2/p1)^(1/n));//Volumetric efficiency in percentage
+nva=((p1/pa)*(ta/t1)*(nvs))*100;//Volumetric efficiency referred to ambient condition in percentage
+wlp=(n/(n-1))*R*t1*(((p2/p1)^((n-1)/n))-1);//work done in Lp cylinder
+whp=(n/(n-1))*R*t5*(((p3/p2)^((n-1)/n))-1);//work done in Hp cylinder
+wt=wlp+whp;//work done in total cylinder
+wiso=R*t1*log(p3/p1);//Isothermal work done per kg of air
+niso=(wiso/wt)*100;//Isothermal efficiency in percentage
+
+//OUTPUT
+printf('(i)The volumetric efficiency referred to ambient condition is %3.2f percentage \n (ii)work done to deliver air by compressor is %3.2f kJ/kg \n (iii)Isothermal efficiency is %3.2f percentage',nva,wt,niso)
+
+
+
diff --git a/1808/CH5/EX5.16/Chapter5_Exampl16.sce b/1808/CH5/EX5.16/Chapter5_Exampl16.sce new file mode 100644 index 000000000..2f205c9d5 --- /dev/null +++ b/1808/CH5/EX5.16/Chapter5_Exampl16.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+n=1.3;//index of expansion
+N=2;//no.of stages
+R=0.287;//gas constant
+m=5;//mass flow rate
+t1=288;//temperature in K
+pd=16;//delivery pressure in bar
+p1=1;//pressure in bar
+cp=0.997;//specific pressure in kJ/kgK
+cv=0.71;//specific volume in kJ/kgK
+g=1.4;//constant
+s=400;//speed in rpm
+c1=0.05;//clearance volume
+c2=0.08;//clearance volume
+
+//CALCULATIONS
+ip=N*(n/(n-1))*(m/60)*R*t1*(((pd/p1)^((n-1)/(n*N)))-1);//indicated power in kW
+ipiso=(m/60)*R*t1*log(pd/p1);//indicated power in isothermal condition
+niso=(ipiso/ip)*100;//Isothermal efficiency in percentage
+t2=t1*((pd/p1)^((n-1)/(n*N)));//temperature in K
+Qlp=cv*(g-n)*(t2-t1)*(m/60)/(n-1);//Heat transferred in LP cylinder per second
+Qic=(m/60)*cp*(t2-t1);//Heat transferred in intercooler per seconds
+va=(m/60)*R*t1/(p1*10^2);//Free air delivered in m^3/s
+nvlp=(1+c1-(c1*((pd/p1)^(1/N*n))));//Volumetric efficiency of LP cylinder in percentage
+nvlp1=nvlp*100;//Volumetric efficiency of LP cylinder in percentage
+nvhp=(1+c2-(c2*(pd/p1)^(1/N*n)));//Volumetric efficiency of HP cylinder in percentage
+nvhp1=nvhp*100;//Volumetric efficiency of HP cylinder in percentage
+vslp=va*60/(nvlp*s);//swept volume of LP cylinder
+vshp=va*60/(sqrt(pd*p1)*nvhp*s);//swept volume of HP cylinder
+
+//OUTPUT
+printf('(i)Power required to run the compressor is %3.2f kW \n (ii)Isothermal efficiency is %3.2f percentage \n (iii)Heat transferred in LP cylinder per second is %3.4f kW \n (iv)Free air delivered is %3.5f m^3/sec \n (v)volumetric efficiency of LP is %3.2f percentage \n volumetric efficiency of HP is %3.2f percentage \n swept volume of LP is %3.6f m^3 \n swept volume of HP is %3.6f m^3 \n',ip,niso,Qlp,va,nvlp1,nvhp1,vslp,vshp)
+
+
+
diff --git a/1808/CH5/EX5.17/Chapter5_Exampl17.sce b/1808/CH5/EX5.17/Chapter5_Exampl17.sce new file mode 100644 index 000000000..f2cbcfbac --- /dev/null +++ b/1808/CH5/EX5.17/Chapter5_Exampl17.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+p1=0.98;//suction pressure in bar
+p3=18;//delivery pressure in bar
+t1=293;//free conduction temperature in K
+t5=303;//suction temperature in K
+pa=1;//ambient temperature in bar
+n=1.3;//index of expansion
+ta=293;//ambient temperature in K
+N=150;//speed in rpm
+c=0.06;//clearence volume
+v1=5;//volume in m^3
+R=0.287;//gas constant
+
+
+//CALCULATIONS
+m=p1*10^2*v1/(R*t1);//mass of air handled/min
+p2=(p1*p3)^(1/2);//pressure in bar
+ip=N*(n/(n-1))*(m/60)*R*t1*(((p3/p1)^((n-1)/(n*N)))-1);//indicated power in kW
+nv=1+c-(c*((p2/p1)^(1/n)));//volumetric efficiency in percentage
+va=4.5*1*t1/(p1*t5);//colume of air in m^2/min
+vs=va/nv;//swept volume in m^2/min
+d=(vs*4/(3.14*N))^(1/3)*100;//dimensions im cm
+
+//OUTPUT
+printf('(i)Indicated power is %3.3f kW \n (ii)The dimensions of the LP cylinder (d=l)%3.2f cm',ip,d)
+
+
+
diff --git a/1808/CH5/EX5.18/Chapter5_Exampl18.sce b/1808/CH5/EX5.18/Chapter5_Exampl18.sce new file mode 100644 index 000000000..3806ecfb6 --- /dev/null +++ b/1808/CH5/EX5.18/Chapter5_Exampl18.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+c=4;//clearance
+p1=1;//pressure in bar
+p5=120;//pressure in bar
+va=15;//volume in m^3/min
+n=1.2;//index of expansion
+
+//CALCULATIONS
+N=log((p5/p1))/log(c);//No.of stages
+//take N=3.5=4 APPROXIMATELY
+C=(p5/p1)^(1/4);//Exact pressure ratio
+p2=C*p1;//Intermediate pressure in bar
+p3=C*p2;//Intermediate pressure in bar
+p4=C*p3;//Intermediate pressure in bar
+P5=C*p4;//Intermediate pressure in bar
+ip=p1*10^2*(va/60)*(n/(n-1))*N*(((p5/p1)^((n-1)/(n*N)))-1);//Minimum power to compress in kW
+
+//OUTPUT
+printf('(i)Number of stages %3.1f \n (ii)Exact pressure ratio %3.2f \n (iii)Intermediate pressure is p2 %3.4f bar \n p3 %3.4f bar \n p4 %3.4f bar \n p5 %3.4f bar \n (iv)Minimum power to compress is %3.2f kW ',N,C,p2,p3,p4,p5,ip)
+
+
+
+
+
+
+
diff --git a/1808/CH5/EX5.19/Chapter5_Exampl19.sce b/1808/CH5/EX5.19/Chapter5_Exampl19.sce new file mode 100644 index 000000000..d1ff880fc --- /dev/null +++ b/1808/CH5/EX5.19/Chapter5_Exampl19.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+va=30;//volume in m^3
+p1=1;//pressure in bar
+p2=16;//pressure in bar
+n=1.32;//index of expansion and compression
+N=320;//speed of the copressor in rpm
+t1=300;//temperature in K
+t2=312;//temperature in K
+c=0.04;//clearance
+nm=0.8;//mech efficiency in percentage
+
+//CALCULATIONS
+va1=p1*va*t2/(t1*p1);//volume in m^3/min
+nv=(1+c-c*(((p2/p1)^(1/n))));//volumetric efficiency in percentage
+vs=va1/nv;//swept volume in m^3/min
+d=((vs*4/(3.14*1.2*N*4))^(1/3))*100;//dimensions of the cylinder in cm
+l=1.2*d;//dimensions of the cylinder in cm
+ip=(n/(n-1))*p1*10^2*(va1/30)*(((p2/p1)^((n-1)/n))-1);//Power required for motor in kW
+mp=ip/(2*nm);//Power required for motor in kW
+
+//OUTPUT
+printf('(i)Dimensions of the cylinder bore %3.2f cm \n stroke %3.2f cm \n (ii)Power required for the motor is %3.2f kW',d,l,mp)
+
diff --git a/1808/CH5/EX5.2/Chapter5_Exampl2.sce b/1808/CH5/EX5.2/Chapter5_Exampl2.sce new file mode 100644 index 000000000..26979136f --- /dev/null +++ b/1808/CH5/EX5.2/Chapter5_Exampl2.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+p1=1;//pressure in bar
+n=1.2;//constant
+N=100;//speed in rpm
+Rp=6;//compression ratio at constant pressure
+aps=150;//average piston speed in m/min
+ip=50;//indicated power in kW
+
+
+//CALCULATIONS
+pm=p1*(n/(n-1))*((Rp^((n-1)/n))-1);//Mean effective pressure in bars
+a=ip*60/(pm*10^2*150);//size of the cylinder in m^2
+d=sqrt(a*4/3.14);//size of the cylinder in m^2
+l=150/(2*N);//size of the cylinder in m^2
+
+//OUTPUT
+printf('(i)size of the cylinder is d %3.4f m \n l %3.2f m',d,l)
+
+
diff --git a/1808/CH5/EX5.20/Chapter5_Exampl20.sce b/1808/CH5/EX5.20/Chapter5_Exampl20.sce new file mode 100644 index 000000000..9d3a2c89e --- /dev/null +++ b/1808/CH5/EX5.20/Chapter5_Exampl20.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+p3=60;//pressure in bar
+p2=7.33;//pressure in bar
+p1=1;//pressure in bar
+n=1.35;//index of expansion and compression
+d1=0.1;//diameter in m
+l1=0.1125;//stroke length in m
+t1=288;//temperature in K
+N=250;//speed in rpm
+Ns=2;//no.of stages
+t5=303;//temperature in K
+R=0.287;//gas constant
+
+//CALCULATONS
+val=(3.14*d1^2*l1*N)/4;//volume of air at atmospheric condition
+m=p1*10^2*val/(R*t1);//mass of air required in kg/min
+ipt=(n/(n-1))*p1*10^2*(val/60)*(((p2/p1)^((n-1)/n))-1)+(n/(n-1))*R*t5*(val/60)*(((p3/p2)^((n-1)/n))-1);//Power required for motor in kW
+d2=sqrt((d1*100)^2*(t5/t1)*(p1/p2));//diameter of the high pressure cylinder in cm
+
+//OUTPUT
+printf('(i)Power of the compressor is %3.3f kW \n (ii)diameter of the high pressure cylinder is %3.2f cm',ipt,d2)
diff --git a/1808/CH5/EX5.21/Chapter5_Exampl21.sce b/1808/CH5/EX5.21/Chapter5_Exampl21.sce new file mode 100644 index 000000000..621ef094a --- /dev/null +++ b/1808/CH5/EX5.21/Chapter5_Exampl21.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+v1=2;//volume in m^3
+R=0.287;//gas constant
+t1=288;//temperature in K
+p2=8;//final pressure in bar
+t2=313;//final temperature in K
+d=14;//displacement in m^3/min
+T=70;//time in seconds
+
+//CALCULATIONS
+m1=p1*10^2*v1/(R*t1);//initial mass in kg
+m2=p2*10^2*v1/(R*t2);//initial mass in kg
+m=m2-m1;//weight of air compressed in kg
+va=m*R*t1/(p1*10^2);//free volume in m^3
+vs=d*T/60;//swept volume in m^3
+nv=(va/vs)*100;//Volumetric efficiency in percentage
+
+//OUTPUT
+printf('(i)Volumetric efficiency is %3.2f percentage ',nv)
+
+
+
+
+
diff --git a/1808/CH5/EX5.22/Chapter5_Exampl22.sce b/1808/CH5/EX5.22/Chapter5_Exampl22.sce new file mode 100644 index 000000000..c0344336d --- /dev/null +++ b/1808/CH5/EX5.22/Chapter5_Exampl22.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+pd=30;//delivery pressure in bar
+t1=288;//temperature in K
+n=1.3;//index of copression
+
+//CALCULATIONS
+p21=sqrt(p1*pd);//Intermediate pressure in bar
+v121=(p21/p1)^(1/n);//volume ratio
+t21=t1*(p21/p1)^((n-1)/n);//temperature in K
+v212=t21/t1;//volume ratio
+v12=v121*v212;//volume ratio
+d12=sqrt(v12);//Ratio of cylinder diameters
+
+//OUTPUT
+printf('(i)Ratio of cylinder diameters is %3.3f ',d12)
+
diff --git a/1808/CH5/EX5.23/Chapter5_Exampl23.sce b/1808/CH5/EX5.23/Chapter5_Exampl23.sce new file mode 100644 index 000000000..02f732aa9 --- /dev/null +++ b/1808/CH5/EX5.23/Chapter5_Exampl23.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+pd=12;//delivery pressure in bar
+R=0.287;//gas constant
+t1=310;//temperature in K
+m=1;//mass of air
+cp=1.005;//specific pressure
+n=1.4;//index of compressor
+
+//CALCULATIONS
+p2=sqrt(p1*pd);//Intermediate pressure in bar
+v1=R*t1/(p1*10^2);//Volume in m^3
+t2=t1*((p2/p1)^((n-1)/n));//temperature in K
+Qc=m*cp*(t2-t1);//Heat rejected in the intercooler per kg of air
+
+//OUTPUT
+printf('(i)Heat rejected in the intercooler per kg of air is %3.2f kW',Qc)
+
+
+
diff --git a/1808/CH5/EX5.24/Chapter5_Exampl24.sce b/1808/CH5/EX5.24/Chapter5_Exampl24.sce new file mode 100644 index 000000000..2a741646b --- /dev/null +++ b/1808/CH5/EX5.24/Chapter5_Exampl24.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+t1=293;//temperature in K
+p2=10;//pressure in bar
+p1=1;//pressure in bar
+n=1.2;//index of compressor
+m=1;//mass pf air
+R=0.287;//gas constant
+g=1.4;//constant
+
+//CALCULATIONS
+t2=t1*((p2/p1)^((n-1)/n));//temperature in K
+wd=m*R*(t2-t1)/(n-1);//workdone during compression per kg of air
+Q=((g-n)/(g-1))*wd;//heat transferred during compression per kg of air
+
+//OUTPUT
+printf('(i)Temperature at the end of the compressor is %3.2f K \n (ii)workdone during compression per kg of air %3.3f kJ/kg \n (iii)heat transferred during compression per kg of air %3.2f kJ/kg',t2,wd,Q)
+
+
+
diff --git a/1808/CH5/EX5.25/Chapter5_Exampl25.sce b/1808/CH5/EX5.25/Chapter5_Exampl25.sce new file mode 100644 index 000000000..8bea75c3f --- /dev/null +++ b/1808/CH5/EX5.25/Chapter5_Exampl25.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+n=1.4;//index of compression
+m=0.1436;//mass of air
+R=0.287;//gas constant
+t1=283;//temperature in K
+t2=303;//temperature in K
+p1=1;//pressure in bar
+pd=30;//delivery pressure in bar
+v1=0.1167;//volume in m^3/s
+nm=0.9;//mechanical efficiency in percentage
+
+//CALCULATIONS
+ip=(n/(n-1))*m*R*((2*sqrt(t1*t2)*((pd/p1)^((n-1)/(2*n))))-(t1+t2));//power required for a compound air compressor
+bp=ip/nm;//Brake power in kW
+
+//OUTPUT
+printf('(i)power required for a compound air compressor %3.4f kW \n (ii)Brake power is %3.2f kW',ip,bp)
+
+
+
+
+
+
+
diff --git a/1808/CH5/EX5.26/Chapter5_Exampl26.sce b/1808/CH5/EX5.26/Chapter5_Exampl26.sce new file mode 100644 index 000000000..d0fc60ebb --- /dev/null +++ b/1808/CH5/EX5.26/Chapter5_Exampl26.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+k=0.05;//clearance
+p1=0.98;//initial pressure in bar
+pd=6.4;//delivery pressure in bar
+n=1.32;//index of compression and expansion
+p0=1;//initial pressure
+t1=305;//temperature in K
+v0=17;//volume in m^3
+t0=288;//teperature in K
+vs=0.02;//volume per stroke in m^3
+
+//CLACULATIONS
+nv=1+k-k*((pd/p1)^(1/n));//volumetric efficiency in percentage
+va=p0*t1*v0/(p1*t0);//volume of air handled at suction condition
+N=va/(vs*nv*2);//speed in rpm
+ip=(n/(n-1))*p1*10^2*(va/60)*((pd/p1)^((n-1)/n)-1);//Indicated power in single stage double acting cylinder in kW
+
+//OUTPUT
+printf('(i)Speed of the compressor is %3.1f rpm \n (ii)Indicated power in single stage double acting cylinder is %3.2f kW',N,ip)
+
+
+
+
+
+
diff --git a/1808/CH5/EX5.27/Chapter5_Exampl27.sce b/1808/CH5/EX5.27/Chapter5_Exampl27.sce new file mode 100644 index 000000000..08f351227 --- /dev/null +++ b/1808/CH5/EX5.27/Chapter5_Exampl27.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+pd=9;//delivery pressure in bar
+n=1.3;//index of compression
+R=0.287;//gas constant
+t1=300;//temperature in K
+m=1;//mass of air
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+p2=sqrt(p1*pd);//intermediate pressure in bar
+wd=2*((n/(n-1))*R*t1*(((pd/p1)^((n-1)/(2*n)))-1));//minimum work done per min
+t2=t1*((p2/p1)^((n-1)/n));//temperature K
+Qr=m*cp*(t2-t1);//heat rejected to intercooler in kJ/kg
+
+//OUTPUT
+printf('(i)minimum work done per min %3.2f kJ/kg \n (ii)%3.4f kJ/kg ',wd,Qr)
diff --git a/1808/CH5/EX5.28/Chapter5_Exampl28.sce b/1808/CH5/EX5.28/Chapter5_Exampl28.sce new file mode 100644 index 000000000..5cd923c56 --- /dev/null +++ b/1808/CH5/EX5.28/Chapter5_Exampl28.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+k=0.05;//clearance
+p1=1;//initial pressure in bar
+pd=5.5;//delivery pressure in bar
+n=1.3;//index of compression
+R=0.287;//gas constant
+N=500;//Speed in rpm
+d=0.2;//diameter in m
+t1=293;//temperature in K
+
+//CALCULATIONS
+nv=1+k-k*((pd/p1)^(1/n));//volumetric efficiency in percentage
+va=nv*(3.14*d^3*1.5*N)/4;//Volume of air in m^3/min
+m=p1*va/(R*t1);//mass of air in kg/min
+ip=(n/(n-1))*p1*10^2*(va/60)*((pd/p1)^((n-1)/n)-1);//Power required to run the compressor in kW
+
+//OUTPUT
+printf('(i)Volumetric efficiency %3.4f percentage \n (ii)Power required to run the compressor is %3.2f kW',nv,ip)
+
+
+
+
diff --git a/1808/CH5/EX5.29/Chapter5_Exampl29.sce b/1808/CH5/EX5.29/Chapter5_Exampl29.sce new file mode 100644 index 000000000..7660d51e0 --- /dev/null +++ b/1808/CH5/EX5.29/Chapter5_Exampl29.sce @@ -0,0 +1,41 @@ +clc
+clear
+//INPUT DATA
+p1=1;//initial pressure in bar
+pd=16;//delivery pressure in bar
+n=1.25;//index of compression
+m=10;//mass flow rate
+R=0.287;//gas constant
+t1=288;//temperature in K
+cp=1.005;//specific pressure
+k1=0.04;//clearance retio
+k2=0.06;//clearance ratio
+N=400;//speed in rpm
+g=1.4;//constant
+
+
+//CALCULATIONS
+p2=sqrt(p1*pd);//intermediate pressure in bar
+ipm=2*(n/(n-1))*(m/60)*R*t1*((p2/p1)^((n-1)/n)-1);//power required in kW
+pi=(m*R*t1/60)*log(pd/p1);//isothermal power
+niso=(pi/ipm)*100;//isothermal efficiency in percentage
+va=m*R*t1/(p1*10^2);//free air delivered in m^3/min
+t2=t1*(p2/p1)^((n-1)/n);//temperature in K
+Qr=(m/60)*cp*(t2-t1);//heat rejected in intercooler in kW
+nvl=1+k1-k1*((p2/p1)^(1/(n*2)));//volumetric efficiency in percentage
+vsl=va/(N*nvl);//swept volume in m^3
+nv2=(1+k2-(k2*((pd/p1)^(1/(n*2)))))*100;//volumetric efficiency in percentage
+vsh=va/(2*((pd/p1)*N*nv2)^(1/2));//swept volume
+Ql=(g-n)*m*R*(t2-t1)/((g-1)*(n-1));//heat transferred in LP
+Qh=(g-n)*m*R*(t2-t1)/((g-1)*(n-1));//heat transferred in HP
+t6=t1*(pd/p1)^((n-1)/n);//temperature in K
+Qi=(m/60)*cp*(t2-t1);//Heat trnsferred in intercooler
+
+//OUTPUT
+printf('(i)The power required is %3.3f kW \n (ii)The isothermal efficiency is %3.3f percentage \n (iii)The free air delivered is %3.4f m^3/min \n (iv)The heat rejected in intercooler is %3.3f kW \n (v)swept volume is %3.5f m^3 \n swept volume is %3.5f m^3 \n (vi)net heat transferred in intercooler is %3.3f kW',ipm,niso,va,Qr,vsl,vsh,Qi)
+
+
+
+
+
+
diff --git a/1808/CH5/EX5.3/Chapter5_Exampl3.sce b/1808/CH5/EX5.3/Chapter5_Exampl3.sce new file mode 100644 index 000000000..0dbacbe00 --- /dev/null +++ b/1808/CH5/EX5.3/Chapter5_Exampl3.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+t1=293;//temperature in K
+p1=1;//pressure in bar
+p2=8;//pressure in bar
+v1=80;//volume in m^3
+g=1.4;//for isentropic compression
+n=1.25;//Adiabatic compression
+
+//CALCULATIONS
+
+//Isothermal compression
+v2=p1*v1/p2;//volume in m^3
+wn=p1*10^2*v1*log(p2/p1);//net work done in kJ/min
+P=wn/60;//Power required in kW
+
+//Adiabatic compression
+v121=(p2/p1)^(1/g);//volume in m^3/min
+v21=v1/v121;//volume in m^3/min
+t211=t1*(p2/v121);//temperature at the end of the compression
+wn1=(g/(g-1))*p1*10^2*v1*(((p2/p1)^((g-1)/g))-1);//net work done in kJ
+P1=wn1/60;//Power in kW
+
+//Polytropic process
+v12=(p2/p1)^(1/n);//volume in m^3/min
+v22=v1/v12;//volume in m^3/min
+t2=t1*(p2/v12);//temperature in K
+wn2=(n/(n-1))*p1*10^2*v1*(((p2/p1)^((n-1)/n))-1);//net work done in kJ
+P2=wn2/60;//power required in kW
+Qc=((g-n)/(g-1))*P2;//Heat transferred in kW
+
+//OUTPUT
+printf('(a)Isothermal compression \n temperature at the end of the compression is 293K \n net work done is %3.2f kJ/min \n Power required is %3.3f kW \n ',wn,P)
+printf('(b)Adiabatic compression \n temperature at the end of the compression is %3.2f K \n net work done is %3.2f kJ/min \n Power required is %3.3f kW \n ',t211,wn1,P1)
+printf('(c)Polytropic compression \n temperature at the end of the compression is %3.2f K \n net work done is %3.2f kJ/min \n Power required is %3.3f kW \n Heat rejected is %3.2f kW\n ',t2,wn2,P2,Qc)
+
diff --git a/1808/CH5/EX5.30/Chapter5_Exampl30.sce b/1808/CH5/EX5.30/Chapter5_Exampl30.sce new file mode 100644 index 000000000..4b06fe769 --- /dev/null +++ b/1808/CH5/EX5.30/Chapter5_Exampl30.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+S=180;//Piston speed in rpm
+N=240;//speed in rpm
+d=0.2;//bore in m
+p1=1;//pressure in bar
+p2=5.67;//compressed pressure in bar
+R=0.287;//gas constant
+t1=288;//entry temperature in K
+n=1.3;//index of compression
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+l=S/(2*N);//Piston speed in m
+vs=(3.14*d^2*l*N)/4;//swept volume in m^3/min
+m=p1*10^2*vs/(R*t1);//mass flow rate in kg/min
+t2=t1*((p2/p1)^((n-1)/n));//exit temperature in K
+wd=(n/(n-1))*(m/60)*R*t1*(((p2/p1)^((n-1)/n))-1);//rate of work done
+wdis=(m/60)*R*t1*log(p2/p1);//Rate of work done by isothermal compression in kW
+
+//OUTPUT
+printf('(i)Mass flow rate %3.2f kg/min \n (ii)rate of work done %3.1f kW \n exit temperature is %3.1f K \n (iii)Rate of work done by isothermal compression is %3.4f kW',m,wd,t2,wdis)
+
+
diff --git a/1808/CH5/EX5.31/Chapter5_Exampl31.sce b/1808/CH5/EX5.31/Chapter5_Exampl31.sce new file mode 100644 index 000000000..1804109af --- /dev/null +++ b/1808/CH5/EX5.31/Chapter5_Exampl31.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+d=0.1;//bore in m
+vc=10*10^-5;//clearance volume
+p1=0.95;//suction pressure
+p2=8;//discharge pressure
+n=1.3;//index of compression
+N=400;//Speed in rpm
+t1=303;//temperature in K
+to=293;//temperature in K
+po=1;//pressure in bar
+
+//CALCULATIONS
+vs=(3.14*(d^3)*1.5)/4;//swept volume in m^3
+k=vc/vs;//clearance ratio
+nv=1+k-(k*((p2/p1)^(1/n)));//volumetric efficiency
+va=vs*nv*N;//volume of air delivered in m^3/min
+vo=p1*va*to/(po*t1);//volume of air delivered in m^3/min
+pm=((n/(n-1))*p1*(va/400)*(((p2/p1)^((n-1)/n))-1))/(vs);//mean effective pressure in bar
+disp(vo)
+//OUTPUT
+printf('(i)The volume of air delivered is %3.4f m^3/min \n (ii)mean effective pressure is %3.3f bar \n ',vo,pm)
diff --git a/1808/CH5/EX5.32/Chapter5_Exampl32.sce b/1808/CH5/EX5.32/Chapter5_Exampl32.sce new file mode 100644 index 000000000..0bec762df --- /dev/null +++ b/1808/CH5/EX5.32/Chapter5_Exampl32.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+v1=10;//volume of air handled in m^3/min
+n=1.4;//index of compression
+m=1;//mass of air
+R=0.287;//gas constant
+t1=293;//initial temperature in K
+p1=1;//pressure in bar
+P2=2;//discharge presuure in bar
+t21=298;//temperature in K
+pd=6;//discharge pressure
+
+//CALCULATIONS
+wd=(n/(n-1))*m*R*t1*(((P2/p1)^((n-1)/n))-1)+(n/(n-1))*R*t21*(((pd/P2)^((n-1)/n))-1);//work done in kJ/kg
+m=p1*10^2*v1/(R*t1);//mass flow rate in kg/min
+ip1=wd*(m/60);//indicated power in kW
+p2=sqrt(p1*pd);//Receiver pressure for best efficiency
+ip2=2*(n/(n-1))*(m/60)*R*t1*(((pd/p1)^((n-1)/(2*n)))-1);//Power required for optimum conditions
+
+//OUTPUT
+printf('(i)power required is %3.3f kW \n (ii)Receiver pressure for the best efficiency %3.4f bar \n (iii)Power required for optimum conditions is %3.3f kW ',ip1,p2,ip2)
diff --git a/1808/CH5/EX5.33/Chapter5_Exampl33.sce b/1808/CH5/EX5.33/Chapter5_Exampl33.sce new file mode 100644 index 000000000..2b4307c83 --- /dev/null +++ b/1808/CH5/EX5.33/Chapter5_Exampl33.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+p1=0.9;//initial pressure in bar
+pd=18;//delivery pressure in bar
+n=1.3;//index of compression
+R=0.287;//gas constant
+t1=303;//temperature in K
+t0=290;//temperature in K
+p11=1;//pressure in bar
+
+//CALCULATIONS
+p2=sqrt(p1*pd);//intermediate pressure in bar
+v0x=(p1/1)*(t0/t1)*0.891743;//volume
+nva=(v0x);//volume of air reduced to atmospheric conditions
+wd=2*((n/(n-1)))*R*t1*(((pd/p1)^((n-1)/(2*n)))-1);//work done required per kg of air delivered in kJ/kg
+wdis=R*t0*log(pd/p11);//isothermal work done in kJ/kg
+niso=(wdis/wd);//isothermal efficiency in percentage
+
+//OUTPUT
+printf('(i)Volumetric efficiency referred to atm conditions %3.5f \n (ii)The work done required to deliver air is %3.2f kJ/kg \n (iii)isothermal efficiency is %3.3f percentage',nva,wd,niso)
+
+
diff --git a/1808/CH5/EX5.34/Chapter5_Exampl34.sce b/1808/CH5/EX5.34/Chapter5_Exampl34.sce new file mode 100644 index 000000000..3875b5ecd --- /dev/null +++ b/1808/CH5/EX5.34/Chapter5_Exampl34.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+R=0.287;//gas constant
+d=0.3;//diameter in m
+l=0.4;//stroke in m
+N=100;//speed in rpm
+p1=1;//pressure in bar
+p2=5;//pressure in bar
+g=1.4;//constant
+n=1.2;//index of copression
+
+//CALCULATIONS
+vs=(3.14*d^2*l*N)/4;//swept volume in m^2/min
+ipis=p1*10^2*(vs/60)*log(p2/p1);//isothermal power required in kW
+niso=(ipis/ipis)*100;//isothermal efficiency in percentage
+pm=ipis/(vs*100);//mean effective pressure in kN/m^2
+wd=(n/(n-1))*p1*10^2*(vs/60)*(((p2/p1)^((n-1)/n))-1);//polytropic work done
+niso1=(ipis/wd)*100;//isothermal efficiency in percentage in polytropic process
+pm1=wd*60/(vs*100);//mean effective pressure in bar
+ipa=(g/(g-1))*p1*10^2*(vs/60)*(((p2/p1)^((g-1)/g))-1);//Adiabatic work done
+niso2=(ipis/ipa)*100;//adiabatic isothermal efficiency in percentage
+pm2=ipa*60/(vs*100);//mean effective pressure in bar
+
+//OUTPUT
+printf('(i)Isothermal \n Isothermal efficiency is %3.1f \n mean effective pressure is %3.4f bar \n (ii)ploytropic process \n Isothermal efficiency %3.2f percentage \n mean effective pressure is %3.4f bar \n (iii)Adiabatic process \n Isothermal efficiency %3.2f percentage \n mean effective pressure is %3.4f bar \n ',niso,pm,niso1,pm1,niso2,pm2)
+
+
diff --git a/1808/CH5/EX5.35/Chapter5_Exampl35.sce b/1808/CH5/EX5.35/Chapter5_Exampl35.sce new file mode 100644 index 000000000..2714794d9 --- /dev/null +++ b/1808/CH5/EX5.35/Chapter5_Exampl35.sce @@ -0,0 +1,25 @@ +clc
+clear
+//INPUT DATA
+p0=1;//suction pressure in bar
+p1=1;//pressure in bar
+p2=6;//delivery pressure in bar
+v0=5;//volume in m^3/min
+t0=288;//suction temperature in K
+t1=300;//initial temperature in K
+k=0.05;//Clearance
+n=1.3;//index of compression
+N=150;//speed in rpm
+
+//CALCULATIONS
+va=(p0/p1)*(t1/t0)*v0;//volume of air delivered in m^3/min
+nv=1+k-k*((p2/p1)^(1/n));//volumetric efficiency in percentage
+vs=va/nv;//stroke volume in m^3/min
+vss=vs/N;//stroke volume per stroke in m^3
+d=(vss*4/(3.14*1.25))^(1/3);//diameter in m
+l=1.25*d;//length in m
+ip=(n/(n-1))*p1*10^2*(va/60)*(((p2/p1)^((n-1)/n))-1);//power required to run the compressor
+
+//OUTPUT
+printf('(i)volumetric efficiency is %3.4f percentage \n (ii)stroke volume of air taken in per stroke is %3.5f m^3 \n (iii)Dimensions of the cylinder stroke %3.2f m \nbore %3.2f m \n (iv)power required to run the compressor is %3.3f kW',nv,vss,l,d,ip)
+
diff --git a/1808/CH5/EX5.4/Chapter5_Exampl4.sce b/1808/CH5/EX5.4/Chapter5_Exampl4.sce new file mode 100644 index 000000000..79f9596e5 --- /dev/null +++ b/1808/CH5/EX5.4/Chapter5_Exampl4.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+n=1.35;//for cylinders
+p1=1;//pressure in bar
+v1=1;//volume in m^3
+p2=7;//pressure in bar
+nm=0.85;//mechanical efficiency in percentage
+nt=0.9;//Turbine efficiency in percentage
+N=300;//speed in rpm
+
+//CALCULATIONS
+//(a)single acting cylinder
+ip1=((n/(n-1))*p1*10^2*v1*(((p2/p1)^((n-1)/n))-1))/60;//indicated power in kW
+bp1=ip1/nm;//brake power in kW
+mp1=bp1/nt;//motor power in kW
+d1=((v1*4/(1.5*N*3.14))^(1/3))*100;//cylinder bore in single acting cylinder in cm
+l1=(1.5)*d1;//stroke in cm
+
+//Double acting cylinder
+d2=((v1*4/(1.5*N*2*3.14))^(1/3))*100;//cylinder bore in double acting cylinder
+l2=1.5*d2;//stroke in cm
+
+//OUTPUT
+printf('(a)Single acting cylinder \n (i)Indicated power is %3.3f kW \n (ii)Power input to the compressor %3.3f kW \n (iii) cylinder bore in single acting cylinder is %3.4f cm \n stroke is %3.2f cm \n',ip1,mp1,d1,l1)
+
+printf('(a)Double acting cylinder \n (i)Indicated power is %3.3f kW \n (ii)Power input to the compressor %3.3f kW \n (iii) cylinder bore in double acting cylinder is %3.4f cm \n stroke is %3.2f cm',ip1,mp1,d2,l2)
+
diff --git a/1808/CH5/EX5.5/Chapter5_Exampl5.sce b/1808/CH5/EX5.5/Chapter5_Exampl5.sce new file mode 100644 index 000000000..7f784992c --- /dev/null +++ b/1808/CH5/EX5.5/Chapter5_Exampl5.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+fad=14;//free air delivered in m^3/min
+N=300;//speed in rpm
+p2=7;//delivery pressure in bar
+p1=1;//pressure in bar
+n=1.3;//index of compression and expansion
+t1=288;//temperature in K
+
+//CALCULATIONS
+
+//Without clearance volume
+vs=fad/(N*2);//swept volume of the cylinder in m^3
+t2=t1*(p2/p1)^((n-1)/n);//Delivery temperature in K
+ip=(n/(n-1))*p1*10^2*(fad/60)*(((p2/p1)^((n-1)/n))-1);//indicated power in kW
+d=((vs*4/(3.14*1.5))^(1/3))*100;//bore of the cylinder in cm
+l=1.5*d;//stroke in cm
+
+//with clearance volume
+vs1=vs/(1.05-vs);//swept volume with clearence volume in m^3
+t2=t1*(p2/p1)^((n-1)/n);//Delivery temperature in K
+nv=(vs/vs1)*100;//volumetric efficiency in percentage
+d1=((vs1*4/(3.14*1.5))^(1/3))*100;//bore of the cylinder in cm
+l1=1.5*d1;//stroke in cm
+//OUTPUT
+printf('(a)Without clearance volume \n (i)swept volume of the cylinder is %3.4f m^3 \n (ii)The delivery temperature is %3.4f K \n (iii)Indicated power is %3.3f kW \n (iv)volumetric efficiency is 100percentage \n (v)bore of the cylinder is %3.2f cm \n stroke %3.4f cm \n',vs,t2,ip,d,l)
+
+printf('(a)With clearance volume \n (i)swept volume of the cylinder is %3.4f m^3 \n (ii)The delivery temperature is %3.4f K \n (iii)Indicated power is %3.3f kW \n (iv)volumetric efficiency is %3.2f percentage \n (v)bore of the cylinder is %3.2f cm \n stroke %3.4f cm \n',vs1,t2,ip,nv,d1,l1)
+
diff --git a/1808/CH5/EX5.6/Chapter5_Exampl6.sce b/1808/CH5/EX5.6/Chapter5_Exampl6.sce new file mode 100644 index 000000000..c263c45c8 --- /dev/null +++ b/1808/CH5/EX5.6/Chapter5_Exampl6.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+vs=0.015;//swept volume in m^3
+vc=0.0008;//clearence volume in m^3
+p3=500;//discharging pressure
+p1=100;//air pressure in kPa
+m=1.4;//isentropic expansion constant
+n=1.3;//polytropic index constant
+
+
+//CALCULATIONS
+v1=vs+vc;//volume in m^3
+v4=vc*((p3/p1)^(1/m));//volume in m^3
+wn=((n/(n-1))*p1*v1*(((p3/p1)^((n-1)/n))-1))-((m/(m-1))*p1*v4*(((p3/p1)^((m-1)/m))-1));//net work done in kJ
+v41=vc*(p3/p1)^(1/n);//volume of absorbing system in m^3
+v14=vs-v41;//volume in m^3
+wn1=(n/(n-1))*p1*(v14)*((((p3/p1)^((n-1)/n))-1));//net work done in kJ
+nd=((wn-wn1)/wn)*100;//percentage in difference in work done
+
+//OUTPUT
+printf('(i)Net cycle work is %3.4f kJ \n (ii)Error evolved is %3.4f ',wn,nd)
+
+
+0
+
+
+
diff --git a/1808/CH5/EX5.7/Chapter5_Exampl7.sce b/1808/CH5/EX5.7/Chapter5_Exampl7.sce new file mode 100644 index 000000000..38a44ebf5 --- /dev/null +++ b/1808/CH5/EX5.7/Chapter5_Exampl7.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+pa=1;//ambient pressure in bar
+ta=15;//temperature in Degree C
+ps1=0.98;//ambient pressure in bar
+ts1=30;//temperature in Degree C
+c=0.04;//clearance
+N=500;//speed in rpm
+p1=1;//ambient pressure
+p2=5;//discharge pressure
+n=1.3;//for cylinders
+
+//CALCULATIONS
+
+//Suction and ambient conditions are same
+vs=0.04*2*N;//swept volume in m^3/min
+nv=1+c-(c*(p2/p1)^(1/n));//volumetric efficiency in percentage
+v14=nv*(vs)/60;//volume in m^3/sec
+ip=(n/(n-1))*p1*100*(v14)*((p2/p1)^((n-1)/n)-1);//indicated power in kJ/min
+
+//Suction and ambient conditions are different
+nv1=1+c-(c*(p2/ps1)^(1/n));//volumetric efficiency in percentage
+v141=nv1*(vs)/60;//volume in m^3/sec
+ip1=(n/(n-1))*ps1*100*(v141)*(((p2/ps1)^((n-1)/n))-1);//indicated power in kJ/min
+vamb=p1*(ta+273)*(v141)/(pa*(ts1+273));//Air discharged in m^3/s
+
+//OUTPUT
+printf('(a)Suction and ambient conditions are same \n (i)Indicated power %3.2f kW \n (ii)air discharged is %3.4f m^3/s \n ',ip,v14)
+printf('(a)Suction and ambient conditions are different \n (i)Indicated power %3.2f kW \n (ii)air discharged is %3.4f m^3/s \n ',ip1,v141)
+
+
diff --git a/1808/CH5/EX5.8/Chapter5_Exampl8.sce b/1808/CH5/EX5.8/Chapter5_Exampl8.sce new file mode 100644 index 000000000..9d29154c6 --- /dev/null +++ b/1808/CH5/EX5.8/Chapter5_Exampl8.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+d=0.3;//bore length in m
+l=0.4;//stroke length in m
+N=300;//speed in rpm
+g=1.4;//constant
+n=1.25;//adiabatic compression constant
+p1=1;//suction pressure
+pd=5;//delivery pressure
+m=1.5;//adiabatic constant
+
+//CALCULATIONS
+vs=(3.14*(d)^2*l*N)/4;//Volume of air compressed per min
+pm=p1*log(pd/p1);//Mean effective pressure in bar
+ip=p1*10^2*(vs/60)*log(pd/p1);//indicated power in kW
+pm1=p1*(n/(n-1))*((((pd/p1)^((n-1)/n)))-1);//Mean effective pressure in bar
+ip1=pm1*vs*100/60;//indicated power in kW
+nso1=(ip/ip1)*100;//Isothermal efficiency in percentage
+pm2=p1*(g/(g-1))*((((pd/p1)^((g-1)/g)))-1);//Mean effective pressure in bar
+ip2=pm2*vs*100/60;//indicated power in kW
+nso2=(ip/ip2)*100;//Isothermal efficiency in percentage
+pm3=p1*(m/(m-1))*((((pd/p1)^((m-1)/m)))-1);//Mean effective pressure in bar
+ip3=pm3*vs*100/60;//indicated power in kW
+nso3=(ip/ip3)*100;//Isothermal efficiency in percentage
+nad=(ip2/ip3)*100;//adiabatic efficiency in percentage
+
+//OUTPUT
+printf('(i)isothermal compression \n Indicated mean effective pressure %3.4f bar \n Ideal power %3.3f kW \n (ii)compression process according to to pv^1.25 \n Indicated mean effective pressure %3.4f bar \n Ideal power %3.3f kW \n isothermal efficiency is %3.3f percentage \n (iii)Compression is reversible adiabatic \n Indicated mean effective pressure %3.4f bar \n Ideal power %3.3f kW \n isothermal efficiency is %3.3f percentage \n (iv)compression is irreversible adiabatic \n Indicated mean effective pressure %3.4f bar \n Ideal power %3.3f kW \n isothermal efficiency is %3.3f percentage \n adiabatic efficiency is %3.3f percentage ',pm,ip,pm1,ip1,nso1,pm2,ip2,nso2,pm3,ip3,nso3,nad)
+
diff --git a/1808/CH5/EX5.9/Chapter5_Exampl9.sce b/1808/CH5/EX5.9/Chapter5_Exampl9.sce new file mode 100644 index 000000000..07249f6ec --- /dev/null +++ b/1808/CH5/EX5.9/Chapter5_Exampl9.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+p1=1;//ambient pressure in bar
+t1=15;//ambient temperature in Degree C
+ps=0.98;//suction pressure in bar
+pd=7;//Delivery pressure in bar
+ts=30;//suction temperature in Degree C
+x=1.25;//ratio of l,d
+c=1/15;//clearance
+va=100;//volume in m^3/min
+n=1.3;//for cylinders
+
+//CALCULATIONS
+
+//(a)If ambient and suction conditions are same
+nv1=(1+c-c*((pd/p1)^(1/n)))*100;//volumetric efficiency in percentage
+vs1=va/nv1;//swept volume in m^3/min
+d1=sqrt(0.260146*4/3.14);//bore length in m
+l1=x*d1;//stroke in m
+N1=500/(2*l1);//Speed in rpm
+ip1=(n/(n-1))*p1*10^2*(va/60)*((pd/p1)^((n-1)/n)-1);//indicated power in kW
+
+//(b)If ambient and suction conditions are different
+nv2=(1+c-c*((pd/ps)^(1/n)))*100;//volumetric efficiency in percentage
+v14=p1*va*(ts+273)/(ps*(t1+273));//volume of air delivered in m^3/min
+vs2=(v14/nv2);//swept volume in m^3/min
+d2=sqrt(vs2*4/(3.14*500));//bore length in m
+l2=x*d2;//stroke length in m
+N2=500/(2*l2);//speed in rpm
+ip2=(n/(n-1))*ps*10^2*(v14/60)*((pd/ps)^((n-1)/n)-1);//indicated power in kW
+
+//OUTPUT
+printf('(a)If ambient and suction conditions are same \n (i)volumetric efficiency %3.3f percentage \n (ii)Bore %3.4f m \n stroke %3.4f m \n speed %3.1f rpm \n (iii)Indicated power is %3.2f kW \n (b)If ambient and suction conditions are different \n (i)volumetric efficiency %3.3f percentage \n (ii)Bore %3.4f m \n stroke %3.4f m \n speed %3.1f rpm \n (iii)Indicated power is %3.2f kW \n ',nv1,d1,l1,N1,ip1,nv2,d2,l2,N2,ip2)
+
+
+
diff --git a/1808/CH6/EX6.1/Chapter6_Exampl1.sce b/1808/CH6/EX6.1/Chapter6_Exampl1.sce new file mode 100644 index 000000000..874963d77 --- /dev/null +++ b/1808/CH6/EX6.1/Chapter6_Exampl1.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+Tmin=263;//lower temperature in K
+Tmax=322;//Higher temperature of refrigerant in K
+Re=10;//capacity in tonnes
+
+
+//CALCULATIONS
+COP=(Tmin/(Tmax-Tmin));//COP
+WD=(Re*210)/(60*COP)*3600;//workdone in kJ/s
+P=WD/3600;//Power required
+Q=(Re*210*60)+WD;//Heat rejected fro the system per hour
+
+//OUTPUT
+printf('(i)COP is %3.2f \n (ii)Heat rejected from the system per hour is %3.1f kJ/hr \n (iii)Power required is %3.3f kW',COP,Q,P)
+
diff --git a/1808/CH6/EX6.10/Chapter6_Exampl10.sce b/1808/CH6/EX6.10/Chapter6_Exampl10.sce new file mode 100644 index 000000000..feba4ba78 --- /dev/null +++ b/1808/CH6/EX6.10/Chapter6_Exampl10.sce @@ -0,0 +1,75 @@ +clc
+clear
+//INPUT DATA
+cp1=1.00;//specific entropy in kJ/kgK
+cpv=0.733;//specific entropy in kJ/kgK
+t21=303;//condenser temperature in K
+t1=265;//evaporator temperature in K
+t31=293;//subcooled temperature in K
+p1=2.354;//pressure in Bar
+p2=7.451;//pressure in Bar
+hf1=28.72;//enthalpy in kJ/kg
+hg1=184.07;//enthalpy in kJ/kg
+hf2=64.59;//enthalpy in kJ/kg
+hg2=199.62;//enthalpy in kJ/kg
+sf1=0.1149;//entropy in kJ/kgK
+sf2=0.24;//entropy in kJ/kgK
+sg1=0.7007;//entropy in kJ/kgK
+sg2=0.6853;//entropy in kJ/kgK
+vg1=0.079;//entropy in kJ/kgK
+vg2=0.0235;//entropy in kJ/kgK
+v1b=0.772;//entropy in kJ/kgK
+t2=309.43;//temperature in K
+
+//CALCULATIONS
+//(i)WET COMPRESSION
+x=((sg2-sf1)/(sg1-sf1));//fraction
+h1b=hf1+x*(hg1-hf1);//enthalpy in kJ/kg
+h2=hg2+cpv*(t2-t21);//enthalpy in kJ/kg
+s1a=sg1+cpv*log(271/t1);//entropy in kJ/kgK
+t2a=(s1a-sg1)/(cpv*t21);//temperature in K
+h2a=hg2+cpv*(t2a-t21);//enthalpy in kJ/kg
+h1a=hg1+cpv*(271-t1);//enthalpy in kJ/kg
+h31=hf2-cpv*(t21-298);//enthalpy in kJ/kg
+Re1=h1b-hf2;//Refrigeration effect in wet condition
+Re2=hg1-hf2;//Refrigeration effect in wet condition
+Re3=h1b-hf2;//Refrigeration effect in wet condition
+Re4=hg1-hf2;//Refrigeration effect in wet condition
+wn1=hg2-h1b;//net workdone in kJ/kg
+wn2=h2-hg1;//net workdone in kJ/kg
+wn3=h2a-hg1;//net workdone in kJ/kg
+wn4=h2-hg1;//net workdone in kJ/kg
+cop1=Re1/wn1;//COP
+cop2=Re2/wn2;//COP
+cop3=Re3/wn3;//COP
+cop4=Re4/wn4;//COP
+m1=2100/Re1;//mass flow rate
+m2=2100/Re2;//mass flow rate
+m3=2100/Re3;//mass flow rate
+m4=2100/Re4;//mass flow rate
+P1=m1*wn1/60;//Power in kW
+P2=m2*wn2/60;//Power in kW
+P3=m3*wn3/60;//Power in kW
+P4=m4*wn4/60;//Power in kW
+Pt1=P1/10;//Power per TR
+Pt2=P2/10;//Power per TR
+Pt3=P3/10;//Power per TR
+Pt4=P4/10;//Power per TR
+d1=((m1*v1b/0.00084883)^(1/3))/100;//displacement in m
+d2=((m2*vg1/0.00084883)^(1/3))/100;//displacement in m
+d3=((m3*vg1/0.00084883)^(1/3))/100;//displacement in m
+d4=((m4*vg1/0.00084883)^(1/3))/100;//displacement in m
+l1=1.5*d1;//stroke in m
+l2=1.5*d2;//stroke in m
+l3=1.5*d3;//stroke in m
+l4=1.5*d4;//stroke in m
+
+
+//OUTPUT
+printf('((i)WET COMPRESSION \n (a)cop is %3.2f \n (b)The power is %3.3f kW/TR \n (c)Bore is %3.5f m \n stroke is %3.4f m \n (d)mass flow rate of refrigerant is %3.1f kg/min \n',cop1,P1,d1,l1,m1)
+
+printf('((ii)DRY COMPRESSION \n (a)cop is %3.2f \n (b)The power is %3.3f kW/TR \n (c)Bore is %3.5f m \n stroke is %3.4f m \n (d)mass flow rate of refrigerant is %3.1f kg/min \n',cop2,P2,d2,l2,m2)
+
+printf('((iii)SUPERHEATED \n (a)cop is %3.2f \n (b)The power is %3.3f kW/TR \n (c)Bore is %3.5f m \n stroke is %3.4f m \n (d)mass flow rate of refrigerant is %3.1f kg/min \n',cop3,P3,d3,l3,m3)
+
+printf('((iv)DRY COMPRESSION AND SUBCOOLED \n (a)cop is %3.2f \n (b)The power is %3.3f kW/TR \n (c)Bore is %3.5f m \n stroke is %3.4f m \n (d)mass flow rate of refrigerant is %3.1f kg/min \n ',cop4,P4,d4,l4,m4)
diff --git a/1808/CH6/EX6.11/Chapter6_Exampl11.sce b/1808/CH6/EX6.11/Chapter6_Exampl11.sce new file mode 100644 index 000000000..6ba1e062f --- /dev/null +++ b/1808/CH6/EX6.11/Chapter6_Exampl11.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+P=11;//Power used to run the compressor in kW
+h1=188.41;//total heat of gas leaving refrigeration in kJ/kg
+h2=213.53;//total heat of gas leaving compressor in kJ/kg
+h4=77.46;//total heat of liquid before throttling in kJ/kg
+
+//CALCULATIONS
+Re=h1-h4;//refrigeration effect
+wd=h2-h1;//work done
+cop=Re/wd;//COP of refrigerant system
+m=P/wd;//mss of refrigerant
+
+//OUTPUT
+printf('(i)COP is %3.3f \n (ii)mass of refrigerant %3.4f kg/s ',cop,m)
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.12/Chapter6_Exampl12.sce b/1808/CH6/EX6.12/Chapter6_Exampl12.sce new file mode 100644 index 000000000..387b835bf --- /dev/null +++ b/1808/CH6/EX6.12/Chapter6_Exampl12.sce @@ -0,0 +1,46 @@ +clc
+clear
+//INPUT DATA
+cp=0.202;//specific pressure in kcal/kg K
+g=1.18;//constant
+t21=303;//condenser temperature in K
+t1=263;//evaporator temperature in K
+t21=313;//subcooled temperature in K
+p1=3.543;//pressure in Bar
+p2=15.335;//pressure in Bar
+hf1=188.4;//enthalpy in kJ/kg
+hg1=401.6;//enthalpy in kJ/kg
+hf2=249.7;//enthalpy in kJ/kg
+hg2=416.6;//enthalpy in kJ/kg
+sf1=0.9573;//entropy in kJ/kgK
+sf2=1.167;//entropy in kJ/kgK
+sg1=1.767;//entropy in kJ/kgK
+sg2=1.699;//entropy in kJ/kgK
+vg1=0.0653;//entropy in kJ/kgK
+vg2=0.0151;//entropy in kJ/kgK
+v1b=0.772;//entropy in kJ/kgK
+hfg1=213.2;//enthalpy in kJ/kg
+hfg2=166.9;//enthalpy in kJ/kg
+vf1=0.759;//sp.volume
+vf2=0.884;//sp.colume
+t22=432.68;//temperature in K
+
+//CALCULATIONS
+t2=t1*((p2/p1)^((g-1)/g));//temperature in K
+v2=vg2*(t2/t21);//sp.volume in m^3/kg
+wc=(g/(g-1))*p1*10^2*vg1*(((p2/p1)^((g-1)/g))-1);//work done in compressor
+h2=hg1+cp*0.202*(t2-t21);//enthalpy in kJ/kg
+Re=hg1-hf2;//Refrigeration effect
+wc1=h2-hg1;//compressor work in kJ/kg
+cop=Re/wc;//COP of the system
+m=(12*210)/(60*Re);//mass of refrigerant in kg/s
+Pc=m*wc;//compressor power
+Pm=Pc/0.75;//Power of the motor in kW
+
+//OUTPUT
+printf('(a)COP of the system is %3.2f \n (b)Mass flow rate oof refrigerant %3.4f kg/s \n (c)Power of the motor is %3.2f kW',cop,m,Pm)
+
+
+
+
+
diff --git a/1808/CH6/EX6.13/Chapter6_Exampl13.sce b/1808/CH6/EX6.13/Chapter6_Exampl13.sce new file mode 100644 index 000000000..54bee5b1a --- /dev/null +++ b/1808/CH6/EX6.13/Chapter6_Exampl13.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+cpv=2.805;//specific pressure in kJ/kg K
+cp1=4.606;//specific pressure in kJ/kg K
+p1=10.01;//pressure in MPa
+p2=1.2;//pressure in MPa
+hf1=298.9;//enthalpy in kJ/kg
+hf2=44.7;//enthalpy in kJ/kg
+hg1=1466;//enthalpy in kJ/kg
+hg2=1406;//enthalpy in kJ/kg
+sf1=1.124;//entropy in kJ/kgK
+sf2=0.188;//entropy in kJ/kgK
+sg1=5.039;//entropy in kJ/kgK
+sg2=5.785;//entropy in kJ/kgK
+vf1=0.128;//volume in m^3/kg
+vf2=0.963;//volume in m^3/kg
+t1=253;//temperature in K
+t11=243.42;//temperature in K
+t21=298;//temperature in K
+t2=404.78;//temperature in K
+t3=293;//temperature in K
+
+//CALCULATIONS
+s1=sg2+cpv*log(t1/t11);//entropy in kJ/kg K
+h1=hg2+cpv*(t1-t11);//enthalpy in kJ/kg
+h2=hg1+cpv*(t2-t21);//enthalpy in kJ/kg
+h3=hf1+cp1*(t21-t3);//enthalpy in kJ/kg
+cop=((h1-h3)/(h2-h1));//COP of the system
+P=1.5*210/(cop*60);//Power of the motor in kW
+
+//OUTPUT
+printf('(a)COP is %3.2f \n (b)Power of the motor is %3.3f kW',cop,P)
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.14/Chapter6_Exampl14.sce b/1808/CH6/EX6.14/Chapter6_Exampl14.sce new file mode 100644 index 000000000..9635931a7 --- /dev/null +++ b/1808/CH6/EX6.14/Chapter6_Exampl14.sce @@ -0,0 +1,38 @@ +clc
+clear
+//INPUT DATA
+hf1=100.4;//enthalpy in kJ/kg
+hf2=-54.56;//enthalpy in kJ/kg
+hg1=1319.22;//enthalpy in kJ/kg
+hg2=1304.99;//enthalpy in kJ/kg
+sf1=0.3473;//entropy in kJ/kgK
+sf2=-0.2134;//entropy in kJ/kgK
+sg1=4.4852;//entropy in kJ/kgK
+sg2=5.0585;//entropy in kJ/kgK
+t1=20;//temperature in Degree C
+t2=-15;//temperature in Degree C
+mi=30;//mass of ice
+hfgi=335;//enthalpy in kJ/kg
+cpw=4.1868;//specific pressure of water in kJ/kg K
+
+//CALCULATIONS
+x1=((sg1-sf2)/(sg2-sf2));//fraction
+h1=hf2+x1*(hg2-hf2);//enthalpy in kJ/kg
+copt=((h1-hf1)/(hg1-h1));//COP of the system
+copa=copt*0.6;//actual cop
+Qa=mi*10^3*(cpw*t1+hfgi)/(24*3600);//heat removed from water in kJ/s
+w=Qa/copa;//Power required to drive compressor in kW
+
+//OUTPUT
+printf('power required to drive the compresor i %3.2f kW',w)
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.15/Chapter6_Exampl15.sce b/1808/CH6/EX6.15/Chapter6_Exampl15.sce new file mode 100644 index 000000000..0b802384a --- /dev/null +++ b/1808/CH6/EX6.15/Chapter6_Exampl15.sce @@ -0,0 +1,42 @@ +clc
+clear
+//INPUT DATA
+p1=1;//cold chamber pressure in bar
+p2=6;//compressor pressure in bar
+g=1.4;//constant
+t1=278;//temperature in K
+t3=298;//temperature in K
+cp=1.005;//specific pressure
+n1=1.3;//index of compression
+n2=1.35;//index of compresion
+R=0.287;//gas constant
+
+//CALCULATIONS
+t2=t1*((p2/p1)^((g-1)/g));//temperature in K
+t4=t3/((p2/p1)^((g-1)/g));//temperature in K
+Qa=cp*(t1-t4);//Refrigeration effect
+Qr=cp*(t2-t3);//heat rejected to the cooling medium in kJ/kg
+wn=Qr-Qa;//net work done in kJ/kg
+copa=Qa/wn;// actual COP
+t21=t1*((p2/p1)^((n1-1)/n1));//temperature in K
+t41=t3/((p2/p1)^((n2-1)/n2));//temperature in K
+Qa1=cp*(t1-t41);//Refrigeration effect
+wn1=(n1/(n1-1))*R*(t21-t1)-(n2/(n2-1))*R*(t3-t41);//net work done in kJ/kg
+copb=Qa1/wn1;// actual COP
+P1=210/(60*copa);//Power per ton of refrigeration in kW/TR
+m1=210*3600/(60*Qa);//air flow rate in kg/hr
+P2=210/(60*copb);//Power per ton of refrigeration in kW/TR
+m2=210*60/(Qa1);//air flow rate in kg/hr
+
+//OUTPUT
+printf('(a) \n (i)Refrigeration effect is %3.2f kJ/kg \n (ii)heat rejected to the cooling medium is %3.2f kJ/kg \n (iii)COPa %3.3f \n (b) \n (I)Refrigeration effect is %3.2f kJ/kg \n CASE A \n (1)Power per ton of refrigeration is %3.3f kW/TR \n (2)air flow rate is %3.2f kg/hr \n CASE B \n (1)Power per ton of refrigeration is %3.3f kW/TR \n (2)air flow rate is %3.2f kg/hr',Qa,Qr,copa,Qa1,P1,m1,P2,m2)
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.16/Chapter6_Exampl16.sce b/1808/CH6/EX6.16/Chapter6_Exampl16.sce new file mode 100644 index 000000000..2602e7f4b --- /dev/null +++ b/1808/CH6/EX6.16/Chapter6_Exampl16.sce @@ -0,0 +1,45 @@ +clc
+clear
+//INPUT DATA
+t1=293;//temperature in K
+t21=363;//temperature in K
+t3=308;//temperature in K
+t41=273;//temperature in K
+p1=1;//compressor pressure in bar
+p2=2;//turbine pressure in bar
+g=1.4;//constant
+cp=1.005;//specific pressure
+m=1;//mass of air
+N=350;//speed in rpm
+R=0.287;//gas constant
+nv=0.9;//volumetric efficiency in percentage
+
+
+//CALCULATIONS
+t2=t1*((p2/p1)^((g-1)/g));//temperature in K
+nc=(((t2-t1)/(t21-t1)))*100;//compressor efficiency in percentage
+t4=t3*((p1/p2)^((g-1)/g));//temperature in K
+nt=(((t3-t41)/(t3-t4)))*100;//turbine efficiency in percentage
+Qa=cp*(t1-t41);//Refrigeration effect
+m1=30*210/Qa;//mass flow rate of air in kg/min
+wn=cp*(m1/60)*((t21-t1)-(t3-t41));//power input in kW
+cop=Qa*m1/(wn*60);//COP
+v1=m*R*t1/(p1*10^2);//volume in m^3/kg
+vs=v1/nv;//swept volume in m^3/kg
+d=(4*vs/(3.14*1.5*N))^(1/3);//diameter of compressor in m
+l=1.5*d;//compressor length in m
+
+//OUTPUT
+printf('(i)The compressor efficiency is %3.2f percentage \n (ii)Turbine efficiency is %3.2f percentage \n (iii)Refrigeration effect %3.2f kJ/kg \n (iv)power input is %3.2f kW \n (v)COP is %3.3f \n (vi)compressor diameter is %3.4f m \n length %3.3f m',nc,nt,Qa,wn,cop,d,l)
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.2/Chapter6_Exampl2.sce b/1808/CH6/EX6.2/Chapter6_Exampl2.sce new file mode 100644 index 000000000..c128bd3f3 --- /dev/null +++ b/1808/CH6/EX6.2/Chapter6_Exampl2.sce @@ -0,0 +1,16 @@ +clc
+clear
+//INPUT DATA
+COP=4;//COP
+WD=20;//workdone of cycle in kW
+
+//CALCULATIONS
+x=1+(1/COP);//Ratio of temperatures
+Re=COP*WD;//Refrigeration effect in kW
+Re1=Re*60;//Refrigeration effect in kJ/min
+Re2=Re1/210;//Refrigeration effect in TR
+Hd=Re+WD;//Heat delivered in kW
+COP1=Hd/WD;//COP of heat pump
+
+//OUTPUT
+printf('(i)Temperature ratio is %3.2f \n (ii)maximum refrigeration effect is %3.2f TR \n (iii)COP of heat pump is %3.2f',x,Re2,COP1)
diff --git a/1808/CH6/EX6.3/Chapter6_Exampl3.sce b/1808/CH6/EX6.3/Chapter6_Exampl3.sce new file mode 100644 index 000000000..086969a03 --- /dev/null +++ b/1808/CH6/EX6.3/Chapter6_Exampl3.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+Re=1;//Heat absorbed in Tonns
+WD=1.25*60;//work done in kJ/min
+Tmin=-40;//low tamperature in Degree C
+Ha=210;//heat absorbed in kJ/min
+
+//CALCULATIONS
+COP=(Re*210)/WD;//COP
+Tmax=((273+Tmin)/COP)+Tmin;//High temperature of the cycle
+Hd=Ha+WD;//Heat rejected in kJ/min
+COPh=Hd/WD;COP;// heat pump
+
+
+//OUTPUT
+printf('(i)COP is %3.2f \n (ii)Tmax is %3.2f Degree C \n (iii)Heat rejected is %3.i kJ/min \n (iv)COP of heat pump is %3.1f ',COP,Tmax,Hd,COPh)
diff --git a/1808/CH6/EX6.4/Chapter6_Exampl4.sce b/1808/CH6/EX6.4/Chapter6_Exampl4.sce new file mode 100644 index 000000000..a71027cd2 --- /dev/null +++ b/1808/CH6/EX6.4/Chapter6_Exampl4.sce @@ -0,0 +1,36 @@ +clc
+clear
+//INPUT DATA
+Tmin=-30;//minimuum temperature in Degree C
+Tmax=35;//maximum temperature in Degree C
+S1=0.6839;//entropy in kJ/kgK from properties of R12 TABLES
+S2=0.6893;//entropy in kJ/kgK from properties of R12 TABLES
+S3=0.2559;//entropy in kJ/kgK from properties of R12 TABLES
+S4=0.2559;//entropy in kJ/kgK from properties of R12 TABLES
+S5=0.0371;//entropy in kJ/kgK from properties of R12 TABLES
+S6=0.7171;//entropy in kJ/kgK from properties of R12 TABLES
+h2=201.5;//enthalpy in kJ/kg from properties of R12 TABLES
+h3=69.5;//enthalpy in kJ/kg from properties of R12 TABLES
+h5=8.9;//enthalpy in kJ/kg from properties of R12 TABLES
+h6=174.2;//enthalpy in kJ/kg from properties of R12 TABLES
+Re=1*210;//Ref.capacity
+
+
+//CALCULATIONS
+x1=(S1-S5)/(S6-S5);//ratio fo entropies
+x2=(S4-S5)/(S6-S5);//ratio fo entropies
+h1=h5+x1*(h6-h5);//enthalpy at point 1
+h4=h5+x2*(h6-h5);//enthalpy at point 4
+Wc=h2-h1;//work of compression
+We=h3-h4;//work of expansion
+Qa=h1-h4;//Heat absorbed in kJ/kg
+Qr=h2-h3;//Heat rejected in kJ/kg
+Wn=Wc-We;//net workdone in kJ/kg
+COP=(Qa/Wn);//COP
+COPc=(Tmin+273)/(Tmax-Tmin);//COP Carnot
+COPa=0.75*COPc;//Actual COP
+P=Re/(COPa*60);//Power consumption per ton
+Hr=(210/60)+P;//Heat rejected per ton
+
+//OUTPUT
+printf('(a)work of compression is %3.2f kJ/kg \n work of expansion %3.1f kJ/kg \n Heat rejected is %3.i kJ/kg \n COP is %3.2f \n (b)Power consumption per ton is %3.2f kW \n heat rejected per ton is %3.2f kW',Wc,We,Qr,COP,P,Hr)
diff --git a/1808/CH6/EX6.5/Chapter6_Exampl5.sce b/1808/CH6/EX6.5/Chapter6_Exampl5.sce new file mode 100644 index 000000000..9013936e3 --- /dev/null +++ b/1808/CH6/EX6.5/Chapter6_Exampl5.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+tmin=293;//minimum temperature in K
+t3=317;//temperature in K
+m=0.008;//mass flow rate in kg/s
+hf1=54.81;//enthalpy in kJ/kg
+hfg1=140.91;//enthalpy in kJ/kg
+hg1=195.78;//enthalpy in kJ/kg
+hf2=78.68;//enthalpy in kJ/kg
+hfg2=125.87;//enthalpy in kJ/kg
+hg2=204.54;//enthalpy in kJ/kg
+vf1=0.2078;//entropy in kJ/kgK
+vf2=0.2845;//entropy in kJ/kgK
+vg1=0.6884;//entropy in kJ/kgK
+vg2=0.6814;//entropy in kJ/kgK
+t2=320.49;//from t-s diagram temperature in K
+cp=0.64;//specific pressure
+
+//CALCULATIONS
+h2=hg2+cp*(t2-t3);//enthalpy in kJ/kg
+wc=m*(h2-hg1);//compressor work in kJ/s
+Rc=m*(hg1-hf2);//Refrigiration capacity in kW
+Rc1=Rc*60/210;//Refrigiration capacity in TR
+copv=Rc/wc;//COP of VCR system
+copc=(tmin/(t3-tmin));//COP of carnot refrigeration cycle in percentage
+
+//OUTPUT
+printf('(a)The compressive work input is %3.5f kJ/s \n (b)Refrigiration capacity is %3.4f TR \n (c)COPvcr is %3.3f \n (d)COPc is %3.3f',wc,Rc,copv,copc)
+
+
diff --git a/1808/CH6/EX6.6/Chapter6_Exampl6.sce b/1808/CH6/EX6.6/Chapter6_Exampl6.sce new file mode 100644 index 000000000..1752e8c6b --- /dev/null +++ b/1808/CH6/EX6.6/Chapter6_Exampl6.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+t21=335.19;//temperature in K
+t3=343.19;//temperature in K
+cp=0.64;//Specific pressure
+m=0.008;//mass flow rate in kg/s
+p1=4.4962;//pressure in MPa
+p2=1.6;//pressure in MPa
+hf1=47.26;//enthalpy in kJ/kg
+hfg1=145.30;//enthalpy in kJ/kg
+hg1=192.56;//enthalpy in kJ/kg
+hf2=98.19;//enthalpy in kJ/kg
+hfg2=111.62;//enthalpy in kJ/kg
+hg2=209.81;//enthalpy in kJ/kg
+sf1=0.1817;//entropy in kJ/kgK
+sf2=0.6913;//entropy in kJ/kgK
+sg1=0.3329;//entropy in kJ/kgK
+sg2=0.6758;//entropy in kJ/kgK
+t2=343;//from t-s diagram temperature in K
+
+//CALCULATIONS
+h2=hg2+cp*(t2-t3);//enthalpy in kJ/kg
+wc=m*(h2-hg1);//compressor work in kJ/s
+Rc=m*(hg1-hf2);//Refrigiration capacity in kW
+Rc1=Rc*60/210;//Refrigiration capacity in TR
+copv=Rc/wc;//COP of VCR system
+
+//OUTPUT
+printf('(a)The compressive work input is %3.2f kJ/s \n (b)Refrigiration capacity is %3.4f TR \n (c)COPvcr is %3.3f \n',wc,Rc1,copv)
+
+
+
+
+
+
+
diff --git a/1808/CH6/EX6.7/Chapter6_Exampl7.sce b/1808/CH6/EX6.7/Chapter6_Exampl7.sce new file mode 100644 index 000000000..6fbce78a3 --- /dev/null +++ b/1808/CH6/EX6.7/Chapter6_Exampl7.sce @@ -0,0 +1,40 @@ +clc
+clear
+//INPUT DATA
+h2=215;//enthalpy in kJ/kg
+cp1=1.05;//Specific pressure
+nc=0.8;//carnot efficiency in percentage
+h1=192.56;//enthalpy in kJ/kg
+h31=98.19;//enthalpy in kJ/kg
+t3=48;//Temperature in K
+t31=62.19;//Temperature in K
+m=0.008;//mass of air in kg
+h5=47.26;//enthalpy in kJ/kg
+t0=317;//temperature in K
+s3=0.2973;//at superheated table entropy in kJ/kgK
+ha1=209.81;//at superheated table enthalpy in kJ/kg
+sa1=0.6758;//at superheated table enthalpy in kJ/kgK
+ha2=225.34;//at superheated table enthalpy in kJ/kg
+sa2=0.7209;//at superheated table entropy in kJ/kgK
+s2a=0.6984;//at superheated table entropy in kJ/kgK
+s5=0.1857;//at superheated table entropy in kJ/kgK
+
+
+//CALCULATIONS
+h2a=((h2-h1)/nc)+h1;//enthalpy in kJ/kg
+h3=h31-cp1*(t31-t3);//enthalpy in kJ/kg
+wc=m*(h2a-h1);//compressor work in kJ/s
+Rc=m*(h1-h3);//Refrigiration capacity in kW
+Rc1=Rc*60/210;//Refrigiration capacity in TR
+copv=Rc/wc;//COP of VCR system
+x4=((h3-h5)/(h1-h5));//fraction
+s4=s5++x4*(s2a-s5);//at superheated table entropy in kJ/kgK
+ic=m*t0*(s4-s3);//Ireeversibility rate in valve in kW
+
+//OUTPUT
+printf('(a)The compressive work input is %3.5f kJ/s \n (b)Refrigiration capacity is %3.2f TR \n (c)COPvcr is %3.3f \n (d)Ireeversibility rate in valve is %3.4f kW',wc,Rc1,copv,ic)
+
+
+11
+
+
diff --git a/1808/CH6/EX6.8/Chapter6_Exampl8.sce b/1808/CH6/EX6.8/Chapter6_Exampl8.sce new file mode 100644 index 000000000..548b1e66b --- /dev/null +++ b/1808/CH6/EX6.8/Chapter6_Exampl8.sce @@ -0,0 +1,46 @@ +clc
+clear
+//INPUT DATA
+cpv2=2.805;//specific pressure kJ/kgk
+cpv3=4.606;//specific pressure kJ/kgK
+t21=303;//condenser temperature in K
+t1=258;//evaporator temperature in K
+t31=283;//subcooled temperature in K
+nv=0.8;//volumetric efficiency in percentage
+p1=2.36;//pressure in MPa
+p2=11.67;//pressure in MPa
+hf1=112.3;//enthalpy in kJ/kg
+hfg1=1313.7;//enthalpy in kJ/kg
+hg1=1426;//enthalpy in kJ/kg
+hf2=323.1;//enthalpy in kJ/kg
+hfg2=1145.9;//enthalpy in kJ/kg
+hg2=1469;//enthalpy in kJ/kg
+sf1=0.457;//entropy in kJ/kgK
+sf2=1.204;//entropy in kJ/kgK
+sg1=5.549;//entropy in kJ/kgK
+sg2=4.984;//entropy in kJ/kgK
+v1=0.509;//volume in m^3/kg
+t2=369.69;//from t-s diagram temperature in K
+
+
+//CALCULATIONS
+h2=hg2+cpv2*(t2-t21);//enthalpy in kJ/kg
+h31=hf2-cpv3*(t21-t31);//enthalpy in kJ/kg
+Re1=hg1-hf2;//refrigeration effect in kJ/kg
+Re2=hg1-h31;//refrigeration effect in kJ/kg
+mt1=210/Re1;//mass flow rate per ton in kg/min
+mt2=210/Re2;//mass flow rate per ton in kg/min
+vsa1=(mt1*v1)/nv;//compressor volume capacity
+vsa2=(mt2*v1)/nv;//compressor volume capacity
+wn=h2-hg1;//net work done
+cop1=Re1/wn;//COP
+cop2=Re2/wn;//COP
+pcop=((cop1-cop2)/cop2)*100;//Percentage COP of dry and subcooled
+pt1=wn*mt1/60;//Power per ton in kW/TR
+pt2=wn*mt2/60;//Power per ton in kW/TR
+
+//OUTPUT
+printf('DRY COMPRESSION \n(a)Refrigeration effect is %3.1f kJ/kg \n (b)The flow rate of refrigerant per ton is %3.4f kg/min \n (c)The compressor volume capacity %3.2f \n (d)COP is %3.2f \n (e)The power per TR is %3.2f kW/TR \n',Re1,mt1,vsa1,cop1,pt1)
+
+printf('DRY AND SUBCOOLED \n(a)Refrigeration effect is %3.1f kJ/kg \n (b)The flow rate of refrigerant per ton is %3.4f kg/min \n (c)The compressor volume capacity %3.2f \n (d)COP is %3.2f \n (e)The power per TR is %3.2f kW/TR',Re2,mt2,vsa2,cop2,pt2)
+
diff --git a/1808/CH6/EX6.9/Chapter6_Exampl9.sce b/1808/CH6/EX6.9/Chapter6_Exampl9.sce new file mode 100644 index 000000000..e84995749 --- /dev/null +++ b/1808/CH6/EX6.9/Chapter6_Exampl9.sce @@ -0,0 +1,52 @@ +clc
+clear
+//INPUT DATA
+cpic=1.94;//specific pressure inkJ/kgK
+cpv2=2.805;//specific pressure in kJ/kgK
+t21=303;//condenser temperature in K
+t1=258;//evaporator temperature in K
+t31=293;//subcooled temperature in K
+p1=2.36;//pressure in MPa
+p2=11.67;//pressure in MPa
+hf1=112.3;//enthalpy in kJ/kg
+hfg1=1313.7;//enthalpy in kJ/kg
+hg1=1426;//enthalpy in kJ/kg
+hf2=323.1;//enthalpy in kJ/kg
+hfg2=1145.9;//enthalpy in kJ/kg
+hg2=1469;//enthalpy in kJ/kg
+sf1=0.457;//entropy in kJ/kgK
+sf2=1.204;//entropy in kJ/kgK
+sg1=5.549;//entropy in kJ/kgK
+sg2=4.984;//entropy in kJ/kgK
+t2=369.7;//from t-s diagram temperature in K
+nac=0.8;//adiabatic efficiency in percentage
+vsa=2.96;//volume in kg/min
+N=1200;//speed in rpm
+
+//CALCULATIONS
+h2=hg2+cpv2*(t2-t21);//enthalpy in kJ/kg
+Rc=10*1000*(4.1868*30+335+1.94*5)/(24*60);//Refrigeration capacity in kJ/min
+Re=hg1-hf2;//Refrigeration effect in kJ/kg
+m=Rc/Re;//mass flow rate of refrigerant in kg/min
+h2a=((h2-hg1)/nac)+hg1;//enthalpy in kJ/kg
+t2a=((t2-t1)/nac)+t1;//Temperature in k
+d=(vsa*0.509*4/(3.14*1.2*N))^(1/3);//piston displacement of compressor in m
+l=1.2*d;//length of piston displacement in m
+w=(h2a-hg1)/0.95;//workdone in kJ/kg
+wac=m*w/60;//Power of the compressor motor in kW
+copa=(Re/wac)*(m/60);//COP of air
+
+//OUTPUT
+printf('(a)Refrigeration capacity is %3.1f kJ/min \n (b)Mass flow rate of refrigerant is %3.2f kg/min \n (c)The discharge temperature is %3.2f K \n (d)Piston displacement of the compressor is d %3.4f m \n l is %3.4f m \n(e)Power of the compressor motor is %3.2f kW \n (f)COPa is %3.3f',Rc,m,t2a,d,l,wac,copa)
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1808/CH7/EX7.1/Chapter7_Exampl1.sce b/1808/CH7/EX7.1/Chapter7_Exampl1.sce new file mode 100644 index 000000000..96ef2a867 --- /dev/null +++ b/1808/CH7/EX7.1/Chapter7_Exampl1.sce @@ -0,0 +1,15 @@ +clc
+clear
+//INPUT DATA
+w=0.016;//specific humidity in kg/kg
+p=760;//pressure in mm of Hg
+ps=31.78;//saturation pressure in mm of Hg
+
+//CALCULATIONS
+pv=(p)*0.02572/1.02572;//Partial pressure of vapour in mm of Hg
+x=(pv/ps)*100;//Relative humidity in percentage
+
+//OUTPUT
+printf('(a)The partial pressure of vapour is %3.4f mm of Hg \n (b)Relative humidity is %3.2f percentage \n (c)According to steam tables Dew point temperature is 21.34 degree c',pv,x)
+
+
diff --git a/1808/CH7/EX7.10/Chapter7_Exampl10.sce b/1808/CH7/EX7.10/Chapter7_Exampl10.sce new file mode 100644 index 000000000..98730cfb1 --- /dev/null +++ b/1808/CH7/EX7.10/Chapter7_Exampl10.sce @@ -0,0 +1,31 @@ +clc
+clear
+//INPUT DATA
+ta1=15;//dry bulb temperature in Degree c
+ta2=25;//dry bulb temperature in Degree c
+tw1=13;//wet bulb temperature in Degree c
+tw2=18;//wet bulb temperature in Degree c
+V1=30;//volume of air in m^3/min
+V2=12;//volume of air in m^3/min
+pva=11.22;//Saturation pressure in mm Hg
+pvb=15.461;//Saturation pressure in mm Hg
+p=760;//pressure in mm of Hg
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+pv1=(pva-((p-pva)*(ta1-tw1)*1.8/(2800-1.3*(1.8*ta1+32))));//Saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+pv2=pvb-((p-pvb)*(ta2-tw2)*1.8/(2800-(1.3*(1.8*ta2+32))));//Saturation pressure in mm Hg
+w2=0.622*(pv2/(p-pv2));//Specific humidity in kg w.v./kg d.a
+h1=cp*ta1+w1*(2500+1.88*ta1);//Enthalpy of air per kg of dry air in kJ/kg d.a
+h2=cp*ta2+w2*(2500+1.88*ta2);//Enthalpy of air per kg of dry air in kJ/kg d.a
+ma1=V1/0.827;//Dry mass flow rate in kg d.a./min
+ma2=V2/0.8574;//Dry mass flow rate in kg d.a./min
+ma3=ma1+ma2;//Dry mass flow rate in kg d.a./min
+w3=((ma1*w1)+(ma2*w2))/ma3;//Specific humidity in kg w.v./kg d.a
+h3=((ma1*h1)+(ma2*h2))/(ma3);//Enthalpy of air per kg of dry air in kJ/kg d.a
+ta3=((ma1*ta1)+(ma2*ta2))/(ma3);//dry bulb temperature in Degree c
+tw3=((ma1*tw1)+(ma2*tw2))/(ma3);//wet bulb temperature in Degree c
+
+//OUTPUT
+printf('(i)The specific humidity of the mixture is %3.4f kg w.v./kg d.a \n (ii)Specific enthalpy of the mixture is %3.2f kJ/kg d.a. \n (iii)DBT corresponds to mixture is %3.3f Degree C \n (iv)WBT corresponds to mixture is %3.3f Degree C ',w3,h3,ta3,tw3)
diff --git a/1808/CH7/EX7.11/Chapter7_Exampl11.sce b/1808/CH7/EX7.11/Chapter7_Exampl11.sce new file mode 100644 index 000000000..96e36988f --- /dev/null +++ b/1808/CH7/EX7.11/Chapter7_Exampl11.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+t1=2;//temperature in Degree C
+t2=30;//temperature in Degree C
+x1=0.8;//realtive humidity in percentage
+td2=10;//Dew point temperature in Degree C
+ps1=5.2854;//Saturation pressure in mm Hg
+pv2=9.196;//Saturation pressure in mm Hg
+ps2=31.8052840;//Saturation pressure in mm Hg
+p=760;//pressure in mm of Hg
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+pv1=x1*ps1;//saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+w2=0.622*(pv2/(p-pv2));//Specific humidity in kg w.v./kg d.a
+ma1=(1/(1+w1));//Mass of dry air per unit mass of moist air in kg/d.a.
+ma2=(3/(1+w2));//Mass of dry air per unit mass of moist air in kg/d.a.
+ma3=ma1+ma2;//Mass of dry air per unit mass of moist air in kg/d.a.
+t3=((ma1*t1)+(ma2*t2))/(ma3);//Temperature of the air after mixture in Degree C
+w3=((ma1*w1)+(ma2*w2))/(ma3);//Specific humidity of air mixture in kg w.v./kg d.a.
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a
+h2=cp*t2+w2*(2500+1.88*t2);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h3=((ma1*h1)+(ma2*h2))/(ma3);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+
+//OUTPUT
+printf('Temperature of the air after mixture is %3.4f Degree c \n (ii)Specific humidity of the air after mixing is %3.7f kg w.v./kg d.a. \n (iii)Specific enthalpy of the air after mixing is %3.2f kJ/kg d.a.',t3,w3,h3)
+
diff --git a/1808/CH7/EX7.12/Chapter7_Exampl12.sce b/1808/CH7/EX7.12/Chapter7_Exampl12.sce new file mode 100644 index 000000000..55f5ab9bf --- /dev/null +++ b/1808/CH7/EX7.12/Chapter7_Exampl12.sce @@ -0,0 +1,22 @@ +clc
+clear
+//INPUT DATA
+tw1=20;//wet bulb temperature in Degree c
+t1=30;//dry bulb temperature in Degree c
+t2=15;//dry bulb temperature in Degree c
+pva=17.0521;//Saturation pressure in mm Hg
+p=760;//pressure in mm of Hg
+ps1=31.81;//pressure in mm of Hg
+ps2=12.77;//pressure in mm of Hg
+
+//CALCULATIONS
+pv1=(pva-((p-pva)*(t1-tw1)*1.8)/(2800-(1.3*(1.8*t1+32))));//Saturation pressure in mm Hg
+x1=(pv1/ps1)*100;//realtive humidity in percentage
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+pv2=12.55;//Saturation pressure in mm Hg
+x2=(pv2/ps2)*100;//realtive humidity in percentage
+
+//OUTPUT
+printf('(a)Initial RH is %3.2f percentage \n (ii)Final RH is %3.2f percentage \n (c) from the chart Final wet bulb temperature according to chart is 14.5 Degree C ',x1,x2)
+
+
diff --git a/1808/CH7/EX7.13/Chapter7_Exampl13.sce b/1808/CH7/EX7.13/Chapter7_Exampl13.sce new file mode 100644 index 000000000..79fc590b1 --- /dev/null +++ b/1808/CH7/EX7.13/Chapter7_Exampl13.sce @@ -0,0 +1,27 @@ +clc
+clear
+//INPUT DATA
+t2=50;//dry bulb temperature in Degree c
+t1=30;//dry bulb temperature in Degree c
+t11=25;//wet bulb temperature in Degree c
+V=300;//volume in m^3
+Ra1=287.3;//rate of flow
+p=760;//pressure in mm of Hg
+pva=23.74;//Saturation pressure in mm Hg
+cp=1.005;//specific pressure
+ps2=92.54;//Saturation pressure in mm Hg
+
+
+//CALCULATIONS
+va1=(Ra1*(273+t1))/((p-21.275)*133.5);//Amount of dry air in m^3/kg d.a.
+pv1=(pva-((p-pva)*(t1-t11)*1.8)/(2800-(1.3*(1.8*t1+32))));//Saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+ma=V/va1;//mass flow rate in kg d.a.
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h2=cp*t2+w1*(2500+1.88*t2);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+pv2=(w1*p/0.6379);//saturation pressure in mm Hg
+Qa=ma*(h2-h1);//Quantity of heat added in kJ
+x2=(pv2/ps2)*100;//Final RH in percentage
+
+//OUTPUT
+printf('(i)Quantity of heat added is %3.2f kJ \n (ii)Final RH is %3.2f percentage \n (iii)from chart Final WBT from the chart is 29 Degree C',Qa,x2)
diff --git a/1808/CH7/EX7.14/Chapter7_Exampl14.sce b/1808/CH7/EX7.14/Chapter7_Exampl14.sce new file mode 100644 index 000000000..5cbd58a7b --- /dev/null +++ b/1808/CH7/EX7.14/Chapter7_Exampl14.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+t1=15;//dry bulb temperature in Degree C
+t3=41;//heating coil temperature in Degree C
+t11=11;//wet bulb temperature in Degree C
+p=760;//pressure in mm of Hg
+x=0.4;//realtive humidity in percentage
+pva=9.83;//Saturation pressure in mm Hg
+ps2=33.68;//Saturation pressure in mm Hg
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+t2=t3-(x*(t3-t1));//dry bulb temperature in Degree c
+pv1=(pva-((p-pva)*(t1-t11)*1.8)/(2800-(1.3*(1.8*t1+32))));//Saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+x2=(pv1/ps2)*100;//realtive humidity in percentage
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h2=cp*t2+w1*(2500+1.88*t2);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+Qa=h2-h1;//Sensible heat addition in kJ/kg d.a.
+
+//OUTPUT
+printf('(a)DBT is %3.1f Degree C \n (b)WBT is from the chart equal to 16.8 Degree C \n (c)RH is %3.2f percentage \n (d)Sensible heat addition is %3.2f kJ/kg d.a ',t2,x2,Qa)
+
+
+
diff --git a/1808/CH7/EX7.15/Chapter7_Exampl15.sce b/1808/CH7/EX7.15/Chapter7_Exampl15.sce new file mode 100644 index 000000000..d1f21df57 --- /dev/null +++ b/1808/CH7/EX7.15/Chapter7_Exampl15.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+V1=200;//volume in m^3/min
+t1=30;//dry bulb temperature in Degree c
+x1=0.8;//realtive humidity in percentage
+t3=14;//Surface temperature in Degree C
+x=0.1;//Coil bypass factor
+ps1=31.81;//Saturation temperature in mm Hg
+pv3=11.97;//Saturation temperature in mm Hg
+cp=1.005;//specific pressure
+R1=287.3;//gas constant
+p=760;//pressure in mm of Hg
+
+//CALCULATIONS
+t2=x*(t1-t3)+t3;//Temperature of air leaving coil in Degree C
+pv1=x1*ps1;//Saturation temperature in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+w3=0.622*(pv3/(p-pv3));//Specific humidity in kg w.v./kg d.a
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h3=cp*t3+w3*(2500+1.88*t3);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h2=(x*(h1-h3))+h3;//Enthalpy of air per kg of dry air in kJ/kg d.a.
+w2=(x*(w1-w3))+w3;//Specific humidity in kg w.v./kg d.a
+v1=R1*(t1+273)/((p-pv1)*133.5);//volume in m^3/kg d.a
+ma=V1/v1;//mass of dry air through the coil in kg d.a./min
+Rc=ma*(h1-h2)/210;//Capacity of the coil in TR
+mw=ma*(w1-w2);//Amount of water vapour removed per minute in kg w.v./kg d.a.
+h4=cp*t1+w2*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+SHF=(h4-h2)/(h1-h2);//Sensible heat factor
+
+//OUTPUT
+printf('(i)The temperature of air leaving the cooling coil is %3.1f Degree C \n (b)Capacity of the cooling coil is %3.2f TR \n (c)Amount of water removed per minute is %3.3f kg w.v./kg d.a. \n (d)Sensible heat factor is %3.4f',t2,Rc,mw,SHF)
diff --git a/1808/CH7/EX7.16/Chapter7_Exampl16.sce b/1808/CH7/EX7.16/Chapter7_Exampl16.sce new file mode 100644 index 000000000..885ea3638 --- /dev/null +++ b/1808/CH7/EX7.16/Chapter7_Exampl16.sce @@ -0,0 +1,37 @@ +clc
+clear
+//INPUT DATA
+t1=30;//dry bulb temperature in Degree c
+t2=25;//Coil cooling temperature in Degree C
+x1=0.6;//realtive humidity in percentage
+t3=10;//Coil cooling temperature in Degree C
+x=0.2;//bypass factor
+Ra=287.3;//gas constant
+p=760;//pressure in mm of Hg
+V1=80;//volume in m^3/kg d.a
+ps1=31.81;//Saturation pressure in mm Hg
+cp=1.005;//specific pressure
+ps2=23.74;//Saturation pressure in mm Hg
+pv3=9.196;//Saturation pressure in mm Hg
+
+
+//CALCULATIONS
+v1=Ra*(273+t1)/((p-19.08)*133.5);//volume in m^3/kg d.a.
+ma=V1/v1;//Mass of dry air entering the coil in kg d.a./min
+pv1=x1*ps1;//Saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h2=cp*t2+w1*(2500+1.88*t2);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+Rc1=ma*(h1-h2);//Capacity of the coil in TR
+x2=(pv1/ps2)*100;//realtive humidity in percentage
+t2r=x*(t1-t3)+t3;//Temperature at refrigeration in Degree c
+w3=0.622*(pv3/(p-pv3));//Specific humidity in kg w.v./kg d.a
+h3=cp*t3+w3*(2500+1.88*t3);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h2r=x*(h1-h3)+h3;//Enthalpy of air per kg of dry air in kJ/kg d.a.
+Rc2=ma*(h1-h2r);//Capacity of the coil in TR
+w2=x*(w1-w3)+w3;//Specific humidity in kg w.v./kg d.a
+mw=ma*(w1-w2);//Condensate flow in kg w.v./min
+
+//OUTPUT
+printf('CASE I \n (a1)Refrigeration required is %3.2f kJ/min \n (b1)Final RH is %3.3f percentage \n CASE II \n (a2)Refrigeration required is %3.2f kJ/min \n (b2)condensate flow is %3.4f kg w.v./min ',Rc1,x2,Rc2,mw)
+
diff --git a/1808/CH7/EX7.17/Chapter7_Exampl17.sce b/1808/CH7/EX7.17/Chapter7_Exampl17.sce new file mode 100644 index 000000000..ebf2296a4 --- /dev/null +++ b/1808/CH7/EX7.17/Chapter7_Exampl17.sce @@ -0,0 +1,29 @@ +clc
+clear
+//INPUT DATA
+t1=40;//dry bulb temperature in Degree c
+t2=30;//dry bulb temperature in Degree c
+x1=0.2;//realtive humidity in percentage
+tw2=20;//wet bulb temperature in Degree c
+ps1=55.31;//Saturation temperature in mm Hg
+ps2=31.81;//Saturation temperature in mm Hg
+pv2a=17.521;//Saturation temperature at WBT in mm Hg
+p=760;//pressure in mm of Hg
+Ra=287.3;//gas constant
+V1=150;//volumme in m^3
+
+
+//CALCULATIONS
+pv1=x1*ps1;//Saturation temperature in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+pv2=(pv2a-((p-pv2a)*(t2-tw2)*1.8)/(2800-(1.3*(1.8*t2+32))));//Saturation pressure in mm Hg
+x2=(pv2/ps2)*100;//realtive humidity in percentage
+w2=0.622*(pv2/(p-pv2));//Specific humidity in kg w.v./kg d.a
+v1=Ra*(273+t1)/((p-11.06)*133.5);//volume
+ma=V1/v1;//Amount of air added in kg d.a./min
+mw=ma*(w2-w1);//Amount of water vapour added in kg d.a./min
+nh=((t1-t2)/(t1-tw2))*100;//humidifier efficiency in percentage
+
+//OUTPUT
+printf('(a)From chart Dew point temperature is corresponds to pv2 is 14.5 Degree c \n (b)Amount of water vapour added is %3.3f kg w.v./min \n (c)Humidifier efficiency is %3.1f percentage',mw,nh)
+
diff --git a/1808/CH7/EX7.18/Chapter7_Exampl18.sce b/1808/CH7/EX7.18/Chapter7_Exampl18.sce new file mode 100644 index 000000000..1f821f3d0 --- /dev/null +++ b/1808/CH7/EX7.18/Chapter7_Exampl18.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+V1=100;//volue in mm^3/min
+t1=6;//dry bulb temperature in Degree c
+t11=3;//dry bulb temperature in Degree c
+Rc=40;//Capacity of the coil in TR
+mw=40;//Amount of water vapour added in kg d.a./min
+pva=5.68;//saturation pressure
+p=760;//pressure in mm of Hg
+Ra=287.3;//gas constant
+cp=1.005;//specific pressure
+
+
+//CALCULATIONS
+pv1=(pva-((p-pva)*(t1-t11)*1.8)/(2800-(1.3*(1.8*t1+32))));//Saturation pressure in mm Hg
+v1=Ra*(273+t1)/((p-pv1)*133.5);//volume
+ma=V1/v1;//Amount of air added in kg d.a./min
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+w3=w1+(mw/(ma*60));//Specific humidity in kg w.v./kg d.a
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h3=h1+33.4;//Enthalpy of air per kg of dry air in kJ/kg d.a.
+t3=h3-(w3*2500)/1.02149;//dry bulb temperature in Degree c
+
+//OUTPUT
+printf('(a)Dry bulb temperature is %3.2f Degree C \n (b)From the psychrometric chart wet bulb temperature is 17.8 Degree C ',t3)
diff --git a/1808/CH7/EX7.19/Chapter7_Exampl19.sce b/1808/CH7/EX7.19/Chapter7_Exampl19.sce new file mode 100644 index 000000000..4e92de16b --- /dev/null +++ b/1808/CH7/EX7.19/Chapter7_Exampl19.sce @@ -0,0 +1,33 @@ +clc
+clear
+//INPUT DATA
+t1=12;//dry bulb temperature in Degree c
+t4=40;//dry bulb temperature in Degree c
+x1=0.9;//realtive humidity in percentage
+t41=25;//wet bulb temperature in Degree c
+x3=0.8;//realtive humidity in percentage
+ps1=10.503;//Saturation temperature in mm Hg
+pv4a=23.74;//Saturation temperature in mm Hg
+t3=22.5;//dry bulb temperature in Degree c
+cp=1.005;//specific pressure
+p=760;//pressure in mm of Hg
+t5=20.8;//dry bulb temperature in Degree c
+
+//CALCULATIONS
+pv1=x1*ps1;//Saturation temperature in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+pv4=(pv4a-((p-pv4a)*(t4-t41)*1.8)/(2800-(1.3*(1.8*t4+32))));//Saturation pressure in mm Hg
+ps3=pv4/x3;//Saturation temperature in mm Hg
+w3=0.622*(pv4/(p-pv4));//Specific humidity in kg w.v./kg d.a
+h3=cp*t3+w3*(2500+1.88*t3);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+t2=(h3-(w1*2500))/1.0191;//dry bulb temperature in Degree c
+h4=cp*t4+w3*(2500+1.88*t4);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+ht=(h4-h1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+mw=(w3-w1);//Additional water required in the air washer in kg w.v./kg d.a.
+nh=((t2-t3)/(t2-t5))*100;//humidifier efficiency in percentage
+
+//OUTPUT
+printf('(a)Temperature at the end of the preheating is %3.2f Degree c \n (b)Total heat required is %3.2f kJ/kg d.a.\n (c)Additional heat required in the air washeer is %3.6f kg w.v./kg d.a.\n (d)humidifier efficiency is %3.2f percentage',t2,ht,mw,nh)
+
+
diff --git a/1808/CH7/EX7.2/Chapter7_Exampl2.sce b/1808/CH7/EX7.2/Chapter7_Exampl2.sce new file mode 100644 index 000000000..55b20da30 --- /dev/null +++ b/1808/CH7/EX7.2/Chapter7_Exampl2.sce @@ -0,0 +1,26 @@ +clc
+clear
+//INPUT DATA
+t=20;//Moist temperature in Degree c
+td=15;//Dew point temperature in Degree c
+pv=12.79;//vapour pressure in mm of Hg
+p2=17.52//pressure of water vapour in mm of Hg
+pa=727.21;//pressure of air in mm of Hg
+hfgt=2454.1;//Specific enthalpy in kJ/kgw.v.
+hfgd=2465.9;//Specific enthalpy in kJ/kgw.v.
+cpa=1.005;//specific pressure
+Ra=287.3;//gas constant
+
+//CALCULATIONS
+pv1=12.79*133.5;//prtial pressure in N/m^2
+x=(pv/p2)*100;//realtive humidity in percentage
+w=0.622*(pv/pa);//Specific humidity in kg w.v./kg d.a
+hv=((4.1868*td)+(hfgd)+(1.88*(t-td)));//Specific enthalpy of water vapour in kJ/kg w.v
+hv1=2500+1.8*(t);//Specific enthalpy of water vapour in kJ/kg w.v
+hv2=4.1868*t+2454.1;//Specific enthalpy of water vapour in kJ/kg w.v
+h=cpa*t+w*hv;//Enthalpy of air per kg of dry air in kJ/kg d.a
+va=(Ra*(t+273))/(pa*133.5);//Specific volume of air per kg of dry air in m^3/kg d.a
+
+//OUTPUT
+printf('(a)From steam tables partial pressure of water is %3.1f N/m^2 \n (b)Relative humidity is %3.2f percentage\n (c)Specific humidity %3.5f kg w.v./kg d.a \n(d)Specific enthalpy of water vapour is %3.3f kJ/kg w.v. \n (e)Enthalpy of air per kg of dry air is %3.2f kJ/kg d.a.\n (f)Specific volume of air per kg of dry air is %3.4f m^3/kg d.a.',pv1,x,w,hv,h,va)
+
diff --git a/1808/CH7/EX7.20/Chapter7_Exampl20.sce b/1808/CH7/EX7.20/Chapter7_Exampl20.sce new file mode 100644 index 000000000..28e5e0b41 --- /dev/null +++ b/1808/CH7/EX7.20/Chapter7_Exampl20.sce @@ -0,0 +1,35 @@ +clc
+clear
+//INPUT DATA
+t1=35;//dry bulb temperature in Degree c
+x1=0.8;//realtive humidity in percentage
+t2=15;//Apparatus dew point in Degree c
+t4=25;//dry bulb temperature in Degree c
+V1=200;//Quantity of moist air in m^3/min
+x=0.3;//bypass factor
+ps1=42.16;//Saturation pressure in mm Hg
+p=760;//pressure in mm of Hg
+Ra=287.3;//gas constant
+cp=1.005;//specific pressure
+ps2=12.77;//Saturation pressure in mm Hg
+ps4=23.74;//Saturation pressure in mm Hg
+
+//CALCULATIONS
+pv1=x1*ps1;//Saturation pressure in mm Hg
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+t3=(x*(t1-t2))+t2;//dry bulb temperature in Degree c
+h1=cp*t1+w1*(2500+1.88*t1);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+w2=0.622*(ps2/(p-ps2));//Specific humidity in kg w.v./kg d.a
+w3=x*(w1-w2)+w2;//Specific humidity in kg w.v./kg d.a
+h3=cp*t3+w3*(2500+1.88*t3);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h4=cp*t4+w3*(2500+1.88*t4);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+v1=Ra*(273+t1)/((p-pv1)*133.5);//volume
+ma=V1/v1;//Amount of air added in kg d.a./min
+Qc=ma*(h1-h3)/210;//Capacity of cooling coil in TR
+Qh=ma*(h4-h3)/60;//Capacity of heating coil in kW
+mw=ma*(w1-w3);//Quantity of water removed in kg w.v./min
+pv4=w3*p/1.01611;//Saturation pressure in mm Hg
+x4=(pv4/ps4)*100;//realtive humidity in percentage
+
+//OUTPUT
+printf('(a)Capacity of cooling coil is %3.2f TR \n (b)Capacity of heating coil is %3.2f kW \n(c)Quantity of water removed is %3.4f kg w.v./min \n (d)realtive humidity is %3.2f percentage',Qc,Qh,mw,x4)
diff --git a/1808/CH7/EX7.21/Chapter7_Exampl21.sce b/1808/CH7/EX7.21/Chapter7_Exampl21.sce new file mode 100644 index 000000000..1dd98ff61 --- /dev/null +++ b/1808/CH7/EX7.21/Chapter7_Exampl21.sce @@ -0,0 +1,34 @@ +clc
+clear
+//INPUT DATA
+t1=35;//dry bulb temperature in Degree c
+t4=25;//dry bulb temperature in Degree c
+V1=40;//Moist air circulation in m^3/min
+x1=0.8;//realtive humidity in percentage
+x4=0.6;//realtive humidity in percentage
+p=760;//pressure in mm of Hg
+Ra=287.3;//gas constant
+cp=1.005;//specific pressure
+ps1=42.157;//Saturation pressure in mm Hg
+ps4=23.74;//Saturation pressure in mm Hg
+t3=16.6;//dry bulb temperature in Degree c
+
+//CALCULATIONS
+pv1=x1*ps1;//Saturation pressure in mm Hg
+v1=Ra*(273+t1)/((p-pv1)*133.5);//volume
+ma=V1/v1;//Amount of air added in kg d.a./min
+w1=0.622*(pv1/(p-pv1));//Specific humidity in kg w.v./kg d.a
+h1=cp*t1+w1*(2500+(1.88*t1));//Enthalpy of air per kg of dry air in kJ/kg d.a.
+pv4=x4*ps4;//Saturation pressure in mm Hg
+x3=(pv4/pv4)*100;//realtive humidity in percentage
+w4=0.622*(pv4/(p-pv4));//Specific humidity in kg w.v./kg d.a
+h4=cp*t4+w4*(2500+1.88*t4);//Enthalpy of air per kg of dry air in kJ/kg d.a.
+h3=cp*t3+w4*(2500+(1.88*t3));//Enthalpy of air per kg of dry air in kJ/kg d.a.
+Qc=ma*(h1-h3)/210;//Capacity of cooling coil in TR
+Qh=ma*(h4-h3)/60;//Capacity of heating coil in kW
+mw=ma*(w1-w4);//Quantity of water removed in kg w.v./min
+
+//OUTPUT
+printf('(a)Capacity of cooling coil is %3.2f kJ/min \n (b)Capacity of heating coil is %3.1f kW \n (c)Quantity of water removed is %3.3f kg w.v./min',Qc,Qh,mw)
+
+
diff --git a/1808/CH7/EX7.3/Chapter7_Exampl3.sce b/1808/CH7/EX7.3/Chapter7_Exampl3.sce new file mode 100644 index 000000000..1538d6d80 --- /dev/null +++ b/1808/CH7/EX7.3/Chapter7_Exampl3.sce @@ -0,0 +1,30 @@ +clc
+clear
+//INPUT DATA
+ta=25;//Dry bulb temperature in Degree c
+tw=15;//Wet bulb temperature in Degree c
+td=7.56;//Dew point temperature in Degree c
+p=760;//Atmospheric air in mm of Hg
+pv1=12.77;//Saturation pressure of water in mm of Hg
+ps=23.74;//Saturation pressure of water in mm of Hg
+Pv=7.788;//Saturation pressure of water in mm of Hg
+Ra=287.3;//gas constant
+Rv=461;//vapour constant
+pa=1015639.698;//air pressure
+cp=1.005;//specific pressure
+
+//CALCULATIONS
+pv=pv1-((p-pv1)*(ta-tw)*1.8/(2800-1.3*(1.8*ta+32)));//Saturation pressure of water in mm of Hg
+x=(pv/ps)*100;//realtive humidity in percentage
+w=0.622*(pv/(p-pv));//Specific humidity in kg w.v./kg d.a
+h=cp*ta+w*(2500+1.88*ta);//Enthalpy of air per kg of dry air in kJ/kg d.a
+Roa=(pa/10)/(Ra*(273+ta));//Density of air in kg/m^3
+Rov=pv*133.5/(Rv*(273+ta));//Density of vapour in kg/m^3
+Ro=Rov+Roa;//Density in kg/m^3
+
+//OUTPUT
+printf('(a)Relative humidity is %3.2f percentage \n (b)Humidity ratio is %3.4f kg w.v./kg d.a \n (c)Dew point temperature is %3.2f Degree c \n (d)Enthalpy of air per kg of dry air is %3.2f kJ/kg d.a.\n (e)Partial pressure of vapour is %3.3f mm Hg \n (f)density is %3.3f kg/m^3',x,w,td=7.56,h,Pv,Ro)
+
+
+
+
diff --git a/1808/CH7/EX7.4/Chapter7_Exampl4.sce b/1808/CH7/EX7.4/Chapter7_Exampl4.sce new file mode 100644 index 000000000..1f363a0de --- /dev/null +++ b/1808/CH7/EX7.4/Chapter7_Exampl4.sce @@ -0,0 +1,19 @@ +clc
+clear
+//INPUT DATA
+p=760;//pressure in mm of Hg
+t=30;//dry bulb temperature in Degree c
+p2=0.04246*10^5;//pressure in N/m^2
+cp=1.005;//specific pressure
+hfg=2500;//Specific enthalpy in kJ/kgw.v.
+cpv=1.88;//specific pressure
+
+//CALCULATIONS
+ps=(p2/133.5);//pressure in mm of Hg
+ws=(0.62*(ps/(p-ps)));//Specific humidity in kg w.v./kg d.a
+h=(cp*t)+ws*(hfg+(cpv*t));//Enthalpy of air per kg of dry air in kJ/kg d.a
+
+//OUTPUT
+printf('(a)Accorrding to steam tables The vapour pressure is %3.2f mm Hg \n (b)Specific humidity %3.4f kg w.v./kg d.a \n (c)Enthalpy of air per kg of dry air is %3.2f kJ/kg d.a ',ps,ws,h)
+
+
diff --git a/1808/CH7/EX7.5/Chapter7_Exampl5.sce b/1808/CH7/EX7.5/Chapter7_Exampl5.sce new file mode 100644 index 000000000..0d8e424d9 --- /dev/null +++ b/1808/CH7/EX7.5/Chapter7_Exampl5.sce @@ -0,0 +1,24 @@ +clc
+clear
+//INPUT DATA
+t=30;//dry bulb temperature in Degree c
+x=30;//realtive humidity in percentage
+p=760;//pressure in mm of Hg
+p2=0.04246*10^5;//pressure in N/m^2
+V=100;//volume in m^3
+Rv=0.461;//vapour constant
+Ra=0.2871;//gas constant
+
+//CALCULATIONS
+ps=(p2/133.5);//pressure in mm of Hg
+pv=0.3*ps;//vapour pressure in mm of Hg
+pa=p-pv;//air pressure in mm of Hg
+w=(0.62*(pv/(p-pv)));//Specific humidity in kg w.v./kg d.a
+ws=(0.62*(ps/(p-ps)));//Specific humidity in kg w.v./kg d.a
+m=w/ws;//Degree of saturation
+mv=pv*133.5*V/(Rv*(t+273)*1000);//Mass of vapour in kg.w.v.
+ma=pa*133.5*V/(Ra*1000*(t+273));//Mass of dry air in kg.d.a.
+
+//OUTPUT
+printf('(a)Partial pressure of dry air vapour is %3.3f mm Hg \n (b)Dew point temperature is 10.62 Degree c \n (c)Specific humidity %3.4f kg w.v./kg d.a \n (d)Degree of saturation is %3.3f \n (e)mass of vapour is %3.4f kg.w.v.\n (f)mass of dry air is %3.4f kg.d.a',pa,w,m,mv,ma)
+
diff --git a/1808/CH7/EX7.6/Chapter7_Exampl6.sce b/1808/CH7/EX7.6/Chapter7_Exampl6.sce new file mode 100644 index 000000000..029754fbf --- /dev/null +++ b/1808/CH7/EX7.6/Chapter7_Exampl6.sce @@ -0,0 +1,23 @@ +clc
+clear
+//INPUT DATA
+t=35;//dry bulb temperature in Degree c
+td=15;//dew point temperature in Degree c
+p=760;//pressure in mm of Hg
+pv1=0.017051*10^5;//saturation pressure
+ps1=0.05628 *10^5;//saturation pressure
+cp=1.005;//specific pressure
+cpv=1.88;//specific volume
+hfg=2500;//Specific enthalpy in kJ/kgw.v.
+
+//CALCULATIONS
+pv=pv1*133.5;//vapour pressure in mm of Hg
+ps=ps1*133.5;//pressure in mm of Hg
+x=(pv/ps)*100;//realtive humidity in percentage
+ws=(0.622*(12.77/(760-12.77)));//Specific humidity in kg w.v./kg d.a
+h=(cp*t)+ws*(hfg+(cpv*t));//Enthalpy of air per kg of dry air in kJ/kg d.a
+
+//OUTPUT
+printf('(a)Relative humidity is %3.2f percentage \n (b)Specific humidity %3.4f kg w.v./kg d.a \n (c)Enthalpy of air per kg of dry air is %3.2f kJ/kg d.a ',x,ws,h)
+
+
diff --git a/1808/CH7/EX7.7/Chapter7_Exampl7.sce b/1808/CH7/EX7.7/Chapter7_Exampl7.sce new file mode 100644 index 000000000..1547af956 --- /dev/null +++ b/1808/CH7/EX7.7/Chapter7_Exampl7.sce @@ -0,0 +1,14 @@ +clc
+clear
+//INPUT DATA
+t=25;//dry bulb temperature in Degree c
+ws=8.6/100;//Specific humidity in kg w.v./kg d.a
+p=760;//pressure in mm of Hg
+ps=23.74;//Saturation pressure in mm of Hg
+
+//CALCULATIONS
+pv=10.508/1.01383;//Partial pressure of water vapour in mm Hg
+x=(pv/ps)*100;//realtive humidity in percentage
+
+//OUTPUT
+printf('(a)Partial pressure of dry air vapour is %3.3f mm Hg \n (b)Relative humidity is %3.2f percentage ',pv,x)
diff --git a/1808/CH7/EX7.8/Chapter7_Exampl8.sce b/1808/CH7/EX7.8/Chapter7_Exampl8.sce new file mode 100644 index 000000000..d075d4fa7 --- /dev/null +++ b/1808/CH7/EX7.8/Chapter7_Exampl8.sce @@ -0,0 +1,32 @@ +clc
+clear
+//INPUT DATA
+p=0.95*10^5;//Atmospheric air
+tw1=20//Wet bulb temperature in Degree c
+t1=31;//Dry bulb temperature in Degree c
+pv2=0.02339*10^5;//vapour pressure in mm of Hg
+hv1=2556.3;//Enthalpy corresponds tovapour inlet in kJ/kg w.v
+hv2=2538.1;//Enthalpy corresponds tovapour outlet in kJ/kg w.v
+h6=83.96;//sensible heat of water in kJ/kg w.v
+m=1;//mass flow rate in kg.d.a.
+cp=1.005;//specific pressure
+t1=30;//temperature in K
+t2=20;//temperature in K
+ps1=0.0425;//Saturation pressure in bar
+
+//CALCULATIONS
+ha1=m*cp*t1;//Enthalpy of air per kg of dry air in kJ/kg d.a
+ha2=m*cp*t2;//Enthalpy of air per kg of dry air in kJ/kg d.a
+w2=0.622*(pv2/(p-pv2));//Specific humidity in kg w.v./kg d.a
+w1=(w2*(hv2-h6)+(ha2-ha1))/(hv1-h6);//Specific humidity in kg w.v./kg d.a
+pv1=0.01759485/1.01759485;//vapour pressure in bar
+x=(pv1/ps1)*100;//realtive humidity in percentage
+
+//OUTPUT
+printf('(a)Humidity ratio is %3.4f kg w.v./kg d.a \n (b)Vapour pressure is %3.5f bar \n relative humidity is %3.2f percentage \n (c) According to steam tables Dew point temperature is 14.5 Degree c ',w1,pv1,x)
+
+
+
+
+
+
diff --git a/1808/CH7/EX7.9/Chapter7_Exampl9.sce b/1808/CH7/EX7.9/Chapter7_Exampl9.sce new file mode 100644 index 000000000..fca84be2e --- /dev/null +++ b/1808/CH7/EX7.9/Chapter7_Exampl9.sce @@ -0,0 +1,28 @@ +clc
+clear
+//INPUT DATA
+m1=1;//mass flow rate in kg
+m2=2;//mass flow rate in kg
+t1=50;//temperature in Degree C
+t2=20;//temperature in Degree C
+x1=0.5;//temperature in Degree C
+td2=20;//Dew point temperature in Degree C
+ps1=0.12354*10^5;//Saturation pressure in N/m^2
+ps2=17.52;//Saturation pressure in mm Hg
+p=760;//pressure in mm of Hg
+m12=0.5;//Ratio of masses
+
+//CALCULATIONS
+pv1=x1*ps1;//vapour pressure in mm Hg
+pv11=pv1/133.5;//vapour pressure in mm Hg
+w1=0.622*(pv11/(p-pv11));//Specific humidity in kg w.v./kg d.a
+w2=0.622*(ps2/(p-ps2));//Specific humidity in kg w.v./kg d.a
+w3=(w1+2*w2)/3;//Specific humidity in kg w.v./kg d.a
+t3=(t1+2*t2)/3;//temperature in Degree C
+
+//OUTPUT
+printf('(i)Temperature of the mixture is %3.1f Degree C \n (ii)Specific humidity of the mixture is %3.5f kg/kg d.a.',t3,w3)
+
+
+
+
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