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authorpriyanka2015-06-24 15:03:17 +0530
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-rwxr-xr-x2168/CH15/EX15.1/Chapter15_example1.sce13
-rwxr-xr-x2168/CH15/EX15.10/Chapter15_example10.sce30
-rwxr-xr-x2168/CH15/EX15.2/Chapter15_example2.sce16
-rwxr-xr-x2168/CH15/EX15.3/Chapter15_example3.sce17
-rwxr-xr-x2168/CH15/EX15.4/Chapter15_example4.sce26
-rwxr-xr-x2168/CH15/EX15.5/Chapter15_example5.sce13
-rwxr-xr-x2168/CH15/EX15.6/Chapter15_example6.sce20
-rwxr-xr-x2168/CH15/EX15.7/Chapter15_example7.sce20
-rwxr-xr-x2168/CH15/EX15.8/Chapter15_example8.sce22
-rwxr-xr-x2168/CH15/EX15.9/Chapter15_example9.sce26
10 files changed, 203 insertions, 0 deletions
diff --git a/2168/CH15/EX15.1/Chapter15_example1.sce b/2168/CH15/EX15.1/Chapter15_example1.sce
new file mode 100755
index 000000000..f3d20efc8
--- /dev/null
+++ b/2168/CH15/EX15.1/Chapter15_example1.sce
@@ -0,0 +1,13 @@
+clc
+clear
+//Input data
+n=6//Number of cylinders
+p=720//Horse power in h.p
+N=180//Speed in r.p.m
+f=250//Fuel rate in gm per horse power hour
+
+//Calculations
+w=(((f/1000)*p)/((N/2)*60*n))*1000//Weight of fuel per cycle in gm/cycle
+
+//Outptut
+printf('The quantity of fuel to be injected per cylinder is %3.2f gm/cycle',w)
diff --git a/2168/CH15/EX15.10/Chapter15_example10.sce b/2168/CH15/EX15.10/Chapter15_example10.sce
new file mode 100755
index 000000000..7f5c74119
--- /dev/null
+++ b/2168/CH15/EX15.10/Chapter15_example10.sce
@@ -0,0 +1,30 @@
+clc
+clear
+//Input data
+n=6//Number of cylinders
+d=11.5//Bore in cm
+l=14//Stroke in cm
+af=16//Air fuel ratio
+pa=1.03//Pressure of air intake in kg/cm^2
+Ta=24+273//Temperature of air intake in K
+nv=76.5//Volumetric efficiency in percent
+R=29.27//Characteristic gas constant in kg.m/kg.K
+N=1500//Speed in r.p.m
+ip=125//Injection pressure in kg/cm^2
+cp=40//Compression pressure in kg/cm^2
+q=18.5//Fuel injection occupies 18.5 degrees of crenk travel
+fsw=760//Fuel specific weight in kg/m^2
+dc=0.94//Orifice discharge coefficient
+
+//Calculations
+Vs=((3.14/4)*d^2*l)//Stroke volume in c.c
+Va=(Vs*(nv/100))//Volume of air supplied in c.c
+wa=((pa*10^4*Va*10^-6)/(R*Ta))//Weight of air supplied per cylinder per cycle in kg
+wf=(wa/af)//Weight of fuel injected per cylinder per cycle in kg
+I=((60*q)/(N*360))//Injection time per cycle in sec
+F=(wf/I)//Fuel injected per cylinder per sec in kg/sec
+Af=(F/(dc*sqrt(2*9.81*fsw*(ip-cp)*10^4)))//Area of orifice in sq.m
+df=sqrt(Af/(3.14/4))*1000//Diameter of orifice in mm
+
+//Output
+printf('Maximum amount of fuel injected per cylinder per sec is %3.2f kg/sec \n Diameter of orifice is %3.3f mm',F,df)
diff --git a/2168/CH15/EX15.2/Chapter15_example2.sce b/2168/CH15/EX15.2/Chapter15_example2.sce
new file mode 100755
index 000000000..5ca12c5b1
--- /dev/null
+++ b/2168/CH15/EX15.2/Chapter15_example2.sce
@@ -0,0 +1,16 @@
+clc
+clear
+//Input data
+n=4//Number of cylinders
+fc=0.215//Brake specific fuel consumption in kg/B.H.P hour
+BHP=400//Brake horse power in B.H.P
+N=250//Speed in r.p.m
+sg=0.9//Specific gravity
+
+//Calculations
+Fc=(fc*BHP)//Fuel consumption per hour in kg/hr
+Fcy=(Fc/n)//Fuel consumption per cylinder in kg/hr
+Fcyc=((Fcy/(60*(N/2)))/(sg*1000))*10^6//Fuel consumption per cycle in kg. In textbook it is given wrong as 0.0287 instead of 3.185
+
+//Output
+printf('The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c',Fcyc)
diff --git a/2168/CH15/EX15.3/Chapter15_example3.sce b/2168/CH15/EX15.3/Chapter15_example3.sce
new file mode 100755
index 000000000..e22c7a16b
--- /dev/null
+++ b/2168/CH15/EX15.3/Chapter15_example3.sce
@@ -0,0 +1,17 @@
+clc
+clear
+//Input data
+n=4//Number of cylinders
+p=450//Brake Horse power in B.H.P
+N=200//Speed in r.p.m
+f=0.2//Fuel rate in kg per horse power hour
+g=0.9//Specific gravity of fuel
+
+//Output
+Fc=(p*f)//Fuel consumption per hour in kg/hr
+Fcy=(Fc/n)//Fuel consumption per cylinder in kg/hr
+Fcyc=(Fcy/(60*(N/2)))//Fuel consumption per cycle in kg
+q=(Fcyc/(g*1000))*10^6//Quantity of fuel injected per cylinder per cycle in c.c
+
+//Output
+printf('The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c',q)
diff --git a/2168/CH15/EX15.4/Chapter15_example4.sce b/2168/CH15/EX15.4/Chapter15_example4.sce
new file mode 100755
index 000000000..e628369ed
--- /dev/null
+++ b/2168/CH15/EX15.4/Chapter15_example4.sce
@@ -0,0 +1,26 @@
+clc
+clear
+//Input data
+//Data from problem 1
+n=6//Number of cylinders
+p=720//Horse power in h.p
+N=180//Speed in r.p.m
+f=250//Fuel rate in gm per horse power hour
+
+Vo=20//Volume of oil in the suction chamber in c.c
+dp=80//Discharge pressure in kg/cm^2
+voi=6//Volume of oil in the injector in c.c
+g=0.9//Specific gravity of oil
+b=78.8*10^-6//Coefficient of compressibility in cm^2/kg when pressure is taken as atmospheric
+
+//Calculations
+w=(((f/1000)*p)/((N/2)*60*n))*1000//Weight of fuel per cycle in gm/cycle
+Va=(w/g)//Volume of air per cycle in c.c
+V1=(Vo+Va)//Initial volume in c.c
+dV12=(b*V1*dp)//Change in volume in c.c
+//Assuming in accordance with average practice that s=2d, nv=0.94 and full load in this pump type x=0.5
+d=((voi+dV12)/((3.14/4)*2*0.94*0.5))^(1/3)//Diameter in cm
+l=(2*d)//Stroke in cm
+
+//Output
+printf('The diameter of the pump is %3.2f cm \n The total stroke is %3.2f cm',d,l)
diff --git a/2168/CH15/EX15.5/Chapter15_example5.sce b/2168/CH15/EX15.5/Chapter15_example5.sce
new file mode 100755
index 000000000..d09aaf328
--- /dev/null
+++ b/2168/CH15/EX15.5/Chapter15_example5.sce
@@ -0,0 +1,13 @@
+clc
+clear
+//Input data
+p=110//Oil pressure in kg/cm^2
+pc=25//Pressure in the combustion chamber in kg/cm^2
+q=0.805//Velocity coefficient. In textbook it is given wrong as 9.805
+d=0.906//Specific gravity
+
+//Calculations
+v=(37.1*q*sqrt((p-pc)/d))//Velocity in m/s
+
+//Output
+printf('The velocity of injection is %3.0f m/s',v)
diff --git a/2168/CH15/EX15.6/Chapter15_example6.sce b/2168/CH15/EX15.6/Chapter15_example6.sce
new file mode 100755
index 000000000..0b1aa0ec0
--- /dev/null
+++ b/2168/CH15/EX15.6/Chapter15_example6.sce
@@ -0,0 +1,20 @@
+clc
+clear
+//Input data
+Vf=6.2//Volume of fuel in c.c
+l=65//Length of fuel line in cm
+di=2.5//Inner diameter in mm
+V=2.75//Volume of fuel in the injector valve in c.c
+Vd=0.15//Volume of fuel to be delivered in c.c. In textbook it is given wrong as 0.047
+p=140//Pressure in kg/cm^2
+pp=1//Pump pressure in kg/cm^2
+patm=1.03//Atmospheric pressure in kg/cm^2
+b=78.8*10^-6//Coefficient of compressibility in cm^2/kg when pressure is taken as atmospheric
+
+//Calculations
+V1=(Vf+(3.14/4)*(di/10)^2*l+V)//Initial volume in c.c
+dV=((b*V1*(p-pp)/patm))//Change in volume in c.c
+d=(dV+Vd)//Total displacement of the plunger in c.c
+
+//Output
+printf('The total displacement of the plunger is %3.3f c.c',d)
diff --git a/2168/CH15/EX15.7/Chapter15_example7.sce b/2168/CH15/EX15.7/Chapter15_example7.sce
new file mode 100755
index 000000000..e1bf64bc6
--- /dev/null
+++ b/2168/CH15/EX15.7/Chapter15_example7.sce
@@ -0,0 +1,20 @@
+clc
+clear
+//Input data
+Vf=6.75//Volume of fuel in c.c
+l=65//Length of fuel line in cm
+di=2.5//Inner diameter in mm
+V=2.45//Volume of fuel in the injector valve in c.c
+Vd=0.15//Volume of fuel to be delivered in c.c.
+p=150//Pressure in kg/cm^2
+pp=1//Pump pressure in kg/cm^2
+patm=1.03//Atmospheric pressure in kg/cm^2
+b=78.8*10^-6//Coefficient of compressibility in cm^2/kg when pressure is taken as atmospheric
+
+//Calculations
+V1=(Vf+(3.14/4)*(di/10)^2*l+V)//Initial volume in c.c
+dV=((b*V1*(p-pp)/patm))//Change in volume in c.c
+d=(dV+Vd)//Total displacement of the plunger in c.c
+
+//Output
+printf('The total displacement of the plunger is %3.3f c.c',d)
diff --git a/2168/CH15/EX15.8/Chapter15_example8.sce b/2168/CH15/EX15.8/Chapter15_example8.sce
new file mode 100755
index 000000000..5823419be
--- /dev/null
+++ b/2168/CH15/EX15.8/Chapter15_example8.sce
@@ -0,0 +1,22 @@
+clc
+clear
+//Input data
+Vf=6.75//Volume of fuel in c.c
+l=65//Length of fuel line in cm
+di=2.5//Inner diameter in mm
+V=2.45//Volume of fuel in the injector valve in c.c
+Vd=0.15//Volume of fuel to be delivered in c.c.
+p=150//Pressure in kg/cm^2
+pp=1//Pump pressure in kg/cm^2
+patm=1.03//Atmospheric pressure in kg/cm^2
+b=78.8*10^-6//Coefficient of compressibility in cm^2/kg when pressure is taken as atmospheric
+dp=0.75//Diameter of the plunger in cm
+
+//Calculations
+V1=(Vf+(3.14/4)*(di/10)^2*l+V)//Initial volume in c.c
+dV=((b*V1*(p-pp)/patm))//Change in volume in c.c
+d=(dV+Vd)//Total displacement of the plunger in c.c
+s=((4/3.14)*(d/dp^2))*10//Stroke in mm
+
+//Output
+printf('The effective plunger stroke is %3.1f mm',s)
diff --git a/2168/CH15/EX15.9/Chapter15_example9.sce b/2168/CH15/EX15.9/Chapter15_example9.sce
new file mode 100755
index 000000000..469004568
--- /dev/null
+++ b/2168/CH15/EX15.9/Chapter15_example9.sce
@@ -0,0 +1,26 @@
+clc
+clear
+//Input data
+n=6//Number of cylinders
+p=300//Horse power in H.P
+N=1200//Speed in r.p.m
+f=0.2//Fuel rate in kg per B.H.P hour
+ip=200//Injection pressure in kg/cm^2
+cp=40//Pressure in the combustion chamber in kg/cm^2
+pic=33//Period of injection of the crank angle in degrees
+g=0.83//Specific gravity of fuel. In textbook, it is given wrong as 0.89
+Cd=0.9//Coefficient of discharge
+
+//Output
+Fc=(p*f)//Fuel consumption per hour in kg/hr
+Fcy=(Fc/n)//Fuel consumption per cylinder in kg/hr
+Fcyc=(Fcy/(60*(N/2)))//Fuel consumption per cycle in kg
+q=(Fcyc/(g*1000))*10^6//Quantity of fuel injected per cylinder per cycle in c.c
+I=((pic/360)*(1/N)*60)//Injection period in sec
+df=(g/1000)//Density of fuel in kg/m^3
+v=sqrt(2*981*((ip-cp)/df))//Velocity of fuel through orifice in m/s
+A=(q/(Cd*v*I))//Area of orifice in cm^2
+d=sqrt(A/(3.14/4))*10//Diameter in mm
+
+//Output
+printf('The diameter of the single orifice injector is %3.2f mm',d)