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-rwxr-xr-x2657/CH12/EX12.1/Ex12_1.sce15
-rwxr-xr-x2657/CH12/EX12.2/Ex12_2.sce24
-rwxr-xr-x2657/CH12/EX12.3/Ex12_3.sce22
-rwxr-xr-x2657/CH12/EX12.4/Ex12_4.sce16
-rwxr-xr-x2657/CH12/EX12.5/Ex12_5.sce21
5 files changed, 98 insertions, 0 deletions
diff --git a/2657/CH12/EX12.1/Ex12_1.sce b/2657/CH12/EX12.1/Ex12_1.sce
new file mode 100755
index 000000000..884b496d1
--- /dev/null
+++ b/2657/CH12/EX12.1/Ex12_1.sce
@@ -0,0 +1,15 @@
+//Calculation of quantity of fuel injected
+clc,clear
+//Given:
+n=6 //Number of cylinders
+bsfc=245 //Brake specific fuel consumption in gm/kWh
+bp=89 //Brake power in kW
+N=2500 //Engine speed in rpm
+s=0.84 //Specific gravity of the fuel
+//Solution:
+m_f=bsfc*bp/(1000) //Fuel consumption in kg/hr
+m_f=m_f/n //Fuel consumption per cylinder in kg/hr
+m_f=(m_f/3600)/(N/(2*60)) //Fuel consumption per cycle in kg
+V_f=m_f*1000/s //Volume of fuel consume per cycle in cc
+//Results:
+printf("\n The quantity of fuel to be injected per cycle per cylinder, V_f = %.4f cc",V_f)
diff --git a/2657/CH12/EX12.2/Ex12_2.sce b/2657/CH12/EX12.2/Ex12_2.sce
new file mode 100755
index 000000000..601291496
--- /dev/null
+++ b/2657/CH12/EX12.2/Ex12_2.sce
@@ -0,0 +1,24 @@
+//Calculation of orifice area
+clc,clear
+//Given:
+n=8 //Number of cylinders
+bp=368 //Brake power in kW
+N=800 //Engine speed in rpm
+bsfc=0.238 //Brake specific fuel consumption in kg/kWh
+P1=35,P2=60 //Beginning pressure and maximum pressure in cylinder in bar
+P1_i=210,P2_i=600 //Expected pressure and maximum pressure at injector in bar
+theta_i=12 //Period of injection in degrees
+Cd=0.6 //Coefficient of discharge for the injector
+s=0.85 //Specific gravity of the fuel
+P_atm=1.013 //Atmospheric pressure in bar
+//Solution:
+m_f=bsfc*bp/(n*60) //Fuel consumption per cylinder in kg/min
+m_f=m_f/(N/2) //Fuel consumption per cycle in kg
+t=theta_i/360*60/N //Time for injection in s
+m_f=m_f/t //Fuel consumption per cycle in kg/s
+deltaP1=P1_i-P1 //Pressure difference at beginning in bar
+deltaP2=P2_i-P2 //Pressure difference at end in bar
+deltaP_av=(deltaP1+deltaP2)/2 //Average pressure difference in bar
+A_f=m_f/(Cd*sqrt(2*s*1000*deltaP_av*10^5)) //Orifice area of fuel injector in m^2
+//Results:
+printf("\n The Orifice area of fuel injector, Af = %.5f cm^2",A_f*10000)
diff --git a/2657/CH12/EX12.3/Ex12_3.sce b/2657/CH12/EX12.3/Ex12_3.sce
new file mode 100755
index 000000000..abafaf56e
--- /dev/null
+++ b/2657/CH12/EX12.3/Ex12_3.sce
@@ -0,0 +1,22 @@
+//Calculation of orifice diameter
+clc,clear
+//Given:
+bp=15 //Brake power per cylinder in kW
+N=2000 //Engine speed in rpm
+bsfc=0.272 //Brake specific fuel consumption in kg/kWh
+API=32 //American Petroleum Institute specific gravity in degreeAPI
+theta_i=30 //Period of injection in degrees
+P_i=120 //Fuel injection pressure in bar
+P_c=30 //Combustion chamber pressure in bar
+Cd=0.9 //Coefficient of discharge for the injector
+function rho=specificgravity(API),rho=141.5/(131.5+API),endfunction //Specific gravity(rho) as a function of API
+//Solution:
+s=specificgravity(API) //Specific gravity of fuel
+m_f=bsfc*bp*2/(N*60) //Fuel consumption in kg
+t=theta_i/360*60/N //Time for injection in s
+m_f=m_f/t //Fuel consumption per cycle in kg/s
+A_f=m_f/(Cd*sqrt(2*s*1000*(P_i-P_c)*10^5)) //Orifice area of fuel injector in m^2
+A_f=A_f*10^6 //Orifice area of fuel injector in mm^2
+d_f=sqrt(4*A_f/%pi) //Diameter of fuel orifice in mm
+//Results:
+printf("\n The diameter of the fuel orifice, d = %.2f mm\n\n",d_f)
diff --git a/2657/CH12/EX12.4/Ex12_4.sce b/2657/CH12/EX12.4/Ex12_4.sce
new file mode 100755
index 000000000..4905cc7b8
--- /dev/null
+++ b/2657/CH12/EX12.4/Ex12_4.sce
@@ -0,0 +1,16 @@
+//Calculations on spray penetration
+clc,clear
+//Given:
+s1=20 //Distance of penetration in cm
+t1=16 //Penetration time in millisec
+P_i1=140 //Injection pressure in bar
+s2=s1 //Same distance of penetration in cm
+P_i2=220 //Injection pressure in bar
+P_c=15 //Combustion chamber pressure in bar
+//Solution:
+deltaP1=P_i1-P_c //Pressure difference for 140 bar injection pressure
+deltaP2=P_i2-P_c //Pressure difference for 220 bar injection pressure
+t2=t1*(s2/s1)*sqrt(deltaP1/deltaP2) //Penetration time for 220 bar injection pressure in millisec
+//Results:
+printf("\n Penetration time for 220 bar injection pressure, t2 = %.1f milli-seconds\n\n",t2)
+//Answer in the book is wrong
diff --git a/2657/CH12/EX12.5/Ex12_5.sce b/2657/CH12/EX12.5/Ex12_5.sce
new file mode 100755
index 000000000..ac4071b6b
--- /dev/null
+++ b/2657/CH12/EX12.5/Ex12_5.sce
@@ -0,0 +1,21 @@
+//Calculations on diesel engine fuel pump
+clc,clear
+//Given:
+V_b=7 //Volume of fuel in the barrel in cc
+D_l=3,L_l=700 //Diameter and length of fuel delivery line in mm
+V_iv=3 //Volume of fuel in the injection valve in cc
+P2=200 //Delivery pressure in bar
+P1=1 //Sump pressure in bar
+V_d=0.15 //Volume to be delivered in cc
+C=78.8D-6 //Coefficient of compressibility
+d=8 //Diameter of the plunger in mm
+//Solution:
+V_l=%pi/4*D_l^2*L_l*10^-3 //Volume of fuel in delivery line in cc
+V1=V_b+V_l+V_iv //Total initial fuel volume in cc
+deltaV=C*(P2-P1)*V1 //Change in volume due to compression in cc
+V_p=deltaV+V_d //Displaced volume by plunger in cc
+A_p=%pi/4*d^2*10^-2 //Area of the plunger in cm^2
+l=V_p/A_p //Effective stroke of plunger in cm
+//Results:
+printf("\n The plunger displacement = %.3f cc",V_p)
+printf("\n The effective stroke of the plunger, l = %.2f mm\n\n",l*10)