diff options
Diffstat (limited to '2090/CH16')
| -rwxr-xr-x | 2090/CH16/EX16.1/Chapter16_example1.sce | 26 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.10/Chapter16_example10.sce | 55 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.11/Chapter16_example11.sce | 31 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.12/Chapter16_example12.sce | 59 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.2/Chapter16_example2.sce | 24 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.3/Chapter16_example3.sce | 14 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.4/Chapter16_example4.sce | 38 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.5/Chapter16_example5.sce | 30 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.6/Chapter16_example6.sce | 41 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.7/Chapter16_example7.sce | 44 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.8/Chapter16_example8.sce | 29 | ||||
| -rwxr-xr-x | 2090/CH16/EX16.9/Chapter16_example9.sce | 42 |
12 files changed, 433 insertions, 0 deletions
diff --git a/2090/CH16/EX16.1/Chapter16_example1.sce b/2090/CH16/EX16.1/Chapter16_example1.sce new file mode 100755 index 000000000..963adf038 --- /dev/null +++ b/2090/CH16/EX16.1/Chapter16_example1.sce @@ -0,0 +1,26 @@ +clc
+clear
+//Input data
+N=3000;//The speed of the engine in rpm
+r=9;//Compression ratio
+l=17.2;//The length of the connecting rod in cm
+t=20;//The combustion ends at a TDC in degrees
+pi=3.141;//Mathematical constant of pi
+k=3;//Three litre spark engine
+n=6;//V-6 Engine
+
+//Calculations
+Vs=(k/n)*10^-3;//Swept volume per cylinder in m^3
+d=[[(Vs*4)/pi]^(1/3)];//The diameter of the bore in m
+L=d*100;//The length of the stroke in cm
+up=2*d*N/60;//Average piston speed in m/s
+Vc=[Vs/(r-1)]*10^6;//Clearence volume in cm^3
+cr=(L)/2;//Crank radius in cm
+R=l/cr;//The ratio of the connecting rod length to crank radius
+up1=up*[(pi/2)*sin(pi/9)*(1+(cos(pi/9)/(R^2-(sin(pi/9)^2))^(1/2)))];//The piston speed at the end of combustion in m/s
+s=(cr*cos(pi/9))+(l^2-(cr^2)*(sin(pi/9))^2)^(1/2);//Distance between crank axis and wrist pin in cm
+x=l+cr-s;//The distance the piston travels from TDC at the end of combustion in cm
+V=Vc+(pi/4)*(d*100)^2*x;//Instantaneous volume in cm^3
+
+//Output
+printf('(a)The cylinder bore and The stroke length (d = L) = %3.1f cm \n (b) The average piston speed = %3.1f m/s \n (c) The clearence volume of one cylinder = %3.1f cm^3 \n (d) The piston speed at the end of combustion = %3.2f m/s \n (e) The distance the piston travels from TDC at the end of combustion = %3.2f cm \n (f) Instantaneous volume = %3.1f cm^3 ',L,up,Vc,up1,x,V)
diff --git a/2090/CH16/EX16.10/Chapter16_example10.sce b/2090/CH16/EX16.10/Chapter16_example10.sce new file mode 100755 index 000000000..1391d26d8 --- /dev/null +++ b/2090/CH16/EX16.10/Chapter16_example10.sce @@ -0,0 +1,55 @@ +clc
+clear
+//Input data
+d=0.18;//The diameter of the cylinder in m
+pi=3.141;//Mathematical constant of pi
+L=0.24;//The length of the stroke in m
+t=30;//Duration trail in min
+N=9000;//Number of revolutions
+Ne=4450;//Total number of explosions
+pmi=5.35;//Gross imep in bar
+pp=0.35;//Pumping imep in bar
+W=40;//Net load on brake wheel in kg
+dd=0.96;//Diameter of the brake wheel drum in m
+dr=0.04;//Diameter of the rope in m
+V=2.6;//Volume of gas used in m^3
+pg=136;//pressure of gas in mmof Hg
+dg=0.655;//The density of gas in kg/m^3
+T=290;//The ambient temperature of air in K
+CV=19000;//The calorific value of the fuel in kJ/m^3
+ta=40;//Total air used in m^3
+p=720;//Pressure of air in mm of Hg
+Te=340;//Temperature of exhaust gas in degree centigrade
+Cpg=1.1;//Specific heat of gas in kJ/kgK
+C=80;//Cooling water circulated in kg
+Tr=30;//Rise in temperature of cooling water in degree centigrade
+R=287;//Real gas constant in J/kgK
+
+//Calculations
+ip=(pmi-pp)*10^5*L*(pi/4)*d^2*(Ne/(30*60))*(1/1000);//The indicated power in kW
+bp=(pi*(N/(30*60))*W*9.81*(dd+dr)*(1/1000));//The brake power in kW
+pgs=760+(pg/13.6);//Pressure of gas supplied in mm of Hg
+Vg=((pgs*V)/290)*(273/760);//The volume of gas in m^3
+Q=(Vg*CV)/30;//Heat supplied by gas used at NTP in kJ/min
+Qbp=bp*60;//Heat equivalent of bp in kJ/min
+Qc=(C/t)*4.18*Tr;//Heat lost to cooling medium in kJ/min
+Va=[((p*ta)/T)*(273/760)]/30;//Volume of air used in kg/min
+da=(1.013*10^5)/(R*273);//The density of air in kg/m^3
+ma=Va*da;//Mass of air used in kg/min
+mg=(Vg/30)*dg;//Mass of gas at NTP in kg/min
+me=ma+mg;//Total mass of exhaust gas in kg/min
+Qe=me*Cpg*(Te-(T-273));//Heat loss to exhaust gas in kJ/min
+Qu=Q-(Qe+Qc+Qbp);//Unaccounted heat loss in kJ/min
+nm=(bp/ip)*100;//Mechanical efficiency in percent
+ni=((ip*60)/Q)*100;//Indicated thermal efficiency in percent
+x=((Qbp/1571)*100);//percentage heat in bp
+y=((Qc/1571)*100);//Percent heat lost to cooling water
+z=((Qe/1571)*100);//Percent heat to exhaust gases
+k=((Qu/1571)*100);//Percent heat unaccounted
+
+//Output
+printf('--------------------------------------------------------------------------------------------\n Heatinput kJ/min percent Heat expenditure kJ/min percent \n ----------------------------------------------------------------------------------------- \n Heat \n supplied 1571 100 (a) Heat in bp %3.1f %3.1f \n (b) Heat loss to cooling water %3.1f %3.1f \n (c) Heat to exhaust gas %3.1f %3.1f \n (d) Unaccounted heat %3.1f %3.1f \n ----------------------------------------------------------------------------------------- \n The mechanical efficiency = %3.2f percent \n The Indicated thermal efficiency = %3.1f percent ',Qbp,x,Qc,y,Qe,z,Qu,k,nm,ni)
+
+
+
+
diff --git a/2090/CH16/EX16.11/Chapter16_example11.sce b/2090/CH16/EX16.11/Chapter16_example11.sce new file mode 100755 index 000000000..db3bb5418 --- /dev/null +++ b/2090/CH16/EX16.11/Chapter16_example11.sce @@ -0,0 +1,31 @@ +clc
+clear
+//Input data
+bp=30;//The brake power in kw
+mf=10;//Mass flow rate of fuel in kg/h
+CV=42000;//Calorific value of the fuel in kJ/kg
+mw=9;//Mass flow rate of water in kg/min
+Tr=60;//Rise in temperature of the cooling water in degree centigrade
+mwe=9.5;//Mass flow rate of water through exhaust gas calorimeter in kg/min
+Tc=40;//Rise in temperature when passing through calorimeter in degree centigrade
+Te=80;//Temperature of exhaust gas leaving the calorimeter in degree centigrade
+a=20;//Air fuel ratio
+T=17;//Ambient temperature in degree centigrade
+Cpw=4.18;//Specific heat of water in kJ/kgK
+Cpg=1;//Mean specific heat of gas in kJ/kgK
+
+//Calculations
+Qf=(mf/60)*CV;//Heat supplied by fuel in kJ/min
+Qbp=bp*60;//Heat equvalent to bp in kJ/min
+Qc=mw*Cpw*Tr;//Heat carried away by the jacket cooling water in kJ/min
+mg=(mf/60)+(mf/60)*a;//Mass of exhaust gas formed in kg/min
+Qe=(mwe*Cpw*Tc)+(mg*Cpg*(Te-T));//Heat carried away by exhaust gas in kJ/min
+Qu=Qf-(Qbp+Qc+Qe);//Unaccounted heat in kJ/min
+x=((Qbp/Qf))*100;//Percentage heat in bp
+y=(Qc/Qf)*100;//Percentage loss of cooling water
+z=(Qe/Qf)*100;//Percentage loaa of heat to exhaust gases
+k=(Qu/Qf)*100;//Percentage heat loss unaccounted
+
+//Output
+printf( '----------------------------------------------------------------------------------------------------\n Heat input kJ/min percent Heat expenditure kJ/min percent \n ---------------------------------------------------------------------------------------------------\n Heat supplied by fuel %3.0f 100 (a) Heat in bp %3.0f %3.2f \n (b) Heat loss to cooling water %3.0f %3.2f \n (c) Heat to exhaust gases %3.0f %3.2f \n (d)Unaccounted heat loss %3.0f %3.2f \n total %3.0f 100 total %3.0f 100 \n --------------------------------------------------------------------------------------------------- ',Qf,Qbp,x,Qc,y,Qe,z,Qu,k,Qf,Qf)
+
diff --git a/2090/CH16/EX16.12/Chapter16_example12.sce b/2090/CH16/EX16.12/Chapter16_example12.sce new file mode 100755 index 000000000..4fae01a27 --- /dev/null +++ b/2090/CH16/EX16.12/Chapter16_example12.sce @@ -0,0 +1,59 @@ +clc
+clear
+//Input data
+n=4;//Number of cylinders
+d=0.085;//The diameter of the bore m
+L=0.095;//The length of the stroke in m
+tr=0.35;//Torque radius in m
+N=3000;//The speed of the engine in rpm
+w=430;//Net brake load in N
+w1=300;//Net brake load produced at the same speed by three cylinders in N
+mf=0.24;//The mass flow rate of fuel in kg/min
+CV=44000;//The calorific value of the fuel in kJ/kg
+mw=65;//Mass flow rate of water in kg/min
+Tw=12;//The rise in temperature in degree centigrade
+a=15;//The air fuel ratio
+Te=450;//The temperature of the exhaust gas in degree centigrade
+Ta=17;//Ambient temperature in degree centigrade
+p=76;//Barometric pressure in cm of Hg
+H=15.5;//The proportion of hydrogen by mass in the fuel in percent
+Cpe=1;//The mean specific heat of dry exhaust gas in kJ/kgK
+Cps=2;//The specific heat of super heated steam in kJ/kgK
+Cpw=4.18;//The specific heat of water in kJ/kgK
+Ts=100;//At 76 cm of Hg The temperature in degree centigrade
+hfg=2257;//The Enthalpy in kJ/kg
+pi=3.141;//Mathematical constant of pi
+R=287;//Real gas constant in J/kgK
+
+//Calculations
+bp=(2*pi*N*w*tr)/(60*1000);//The brake power in kW
+bp1=(2*pi*N*w1*0.35)/(60*1000);//The brake power when each cylinder is cut off in kW
+ip=bp-bp1;//Indicated power per cycle in kW
+ip1=n*ip;//Indicated power of the engine in kW
+imep=[(ip1*60*1000)/(L*(pi/4)*d^2*(N/2)*n)]/10^5;//The indicated mean effective pressure in bar
+ni=[(ip1*60)/(mf*CV)]*100;//Indicated thermal efficiency in percent
+bsfc=(mf*60)/bp;//Brake specific fuel consumption in kg/kWh
+Vs=(pi/4)*d^2*L*(N/2)*n;//Swept volume in m^3/min
+ma=a*mf;//Mass flow rate of air in kg/min
+da=(1*10^5)/(R*(Ta+273));//The density of air in kg/m^3
+Va=ma/da;//Volume of air flow in m^3/min
+nv=[Va/Vs]*100;//Volumetric efficiency in percent
+Qf=mf*CV;//Heat supplied by fuel in kJ/min
+Qbp=bp*60;//The heat equivalent to bp in kJ/min
+Qc=mw*Cpw*Tw;//Heat lost to cooling water in kJ/min
+mv=9*(H/100)*mf;//Mass of water vapour in kg/min
+me=ma+mf-mv;//Mass of dry exhaust gas in kg/min
+Qe=me*Cpe*(Te-Ta);//Heat carried away by the exhaust gas in kJ/min
+Qs=(mv*([Cpw*(Ts-Ta)]+hfg+(Cps*(Te-Ts))));//Heat lost in steam in kJ/min
+Qu=Qf-(Qbp+Qc+Qe+Qs);//Unaccounted heat loss in kJ/min
+x=(Qbp/Qf)*100;//Percentage of heat in bp
+y=(Qc/Qf)*100;//Percentage of heat loss in colling water
+z=(Qe/Qf)*100;//Percentage heat loss in dry exhaust gas
+k=(Qs/Qf)*100;//Percentage heat lost to steam
+l=(Qu/Qf)*100;//Percentage of unaccounted heat lost
+
+//Output
+printf('---------------------------------------------------------------------------------------------------\n Heat input kJ/min percent Heat expenditure kJ/min percent \n -------------------------------------------------------------------------------------------------------\n Heat supplied by fuel %3.0f 100 (a) Heat in bp %3.0f %3.2f \n (b) Heat lost to cooling water %3.0f %3.2f \n (c) Heat to dry exhaust %3.0f %3.2f \n (d) Heat lost in steam %3.0f %3.2f \n (e) Unaccounted heat loss %3.0f %3.2f \n total %3.0f 100 Total %3.0f 100 \n --------------------------------------------------------------------------------------------------------\n \n The indicated mean effective pressure = %3.2f bar \n The indicated thermal efficiency = %3.1f percent \n The brake specific fuel comsumption = %3.4f kg/kWh \n The volumetric efficiency = %3.1f percent ',Qf,Qbp,x,Qc,y,Qe,z,Qs,k,Qu,l,Qf,Qf,imep,ni,bsfc,nv)
+
+
+
diff --git a/2090/CH16/EX16.2/Chapter16_example2.sce b/2090/CH16/EX16.2/Chapter16_example2.sce new file mode 100755 index 000000000..4c006fe68 --- /dev/null +++ b/2090/CH16/EX16.2/Chapter16_example2.sce @@ -0,0 +1,24 @@ +clc
+clear
+//Input data
+d=0.175;//The diameter of the bore in m
+pi=3.141;//The mathematical constant of pi
+L=0.32;//The length of the stroke in m
+p=6.5;//Mean effective pressure in bar
+pp=0.4;//Pumping loop mean effective pressure in bar
+N=510;//The speed of the engine in rpm
+pm=0.65;//Diagrams from the dead cycle give a mep in bar
+n=55;//Firing strokes per minute
+
+//Calculations
+pmi=p-pp;//The net imep at full load in bar
+c=((N/2)-n);//Dead cycles per minute at no load
+ipw=pmi*10^5*L*(pi/4)*d^2*(n/60)*(1/1000);//Indicating power for working cycles in kW
+Pp=pm*10^5*L*(pi/4)*d^2*(c/60)*(1/1000);//Pumping power of dead cycles in kW
+fp=ipw-Pp;//Power in kW
+fip=pmi*10^5*L*(pi/4)*d^2*(N/(2*60))*(1/1000);//Full load indicated power in kW
+fbp=fip-fp;//Full load break power in kW
+nm=(fbp/fip)*100;//Mechanical efficiency in percent
+
+//Output
+printf(' The full load break power = %3.2f kW \n The mechanical efficiency of the engine = %3.1f percent ',fbp,nm)
diff --git a/2090/CH16/EX16.3/Chapter16_example3.sce b/2090/CH16/EX16.3/Chapter16_example3.sce new file mode 100755 index 000000000..ef657ddcb --- /dev/null +++ b/2090/CH16/EX16.3/Chapter16_example3.sce @@ -0,0 +1,14 @@ +clc
+clear
+//Input data
+d=0.09;//The diameter of the bore in m
+L=0.1;//The length of the stroke in m
+T=120;//The torque measured in Nm
+pi=3.141;//Mathematical constant of pi
+n=4;//Number of cylinders
+
+//Calculations
+pmb=[(4*pi*T)/(L*(pi/4)*d^2*n)]/10^5;//The brake mean effective pressure in bar
+
+//Output
+printf('The brake mean effective pressure = %3.2f bar',pmb)
diff --git a/2090/CH16/EX16.4/Chapter16_example4.sce b/2090/CH16/EX16.4/Chapter16_example4.sce new file mode 100755 index 000000000..61c9f91b8 --- /dev/null +++ b/2090/CH16/EX16.4/Chapter16_example4.sce @@ -0,0 +1,38 @@ +clc
+clear
+//Input data
+d=0.06;//The diameter of the bore in m
+L=0.085;//The length of the stroke in m
+N=3000;//The speed of the engine in rpm
+r=0.35;//Torque arm radius in m
+W=160;//Weight in N
+pi=3.141;//Mathemetical constant
+f=6.6;//Fuel consumption in l/h
+g=0.78;//specific gravity of the fuel
+CV=44000;//The calorific value of the fuel in kJ/kg
+w1=114;//Brake load for cylinder 1 in N
+w2=110;//Brake load for cylinder 2 in N
+w3=112;//Brake load for cylinder 3 in N
+w4=116;//Brake load for cylinder 4 in N
+n=4;//Number of cylinders
+
+//Calculations
+Vf=(f*10^-3)/3600;//Volume flow rate of fuel in m^3/s
+df=g*1000;//The density of the fuel in kg/m^3
+mf=df*Vf;//Mass flow rate of fuel in kg/s
+T=W*r;//Torque in Nm
+bp=(2*pi*N*T)/(60*1000);//The brake power in kW
+pmb=[(120*bp*1000)/(L*(pi/4)*d^2*N*n)]/10^5;//Brake mean effective pressure in bar
+nb=[(bp)/(mf*CV)]*100;//The brake thermal efficiency in percent
+bsfc=(mf*3600)/bp;//Brake specific fuel consumption in kg/kWh
+bp1=[(2*pi*N*w1*r)/(60*1000)];//Brake power from morse test in kW
+ip1=bp-bp1;//Indicated power in kW
+ip2=bp-[(2*pi*N*w2*r)/(60*1000)];//Indicated power in kW
+ip3=bp-[(2*pi*N*w3*r)/(60*1000)];//Indicated power in kW
+ip4=bp-[(2*pi*N*w4*r)/(60*1000)];//Indicated power in kW
+ip=ip1+ip2+ip3+ip4;//Total indicated power in kW
+nm=(bp/ip)*100;//Mechanical efficiency in %
+pmi=pmb/(nm/100);//The imep in bar
+
+//Output
+printf('The brake power = %3.2f kW \n The brake mean effective pressure = %3.2f bar \n The brake thermal efficiency = %3.0f percent \n The brake specific fuel consumption = %3.3f kg/kWh \n The indicated power = %3.2f kW \n The mechanical efficiency = %3.1f percent \n The indicated mean effective pressure = %3.1f bar ',bp,pmb,nb,bsfc,ip,nm,pmi)
diff --git a/2090/CH16/EX16.5/Chapter16_example5.sce b/2090/CH16/EX16.5/Chapter16_example5.sce new file mode 100755 index 000000000..c8cda2adb --- /dev/null +++ b/2090/CH16/EX16.5/Chapter16_example5.sce @@ -0,0 +1,30 @@ +clc
+clear
+//Input data
+d=0.15;//The diameter of the bore in m
+L=0.16;//The length of the stroke in m
+N=500;//The speed of the engine in rpm
+mf=0.0475;//Fuel consumption in kg/min
+CV=42000;//The calorific value in kJ/kg
+w=400;//The tension on either side of the pulley in N
+c=2.2;//Brake circumference in m
+l=50;//Length of the indicator diagram in mm
+ap=475;//Area of the positive loop of indicator diagram in mm^2
+an=25;//Area of the negative loop of indicator diagram in mm^2
+s=0.8333;//Spring constant in bar/mm
+pi=3.141;//Mathematical constant of pi
+
+//Calculations
+r=c/(2*pi);//Arm length in m
+T=w*r;//Torque in Nm
+bp=(2*pi*N*T)/(60*1000);//Brake power in kW
+M=(ap-an)/l;//Mean height of indicator diagram in mm
+imep=M*s;//Indicated mean effective pressure in bar
+ip=(imep*10^5*L*(pi/4)*d^2*(N/(2*60))*(1/1000));//Indicated power in kW
+nm=(bp/ip)*100;//The mechanical efficiency in percent
+nb=[(bp*60)/(mf*CV)]*100;//The brake thermal efficiency in percent
+ni=[(nb/100)/(nm/100)]*100;//The indicated thermal efficiency in percent
+bsfc=(mf*60)/bp;//Brake specific fuel consumption in kg/kWh
+
+//Output
+printf('(a) The brake power = %3.2f kW \n (b) The indicated power = %3.3f kW \n (c) The mechanical efficiency = %3.0f percent \n (d) The brake thermal efficiency = %3.2f percent \n (e) The indicated thermal efficiency = %3.1f percent \n (f) The brake specific fuel consumption = %3.3f kg/kWh ',bp,ip,nm,nb,ni,bsfc)
diff --git a/2090/CH16/EX16.6/Chapter16_example6.sce b/2090/CH16/EX16.6/Chapter16_example6.sce new file mode 100755 index 000000000..f4f27ea7a --- /dev/null +++ b/2090/CH16/EX16.6/Chapter16_example6.sce @@ -0,0 +1,41 @@ +clc
+clear
+//Input data
+n=8;//Number of cylinders
+d=0.08;//The diameter of the bore in m
+L=0.1;//The length of the stroke in m
+N=4500;//The speed of the engine in rpm
+dy=0.55;//The dynamometer readings in m
+w=40;//The weight of the dynamometer scale reading in kg
+c=100;//Fuel consumption in cc
+t=9.5;//Time taken for fuel consumption in s
+CV=44000;//The calorific value of the fuel in kJ/kg
+p=1;//The atmospheric air pressure in bar
+T=300;//The atmospheric air temperature in K
+pi=3.141;//Mathematical constant of pi
+ma=6;//Mass flow rate of air in kg/min
+g=0.7;//Specific gravity of the fuel
+Vc=65;//The clearance volume of each cylinder in cc
+R=287;//Real gas constant in J/kgK
+g=1.4;//Isentropic index
+
+//Calculations
+bp=(2*pi*N*dy*w*9.81)/(60*1000);//The brake power in kW
+bmep=[(bp*1000*60)/(L*(pi/4)*d^2*(N/2)*n)]/10^5;//The brake mean effective pressure in bar
+mf=(c*g*3600)/(t*2*1000);//The mass flow rate of fuel in kg/h
+bsfc=(mf/bp);//Brake specific fuel consumption in kg/kWh
+bsac=(ma*60)/bp;//Brake specific air consumption in kg/kWh
+a=bsac/bsfc;//Air fuel ratio
+nb=((bp*3600)/(mf*CV))*100;//The brake thermal efficiency in percent
+Va=(ma*R*T)/(p*10^5);//The volume flow rate of air at intake condition in m^3/min
+Vs=(pi/4)*d^2*L*(N/2)*n;//The swept volume per minute in m^3/min
+nv=(Va/Vs)*100;//Volumetric efficiency in percent
+Vs1=[(pi/4)*d^2*L]*10^6;//Swept volume per cylinder in cc
+cr=(Vs1+Vc)/Vc;//Compression ratio
+na=[1-(1/cr)^(g-1)]*100;//Air standard efficiency in percent
+re=[(nb)/(na)]*100;//Relative efficiency in percent
+
+//Output
+printf( 'The brake power = %3.1f kW \n The brake mean effective pressure = %3.3f bar \n The brake specific fuel consumption = %3.3f kg/kWh \n The brake specific air consumption = %3.2f kg/kWh \n The air fuel ratio = %3.2f \n The brake thermal efficiency = %3.1f percent \n The volumetric efficiency = %3.1f percent \n The relative efficiency = %3.1f percent ',bp,bmep,bsfc,bsac,a,nb,nv,re)
+
+
diff --git a/2090/CH16/EX16.7/Chapter16_example7.sce b/2090/CH16/EX16.7/Chapter16_example7.sce new file mode 100755 index 000000000..29ccb964c --- /dev/null +++ b/2090/CH16/EX16.7/Chapter16_example7.sce @@ -0,0 +1,44 @@ +clc
+clear
+//Input data
+n=6;//Number of cylinders
+Do=0.03;//Orifice diameter in m
+Cd=0.6;//Coefficient of discharge
+H=0.14;//Pressure drop across the orifice
+d=0.1;//The diameter of the bore in m
+L=0.11;//The length of the stroke in m
+W=540;//Brake load in N
+N=2500;//Engine speed in rpm
+ch=83/17;//C/H ratio by mass
+p=1;//Ambient pressure in bar
+t=18;//Time taken for fuel consumption in s
+f=100;//The amount of fuel consumption in cc
+T=300;//Ambient air temperature in K
+df=780;//The density of the fuel in kg/m^3
+R=287;//Real gas constant in J/kgK
+g=9.81;//Gravitational force constant in m/s^2
+pi=3.141;//Mathematical constant
+dhg=13600;//Density of Hg in kg/m^3
+
+//Calculations
+da=(p*10^5)/(R*T);//The density of air in kg/m^3
+Va=(Cd*(pi/4)*Do^2*[2*g*H*(dhg/da)]^(1/2));//Volume flow rate of air in m^3/s
+Vs=(pi/4)*d^2*L*(N/(2*60))*n;//Swept volume per second in m^3/s
+nv=(Va/Vs)*100;//Volumetric efficiency in percent
+bp=(W*N)/(20000);//The brake power in kW
+bmep=[(bp*1000)/(L*(pi/4)*d^2*(N/(2*60))*n)]/10^5;//The brake mean effective pressure in bar
+T=(60*bp*1000)/(2*pi*N);//Torque in Nm
+mf=(f/18)*(780/1000)*(1/1000)*3600;//Mass flow rate of fuel in kg/h
+bsfc=mf/bp;//The brake specific fuel consumption in kg/kWh
+so=(0.83*(32/12))+(0.17*(8/1));//Stoichiometric oxygen required per kg of fuel in kg/kg fuel
+sa=so/bsfc;//Stoichiometric air required in kg/kg fuel
+maa=Va*da;//Actual mass flow rate of air in kg/s
+af=(maa*3600)/mf;//Actual air fuel ratio
+pea=[(af-sa)/sa]*100;//Percentage of excess air in percent
+
+//Output
+printf('The volumetric efficiency = %3.1f percent \n The brake mean effective pressure = %3.2f bar \n The brake power = %3.1f kW \n The Torque = %3.1f Nm \n The brake specific fuel consumption = %3.3f kg/kWh \n The percentage of excess air = %3.1f percent ',nv,bmep,bp,T,bsfc,pea)
+
+
+
+
diff --git a/2090/CH16/EX16.8/Chapter16_example8.sce b/2090/CH16/EX16.8/Chapter16_example8.sce new file mode 100755 index 000000000..89b541dc7 --- /dev/null +++ b/2090/CH16/EX16.8/Chapter16_example8.sce @@ -0,0 +1,29 @@ +clc
+clear
+//Input data
+d=0.2;//The diameter of bore in m
+L=0.3;//The length of the stroke in m
+r=5.5;//The compression ratio of the engine
+N=400;//The speed of the engine in rpm
+imep=4.5;//The indicative mean effective pressure in bar
+a=6;//Air to gas by volume
+CV=12000;//The calorific value of the gas in kJ/m^3
+T=340;//The temperature at the beginning of the compression stroke in K
+p=0.97;//The pressure at the beginning of the compression stroke in bar
+pi=3.141;//The mathematical constant of pi
+g=1.4;//Adiabatic index
+
+//Calculations
+Vs=(pi/4)*d^2*L;//The swept volume in m^3
+Vc=Vs/(r-1);//The clearance volume in m^3
+V=Vs+Vc;//Total cylinder volume in m^3
+Vg=V/7;//Volume of the gas in total cylinder volume in m^3
+Vntp=((p*Vg)/T)*(273/1.013);//Volume of gas at NTP in m^3
+Q=Vntp*CV*(N/(2*60));//Heat supplied by the fuel in kJ/s
+ip=(imep*10^5*L*(pi/4)*d^2*(N/(2*60))*(1/1000));//Indicated power in kW
+ni=(ip/Q)*100;//Indicated thermal efficincy in percent
+na=[1-(1/r)^(g-1)]*100;//Air standard efficiency in percent
+nr=(ni/na)*100;//Relative efficiency based on indicated thermal efficiency in percent
+
+//Output
+printf('The indicated power = %3.2f kW \n The thermal efficiency = %3.1f percent \n The relative efficiency = %3.1f percent ',ip,ni,nr)
diff --git a/2090/CH16/EX16.9/Chapter16_example9.sce b/2090/CH16/EX16.9/Chapter16_example9.sce new file mode 100755 index 000000000..39c871a64 --- /dev/null +++ b/2090/CH16/EX16.9/Chapter16_example9.sce @@ -0,0 +1,42 @@ +clc
+clear
+//Input data
+n=6;//Number of cylinder
+bp=130;//Brake power in kW
+N=1800;//The speed of the engine in rpm
+CV=42000;//The calorific value of the fuel in kJ/kg
+C=86;//The composition of carbon in the fuel in percent
+H=13;//The composition of Hydrogen in the fuel in percent
+NC=1;//The non combustibles present in the fuel in percent
+na=85;//The absolute volumetric efficiency in percent
+ni=38;//The indicated thermal efficiency in percent
+nm=80;//The mechanical efficiency in percent
+ac=110;//The excess consumption of air in percent
+sb=1.2;//The stroke to the bore ratio
+da=1.3;//The density of air in kg/m^3
+pi=3.141;//Mathematical constant of pi
+
+//Calculations
+saf=(((C/100)*(32/12))+((H/100)*(8/1)))*(1/0.23);//The stoichiometric air fuel ratio
+aaf=saf*(1+1.1);//The actual air fuel ratio
+Ma=(0.23*32)+(0.77*28);//The molecular weight of air in kg/kmol
+a=(C/100)/12;//For carbon balance
+b=(H/100)/2;//For hydrogen balace
+x=aaf/Ma;//Number of kmol of air per kg of fuel
+c=(0.21*x)-a-(b/2);//For oxygen balance
+d1=0.79*x;//For nitrogen balance
+ip=bp/(nm/100);//The indicated power in kW
+mf=ip/[(ni/100)*CV];//The mass flow rate of fuel in kg/s
+ma=mf*aaf;//The mass flow rate of air in kg/s
+Va=ma/da;//Actual volume flow rate in m^3/s
+Vs=Va/(na/100);//The swept volume per second in m^3/s
+d=[[Vs*(4/pi)*(1/1.2)*((2*60)/N)*(1/n)]^(1/3)]*1000;//The diameter of the bore in mm
+L=1.2*d;//The length of the stroke in mm
+T=a+c+d1;//The total composition in kmol
+CO2=(a/T)*100;//The volume of CO2 in %
+O2=(c/T)*100;//The volume of O2 in %
+N2=(d1/T)*100;//The volume of N2 in %
+
+//Output
+printf(' The volumetric composition of dry exhaust gas : \n 1) CO2 = %3.5f kmol and volume = %3.2f percent \n 2) O2 = %3.5f kmol and volume = %3.2f percent \n 3) N2 = %3.5f kmol and volume = %3.2f percent \n The bore of the engine = %3.0f mm \n The stroke of the engine = %3.1f mm ',a,CO2,c,O2,d1,N2,d,L)
+
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