diff options
Diffstat (limited to '2657/CH12')
-rwxr-xr-x | 2657/CH12/EX12.1/Ex12_1.sce | 15 | ||||
-rwxr-xr-x | 2657/CH12/EX12.2/Ex12_2.sce | 24 | ||||
-rwxr-xr-x | 2657/CH12/EX12.3/Ex12_3.sce | 22 | ||||
-rwxr-xr-x | 2657/CH12/EX12.4/Ex12_4.sce | 16 | ||||
-rwxr-xr-x | 2657/CH12/EX12.5/Ex12_5.sce | 21 |
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) |