diff options
| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Fundamental_of_internal_combustion_engines_by_H_N_Gupta | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
| download | Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.tar.gz Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.tar.bz2 Python-Textbook-Companions-6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d.zip | |
Removed duplicates
Diffstat (limited to 'Fundamental_of_internal_combustion_engines_by_H_N_Gupta')
13 files changed, 5181 insertions, 0 deletions
diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap13.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap13.ipynb new file mode 100755 index 00000000..d404f5f6 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap13.ipynb @@ -0,0 +1,124 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13: Engine friction and lubrication"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.1 Page No 423"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.08\t\t #The diameter of bore in m\n",
+ "L=0.075\t\t #The length of the stroke in m\n",
+ "l=0.152\t\t #The connecting rod length in m\n",
+ "h=0.062\t\t #Skirt length of the piston in m\n",
+ "Fr=8000\t\t #Compressive force in the connecting rod in N\n",
+ "p=3000\t\t #The pressure in the cylinder kPa\n",
+ "y=0.004*10**-3\t #The clearence between piston and cylinder wall in m\n",
+ "U=0.006\t\t #The dynamic viscosity of the lubricating oil in pa.s\n",
+ "u=8.2\t\t #The piston speed in m/s\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "ts=(U*u)/y\t #The shear stress in N/m**2\n",
+ "A=pi*d*h\t #Contact area between the piston and the cylinder in m**2\n",
+ "Ff=ts*A\t\t #Friction force on the piston inN\n",
+ "r=L/2.0\t\t #Crank length in m\n",
+ "A=math.atan(r/l)\t #The angle made by the crank in radians\n",
+ "Ft=Fr*sin(A)\t#The side thrust in N\n",
+ "\n",
+ "#Output \n",
+ "print\"The friction force on the piston \",round(Ff,0),\"N\"\n",
+ "print\"The thrust force on the cylinder wall is\",round(Ft,0),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The friction force on the piston 192.0 N\n",
+ "The thrust force on the cylinder wall is 1916.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13.2 Page No 424"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.065\t\t#The cylinder bore diameter in m\n",
+ "L=6.0\t\t#The length of the stroke in cm\n",
+ "l=12.0\t\t#The length of the connecting rod in cm\n",
+ "p=50.0\t\t#The pressure in the cylinder in bar\n",
+ "q=90.0\t\t#The crank position in power stroke of the cycle for one cylinder in degrees\n",
+ "Ff=900.0\t\t#Friction force in N\n",
+ "o=0.2\t\t#Wrist pin off set in cm\n",
+ "\n",
+ "#Calculations\n",
+ "r=L/2.0\t\t#The crank length in cm\n",
+ "sineA=r/l\t \n",
+ "cosA=(1-(sineA)**2)**(1/2.0)\t\n",
+ "Fr=(((p*10**5*(pi/4.0)*d**2)-Ff)/cosA)/1000.0 \n",
+ "Ft=Fr*sineA\t #The side thrust on the piston in kN\n",
+ "sineA1=(r-o)/l\t#The value of sine\n",
+ "cosA1=(1-(sineA1)**2)**(1/2.0)\t\t\t\n",
+ "Fr1=(((p*10**5*(pi/4.0)*d**2)-Ff)/cosA1)/1000.0\t\n",
+ "Ft1=Fr1*sineA1 #The side thrust in kN\n",
+ "\n",
+ "#Output \n",
+ "print\"(a) The force in the connecting rod \",round(Fr,3),\" kN\"\n",
+ "print\"The side thrust on the piston =\",round(Ft,2),\"kN\"\n",
+ "print\"(b) The side thrust on the piston =\",round(Ft1,3),\" kN\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The force in the connecting rod 16.206 kN\n",
+ "The side thrust on the piston = 4.05 kN\n",
+ "(b) The side thrust on the piston = 3.765 kN\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap2.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap2.ipynb new file mode 100755 index 00000000..5baf7b4f --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/Chap2.ipynb @@ -0,0 +1,542 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 Test"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 Page No: 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=20.0 #Cylinder bore diameter in cm\n",
+ "L=25.0 #Stroke length in cm\n",
+ "Vc=1570.0 #The clearance volume in cm**3\n",
+ "P1=1.0 #Pressure at the beginning of the compression in bar\n",
+ "T1=300.0 #Temperature at the beginning of the compression in K\n",
+ "T3=1673 #The maximum temperature of the cycle in K\n",
+ "Cv=0.718 #specific heat at constant volume for air in kJ/kgK\n",
+ "R=0.287 #Real gas constant in kJ/kgK\n",
+ "g=1.4 #Isentropic index\n",
+ "c=500.0 #Number of cycles per minute\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(math.pi/4.0)*d**2*L\n",
+ "V1=Vs+Vc\n",
+ "V2=Vc #cm**3\n",
+ "r=V1/V2 #Compression ratio\n",
+ "T2=T1*r**(g-1)\n",
+ "P2=P1*r**g\n",
+ "P3=P2*(T3/T2) #In constant volume, process Pressure at point 3 in bar\n",
+ "T4=T3*(1/r)**(g-1) #In isentropic process, Temperature at point 4 in degree centigrade\n",
+ "P4=P3*(1/r)**(g) #In isentropic process, Pressure at point 4 in bar\n",
+ "no=(1-(1/r)**(g-1))*100 #Air standard efficiency of otto cycle\n",
+ "Q1=Cv*(T3-T2) #Heat supplied in kJ/kg\n",
+ "Q2=Cv*(T4-T1) #Heat rejected in kJ/kg\n",
+ "W=Q1-Q2 #Work done per unit mass in kJ/kg\n",
+ "m=((P1*10**5*V1*10**-6)/(R*T1))/1000.0 #The amount of mass in kg\n",
+ "W1=W*m #Work done in kJ\n",
+ "pm=((W1*10**3)/(Vs*10.0**-6))/10.0**5 #Mean effective pressure in N/m**2\n",
+ "P=W1*(c/60.0) #Power developed in kW\n",
+ "\n",
+ "#Output\n",
+ "print\"Temperature at point 2 = \",round(T2-273.15,1),\"C\"\n",
+ "print\"Pressure at point 2 =\" ,round(P2,2),\" bar\"\n",
+ "print\"Pressure at point 3 = \",round(P3,2),\"bar\" \n",
+ "print\"Temperature at point 4 = \",round(T4-273.15),\" C \\nPressure at point 4 = \",round(P4,3),\"bar\"\n",
+ "print\"Air standard efficiency of otto cycle = \",round(no,2),\" percent \"\n",
+ "print\"Work done = \",round(W1,2),\" kJ\"\n",
+ "print\"Mean effective pressure = \",round(pm,2),\"bar \"\n",
+ "print\"Power developed = \",round(P,1),\"kW \"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature at point 2 = 341.3 C\n",
+ "Pressure at point 2 = 12.29 bar\n",
+ "Pressure at point 3 = 33.47 bar\n",
+ "Temperature at point 4 = 544.0 C \n",
+ "Pressure at point 4 = 2.723 bar\n",
+ "Air standard efficiency of otto cycle = 51.17 percent \n",
+ "Work done = 4.26 kJ\n",
+ "Mean effective pressure = 5.42 bar \n",
+ "Power developed = 35.5 kW \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page No: 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "CV=42000.0 #The calorific value of the fuel in kJ/kg\n",
+ "pa=5.0 #Percentage of compression\n",
+ "Pa=1.2 #Pressure in the cylinder at 5% compression stroke\n",
+ "pb=75 #Percentage of compression\n",
+ "Pb=4.8 #Pressure in the cylinder at 75% compression stroke\n",
+ "g=1.3 #polytropic index\n",
+ "g1=1.4 #Isentropic index\n",
+ "n=0.6 #Air standard efficiency\n",
+ "\n",
+ "#Calculations\n",
+ "V=(Pb/Pa)**(1/1.3)#Ratio of volumes\n",
+ "r=(V*(pb/100.0)-(pa/100.0))/((1-(pa/100.0))-(V*(1-(pb/100.0)))) #Compression ratio\n",
+ "n1=((1-(1/r)**(g1-1)))*100 #Relative efficiency\n",
+ "nthj=n*(n1/100.0) #Indicated thermal efficiency\n",
+ "x=(1/(CV*nthj))*3600 #Specific fuel consumption in kg/kW.h\n",
+ "\n",
+ "#Output\n",
+ "print\"The compression ratio of the engine is \",round(r,1)\n",
+ "print\"The specific fuel consumption is \",round(x,3),\"kg/kwh\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compression ratio of the engine is 9.5\n",
+ "The specific fuel consumption is 0.241 kg/kwh\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page No: 48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.2 #The diameter of the cylinder bore in m\n",
+ "L=0.3 #The length of the stroke in m\n",
+ "P1=1 #The pressure at the beginning of the compression in bar\n",
+ "T1=300.0 #The temperature at the beginning of the compression in K\n",
+ "r=16.0 #Compression ratio\n",
+ "V=0.08 #Cutt off takes place at 8& of the stroke\n",
+ "R=0.287 #Real gas constant in kJ/kgK\n",
+ "g=1.4 #Isentropic index\n",
+ "Cp=1.005 #Specific heat at constant prassure in kJ/kgK\n",
+ "Cv=0.718 #specific heat at constant volume for air in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(math.pi/4.0)*d**2*L #Swept volume in m**3\n",
+ "Vc=Vs/(r-1) #Clearance volume in m**3\n",
+ "V2=Vc #Volume at point 2 in m**3\n",
+ "V1=Vs+Vc #Volume at point 1 in m**3\n",
+ "m=(P1*10**5*V1)/(R*T1) #The amount of mass in kg\n",
+ "P2=P1*(r**g) #Pressure at point 2 in bar\n",
+ "P3=P2 #Pressure at point 3 in bar\n",
+ "T2=T1*r**(g-1) #Temperature at point 2 in K\n",
+ "V3=(V*Vs)+V2 #Volume at point 3 in m**3\n",
+ "C=V3/V2 #Cut off ratio\n",
+ "T3=C*T2 #Temperature at point 3 in K\n",
+ "P4=P3*(C/r)**g #Pressure at the point 4 in bar\n",
+ "T4=T3*(C/r)**(g-1) #Temperature at point 4 in K\n",
+ "V4=V1 #Volume at point 4 in m**3\n",
+ "Q1=(m*Cp*(T3-T2))/1000.0 #Heat supplied in kJ\n",
+ "Q2=(m*Cv*(T4-T1))/1000.0 #Heat rejected in kJ\n",
+ "W=(Q1-Q2) #Work done per cycle in kJ\n",
+ "na=(W/Q1)*100 #Air standard efficiency\n",
+ "Pm=(W*1000/Vs)/10.0**5 #Mean effective pressure in bar\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) Volume at point 2 = \",round(V2,5),\" m**3 \\nVolume at point 1 = \",round(V1,5),\"m**3 \"\n",
+ "print\"Pressure at point 2 = \",round(P2,1),\" bar\" \n",
+ "print\"Temperature at point 2 = \",round(T2,1),\"K\"\n",
+ "print\"beeta is\",C,\"\\nTemperature at point 3 = \",round(T3,0),\" K \\nPressure at point 4 =\",round(P4,3),\" bar\"\n",
+ "print\"Temperature at point 4 = \",round(T4,1),\" K \\nVolume at point 4 = \",round(V4,5),\"m**3\" \n",
+ "print\"(b) cut off ratio = \",round(c,1)\n",
+ "print\"(c) Work done per cycle = \",round(W,3),\"kJ\" \n",
+ "print\"(d) air smath.tandard efficiency = \",round(na,1),\" percent\" \n",
+ "print\"(e)Mean effective pressure = \",round(Pm,2),\" bar \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Volume at point 2 = 0.00063 m**3 \n",
+ "Volume at point 1 = 0.01005 m**3 \n",
+ "Pressure at point 2 = 48.5 bar\n",
+ "Temperature at point 2 = 909.4 K\n",
+ "beeta is 2.2 \n",
+ "Temperature at point 3 = 2001.0 K \n",
+ "Pressure at point 4 = 3.016 bar\n",
+ "Temperature at point 4 = 904.7 K \n",
+ "Volume at point 4 = 0.01005 m**3\n",
+ "(b) cut off ratio = 500.0\n",
+ "(c) Work done per cycle = 7.736 kJ\n",
+ "(d) air smath.tandard efficiency = 60.4 percent\n",
+ "(e)Mean effective pressure = 8.21 bar \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page No:50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "P1=7\n",
+ "g=1.4\n",
+ "r=12\n",
+ "import numpy as np\n",
+ "from scipy.optimize import fsolve\n",
+ "\n",
+ "def f(b):\n",
+ " return P1-1/((g-1)*(r-1))*(g*r**g*(b-1)-r*(b**1.4-1))\n",
+ "b = fsolve(f, 1)\n",
+ "f(b)\n",
+ "na=(1-(1/(r**(g-1)))*(((b**g)-1)/(g*(b-1))))*100 #Air standard efficiency\n",
+ "\n",
+ "#Output \n",
+ "print\"The cut off ratio = \",round(b,1),\" \\n The air standard efficiency =\",na,\"percent\"\n",
+ "#NOTE:In the book Answer is wrong for Air standard efficiency .\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cut off ratio = 2.2 \n",
+ " The air standard efficiency = [ 55.47110058] percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no 51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "m=30.0 #The air fuel ratio by mass\n",
+ "T1=300 #The temperature of air at the beginning of the compression in K\n",
+ "r=16 #The compression ratio\n",
+ "CV=42000 #The calorific value of the fuel in kJ/kg\n",
+ "g=1.4 #Isentropic index\n",
+ "Cp=1.005 #Specific heat at constant prassure in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "T2=T1*(r**(g-1)) #Temperature at point 2 in K\n",
+ "T3=((1/m)*(CV/Cp))+T2 #Temperature at point 3 in K\n",
+ "C=T3/T2 #The cut off ratio\n",
+ "n=(1-((1/r**(g-1))*(((C**g)-1)/(g*(C-1)))))*100#The ideal efficiency of the engine based on the air standard cycle\n",
+ "\n",
+ "#Output\n",
+ "print\" The ideal efficiency of the engine based on the air standard cycle = \",round(n,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The ideal efficiency of the engine based on the air standard cycle = 58.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page No: 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "p1=1.0 #Inlet pressure in bar\n",
+ "p2=32.425 #Pressure at the end of isentropic compression in bar\n",
+ "r=6.0 #Ratio of expansion\n",
+ "r1=1.4 #Isentropic index\n",
+ "\n",
+ "#Calculations\n",
+ "rc=(p2/p1)**(1/r1) #Compression ratio\n",
+ "b=(rc/r) #cut-off ratio\n",
+ "n=(1-((b**r1-1)/(rc**(r1-1)*r1*(b-1))))*100 \n",
+ "pm=((p1*rc**r1*n/100.0*r1*(b-1))/((r1-1)*(rc-1))) \n",
+ "\n",
+ "#Output\n",
+ "print\"Air-smath efficiency is \",round(n,3),\"percent\"\n",
+ "print\"Mean effective pressure is \",round(pm,3),\"bar\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Air-smath efficiency is 56.671 percent\n",
+ "Mean effective pressure is 5.847 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8 Page No: 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "rc=15.0 #Compression ratio\n",
+ "p1=1 #Pressure at which compression begins in bar\n",
+ "T1=27.0+273.0 #Temperature in K\n",
+ "pm=60 #Maximum pressure in bar\n",
+ "h=2 #Heat transfered to air at constant volume is twice that at consmath.tant pressure\n",
+ "g=1.4 #Isentropic index\n",
+ "Cv=0.718 #specific heat at constant volume for air in kJ/kgK\n",
+ "Cp=1.005 #specific heat at constant pressure for air in kJ/kgK\n",
+ "R=0.287 #Real gas constant in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "T2=(T1*rc**(g-1)) #Temperature in K\n",
+ "p2=(p1*rc**g) #Pressure in bar\n",
+ "T3=(T2*(pm/p2)) #Temperature in K\n",
+ "T4=(Cv*(T3-T2))/(2*Cp)+T3 #Temperature in K\n",
+ "b=(T4/T3) #Cut-off ratio\n",
+ "T5=(T4*(b/rc)**(g-1)) #Temperature in K\n",
+ "p5=(p1*(T5/T1)) #Pressure in bar\n",
+ "Q1=(Cv*(T3-T2))+(Cp*(T4-T3))#Heat supplied per unit mass in kJ/kg\n",
+ "Q2=Cv*(T5-T1) #Heat rejected per unit mass in kJ/kg\n",
+ "W=(Q1-Q2) #Workdone in kJ/kg\n",
+ "n=(W/Q1)*100 #Air standard efficiency\n",
+ "Vs=((1*R*1000*T1)/(p1*10**5))*(1-1/rc) #Swept volume in m**3/kg\n",
+ "pmean=((W*1000)/Vs)/10.0**5 #Mean-effective pressure in bar\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The pressures and temperatures at the cardinal points of the cycle are \"\n",
+ "print\"T2 =\",round(T2,1),\" K \\np2 =\", round(p2,1),\" bar \\nT3 = \",round(T3,1), \"K \\np3 =\",round(p3,1),\"bar\"\n",
+ "print\"T4 =\",round(T4,1),\"K \\nT5 = \",round(T5,1), \"K \\np5 = \",round(p5,1),\"bar\"\n",
+ "print\"(b) The cycle efficiency is\",round(n,0),\"percent\" \n",
+ "print\"(c) The mean effective pressure of the cycle is \",round(pmean,1),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The pressures and temperatures at the cardinal points of the cycle are \n",
+ "T2 = 886.3 K \n",
+ "p2 = 44.3 bar \n",
+ "T3 = 1200.0 K \n",
+ "p3 = 52.2 bar\n",
+ "T4 = 1312.1 K \n",
+ "T5 = 460.3 K \n",
+ "p5 = 1.5 bar\n",
+ "(b) The cycle efficiency is 66.0 percent\n",
+ "(c) The mean effective pressure of the cycle is 2.8 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page No: 55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "r=12.0 #Compression ratio\n",
+ "B=1.615 #Cut off ratio\n",
+ "p3=52.17 #Maximum pressure in bar\n",
+ "p4=p3 #Maximum pressure in bar\n",
+ "p1=1 #Initial pressure in bar\n",
+ "T1=(62+273) #Initial temperature in K\n",
+ "n=1.35 #Indices of compression and expansion\n",
+ "g=1.4 #Adiabatic exponent\n",
+ "mR=0.287 #Real gas constant in kJ/kgK\n",
+ "Cv=0.718 #specific heat at constant volume for air in kJ/kgK\n",
+ "Cp=1.005 #specific heat at constant pressure for air in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "T2=T1*r**(n-1) #The temperature at point 2 in K\n",
+ "p2=p1*(r)**n #The pressure at point 2 in bar\n",
+ "T3=T2*(p3/p2) #The temperature at point 3 in K\n",
+ "T4=T3*B #The temperature at point 4 in K\n",
+ "T5=T4*(B/r)**(n-1) #The temperature at point 5 in K\n",
+ "Q12=((g-n)/(g-1))*mR*((T1-T2)/(n-1)) # kJ/kg\n",
+ "Q23=Cv*(T3-T2) \n",
+ "Q34=Cp*(T4-T3) \n",
+ "Q45=((g-n)/(g-1))*mR*((T4-T5)/(n-1)) \n",
+ "Q51=Cv*(T1-T5) \n",
+ "Q1=Q23+Q34+Q45 \n",
+ "Q2=-Q12+(-Q51) \n",
+ "W=Q1-Q2 \n",
+ "E=(W/Q1)*100 \n",
+ "Vs=((mR*T1)/p1)*(r-1)/r # m**3/kg\n",
+ "pm=(W*1000/Vs)/10.0**3 #Mean effective pressure in bar\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The temperature at cardinal points \\nT2 =\",round(T2,0),\" K\\nT3 = \",round(T3,0),\"K \\nT4 = \",round(T4,0),\"K \\nT5 = \",round(T5,0),\"K \" \n",
+ "print\"(b) The cycle efficiency = \",round(E,1),\" percent\"\n",
+ "print\"(c) The mean effective pressure of the cycle = \",round(pm,3),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The temperature at cardinal points \n",
+ "T2 = 799.0 K\n",
+ "T3 = 1456.0 K \n",
+ "T4 = 2352.0 K \n",
+ "T5 = 1166.0 K \n",
+ "(b) The cycle efficiency = 56.9 percent\n",
+ "(c) The mean effective pressure of the cycle = 9.638 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page No: 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "p1=1.0 #Inlet pressure in bar\n",
+ "T1=27.0+273.0 #Temperature in K\n",
+ "p2=4.0 #pressure at point 2 in bar\n",
+ "p3=16.0 #Maximum pressure in bar\n",
+ "Cv=0.573 #specific heat at constant volume for gas in kJ/kgK\n",
+ "Cp=0.761 #specific heat at constant pressure for gas in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "g=(Cp/Cv) \n",
+ "T2=(T1*(p2/p1)**((g-1)/g)) # K\n",
+ "T3=(p3/p2)*T2 \n",
+ "T4=T3*(p1/p3)**((g-1)/g) \n",
+ "Q1=Cv*(T3-T2) #kJ/kg\n",
+ "Q2=Cp*(T4-T1) \n",
+ "W=Q1-Q2 \n",
+ "n=(W/Q1)*100 \n",
+ "r=(p2/p1)**(1/g)\n",
+ "R=(Cp-Cv) #kJ/kg.K\n",
+ "Vs=(R*1000*T1*(r-1))/(p1*10.0**5*r) #m**3/kg\n",
+ "pm=(W/(Vs*100.0)) \n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The work done per kg of gas is \",round(W,1),\"kJ/kg\"\n",
+ "print\"(b) The efficiency of the cycle is \",round(n,1),\"percent \"\n",
+ "print\"(c) Mean effective pressure is \",round(pm,1),\"bar\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The work done per kg of gas is 306.2 kJ/kg\n",
+ "(b) The efficiency of the cycle is 42.2 percent \n",
+ "(c) Mean effective pressure is 8.4 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/README.txt b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/README.txt new file mode 100755 index 00000000..67ddb01f --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/README.txt @@ -0,0 +1,10 @@ +Contributed By: Tamanna Mittal +Course: others +College/Institute/Organization: NTPC +Department/Designation: commerce +Book Title: Fundamental of internal combustion engines +Author: H N Gupta +Publisher: Prentice Hall of India Pvt Ltd, New Delhi +Year of publication: 2006 +Isbn: 812032854X +Edition: 1st edition
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap10.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap10.ipynb new file mode 100755 index 00000000..880733d2 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap10.ipynb @@ -0,0 +1,391 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10:CI Engines-Fuel Injection System"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 page no: 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "bsfc=0.3 #The brake specific fuel consumption in kg/kWh\n",
+ "bp=250 #The brake power in kW\n",
+ "N=1500 #Number of cycles per min in rpm\n",
+ "CA=15 #Crank angle in degrees\n",
+ "pi1=30 #The pressure of air in the cylinder at the beginning of the injection in bar\n",
+ "pi2=60 #The pressure of air in the cylinder at the end of the injection in bar\n",
+ "pf1=220 #The fuel injection pressure at the beginning in bar\n",
+ "pf2=550 #The fuel injection pressure at the end in bar\n",
+ "Cd=0.65 #The coefficient of discharge for the injector \n",
+ "df=850 #The density of the fuel in kg/m**3\n",
+ "p1=1.013 #The atmospheric pressure in bar\n",
+ "n=4.0 #The number of orifices used in the nozzle\n",
+ "x=6.0 #Number of cylinders\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "mf=bsfc*bp/60.0 \n",
+ "F=(mf/(N/2.0))*(1/x) \n",
+ "s=(CA/360.0)/(N/60.0) \n",
+ "mf1=F/s \n",
+ "p1=pf1-pi1 \n",
+ "p2=pf2-pi2\n",
+ "pa=(p1+p2)/2.0\n",
+ "Af=(mf1/(Cd*(2*df*pa*10.0**5)**(1/2.0)))*10**6\n",
+ "do=((Af/n)*(4/math.pi))**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The nozzle area required per injection = \" ,round(Af,3),\"mm**2\"\n",
+ "print\"The diameter of the orifice = \",round(do,3), \"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The nozzle area required per injection = 1.067 mm**2\n",
+ "The diameter of the orifice = 0.583 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 page no: 333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "bp=30 #The brake power of the engine in kW\n",
+ "N=3000 #The engine speed in rpm \n",
+ "bsfc=0.28 #The brake specific fuel consumption in kg/kWh \n",
+ "Api=35 \n",
+ "p2=160 #The pressure at which fuel is injected in bar\n",
+ "CA=28 #The crank angle in degrees\n",
+ "p1=35 #The pressure in the combustion chamber in bar\n",
+ "Cv=0.92 #The coefficient of velocity \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "S=141.5/(131.5+Api) \n",
+ "df=S*1000 \n",
+ "D=(CA/360.0)/(N/60.0)\n",
+ "F=(bsfc*bp)/((N/2.0)*60)\n",
+ "mf=F/D\n",
+ "Cf=Cv*((2*(p2-p1)*10**5)/df)**(1/2.0)\n",
+ "Af=(mf/(df*Cf))*10**6\n",
+ "d=(4*Af/math.pi)**(1/2.0) \n",
+ "\n",
+ "#Output\n",
+ "print\"The velocity of injection of the fuel = \",round(Cf,1),\"m/s \"\n",
+ "print\"The diameter of the fuel orifice = \",round(d,3),\" mm \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of injection of the fuel = 157.8 m/s \n",
+ "The diameter of the fuel orifice = 0.755 mm \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3 Page no 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.8*10**-3 #The diameter of an orifice in m\n",
+ "A=1.65*10**-6 #The cross sectional area in m**2\n",
+ "Cd=0.9 #The discharge coefficient of the orifice \n",
+ "Cp=0.85 #The coefficient of the passage\n",
+ "p1=170 #The injection pressure in bar\n",
+ "p2=25 #The compression pressure of the discharge in bar\n",
+ "df=850 #The density of the fuel in kg/m**3\n",
+ "\n",
+ "#Calculations\n",
+ "Q=((145/(22.931*10.0**9))**(1/2.0))*10**6 \n",
+ "p=170-(2.161*10**9*(Q/10.0**6)**2)\n",
+ "Cf=Cd*((2*(p-p2)*10**5)/df)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The discharge of fuel through the injector = \",round(Q,1),\"cm**2/s\" \n",
+ "print\"The jet velocity through the orifice = \",round(Cf,1),\" m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The discharge of fuel through the injector = 79.5 cm**2/s\n",
+ "The jet velocity through the orifice = 158.2 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4 page no: 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "s=20 #Spray penetration in cm\n",
+ "t1=15.7 #The spray penetration of 20 cm in ms\n",
+ "pi1=150 #The injection pressure in bar\n",
+ "pi2=450.0 #The injection pressure to be used in bar\n",
+ "p2=15 #The combustion chamber pressure in bar\n",
+ "d1=0.34 #The diameter of the orifice in mm\n",
+ "s1=20 #The penetration for an orifice in cm\n",
+ "d2=0.17 #If the diameter of the orifice in cm\n",
+ "t11=12 #The spray penetration in ms\n",
+ "\n",
+ "#Calculations\n",
+ "t2=(t1*(pi1-p2)**(1/2.0))/(pi2-p2)**(1/2.0)\n",
+ "s2=d2*(s1/d1)\n",
+ "t21=t11*(d2/d1)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The time required for the spray to penetrate = \",round(t2,3),\"ms\"\n",
+ "print\"(b) The spray penetration of the orifice = \",round(s2,3),\"cm\"\n",
+ "print\"The time required for the spray to penetrate = \",round(t21,3),\"ms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The time required for the spray to penetrate = 8.746 ms\n",
+ "(b) The spray penetration of the orifice = 10.0 cm\n",
+ "The time required for the spray to penetrate = 6.0 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 page no: 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "v=6.5 #The volume of fuel in the barrel in cc\n",
+ "d=0.3 #The dimeter of fuel pipe line in cm\n",
+ "l=65 #The length of the fuel pipe line in cm \n",
+ "vi=2.5 #The volume of fuel in the injection valve in cc\n",
+ "K=78.5*10**-6 #The coefficient of compressibility of the oil per bar\n",
+ "p1=1 #The atmospheric pressure in bar\n",
+ "p2=180 #The pressure due to pump in bar\n",
+ "v3=0.1 #The pump displacement necessary for the fuel in cc\n",
+ "e=0.75 #The effective stroke of the plunger in cm\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "V1=v+((math.pi*d**2)/4.0)*l+vi\n",
+ "V=K*V1*(p2-p1)\n",
+ "T=(V)+v3\n",
+ "L=T*(4/math.pi)*(1/(e**2))\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The total displacement of the plunger = \",round(T,3),\"cc\" \n",
+ "print\"(b) The effective stroke of the plunger = \",round(L,3),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The total displacement of the plunger = 0.291 cc\n",
+ "(b) The effective stroke of the plunger = 0.659 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 page no: 337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=4.0 #Number of cylinders \n",
+ "N=2500 #The engine speed in rpm \n",
+ "P=90 #The power produced by the engine in kW\n",
+ "bsfc=0.28 #The brake specific fuel consumption in kg/kWh\n",
+ "v=3.5 #The volume of fuel in the barrel in cc\n",
+ "vp=2.5 #Volume of fuel in the pipe line in cc\n",
+ "vi=2.0 #The fuel inside the injector in cc\n",
+ "p1=280.0 #The average injection pressure in bar\n",
+ "p2=30.0 #The compression pressure of air during injection in bar\n",
+ "df=850.0 #The density of the fuel in kg/m**3\n",
+ "K=80*10**-6 #The coefficient of compressibility of fuel per bar\n",
+ "pi=1.0 #The pressure with which fuel enter into the barrel in bar\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "F=(bsfc*P)/((N/2.0)*60)\n",
+ "F1=F/n\n",
+ "Vf=(F1/df)*10**6\n",
+ "V1=v+vp+vi\n",
+ "V=K*V1*(p1-math.pi)\n",
+ "Vp=Vf+V\n",
+ "W=((1/2.0)*(p1-math.pi)*10**5*V*10**-6)+((p1-p2)*10**5*Vf*10**-6)\n",
+ "P1=(W*N)/(2*60*1000) #Power lost per cylinder in kW\n",
+ "P2=P1*4 #Total power lost for pumping the fuel in kW\n",
+ "\n",
+ "#Output \n",
+ "print\"The displacement volume of one plunger per cycle = \",round(Vp,3),\"cc\" \n",
+ "print\"Total power lost for pumping the fuel = \",round(P2,3),\"kW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The displacement volume of one plunger per cycle = 0.276 cc\n",
+ "Total power lost for pumping the fuel = 0.41 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 page no: 339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "v1=0.3 #Velocity of the pump plunger in m/s\n",
+ "l=0.575 #The length of the fuel pipe in m\n",
+ "A=1/20.0 #The cross sectional area of pipe to the plunger cylinder\n",
+ "a=1/40.0 #The area of nozzle hole to the pipe \n",
+ "p1=27.6 #Initial pressure in the line in bar \n",
+ "p2=27.6 #The compression pressure of the engine\n",
+ "K=17830*10**5 #The bulk modulus of fuel in N/m**2\n",
+ "df=860.0 #The density of the fuel in kg/m**3\n",
+ "\n",
+ "#Calculations\n",
+ "Vs=(K/df)**(1/2.0)\n",
+ "t=l/Vs\n",
+ "Vp=(1/A)*v1\n",
+ "p=((K/Vs)*Vp)/10.0**5\n",
+ "pi=p+p1\n",
+ "po=p1+p\n",
+ "vc=Vp-(a*((2*(po-p2))/df)**(1/2.0))\n",
+ "pr=26.8 #By trial , Pressure\n",
+ "Vc=pr*(Vs/(K/10.0**5))\n",
+ "po1=p1+p+pr\n",
+ "vo=a*((2*(po1-p2)*10**5)/df)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The velocity of the pressure disturbance = \",round(Vs,0),\"m/s\"\n",
+ "print\"(b) The time taken by the disturbance to travel through the pipe line = \",round(t,4),\" s\" \n",
+ "print\"(c) The velocity at the pump end of the pipe line as the plunger moves = \",round(Vp,2),\" m/s\"\n",
+ "print\"The pressure at the pump end of pipe line as the plunger moves = \",round(pi,2),\"bar\"\n",
+ "print\"(d)The magnitude of the first reflected pressure = \",round(pr,2),\"bar\" \n",
+ "print\"The magnitude of the first reflected velocity wave =\",round(Vc,2),\"m/s\" \n",
+ "print\"(e)The pressure at the oriface end of the pipe line after the first reflection = \",round(po1,1),\"bar\"\n",
+ "print\"The velocity at the oriface end of the pipe line after the first reflection = \",round(vo,2),\" m/s \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The velocity of the pressure disturbance = 1440.0 m/s\n",
+ "(b) The time taken by the disturbance to travel through the pipe line = 0.0004 s\n",
+ "(c) The velocity at the pump end of the pipe line as the plunger moves = 6.0 m/s\n",
+ "The pressure at the pump end of pipe line as the plunger moves = 101.9 bar\n",
+ "(d)The magnitude of the first reflected pressure = 26.8 bar\n",
+ "The magnitude of the first reflected velocity wave = 2.16 m/s\n",
+ "(e)The pressure at the oriface end of the pipe line after the first reflection = 128.7 bar\n",
+ "The velocity at the oriface end of the pipe line after the first reflection = 3.83 m/s \n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap11.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap11.ipynb new file mode 100755 index 00000000..4d28299a --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap11.ipynb @@ -0,0 +1,337 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11:Two Stroke Engines"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1 page no:367"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "nsc=75 #The scavenging efficiency of the two stroke engine in percent \n",
+ "ns=20 #The scavenging efficiency is increased by in percent\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Rsc=math.log(1/(1-(nsc/100.0)))\n",
+ "nsc1=(nsc/100.0)+((nsc/100.0)*(ns/100.0))\n",
+ "Rsc1=math.log(1/(1-(nsc1)))\n",
+ "Rscr=((Rsc1-Rsc)/Rsc)*100 #Percentage increase in scavenging ratio in persent\n",
+ "\n",
+ "#Output\n",
+ "print\"The percentage change in the scavenging ratio = \",round(Rscr,1),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage change in the scavenging ratio = 66.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2 page no:367"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.12 #The bore diameter of the engine in m\n",
+ "l=0.15 #The stroke length of the engine in m\n",
+ "r=16.0 #The compression ratio \n",
+ "N=2000.0 #The speed of the engine in rpm\n",
+ "mf=(240/60.0) #Actual air flow per min in kg/min\n",
+ "T=300.0 #Air inlet temperature in K\n",
+ "p=1.025 #Exhaust pressure in bar\n",
+ "R=287 #Real gas constant in J/kg\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "da=(p*10**5)/(R*T)\n",
+ "Vs=((math.pi)*(d**2)*l)/4.0\n",
+ "V=(r/(r-1))*Vs\n",
+ "m=da*V\n",
+ "m1=m*N\n",
+ "Rsc=mf/m1 #Scavenging ratio\n",
+ "nsc=((1-math.exp(-Rsc))*100)\n",
+ "ntr=((nsc/100.0)/Rsc)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The scavenging ratio = \",round(Rsc,2)\n",
+ "print\"(b) The scavenging efficiency = \",round(nsc,1),\"percent \"\n",
+ "print\"(c) The trapping efficiency = \",round(ntr,1),\" percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The scavenging ratio = 0.93\n",
+ "(b) The scavenging efficiency = 60.5 percent \n",
+ "(c) The trapping efficiency = 65.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3 page no: 368"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "mf=6.5 #Mass flow rate of fuel in kg/h\n",
+ "N=3000.0 #The speed of the engine in rpm\n",
+ "a=15 #The air fuel ratio\n",
+ "CV=44000 #The calorific value of the fuel in kJ/kg\n",
+ "pm=9 #The mean piston speed in m/s\n",
+ "pmi=4.8 #The mean pressure in bar\n",
+ "nsc=85 #The scavenging efficiency in percent\n",
+ "nm=80 #The mechanical efficiency in percent\n",
+ "R=290.0 #Real gas constant in J/kgK\n",
+ "p=1.03 #The pressure of the mixture in bar\n",
+ "T=288.0 #The temperature of the mixture in K\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "ma=a*mf\n",
+ "L=((pm*60)/(2*N))*100\n",
+ "mac=mf+ma\n",
+ "mi=(mac)/(nsc/100.0)\n",
+ "da=(p*10**5)/(R*T)\n",
+ "d=(((mi/da)*(4/math.pi)*(1/(L/100.0))*(1/(60*N)))**(1/2.0))*100\n",
+ "ip=(pmi*10**5*(L/100)*((math.pi/4.0)*(d/100)**2)*N)/(60*1000)\n",
+ "bp=ip*(nm/100.0)\n",
+ "nth=(bp/((mf/3600.0)*CV))*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The diameter of the bore = \",round(d,2),\"cm\"\n",
+ "print\"The length of the stroke = \",L,\" cm\" \n",
+ "print\"The brake power = \",round(bp,2),\" kW\"\n",
+ "print\"The brake thermal efficiency =\",round(nth,2),\" percent \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diameter of the bore = 8.83 cm\n",
+ "The length of the stroke = 9.0 cm\n",
+ "The brake power = 10.58 kW\n",
+ "The brake thermal efficiency = 13.32 percent \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4 page no: 369"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.08 #The diameter of the bore in m\n",
+ "L=0.1 #The length of the stroke in m\n",
+ "r=8.0 #The compression ratio \n",
+ "o=60.0 #The exhaust port open before BDC in degrees\n",
+ "v=60.0 #The exhaust port closes after BDC in degrees\n",
+ "a=15.0 #Air fuel ratio \n",
+ "T=300.0 #The temperature of the mixture entering into the engine in K\n",
+ "p=1.05 #The pressure in the cylinder at the time of clomath.sing\n",
+ "R=290.0 #Real gas constant in J/kgK\n",
+ "ma=150.0 #Mass flow rate of air in kg/h\n",
+ "N=4000.0 #The speed of the engine in rpm\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "mf=ma/a\n",
+ "mac=ma+mf\n",
+ "r=(L*100)/2.0\n",
+ "Le=(r+(r*math.sin (math.pi/6.0)))/100.0\n",
+ "Vse=(math.pi*d**2*Le)/4.0\n",
+ "V=(r/(r-1))*Vse\n",
+ "V=0.00043 #Value in book after approximation\n",
+ "da=(p*10**5)/(R*T)\n",
+ "m=V*da\n",
+ "mi=m*60*N\n",
+ "Rsc=mac/mi \n",
+ "nsc=(1-(exp(-Rsc)))*100\n",
+ "ntr=nsc/Rsc\n",
+ "\n",
+ "#Output\n",
+ "print\"The scavenging ratio = \",round(Rsc,2)\n",
+ "print\"The scavenging efficiency =\",round(nsc,2),\" percent \"\n",
+ "print\"The trapping efficiency = \",round(ntr,2),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The scavenging ratio = 1.28\n",
+ "The scavenging efficiency = 72.32 percent \n",
+ "The trapping efficiency = 56.3 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5 page no: 371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=8.25 #The diameter of the bore in cm\n",
+ "L=11.25 #The length of the stroke in cm\n",
+ "r=8.0 #The compression ratio \n",
+ "N=2500.0 #The speed of the engine in rpm\n",
+ "ip=17.0 #Indicated power in kW\n",
+ "a=0.08 #Fuel air ratio \n",
+ "T=345.0 #Inlet temperature mixture in K\n",
+ "p=1.02 #Exhaust pressure in bar\n",
+ "CV=44000.0 #The calorific value of the fuel in kJ/kg\n",
+ "nth=0.29 #Indicated thermal efficiency\n",
+ "M=114.0 #Molar mass of fuel \n",
+ "R=8314.0 #Universal Gas constant in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(math.pi*d**2*L)/4 #Displacement volume in cm**3\n",
+ "V=(r/(r-1))*Vs #Total cylinder volume in m**3\n",
+ "ps=((29*p*10**5)/(R*T))*(1/(1+a*(29/M))) #The density of dry air in kg/m**3\n",
+ "nsc=((ip*1000)/((N/60)*V*10**-6*ps*a*CV*1000*nth))*100 #The scavenging efficiency in percent\n",
+ "\n",
+ "#Output\n",
+ "print\"The scavenging efficiency = \",round(nsc,2),\" percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The scavenging efficiency = 57.54 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6 page no: 372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "S=15.0 #The speed of the math.piston in m/s\n",
+ "ps=0.35 #The scavenging pressure in bar\n",
+ "pa=1.03 #Atmospheric pressure in bar\n",
+ "r=18.0 #The compression ratio \n",
+ "t=35.0 #The inlet temperature in degree centigrade\n",
+ "Rsc=0.9 #The scavenging ratio \n",
+ "ta=15.0 #The atmospheric temperature in degree centigrade\n",
+ "nc=0.75 #Compressor efficiency \n",
+ "g=1.4 #Adiabatic index\n",
+ "R=287.0 #Real gas constant in J/kgK\n",
+ "Cp=1005.0 #Specific heat of gas in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "pi=ps+pa #The scavenging pressure in bar\n",
+ "Ti=(273+ta)+t #The inlet temperature in K\n",
+ "pr=pa/math.pi #The ratio of the pressure for calculations\n",
+ "di=(pi*10**5)/(R*Ti) #The density in kg/m**3\n",
+ "ai=(g*R*Ti)**(1/2.0) #The sonic velocity in m/s\n",
+ "C=(Rsc)/(2*((r-1)/r)*(ai/S)*(pi/pa)*((2/(g-1))*(((pr)**(2/g))-((pr)**((g+1)/g))))**(1/2.0))\n",
+ "ds=(pa*10**5)/(R*Ti) #The density in kg/m**3\n",
+ "mep=(ds*Rsc*Cp*Ti*(((pi/pa)**((g-1)/g))-1))/((nc*((r-1)/r))*10**5) #Mean effective pressure in bar\n",
+ "\n",
+ "#Output\n",
+ "print\"The flow coefficient = \",round(C,3) \n",
+ "print\"The compressor mean effective pressure = \",round(mep,1),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The flow coefficient = 0.028\n",
+ "The compressor mean effective pressure = 0.4 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap15.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap15.ipynb new file mode 100755 index 00000000..8df611a8 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap15.ipynb @@ -0,0 +1,434 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter15:Air Capacity and SuperCharging"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.1 page no: 474"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "Vs=0.0028 #Swept volume in m**3\n",
+ "N=3000 #Speed of the engine in rpm\n",
+ "ip=12.5 #The average indicated power developed in kW/m**3\n",
+ "nv=85 #Volumetric efficiency in percent\n",
+ "p1=1.013 #The atmospheric pressure in bar\n",
+ "T1=288 #The atmospheric temperature in K\n",
+ "ni=74 #Isentropic efficiency in percent\n",
+ "pr=1.6 #The pressure ratio\n",
+ "nm=78 #All mechanical efficiencies in percent\n",
+ "g=1.4 #Adiabatic index\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "Cp=1.005 #The specific heat of gas in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "Vs1=(Vs*(N/2.0)) #Volume swept by the piston per minute in m**3/min\n",
+ "Vi=(nv/100.0)*Vs1 #Unsupercharged induced volume in m**3/min\n",
+ "p2=pr*p1 #Blower delivery pressure in bar\n",
+ "T21=T1*(p2/p1)**((g-1)/g) #Temperature after isentropic compression in K\n",
+ "T2=T1+((T21-T1)/((ni/100.0))) #Blower delivery temperature in K\n",
+ "Ve=(Vs1*p2*T1)/(T2*p1) #Equivalent volume at 1.013 bar and 15 degree centigrade in m**3/min\n",
+ "nv1=(Ve/Vs1)*100 #Volumetric efficiency of supercharged engine in percent\n",
+ "Vii=Ve-Vi #Increase in induced volume in m**3/min\n",
+ "ipa=ip*Vii #Increase in ip from air induced in kW\n",
+ "ipi=((p2-p1)*10**5*Vs1)/(60*1000) #Increase in ip due to increased induction pressure in kW\n",
+ "ipt=ipa+ipi #Total increase in ip in kW\n",
+ "bp=ipt*(nm/100.0) #Increase in engine bp in kW\n",
+ "ma=(p2*(Vs1/60.0)*10**5)/(R*T2) #Mass of air delivered per second by blower in kg/s\n",
+ "P=ma*Cp*(T2-T1) #Power input to blower in kW\n",
+ "Pd=P/(nm/100.0) #Power required to drive the blower in kW\n",
+ "bpn=bp-Pd #Net increase in bp in kW\n",
+ "bpu=ip*Vi*(80/100.0) #The bp of unsupercharged engine in kW\n",
+ "bpp=(bpn/(bpu))*100 #Percentage increase in bp in percent\n",
+ "\n",
+ "#Output\n",
+ "print\"The volumetric efficiency of supercharged engine = \",round(nv1,0),\"percent\"\n",
+ "print\"The increase in brake power by supercharging = \",round(bpn,1),\" kW \"\n",
+ "print\"The percentage increase in brake power = \",round(bpp,1),\" percent \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volumetric efficiency of supercharged engine = 134.0 percent\n",
+ "The increase in brake power by supercharging = 15.1 kW \n",
+ "The percentage increase in brake power = 42.3 percent \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.2 page no: 477"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "p=1.013 #The pressure at the sea level in bar\n",
+ "T=283 #The temperature at the sea level in K\n",
+ "bp=275.0 #Brake power in kW\n",
+ "N=1800.0 #The speed of the engine in rpm\n",
+ "a=20 #Air fuel ratio \n",
+ "R=287 #The real gas constant in J/kgK\n",
+ "bsfc=0.24 #Brake specific fuel consumption in kg/kWh\n",
+ "nv=80 #Volumetric efficiency in percent\n",
+ "p2=0.75 #The atmospheric pressure at altitude in bar\n",
+ "P=9 #The power consumed by supercharger of the total power produced by the engine in percent\n",
+ "T2=303 #The temperature of air leaving the supercharger in K\n",
+ "\n",
+ "#Calculations\n",
+ "mf=(bsfc*bp)/60.0 \n",
+ "ma1=mf*a \n",
+ "ma=(2/N)*ma1 \n",
+ "dai=(p*10**5)/(R*T) \n",
+ "Vd=(ma/(dai*(nv/100.0))) \n",
+ "pmb=(bp*2*60*1000)/(Vd*N*10**5) \n",
+ "GP=bp/(1-0.09) \n",
+ "ma2=(ma1/bp)*GP \n",
+ "ma1=(ma2*2)/N \n",
+ "p21=((R*T2*ma1)/((nv/100.0)*Vd))/10.0**5 \n",
+ "pi=p21-p2 \n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The engine capacity Vd = \",round(Vd,3),\"m**3\" \n",
+ "print\"The bmep of the unsupercharged engine = \",round(pmb,3),\"bar\" \n",
+ "print\"(b) Increase in air pressure required in the supercharged = \",round(pi,3),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The engine capacity Vd = 0.024 m**3\n",
+ "The bmep of the unsupercharged engine = 7.483 bar\n",
+ "(b) Increase in air pressure required in the supercharged = 0.442 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.3 page no: 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "Vs=0.003 #Swept volume in m**3\n",
+ "bmep=9 #Brake mean effective pressure in bar\n",
+ "N=4000 #The speed of the engine in rpm\n",
+ "ni=30.0 #Indicated thermal efficiency in percent\n",
+ "nm=90 #Mechanical efficiency in percent\n",
+ "bmep1=12 #The brake mean effective pressure of other engine in bar\n",
+ "N1=4000 #The speed of other engine in rpm\n",
+ "ni1=25 #The indicated thermal efficiency of other engine in percent\n",
+ "nm1=91 #The mechanical efficiency of other engine in percent\n",
+ "m=200 #The mass of naturally aspired engine in kg\n",
+ "m1=220 #The mass of supercharged engine in kg\n",
+ "CV=44000 #The calorific value of the fuel in kJ/kg\n",
+ "\n",
+ "#Calculations\n",
+ "bp=(bmep*10**5*Vs*N)/(2.0*60.0*1000) \n",
+ "ip=bp/(nm/100.0) \n",
+ "mf=(ip)/((ni/100.0)*CV) \n",
+ "bp1=(bmep1*10**5*Vs*N1)/(2.0*60.0*1000) \n",
+ "ip1=bp1/(nm1/100.0) \n",
+ "mf1=ip1/((ni1/100.0)*CV) \n",
+ "mf2=mf*3600 \n",
+ "mf3=mf1*3600 \n",
+ "x=((200/90.0)-(220/120.0))/((43.2/120.0)-(27.27/90.0)) \n",
+ "\n",
+ "#Output\n",
+ "print\"The maximum hours required for supply of sufficient fuel = \",round(x,3),\"hr\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum hours required for supply of sufficient fuel = 6.823 hr\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.4 Page no 480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.1 #The diameter of the bore in m\n",
+ "L=0.12 #The length of the stroke in m\n",
+ "N=3000 #The speed of the engine in rpm\n",
+ "n=4 #Number of cylinders\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "t=120 #Output Torque in Nm\n",
+ "nm=85 #The mechanical efficiency of the engine in percent\n",
+ "T1=288 #The inlet temperature of air into compressor in K\n",
+ "p1=1 #The inlet pressure of air into compressor in bar\n",
+ "Q=1200 #Heat rejected rate in kJ/min\n",
+ "T=328 #The outlet temperature of air in K\n",
+ "p=1.7 #The outlet pressure of air in bar\n",
+ "nv=90 #Volumetric efficiency in percent\n",
+ "Cp=1.005 #Specific heat of gas in kJ/kg\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "bp=(2*math.pi*N*t)/(60.0*1000.0) #The brake power in kW\n",
+ "ip=bp/(nm/100.0) #The indicated power in kW\n",
+ "pmi=((ip*2*60*1000*4)/(L*(math.pi*d**2)*N*n))/10.0**5 #The mean effective pressure in bar\n",
+ "Vs=(math.pi/4.0)*d**2*L #Swept volume in m**3\n",
+ "Vs1=Vs*(N/2.0)*n #Volume swept by the piston per min \n",
+ "V1=(nv/100.0)*Vs1 #Rate of volume flow of air into the engine in m**3/min\n",
+ "me=((p*10**5*V1)/(R*T))*60 #Rate of mass flow of air into the engine in kg/h\n",
+ "E=Q/60.0 #Energy balance in the after cooling in kJ/s\n",
+ "T2=((bp/E)*T-T1)/((bp/E)-1) #The outlet temperature of air in K\n",
+ "mc=((bp)/(Cp*(T2-T1)))*3600 #Mass flow rate in kg/h\n",
+ "maf=mc-me #Rate of air flow available to the consumer in kg/h\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The imep of the supercharged engine = \",round(pmi,3),\"bar\"\n",
+ "print\"(b) The rate of air consumed by the engine = \",round(me,1),\"kg/h\" \n",
+ "print\"(c) The rate of air flow available to the consumer = \",round(maf,1),\"kg/h\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The imep of the supercharged engine = 4.706 bar\n",
+ "(b) The rate of air consumed by the engine = 551.5 kg/h\n",
+ "(c) The rate of air flow available to the consumer = 1033.5 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.5 page no: 482"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "Vs=0.0045 #Swept volume in m**3\n",
+ "N=4000.0 #The speed of the engine in rpm \n",
+ "nv=150.0 #Overall volumetric efficiency in percent\n",
+ "ni=90.0 #Isentropic efficiency of the compressor in percent\n",
+ "nm=85.0 #Mechanical efficiency in percent\n",
+ "T=330.0 #The temperature of compressed air after cooler in K\n",
+ "p2=1.8 #The pressure of the compressed air in bar\n",
+ "T1=290.0 #The ambient temperature of air in K\n",
+ "p1=1.0 #The pressure of the ambient condition in bar\n",
+ "R=287.0 #The real gas constant in J/kgK\n",
+ "g=1.4 #Adiabatic index\n",
+ "Cp=1.005 #The specific heat of gas in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "T21=T1*(p2/p1)**((g-1)/g) \n",
+ "T2=T1+((T21-T1)/(ni/100.0)) \n",
+ "Vs1=Vs*(N/(2*60)) # m**3/s\n",
+ "Va=(nv/100)*Vs1 \n",
+ "d=(p1*10**5)/(R*T1) # kg/m**3\n",
+ "ma=d*Va # kg/s\n",
+ "Q=ma*Cp*(T2-T) # kJ/s\n",
+ "P=ma*Cp*(T2-T1) # kW\n",
+ "Pa=P/(nm/100.0) \n",
+ "\n",
+ "#Output\n",
+ "print \"(a) The rate of heat rejected from the engine after cooler = \",round(Q,2),\"kJ/s\" \n",
+ "print\"(b) The power absorbed by the supercharger from the engine = \",round(Pa,1),\"kW\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The rate of heat rejected from the engine after cooler = 5.14 kJ/s\n",
+ "(b) The power absorbed by the supercharger from the engine = 18.8 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.6 page no: 483"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "p1=0.98 #The inlet pressure of air in bar\n",
+ "T1=290.0 #The inlet temperature of air in K\n",
+ "p2=1.8 #The pressure of air delivered to the engine in bar\n",
+ "a=20.0 #The air fuel ratio \n",
+ "T3=850.0 #The temperature of the exhaust gases leaving the engine in K\n",
+ "p3=1.6 #The pressure of the exhaust gases leaving the engine in bar\n",
+ "p4=1.03 #The turbine exhaust pressure in bar\n",
+ "nc=80.0 #The isentropic efficiency of compressor in percent\n",
+ "nt=85.0 #The isentropic efficiency of turbine in percent\n",
+ "Cpa=1.005 #The specific heat of air in kJ/kgK\n",
+ "Cpg=1.15 #The specific heat of gas in kJ/kgK\n",
+ "g=1.33 #isentropic index\n",
+ "h=1.0 #Adiabatic index\n",
+ "\n",
+ "#Calculations\n",
+ "T21=T1*(p2/p1)**(0.286) #value taken in book (g-1/g)=0.286 \n",
+ "T2=T1+((T21-T1)/(nc/100.0)) \n",
+ "T22=T2-273 \n",
+ "T41=T3*(p4/p3)**((g-1)/g) \n",
+ "T4=T3-((nt/100.0)*(T3-T41)) \n",
+ "T44=T4-273 \n",
+ "mf=1.0 # kg/s\n",
+ "ma=mf*a # kg/s\n",
+ "Wc=ma*Cpa*(T2-T1) # kW\n",
+ "mg=ma+mf #Mass flow rate of gas in kg/s\n",
+ "Wt=mg*Cpg*(T3-T4) \n",
+ "Pt=(Wc/Wt)*100 \n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The temperature of the air leaving the compressor = \",round(T22,0),\"degree centigrade\" \n",
+ "print\"(b) The temperature of gases leaving the turbine = \",round(T44,0),\"degree centigrade\" \n",
+ "print\"(c) The mechanical power used to run the turbocharger = \",round(Pt,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The temperature of the air leaving the compressor = 86.0 degree centigrade\n",
+ "(b) The temperature of gases leaving the turbine = 502.0 degree centigrade\n",
+ "(c) The mechanical power used to run the turbocharger = 76.6 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15.7 page no: 485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "a=14.0 #Air fuel ratio \n",
+ "T1=288 #The ambient temperature of air in K\n",
+ "T2=(288-23) #The evaporation of fuel cause 23 degree C drop in mixture temperature in K\n",
+ "p=1.3 #Pressure ratio \n",
+ "nc=75 #The isentropic efficiency of the compressor in percent\n",
+ "Cpm=1.05 #The specific heat of the mixture in kJ/kgK\n",
+ "Cpa=1.005 #The specific heat of air in kJ/kgK\n",
+ "g=1.33 #Adiabatic index\n",
+ "h=1.4 #Isentropic index\n",
+ "ma=1 #Mass flow rate of air in kg/s\n",
+ "\n",
+ "#Calculations\n",
+ "T31=T2*p**((g-1)/g) \n",
+ "T3=T2+((T31-T2)/(nc/100.0))\n",
+ "mm=1+(1/a)\n",
+ "Wc1=mm*Cpm*(T3-T2)\n",
+ "T21=T1*p**((h-1)/h)\n",
+ "T4=T1+((T21-T1)/(nc/100.0))\n",
+ "T4_=317 #approx value taken in book of T4=317\n",
+ "Wc2=ma*Cpa*(T4_-T1) \n",
+ "T5=T4-23\n",
+ "Ps=((Wc2-round(Wc1,0))*100)/Wc2\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The power required by the compressor before the supercharger = \",round(Wc1,0),\"kW/kg of air per second\"\n",
+ "print\"(b) The power required by the compressor after the supercharger = \",round(Wc2,1),\"kW/kg of air per second\" \n",
+ "print\"Percentage of turbine power used to run the compressor = \",round(Ps,3),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The power required by the compressor before the supercharger = 27.0 kW/kg of air per second\n",
+ "(b) The power required by the compressor after the supercharger = 29.1 kW/kg of air per second\n",
+ "Percentage of turbine power used to run the compressor = 7.36 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap16.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap16.ipynb new file mode 100755 index 00000000..158b33df --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap16.ipynb @@ -0,0 +1,874 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter16:Engine Testing and Performance"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.1 page no: 519"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=3000 #The speed of the engine in rpm \n",
+ "r=9 #Compression ratio \n",
+ "l=17.2 #The length of the connecting rod in cm\n",
+ "t=20 #The combustion ends at a TDC in degrees\n",
+ "k=3 #Three litre spark engine\n",
+ "n=6.0 #V-6 Engine\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(k/n)*10**-3\n",
+ "d=(((Vs*4)/math.pi)**(1/3.0))\n",
+ "L=d*100\n",
+ "up=2*d*N/60.0\n",
+ "Vc=(Vs/(r-1))*10**6\n",
+ "cr=L/2.0\n",
+ "R=l/cr\n",
+ "up1=up*((math.pi/2.0)*math.sin(math.pi/9.0)*(1+(math.cos(math.pi/9.0)/(R**2-(math.sin(math.pi/9.0)**2))**(1/2.0))))\n",
+ "s=(cr*math.cos(math.pi/9.0))+(l**2-(cr**2)*(math.sin(math.pi/9.0))**2)**(1/2.0)\n",
+ "x=l+cr-s\n",
+ "V=Vc+(math.pi/4.0)*(d*100)**2*x\n",
+ "\n",
+ "#Output \n",
+ "print\"(a)The cylinder bore and The stroke length (d = L) = \",round(L,2),\"cm\"\n",
+ "print\"(b) The average piston speed = \",round(up,2),\"m/s\"\n",
+ "print\"(c) The clearence volume of one cylinder = \",round(Vc,2),\"cm**3\"\n",
+ "print\"(d) The piston speed at the end of combustion = \",round(up1,2),\"m/s\"\n",
+ "print\"(e) The distance the piston travels from TDC at the end of combustion = \",round(x,2),\"cm\" \n",
+ "print\"(f) Instantaneous volume = \",round(V,2),\"cm**3\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The cylinder bore and The stroke length (d = L) = 8.6 cm\n",
+ "(b) The average piston speed = 8.6 m/s\n",
+ "(c) The clearence volume of one cylinder = 62.5 cm**3\n",
+ "(d) The piston speed at the end of combustion = 5.71 m/s\n",
+ "(e) The distance the piston travels from TDC at the end of combustion = 0.32 cm\n",
+ "(f) Instantaneous volume = 81.24 cm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.2 page no: 521"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.175 #The diameter of the bore in m\n",
+ "L=0.32 #The length of the stroke in m\n",
+ "p=6.5 #Mean effective pressure in bar\n",
+ "pp=0.4 #Pumping loop mean effective pressure in bar\n",
+ "N=510.0 #The speed of the engine in rpm\n",
+ "pm=0.65 #Diagrams from the dead cycle give a mep in bar\n",
+ "n=55.0 #Firing strokes per minute \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "pmi=p-pp \n",
+ "c=((N/2.0)-n) \n",
+ "ipw=pmi*10**5*L*(math.pi/4.0)*d**2*(n/60.0)*(1/1000.0) \n",
+ "Pp=pm*10**5*L*(math.pi/4.0)*d**2*(c/60.0)*(1/1000.0) \n",
+ "fp=ipw-Pp #Power in kW\n",
+ "fip=pmi*10**5*L*(math.pi/4.0)*d**2*(N/(2*60))*(1/1000.0)\n",
+ "fbp=fip-fp\n",
+ "nm=(fbp/fip)*100\n",
+ "\n",
+ "#Output \n",
+ "print\" The full load break power = \",round(fbp,2),\"kW\" \n",
+ "print\"The mechanical efficiency of the engine = \",round(nm,2),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The full load break power = 17.32 kW\n",
+ "The mechanical efficiency of the engine = 86.79 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.3 page no: 522"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.09 #The diameter of the bore in m\n",
+ "L=0.1 #The length of the stroke in m\n",
+ "T=120 #The torque measured in Nm\n",
+ "n=4 #Number of cylinders \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "pmb=((4*math.pi*T)/(L*(math.pi/4)*d**2*n))/10.0**5\n",
+ "\n",
+ "#Output \n",
+ "print\"The brake mean effective pressure = \",round(pmb,2),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The brake mean effective pressure = 5.93 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.4 page no: 522"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Input data\n",
+ "d=0.06 #The diameter of the bore in m \n",
+ "L=0.085 #The length of the stroke in m\n",
+ "N=3000 #The speed of the engine in rpm\n",
+ "r=0.35 #Torque arm radius in m\n",
+ "W=160 #Weight in N\n",
+ "f=6.6 #Fuel consumption in l/h\n",
+ "g=0.78 #specific gravity of the fuel \n",
+ "CV=44000 #The calorific value of the fuel in kJ/kg\n",
+ "w1=114 #Brake load for cylinder 1 in N\n",
+ "w2=110 #Brake load for cylinder 2 in N\n",
+ "w3=112 #Brake load for cylinder 3 in N\n",
+ "w4=116 #Brake load for cylinder 4 in N\n",
+ "n=4 #Number of cylinders\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vf=(f*10**-3)/3600.0\n",
+ "df=g*1000\n",
+ "mf=df*Vf\n",
+ "T=W*r\n",
+ "bp=(2*math.pi*N*T)/(60.0*1000.0)\n",
+ "pmb=((120*bp*1000)/(L*(math.pi/4.0)*d**2*N*n))/10.0**5\n",
+ "nb=((bp)/(mf*CV))*100\n",
+ "bsfc=(mf*3600)/bp\n",
+ "bp1=((2*math.pi*N*w1*r)/(60.0*1000.0))\n",
+ "ip1=bp-bp1 \n",
+ "ip2=bp-((2*math.pi*N*w2*r)/(60*1000))\n",
+ "ip3=bp-((2*math.pi*N*w3*r)/(60*1000))\n",
+ "ip4=bp-((2*math.pi*N*w4*r)/(60*1000))\n",
+ "ip=ip1+ip2+ip3+ip4\n",
+ "nm=(bp/ip)*100\n",
+ "pmi=pmb/(nm/100.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The brake power =\",round(bp,3),\"kW \" \n",
+ "print\"The brake mean effective pressure = \",round(pmb,3),\"bar\"\n",
+ "print\"The brake thermal efficiency =\" ,round(nb,0),\"percent\"\n",
+ "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\"\n",
+ "print\"The indicated power = \",round(ip,2),\"kW \"\n",
+ "print\"The mechanical efficiency =\", round(nm,1),\"percent \" \n",
+ "print\"The indicated mean effective pressure = \",round(pmi,1),\"bar\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The brake power = 17.593 kW \n",
+ "The brake mean effective pressure = 7.32 bar\n",
+ "The brake thermal efficiency = 28.0 percent\n",
+ "The brake specific fuel consumption = 0.293 kg/kWh\n",
+ "The indicated power = 20.67 kW \n",
+ "The mechanical efficiency = 85.1 percent \n",
+ "The indicated mean effective pressure = 8.6 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.5 page no: 523"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.15 #The diameter of the bore in m\n",
+ "L=0.16 #The length of the stroke in m\n",
+ "N=500 #The speed of the engine in rpm\n",
+ "mf=0.0475 #Fuel consumption in kg/min\n",
+ "CV=42000 #The calorific value in kJ/kg\n",
+ "w=400 #The tension on either side of the pulley in N\n",
+ "c=2.2 #Brake circumference in m\n",
+ "l=50 #Length of the indicator diagram in mm\n",
+ "ap=475 #Area of the positive loop of indicator diagram in mm**2\n",
+ "an=25 #Area of the negative loop of indicator diagram in mm**2\n",
+ "s=0.8333 #Spring constant in bar/mm\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "r=c/(2.0*math.pi)\n",
+ "T=w*r\n",
+ "bp=(2*math.pi*N*T)/(60.0*1000.0)\n",
+ "M=(ap-an)/l\n",
+ "imep=M*s\n",
+ "ip=(imep*10**5*L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*(1/1000.0))\n",
+ "nm=(bp/ip)*100\n",
+ "nb=((bp*60)/(mf*CV))*100\n",
+ "ni=((nb/100.0)/(nm/100.0))*100\n",
+ "bsfc=(mf*60)/bp\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The brake power = \",round(bp,3),\"kW\"\n",
+ "print \"(b) The indicated power = \",round(ip,3),\"kW\"\n",
+ "print\"(c) The mechanical efficiency = \",round(nm,0),\"percent\"\n",
+ "print\"(d) The brake thermal efficiency = \",round(nb,3)\n",
+ "print\"(e) The indicated thermal efficiency = \",round(ni,3)\n",
+ "print\"(f) The brake specific fuel consumption = \",round(bsfc,3),\" kg/kWh\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The brake power = 7.333 kW\n",
+ "(b) The indicated power = 8.835 kW\n",
+ "(c) The mechanical efficiency = 83.0 percent\n",
+ "(d) The brake thermal efficiency = 22.055\n",
+ "(e) The indicated thermal efficiency = 26.573\n",
+ "(f) The brake specific fuel consumption = 0.389 kg/kWh\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.6 page no: 524"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=8 #Number of cylinders\n",
+ "d=0.08 #The diameter of the bore in m\n",
+ "L=0.1 #The length of the stroke in m\n",
+ "N=4500 #The speed of the engine in rpm \n",
+ "dy=0.55 #The dynamometer readings in m\n",
+ "w=40 #The weight of the dynamometer scale reading in kg\n",
+ "c=100 #Fuel consumption in cc\n",
+ "t=9.5 #Time taken for fuel consumption in s\n",
+ "CV=44000 #The calorific value of the fuel in kJ/kg\n",
+ "p=1 #The atmospheric air pressure in bar\n",
+ "T=300 #The atmospheric air temperature in K\n",
+ "ma=6 #Mass flow rate of air in kg/min\n",
+ "g=0.7 #Specific gravity of the fuel \n",
+ "Vc=65 #The clearance volume of each cylinder in cc\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "g=1.4 #Isentropic index\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "bp=(2*math.pi*N*dy*w*9.81)/(60.0*1000.0)\n",
+ "bmep=((bp*1000*60)/(L*(math.pi/4.0)*d**2*(N/2.0)*n))/10.0**5\n",
+ "mf=(c*g*3600)/(t*2.0*1000.0)\n",
+ "bsfc=(mf/bp)\n",
+ "bsac=(ma*60)/bp\n",
+ "a=bsac/bsfc\n",
+ "nb=((bp*3600)/(mf*CV))*100\n",
+ "Va=(ma*R*T)/(p*10.0**5)\n",
+ "Vs=(math.pi/4.0)*d**2*L*(N/2.0)*n\n",
+ "nv=(Va/Vs)*100\n",
+ "Vs1=((math.pi/4.0)*d**2*L)*10**6\n",
+ "cr=(Vs1+Vc)/Vc\n",
+ "na=(1-(1/cr)**(g-1))*100\n",
+ "re=((nb)/(na))*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The brake power = \",round(bp,1),\"kW\" \n",
+ "print\"The brake mean effective pressure = \",round(bmep,3),\"bar\"\n",
+ "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\"\n",
+ "print\"The brake specific air consumption = \",round(bsac,3),\"kg/kWh\"\n",
+ "print\"The air fuel ratio = \",round(a,2)\n",
+ "print\"The brake thermal efficiency = \",round(nb,3),\"percent\" \n",
+ "print\"The volumetric efficiency = \",round(nv,1),\"percent\"\n",
+ "print\"The relative efficiency = \",round(re,1),\"percent\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The brake power = 101.7 kW\n",
+ "The brake mean effective pressure = 6.744 bar\n",
+ "The brake specific fuel consumption = 0.261 kg/kWh\n",
+ "The brake specific air consumption = 3.54 kg/kWh\n",
+ "The air fuel ratio = 13.57\n",
+ "The brake thermal efficiency = 31.369 percent\n",
+ "The volumetric efficiency = 57.1 percent\n",
+ "The relative efficiency = 54.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.7 page no: 526"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=6 #Number of cylinders\n",
+ "Do=0.03 #Orifice diameter in m\n",
+ "Cd=0.6 #Coefficient of discharge \n",
+ "H=0.14 #Pressure drop across the orifice\n",
+ "d=0.1 #The diameter of the bore in m\n",
+ "L=0.11 #The length of the stroke in m\n",
+ "W=540 #Brake load in N\n",
+ "N=2500 #Engine speed in rpm\n",
+ "ch=83/17 #C/H ratio by mass\n",
+ "p=1 #Ambient pressure in bar\n",
+ "t=18 #Time taken for fuel consumption in s\n",
+ "f=100 #The amount of fuel consumption in cc\n",
+ "T=300.0 #Ambient air temperature in K\n",
+ "df=780 #The density of the fuel in kg/m**3\n",
+ "R=287.0 #Real gas constant in J/kgK\n",
+ "g=9.81 #Gravitational force constant in m/s**2\n",
+ "dhg=13600 #Density of Hg in kg/m**3\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "da=(p*10**5)/(R*T)\n",
+ "Va=(Cd*(math.pi/4.0)*Do**2*(2*g*H*(dhg/da))**(1/2.0))\n",
+ "Vs=(math.pi/4.0)*d**2*L*(N/(2.0*60.0))*n\n",
+ "nv=(Va/Vs)*100\n",
+ "bp=(W*N)/(20000.0)\n",
+ "bmep=((bp*1000)/(L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*n))/10.0**5\n",
+ "T=(60*bp*1000)/(2.0*math.pi*N)\n",
+ "mf=(f/18.0)*(780/1000.0)*(1/1000.0)*3600\n",
+ "bsfc=mf/bp\n",
+ "so=(0.83*(32/12.0))+(0.17*(8/1.0))\n",
+ "sa=so/bsfc\n",
+ "maa=Va*da\n",
+ "af=(maa*3600)/mf\n",
+ "pea=((af-sa)/sa)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The volumetric efficiency = \",round(nv,3),\"percent\" \n",
+ "print\"The brake mean effective pressure = \",round(bmep,3),\"bar\" \n",
+ "print\"The brake power = \",bp,\"kW\" \n",
+ "print\"The Torque = \",round(T,3),\"Nm\"\n",
+ "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\" \n",
+ "print\"The percentage of excess air = \",round(pea,1),\"percent\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volumetric efficiency = 70.433 percent\n",
+ "The brake mean effective pressure = 6.25 bar\n",
+ "The brake power = 67.5 kW\n",
+ "The Torque = 257.831 Nm\n",
+ "The brake specific fuel consumption = 0.231 kg/kWh\n",
+ "The percentage of excess air = 31.9 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.8 page no: 528"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.2 #The diameter of bore in m\n",
+ "L=0.3 #The length of the stroke in m\n",
+ "r=5.5 #The compression ratio of the engine\n",
+ "N=400 #The speed of the engine in rpm\n",
+ "imep=4.5 #The indicative mean effective pressure in bar\n",
+ "a=6 #Air to gas by volume \n",
+ "CV=12000 #The calorific value of the gas in kJ/m**3\n",
+ "T=340 #The temperature at the beginning of the compression stroke in K\n",
+ "p=0.97 #The pressure at the beginning of the compression stroke in bar\n",
+ "g=1.4 #Adiabatic index\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(math.pi/4.0)*d**2*L\n",
+ "Vc=Vs/(r-1)\n",
+ "V=Vs+Vc\n",
+ "Vg=V/7.0\n",
+ "Vntp=((p*Vg)/T)*(273/1.013)\n",
+ "Q=Vntp*CV*(N/(2.0*60.0))\n",
+ "ip=(imep*10**5*L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*(1/1000.0))\n",
+ "ni=(ip/Q)*100\n",
+ "na=(1-(1/r)**(g-1))*100\n",
+ "nr=(ni/na)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The indicated power = \",round(ip,2),\"kW\" \n",
+ "print\"The thermal efficiency = \",round(ni,1),\"percent\" \n",
+ "print\"The relative efficiency = \",round(nr,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The indicated power = 14.14 kW\n",
+ "The thermal efficiency = 27.9 percent\n",
+ "The relative efficiency = 56.5 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.9 page no: 529"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=6.0 #Number of cylinder\n",
+ "bp=130.0 #Brake power in kW\n",
+ "N=1800.0 #The speed of the engine in rpm \n",
+ "CV=42000.0 #The calorific value of the fuel in kJ/kg\n",
+ "C=86.0 #The composition of carbon in the fuel in percent\n",
+ "H=13.0 #The composition of Hydrogen in the fuel in percent\n",
+ "NC=1.0 #The non combustibles present in the fuel in percent\n",
+ "na=85.0 #The absolute volumetric efficiency in percent\n",
+ "ni=38.0 #The indicated thermal efficiency in percent\n",
+ "nm=80.0 #The mechanical efficiency in percent\n",
+ "ac=110.0 #The excess consumption of air in percent\n",
+ "sb=1.2 #The stroke to the bore ratio \n",
+ "da=1.3 #The density of air in kg/m**3\n",
+ "\n",
+ "#Calculations \n",
+ "import math\n",
+ "saf=(((C/100.0)*(32/12.0))+((H/100.0)*(8/1.0)))*(1/0.23)\n",
+ "aaf=saf*(1+1.1)\n",
+ "Ma=(0.23*32)+(0.77*28)\n",
+ "a=(C/100)/12.0\n",
+ "b=(H/100.0)/2.0\n",
+ "x=aaf/Ma\n",
+ "c=(0.21*x)-a-(b/2.0)\n",
+ "d1=0.79*x\n",
+ "ip=bp/(nm/100.0)\n",
+ "mf=ip/((ni/100.0)*CV)\n",
+ "ma=mf*aaf\n",
+ "Va=ma/da\n",
+ "Vs=Va/(na/100.0)#The swept volume per second in m**3/s\n",
+ "d=((Vs*(4/math.pi)*(1/1.2)*((2*60)/N)*(1/n))**(1/3.0))*1000\n",
+ "L=1.2*d\n",
+ "T=a+c+d1\n",
+ "CO2=(a/T)*100\n",
+ "O2=(c/T)*100\n",
+ "N2=(d1/T)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The volumetric composition of dry exhaust gas :\" \n",
+ "print\"1) CO2 = \",round(a,2),\"kmol\",\"and\" \n",
+ "print\"volume = \",round(CO2,2),\"percent\" \n",
+ "print\"2) O2 = \",round(c,3),\"kmol\",\"and\" \n",
+ "print\"volume = \",round(O2,3),\"percent\"\n",
+ "print\"3) N2 = \",round(d1,3),\"kmol\",\"and\" \n",
+ "print\"volume = \",round(N2,3),\"percent\"\n",
+ "print\"The bore of the engine = \",round(d,0),\"mm\" \n",
+ "print\"The stroke of the engine = \",round(L,1),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volumetric composition of dry exhaust gas :\n",
+ "1) CO2 = 0.07 kmol and\n",
+ "volume = 7.03 percent\n",
+ "2) O2 = 0.117 kmol and\n",
+ "volume = 11.456 percent\n",
+ "3) N2 = 0.831 kmol and\n",
+ "volume = 81.517 percent\n",
+ "The bore of the engine = 149.0 mm\n",
+ "The stroke of the engine = 178.8 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.10 page no: 531"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.18 #The diameter of the cylinder in m\n",
+ "L=0.24 #The length of the stroke in m\n",
+ "t=30.0 #Duration trail in min \n",
+ "N=9000.0 #Number of revolutions \n",
+ "Ne=4450.0 #Total number of explosions\n",
+ "pmi=5.35 #Gross imep in bar\n",
+ "pp=0.35 #Pumping imep in bar\n",
+ "W=40.0 #Net load on brake wheel in kg\n",
+ "dd=0.96 #Diameter of the brake wheel drum in m\n",
+ "dr=0.04 #Diameter of the rope in m\n",
+ "V=2.6 #Volume of gas used in m**3\n",
+ "pg=136.0 #pressure of gas in mmof Hg\n",
+ "dg=0.655 #The density of gas in kg/m**3\n",
+ "T=290.0 #The ambient temperature of air in K\n",
+ "CV=19000.0 #The calorific value of the fuel in kJ/m**3\n",
+ "ta=40.0 #Total air used in m**3\n",
+ "p=720.0 #Pressure of air in mm of Hg\n",
+ "Te=340.0 #Temperature of exhaust gas in degree centigrade \n",
+ "Cpg=1.1 #Specific heat of gas in kJ/kgK\n",
+ "C=80.0 #Cooling water circulated in kg\n",
+ "Tr=30.0 #Rise in temperature of cooling water in degree centigrade \n",
+ "R=287.0 #Real gas constant in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "#import math\n",
+ "ip=(pmi-pp)*10**5*L*(math.pi/4.0)*d**2*(Ne/(30.0*60.0))*(1/1000.0)\n",
+ "bp=(math.pi*(N/(30.0*60.0))*W*9.81*(dd+dr)*(1/1000.0))\n",
+ "pgs=760+(pg/13.6)\n",
+ "Vg=((pgs*V)/290.0)*(273/760.0)\n",
+ "Q=(Vg*CV)/30.0\n",
+ "Qbp=bp*60\n",
+ "Qc=(C/t)*4.18*Tr\n",
+ "Va=(((p*ta)/T)*(273/760.0))/30.0\n",
+ "da=(1.013*10**5)/(R*273)\n",
+ "ma=Va*da\n",
+ "mg=(Vg/30)*dg\n",
+ "me=ma+mg\n",
+ "Qe=me*Cpg*(Te-(T-273))\n",
+ "Qu=Q-(Qe+Qc+Qbp)\n",
+ "nm=(bp/ip)*100\n",
+ "ni=((ip*60)/Q)*100 \n",
+ "x=((Qbp/1571.0)*100)\n",
+ "y=((Qc/1571.0)*100)\n",
+ "z=((Qe/1571.0)*100)\n",
+ "k=((Qu/1571.0)*100)\n",
+ "Qf=Qbp+Qc+Qe+Qu\n",
+ "\n",
+ "#Output\n",
+ "print Qe\n",
+ "print\" HEAT BALANCE SEAT\"\n",
+ "\n",
+ "print \" \\nHeat input kj/min % heat expenditure Kj/min %\"\n",
+ "print\"--------------------------------------------------------------------------------------------------\"\n",
+ "print\"heat supplied by fuel \",round(Q,0),\" 100 (a)Heat in bp \", round(Qbp,1),\" \",round(x,1)\n",
+ "print\" (b)Heat loss to cooling tower \",round(Qc,0),\" \",round(y,1)\n",
+ "print\" (c) Heat to exhaust gases \",round(Qe,0),\" \",round(z,0)\n",
+ "print\" (d)unaccounted losses \",round(Qu,0),\" \",round(k,1)\n",
+ "print\"Total \",round(Q,0),\" 100 total \",round(Qf,0), \" 100.0\"\n",
+ " \n",
+ "print\"\\nMechanical efficiency is\",round(nm,2),\"percent\"\n",
+ "print\"Thermal efficiency is\",round(ni,1),\"percent\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "565.475307496\n",
+ " HEAT BALANCE SEAT\n",
+ " \n",
+ "Heat input kj/min % heat expenditure Kj/min %\n",
+ "--------------------------------------------------------------------------------------------------\n",
+ "heat supplied by fuel 1571.0 100 (a)Heat in bp 369.8 23.5\n",
+ " (b)Heat loss to cooling tower 334.0 21.3\n",
+ " (c) Heat to exhaust gases 565.0 36.0\n",
+ " (d)unaccounted losses 301.0 19.1\n",
+ "Total 1571.0 100 total 1571.0 100.0\n",
+ "\n",
+ "Mechanical efficiency is 81.65 percent\n",
+ "Thermal efficiency is 28.8 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.11 page no: 533"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given \n",
+ "bp=30 #The brake power in kw\n",
+ "mf=10 #Mass flow rate of fuel in kg/h\n",
+ "CV=42000 #Calorific value of the fuel in kJ/kg\n",
+ "mw=9 #Mass flow rate of water in kg/min\n",
+ "Tr=60 #Rise in temperature of the cooling water in degree centigrade\n",
+ "mwe=9.5 #Mass flow rate of water through exhaust gas calorimeter in kg/min\n",
+ "Tc=40 #Rise in temperature when passing through calorimeter in degree centigrade\n",
+ "Te=80 #Temperature of exhaust gas leaving the calorimeter in degree centigrade\n",
+ "a=20 #Air fuel ratio\n",
+ "T=17 #Ambient temperature in degree centigrade\n",
+ "Cpw=4.18 #Specific heat of water in kJ/kgK\n",
+ "Cpg=1 #Mean specific heat of gas in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "Qf=(mf/60.0)*CV \n",
+ "Qbp=bp*60\n",
+ "Qc=mw*Cpw*Tr\n",
+ "mg=(mf/60.0)+(mf/60.0)*a\n",
+ "Qe=(mwe*Cpw*Tc)+(mg*Cpg*(Te-T))\n",
+ "Qu=Qf-(Qbp+Qc+Qe)\n",
+ "x=((Qbp/Qf))*100\n",
+ "y=(Qc/Qf)*100\n",
+ "z=(Qe/Qf)*100\n",
+ "k=(Qu/Qf)*100\n",
+ "\n",
+ "#Output\n",
+ "print\" HEAT BALANCE SEAT\"\n",
+ "print \"\\n Heat input kj/min % heat expenditure Kj/min %\"\n",
+ "print\"--------------------------------------------------------------------------------------------------\"\n",
+ "print\"heat supplied by fuel \",Qf,\" 100 (a)Heat in bp \", Qbp,\" \",round(x,0)\n",
+ "print\" (b)Heat loss to cooling tower \",round(Qc,0),\" \",round(y,0)\n",
+ "print\" (c) Heat to exhaust gases \",round(Qe,0),\" \",round(z,0)\n",
+ "print\" (d)unaccounted losses \",round(Qu,0),\" \",round(k,0)\n",
+ "print\"Total \",Qf,\" 100 total \",Qf, \" 100\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " HEAT BALANCE SEAT\n",
+ "\n",
+ " Heat input kj/min % heat expenditure Kj/min %\n",
+ "--------------------------------------------------------------------------------------------------\n",
+ "heat supplied by fuel 7000.0 100 (a)Heat in bp 1800 26.0\n",
+ " (b)Heat loss to cooling tower 2257.0 32.0\n",
+ " (c) Heat to exhaust gases 1809.0 26.0\n",
+ " (d)unaccounted losses 1134.0 16.0\n",
+ "Total 7000.0 100 total 7000.0 100\n"
+ ]
+ }
+ ],
+ "prompt_number": 85
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16.12 page no: 534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "n=4.0 #Number of cylinders\n",
+ "d=0.085 #The diameter of the bore m\n",
+ "L=0.095 #The length of the stroke in m\n",
+ "tr=0.35 #Torque radius in m\n",
+ "N=3000.0 #The speed of the engine in rpm\n",
+ "w=430.0 #Net brake load in N\n",
+ "w1=300.0 #Net brake load produced at the same speed by three cylinders in N\n",
+ "mf=0.24 #The mass flow rate of fuel in kg/min\n",
+ "CV=44000.0 #The calorific value of the fuel in kJ/kg\n",
+ "mw=65.0 #Mass flow rate of water in kg/min\n",
+ "Tw=12.0 #The rise in temperature in degree centigrade\n",
+ "a=15.0 #The air fuel ratio \n",
+ "Te=450.0 #The temperature of the exhaust gas in degree centigrade \n",
+ "Ta=17.0 #Ambient temperature in degree centigrade\n",
+ "p=76.0 #Barometric pressure in cm of Hg\n",
+ "H=15.5 #The proportion of hydrogen by mass in the fuel in percent\n",
+ "Cpe=1.0 #The mean specific heat of dry exhaust gas in kJ/kgK\n",
+ "Cps=2.0 #The specific heat of super heated steam in kJ/kgK\n",
+ "Cpw=4.18 #The specific heat of water in kJ/kgK\n",
+ "Ts=100.0 #At 76 cm of Hg The temperature in degree centigrade \n",
+ "hfg=2257 #The Enthalpy in kJ/kg\n",
+ "R=287.0 #Real gas cons tant in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "bp=(2*math.pi*N*w*tr)/(60.0*1000.0)\n",
+ "bp1=(2*math.pi*N*w1*0.35)/(60.0*1000.0)\n",
+ "ip=bp-bp1\n",
+ "ip1=n*ip\n",
+ "imep=((ip1*60*1000)/(L*(math.pi/4.0)*d**2*(N/2.0)*n))/10.0**5\n",
+ "ni=((ip1*60)/(mf*CV))*100\n",
+ "bsfc=(mf*60)/bp\n",
+ "Vs=(math.pi/4.0)*d**2*L*(N/2.0)*n\n",
+ "ma=a*mf\n",
+ "da=(1*10**5)/(R*(Ta+273))\n",
+ "Va=ma/da\n",
+ "nv=(Va/Vs)*100\n",
+ "Qf=mf*CV\n",
+ "Qbp=bp*60\n",
+ "Qc=mw*Cpw*Tw\n",
+ "mv=9*(H/100.0)*mf\n",
+ "me=ma+mf-mv\n",
+ "Qe=me*Cpe*(Te-Ta)\n",
+ "Qs=(mv*((Cpw*(Ts-Ta))+hfg+(Cps*(Te-Ts))))\n",
+ "Qu=Qf-(Qbp+Qc+Qe+Qs)\n",
+ "x=(Qbp/Qf)*100\n",
+ "y=(Qc/Qf)*100\n",
+ "z=(Qe/Qf)*100\n",
+ "k=(Qs/Qf)*100\n",
+ "l=(Qu/Qf)*100 \n",
+ "\n",
+ "#Output\n",
+ "print\" HEAT BALANCE SEAT\"\n",
+ "print \" Heat input kj/min % heat expenditure Kj/min %\"\n",
+ "print\"----------------------------------------------------------------------------------------------------\"\n",
+ "print\"heat supplied by fuel \",Qf,\" 100 (a)Heat in bp \", Qbp,\" \",round(x,2)\n",
+ "print\" (b)Heat loss to cooling tower \",round(Qc,0),\" \",round(y,2)\n",
+ "print\" (c) Heat to dry exhaust gases \",round(Qe,0),\" \",round(z,1)\n",
+ "print\" (d)Heat loss in steam \",round(Qs,0),\" \",round(k,1)\n",
+ "print\" (e)unaccounted losses \",round(Qu,0),\" \",round(l,2)\n",
+ "print\"Total \",Qf,\" 100 total \",Qf, \" 100\"\n",
+ "\n",
+ "print\"\\nimpep \",round(imep,2),\"bar\"\n",
+ "print\"Thermal efficiency \",round(ni,1),\"percent\"\n",
+ "print\"bsfc=\",round(bsfc,4),\"kg/kwh\"\n",
+ "print\"Volumetric efficiency=\",round(nv,1),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " HEAT BALANCE SEAT\n",
+ " Heat input kj/min % heat expenditure Kj/min %\n",
+ "----------------------------------------------------------------------------------------------------\n",
+ "heat supplied by fuel 10560.0 100 (a)Heat in bp 2836.85816619 26.86\n",
+ " (b)Heat loss to cooling tower 3260.0 30.87\n",
+ " (c) Heat to dry exhaust gases 1518.0 14.4\n",
+ " (d)Heat loss in steam 1106.0 10.5\n",
+ " (e)unaccounted losses 1839.0 17.41\n",
+ "Total 10560.0 100 total 10560.0 100\n",
+ "\n",
+ "impep 10.61 bar\n",
+ "Thermal efficiency 32.5 percent\n",
+ "bsfc= 0.3046 kg/kwh\n",
+ "Volumetric efficiency= 92.6 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap3.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap3.ipynb new file mode 100755 index 00000000..a0a9064e --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap3.ipynb @@ -0,0 +1,1164 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Reactive Sysyem"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=20.0 #Methanol burned with excess air in percentage \n",
+ "p=1.0 #Pressure of air in bar\n",
+ "t=27.0 #Temperature of air in degree centigrade\n",
+ "O=32.0 #The molecular weight of oxygen\n",
+ "N=28.0 #The molecular weight of nitrogen\n",
+ "R=8314.0 #Universal gas constant in Nm/kmolK\n",
+ "C=32.0 #Molecular weight of methanol\n",
+ "CO=44.0 #Molecular weight of the carbondioxide \n",
+ "H=18.0 #Molecular weight of the water\n",
+ "\n",
+ "#Calculations\n",
+ "S=((1.8*O)+(6.768*N))/C\n",
+ "A=((1.8*O)+(6.768*N))/C\n",
+ "M=1.8+6.768\n",
+ "V=(M*R*(t+273))/(p*10**5)\n",
+ "T=(1+1.8+6.768)\n",
+ "Cm=(1/T)\n",
+ "Om=(1.8/T)\n",
+ "Nm=(6.768/T)\n",
+ "Mr=(Cm*C)+(Om*O)+(Nm*N)\n",
+ "Tp=(1+2+6.768+0.3)\n",
+ "COm=(1/Tp)\n",
+ "Hp=(2/Tp)\n",
+ "Np=(6.768/Tp)\n",
+ "Op=(0.3/Tp)\n",
+ "Mp=(COm*CO)+(Hp*H)+(Np*N)+(Op*O)\n",
+ "Pp=(Hp*p)\n",
+ "D=60\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The volume of air supplied per kmole of fuel = \",round(V,3),\"m**3/kmole fuel\"\n",
+ "print\"(b) The molecular weight of the reactants = \",round(Mr,3)\n",
+ "print\"The molecular weight of the products =\",round(Mp,3)\n",
+ "print\"(c) The dew point of the products = \",D,\"degree centigrade\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The volume of air supplied per kmole of fuel = 213.703 m**3/kmole fuel\n",
+ "(b) The molecular weight of the reactants = 29.171\n",
+ "The molecular weight of the products = 27.722\n",
+ "(c) The dew point of the products = 60 degree centigrade\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C1=40.0 #The content of C7H16 in the fuel in percentage\n",
+ "C2=60.0 #The content of C8H18 in the fuel in percentage\n",
+ "d=0.12 #The diameter of the bore in m\n",
+ "l=0.145 #The length of the bore in m\n",
+ "r=8.5 #Compression ratio \n",
+ "p=1.1 #Pressure at exhaust stroke in bar\n",
+ "T=720.0 #The temperature at the exhaust stroke in K\n",
+ "O=32.0 #The molecular weight of oxygen\n",
+ "N=28.0 #The molecular weight of nitrogen\n",
+ "C3=100.0 #Molecular weight of C7H16\n",
+ "C4=114.0 #The molecular weight of C8H18\n",
+ "R=8314.0 #Universal gas constant in Nm/kmolK\n",
+ "CO2=44.0 #Molecular weight of the carbondioxide \n",
+ "C5=28.0 #Molecular weight of the carbonmonoxide\n",
+ "H=18.0 #Molecular weight of the water\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "N2=100-(12+1.5+2.5)\n",
+ "Y=84/3.76\n",
+ "X=13.5/7.6\n",
+ "Z=(22.34-15.25)*2\n",
+ "Hl=(6.4+10.8)/2.0\n",
+ "Hr=7.98\n",
+ "Hd=Hl-Hr\n",
+ "A=((12.58*(O+(3.76*N)))/(((C1/100.0)*C3)+((C2/100.0)*C4)))\n",
+ "Vs=(math.pi/4.0)*d**2*l\n",
+ "Vc=Vs/(r-1)\n",
+ "M=((6.757*CO2)+(0.8446*C5)+(1.408*O)+(47.3*N)+(8.6*H))/(6.757+0.8446+1.408+47.3+8.6)\n",
+ "R1=R/M\n",
+ "m=((p*10**5)*Vc)/(R1*T)\n",
+ "\n",
+ "#Output \n",
+ "print\"(a)The air/fuel ratio = \",round(A,2)\n",
+ "print\"(b)The mass of the exhaust gases in the clearance space = \",round(m*1000,3),\"*10**-3kg\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The air/fuel ratio = 15.93\n",
+ "(b)The mass of the exhaust gases in the clearance space = 0.114 *10**-3kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C=0.86 #The amount of carbon content in the 1kg of fuel by weight in kg\n",
+ "H=0.05 #The amount of hydrogen content in the 1kg of fuel by weight in kg\n",
+ "O=0.02 #The amount of oxygen content in the 1kg of fuel by weight in kg\n",
+ "S=0.005 #The amount of sulphur content in the 1kg of fuel by weight in kg\n",
+ "N=0.065 #The amount of nitrogen content in the 1kg of fuel by weight in kg\n",
+ "E=25.0 #The amount of excess air supplied in percentage\n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n=28.0 #Molecular weight of the nitrogen\n",
+ "\n",
+ "#Calculations\n",
+ "o1=(o/c)*C\n",
+ "coa=(co/c)*C\n",
+ "o2=(o/4.0)*H\n",
+ "h2=(36/4.0)*H\n",
+ "o3=(o/s)*S\n",
+ "s1=(so/s)*S\n",
+ "To=o1+o2+o3\n",
+ "Tt=To-O\n",
+ "As=(Tt*100)/23.0\n",
+ "as_=As*(1+(E/100.0))\n",
+ "o2a=0.23*(E/100)*As\n",
+ "n2a=0.77*(1+(E/100))*As\n",
+ "n2e=n2a+N\n",
+ "Tw=coa+n2e+o2a\n",
+ "pco=(coa/Tw)*100\n",
+ "pn=(n2e/Tw)*100\n",
+ "po=(o2a/Tw)*100\n",
+ "mco=(coa/co)\n",
+ "mn=(n2e/n)\n",
+ "mo=(o2a/o)\n",
+ "Tm=mco+mn+mo\n",
+ "vco=(mco/Tm)*100\n",
+ "vn=(mn/Tm)*100\n",
+ "vo=(mo/Tm)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)Stoichiometric air/fuel ratio = \",round(As,2) \n",
+ "print\"(b)The percentage of dry products of combustion by weight :\"\n",
+ "print\" CO2 = \",round(pco,2),\"percent\"\n",
+ "print\"N2 = \",round(pn,2),\"percent\"\n",
+ "print\"O2 = \",round(po,2),\"percent\"\n",
+ "print\"(c)The percentage of dry products of combustion by volume : \"\n",
+ "print\"CO2 = \",round(vco,2),\"percent\"\n",
+ "print\"N2 = \",round(vn,2),\"percent\"\n",
+ "print\"O2= \",round(vo,2),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Stoichiometric air/fuel ratio = 11.64\n",
+ "(b)The percentage of dry products of combustion by weight :\n",
+ " CO2 = 20.89 percent\n",
+ "N2 = 74.68 percent\n",
+ "O2 = 4.44 percent\n",
+ "(c)The percentage of dry products of combustion by volume : \n",
+ "CO2 = 14.47 percent\n",
+ "N2 = 81.3 percent\n",
+ "O2= 4.23 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "CO=12.0 #The composition of carbondioxide of combustion by volume in percentage \n",
+ "C=0.5 #The composition of carbonmoxide of combustion by volume in percentage \n",
+ "O=4.0 #The composition of oxygen of combustion by volume in percentage \n",
+ "N=83.5 #The composition of nitrogen of combustion by volume in percentage \n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n1=28.0 #Molecular weight of the nitrogen\n",
+ "h=2.0 #Molecular weight of the hydrogen\n",
+ "\n",
+ "#Calculations\n",
+ "m=12+0.5\n",
+ "x=N/3.76\n",
+ "z=(x-(CO+(C/2)+O))*2\n",
+ "n=z*h\n",
+ "Af=((x*o)+(N*n1))/((m*c)+(n))\n",
+ "As=((18.46*o)+(69.41*n1))/173.84\n",
+ "Ta=(Af/As)*100\n",
+ "mc=((m*c)/173.84)*100\n",
+ "mh=(n/173.84)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The air/fuel ratio = \",round(Af,1)\n",
+ "print\"(b)The percent theoretical air = \",round(Ta,1),\"%\"\n",
+ "print\"(c)The percentage composition of fuel on a mass basis : \" \n",
+ "print\"C = \",round(mc,1),\"%\"\n",
+ "print\"H = \",round(mh,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The air/fuel ratio = 17.5\n",
+ "(b)The percent theoretical air = 120.3 %\n",
+ "(c)The percentage composition of fuel on a mass basis : \n",
+ "C = 86.3 %\n",
+ "H = 13.7 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C=86.0 #The composition of carbon in the fuel by weight in percentage\n",
+ "H=14.0 #The composition of hydrogen in the fuel by weight in percentage\n",
+ "e=1.25 #Equivalent ratio\n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n=28.0 #Molecular weight of the nitrogen\n",
+ "h2=2.0 #Molecular weight of the hydrogen\n",
+ "Fc=0.86 #Fraction of C\n",
+ "\n",
+ "#Calculations\n",
+ "Ra=1/Fc\n",
+ "x=2*(1+(0.9765/2.0)-(1.488*0.8))\n",
+ "Tm=0.5957+0.4043+4.476\n",
+ "vc=(0.5957/Tm)*100\n",
+ "vco=(0.4043/Tm)*100\n",
+ "vn=(4.476/Tm)*100\n",
+ "\n",
+ "#Calculations\n",
+ "print\"The percentage analysis of dry exhaust gas by volume : \"\n",
+ "print\"CO = \",round(vc,2),\"percent\" \n",
+ "print\"CO2 = \",round(vco,2),\"percent\" \n",
+ "print\"N2 = \",round(vn,2),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage analysis of dry exhaust gas by volume : \n",
+ "CO = 10.88 percent\n",
+ "CO2 = 7.38 percent\n",
+ "N2 = 81.74 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page no 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of both reactants and products in degree centigrade\n",
+ "p=1.0 #The pressure of both reactants and products in bar\n",
+ "\n",
+ "#Calculations \n",
+ "h=0\n",
+ "hf1=-103.85\n",
+ "hf2=-393.52\n",
+ "hf3=-285.8\n",
+ "hf4=(3*hf2)+(4*hf3)\n",
+ "Q=hf4-hf1\n",
+ "\n",
+ "#Output\n",
+ "print\" The heat transfer per mole of fuel = \",Q,\"kJ/mol fuel\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The heat transfer per mole of fuel = -2219.91 kJ/mol fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page no 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of the air entering the diesel engine in degree centigrade \n",
+ "T=600.0 #The temperature at which the products are released in K\n",
+ "Ta=200.0 #Theoretical air used in percentage \n",
+ "Q=-93.0 #Heat loss from the engine in MJ/kmol fuel\n",
+ "f=1.0 #The fuel rate in kmol/h\n",
+ "\n",
+ "#Calculations \n",
+ "hfr=-290.97\n",
+ "h1=-393.52\n",
+ "h11=12.916\n",
+ "hfc=h1+h11\n",
+ "h2=-241.82\n",
+ "h22=10.498\n",
+ "hfh=h2+h22\n",
+ "h3=0 \n",
+ "h33=9.247\n",
+ "hfo=h3+h33\n",
+ "h4=0\n",
+ "h44=8.891\n",
+ "hfn=h4+h44\n",
+ "hfp=(12*hfc)+(13*hfh)+(18.5*hfo)+(139.12*hfn)\n",
+ "W=Q+hfr-hfp\n",
+ "W1=(f*W*10**3)/3600.0\n",
+ "\n",
+ "#Output\n",
+ "print\"The work for a fuel rate of 1 kmol/h is \",round(W1,0),\"kW\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The work for a fuel rate of 1 kmol/h is 1606.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.8 Page no 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "P=600 #Power of an engine in kW\n",
+ "t=25 #Temperature at which fuel is used in degree centigrade\n",
+ "Ta=150 #Theoretical air used in percentage\n",
+ "T1=400 #The temperature at which air enters in K\n",
+ "T2=700 #The temperature at which the products of combustion leave in K\n",
+ "Q=-150 #The heat loss from the engine in kW\n",
+ "C=12 #Molecular weight of carbon\n",
+ "h=1 #Molecular weight of hydrogen\n",
+ "\n",
+ "#Calculations\n",
+ "hfc=-259.28\n",
+ "hfo1=3.029\n",
+ "hfn1=2.971\n",
+ "HR=(hfc)+(1.5*12.5*hfo1)+(1.5*12.5*3.76*hfn1)\n",
+ "hfco=-393.52\n",
+ "hfco1=17.761\n",
+ "hfh=-241.82\n",
+ "hfh1=14.184\n",
+ "hfo2=12.502\n",
+ "hfn2=11.937\n",
+ "HP=(8*(hfco+hfco1))+(9*(hfh+hfh1))+(6.25*hfo2)+(70.5*hfn2)\n",
+ "H=HP-HR\n",
+ "nf=((Q-P)*3600)/(H*10.0**3)\n",
+ "M=(8*C)+(18*h)\n",
+ "mf=nf*M\n",
+ "\n",
+ "#Output\n",
+ "print\" The fuel consumption for complete combustion is \",round(mf,1),\"kg/h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The fuel consumption for complete combustion is 74.3 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25 #Temperature at which fuel is used for combustion in degree centigrade \n",
+ "p=1 #The pressure at which fuel is used in bar\n",
+ "T=400 #The temperature of the products of combustion in K\n",
+ "R=8.314*10**-3 #Universal gas constant\n",
+ "hfco=-393.52 #The enthalpy of the carbondioxide in MJ/kmol fuel\n",
+ "hfco1=4.008 #The change in enthalpy of the carbondioxide for the given conditions in MJ/kmol fuel\n",
+ "hfh=-241.82 #The enthalpy of the water in MJ/kmol fuel\n",
+ "hfh1=3.452 #The change in enthalpy of the water for the given conditions in MJ/kmol fuel\n",
+ "\n",
+ "#Calculations\n",
+ "hfc=-103.85 \n",
+ "HR=(1*(hfc-(R*(t+273))))+(5*(-R*(t+273)))\n",
+ "HP=(3*(hfco+hfco1-(R*T)))+(4*(hfh+hfh1-(R*T)))\n",
+ "Q=HP-HR\n",
+ "Q1=-Q\n",
+ "\n",
+ "#Output\n",
+ "print\"The heat transfer per mole of propane = \",round(Q1,1),\"kJ/mol propane\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat transfer per mole of propane = 2026.6 kJ/mol propane\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=1500 #The given temperature in K\n",
+ "hfco=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hf1=61.714 #The change in enthalpy for actual state and reference state in MJ/kmol\n",
+ "hfc=-110.52 #The enthalpy of formation for carbonmonoxide in MJ/kmol\n",
+ "hf2=38.848 #The change in enthalpy of CO for actual and reference state in MJ/kmol\n",
+ "hfo=0 #The enthalpy of formation for oxygen gas\n",
+ "hf3=40.61 #The change in enthalpy of oxygen for different states in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "HP=hfco+hf1\n",
+ "HR=(hfc+hf2)+(0.5*(hfo+hf3))\n",
+ "H=HP-HR\n",
+ "\n",
+ "#Output\n",
+ "print\" The enthalpy of combustion is \",round(H,1),\"MJ/kmol CO\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The enthalpy of combustion is -280.4 MJ/kmol CO\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page no 83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=30 #The amount of excess air in percentage\n",
+ "tp=400 #The temperature at which propane enters in K\n",
+ "ta=300 #The temperature at which air enters in K\n",
+ "T=900 #The temperature at which products leave in K\n",
+ "m=83.7 #The average molar specific heat of propane at consmath.tant pressure in kJ/kmolK\n",
+ "Mp=44 #The molecular weight of propane\n",
+ "hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hf1=28.041 #The change in enthalpy of CO2 for actual and reference state in MJ/kmol\n",
+ "hfh=-241.82 #The enthalpy of formation for water in MJ/kmol\n",
+ "hf2=21.924 #The change in enthalpy of water for actual and reference state in MJ/kmol\n",
+ "hfn=0 #The enthalpy of nitrogen gas \n",
+ "hf3=18.221 #The change in enthalpy of nitrgen for actual and reference state in MJ/kmol\n",
+ "hfo=0 #The enthalpy of oxygen gas \n",
+ "hf4=19.246 #The change in enthalpy of oxygen for actual and reference state in MJ/kmol\n",
+ "hfp=-103.85 #The enthalpy of formation for propane in MJ/kmol\n",
+ "R=0.0837 #Universal gas constant \n",
+ "hfo1=0 #The enthalpy of oxygen gas \n",
+ "hf11=0.054 #The change in enthalpy of oxygen gas for actual and reference state in MJ/kmol\n",
+ "hfn1=0 #The enthalpy of nitrogen gas\n",
+ "hfn22=0.054 #The change in enthalpy of nitrogen for actual and reference state in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "HP=(3*(hfc+hf1))+(4*(hfh+hf2))+(24.44*(hfn+hf3))+(1.5*(hfo+hf4))\n",
+ "HR=(1*(hfp+(R*(tp-ta))))+(6.5*(hfo1+hf11))+(24.44*(hfn1+hfn22))\n",
+ "Q=HP-HR\n",
+ "Q1=(-Q/Mp)\n",
+ "\n",
+ "#Output\n",
+ "print\" The amount of heat transfer per kg of fuel is \",round(Q1,3),\"MJ/kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The amount of heat transfer per kg of fuel is 32.0 MJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Ta=150 #The presence of Theoretical air\n",
+ "hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hfh=-285.8 #The enthalpy of formation for water in MJ/kmol\n",
+ "hfon=0 #The enthalpy of formation for oxygen and nitrogen gas \n",
+ "hfch=-74.87 #The enthalpy of formation for methane in MJ/kmol\n",
+ "np=2 #Number of moles of product\n",
+ "nr=4 #Number of moles of reactant\n",
+ "R=8.314*10**-3 #Universal gas constant \n",
+ "t=298 #The temperature in K\n",
+ "hfh1=-241.82 #The enthalpy of formation for water in MJ/kmol\n",
+ "np1=4 #Number of moles of product\n",
+ "nr1=4 #Number of moles of reactant\n",
+ "\n",
+ "#Calculations\n",
+ "HP=(hfc)+(2*hfh)\n",
+ "HR=1*hfch\n",
+ "H=HP-HR\n",
+ "n=np-nr\n",
+ "U1=H-(n*R*t)\n",
+ "HP1=(1*hfc)+(2*hfh1)\n",
+ "H1=HP1-HR\n",
+ "n1=np1-nr1\n",
+ "U2=H1-(n1*R*t)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The water as liquid\" \n",
+ "print\"The standard enthalpy of combustion is \",H, \"MJ/kmol\"\n",
+ "print\"The standard internal energy of combustion is \",round(U1,3),\"MJ/kmol\"\n",
+ "print\"(b)The water as a gas \"\n",
+ "print\"The standard enthalpy of combustion is \",H1,\"MJ/kmol\"\n",
+ "print\"The standard internal energy of combustion is \",U2,\"MJ/kmol\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The water as liquid\n",
+ "The standard enthalpy of combustion is -890.25 MJ/kmol\n",
+ "The standard internal energy of combustion is -885.295 MJ/kmol\n",
+ "(b)The water as a gas \n",
+ "The standard enthalpy of combustion is -802.29 MJ/kmol\n",
+ "The standard internal energy of combustion is -802.29 MJ/kmol\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "cv=44000 #The lower calorific value of liquid fuel in kJ/kg\n",
+ "C=84 #The carbon content present in the fuel in percentage\n",
+ "H=16 #The hydrogen content present in the fuel in percentage\n",
+ "t=25 #The temperature in degree centigrade\n",
+ "hfg=2442 #The enthalpy of vaporization for water in kJ/kg\n",
+ "c=12.0 #Molecular weight of carbon \n",
+ "h=2 #Molecular weight of hydrogen\n",
+ "co2=44.0 #Molecular weight of carbondioxide\n",
+ "h2o=18 #Molecular weight of water \n",
+ "o2=32.0 #Molecular weight of oxygen\n",
+ "R=8.314 #Universal gas constant in J/molK\n",
+ "\n",
+ "#Calculations\n",
+ "CO2=(0.84*(co2/c))\n",
+ "H2O=(0.16*(h2o/h))\n",
+ "cvd=H2O*hfg\n",
+ "HHV=cv+cvd\n",
+ "np=3.08/co2\n",
+ "nr=3.52/o2\n",
+ "n=np-nr\n",
+ "HHVv=HHV+(n*R*(t+273))\n",
+ "LHVv=cv+(n*R*(t+273))\n",
+ "\n",
+ "#Output\n",
+ "print\" The higher calorific value at constant pressure = \",HHV,\"kJ/kg fuel\" \n",
+ "print\"The higher calorific value at constant volume = \",round(HHVv,0),\"kJ/kg fuel\"\n",
+ "print\"The lower calorific value at constant volume = \",round(LHVv,0),\"kJ/kg fuel\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The higher calorific value at constant pressure = 47516.48 kJ/kg fuel\n",
+ "The higher calorific value at constant volume = 47417.0 kJ/kg fuel\n",
+ "The lower calorific value at constant volume = 43901.0 kJ/kg fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=100 #The amount of excess air in percent\n",
+ "T=298 #The temperature of reactants in K\n",
+ "nc=1 #Number of moles of propane\n",
+ "hfch=-103.85 #Enthalpy of formation for propane in MJ/kmol fuel\n",
+ "hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+ "hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel\n",
+ "hfon=0 #Enthalpy of formation for both oxygen and nitrogen gas\n",
+ "T1=1500 #Assuming the products temperature for fist trail in K\n",
+ "hfc1=61.714 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+ "hfh1=48.095 #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+ "hfo1=40.61 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+ "hfn1=38.405 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+ "T2=1600 #Assuming the products temperature for second trail in K\n",
+ "hfc2=67.58 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+ "hfh2=52.844 #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+ "hfo2=44.279 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+ "hfn2=41.903 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+ "\n",
+ "#Calculations\n",
+ "HR=nc*hfch\n",
+ "x=HR-((3*hfc)+(4*hfh)+(5*hfon)+(37.6*hfon))\n",
+ "hfn=x/37.6\n",
+ "HP1=(HR-x)+(3*hfc1)+(4*hfh1)+(5*hfo1)+(37.6*hfn1)\n",
+ "HP2=(HR-x)+(3*hfc2)+(4*hfh2)+(5*hfo2)+(37.6*hfn2)\n",
+ "Te=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+ "\n",
+ "#Output\n",
+ "print\" The adiabatic flame temperature for steady-flow process is \",round(Te,0),\"K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The adiabatic flame temperature for steady-flow process is 1510.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=600 #The initial temperature of air in K\n",
+ "p=1 #The initial pressure of air in atm\n",
+ "R=8.314 #Universal gas constant in J/molK\n",
+ "Tr=298 #The temperature of reactants in K\n",
+ "a=4.503 #Given Constants \n",
+ "b=-8.965*10**-3\n",
+ "c=37.38*10**-6\n",
+ "d=-36.49*10**-9\n",
+ "e=12.22*10**-12\n",
+ "hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+ "hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel\n",
+ "hfn=0 #Enthalpy of formation for nitrogen gas\n",
+ "hfc1=-74.87 #The enthalpy of formation for methane in MJ/kmol fuel \n",
+ "hfh1=9.247 #The change in enthalpy of the water in MJ/kmol\n",
+ "hfn1=8.891 #The change in enthalpy of nitrogen in MJ/kmol\n",
+ "Tc=3700 #The corresponding temperature for the enthalpy of guess nitrogen in K\n",
+ "T1=2800 #The temperature assumed for the first trail in K\n",
+ "hco1=140.444 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh1=115.294 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn1=85.345 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "T2=2500 #The temperature assumed for the second trail in K\n",
+ "hco2=121.926 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh2=98.964 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn2=74.312 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "T3=2600 #The temperature fo the third trail in K\n",
+ "hco3=128.085 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh3=104.37 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn3=77.973 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "Tc1=3000 #Assume temperature for first trail in K\n",
+ "hcoa1=146.645 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hha1=120.813 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hna1=89.036 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "Tc2=3200 #Assume temperature for the second trail in K\n",
+ "hcoa2=165.331 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hha2=137.553 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hna2=100.161 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "HP=(1*hfc)+(2*hfh)+(7.52*hfn)\n",
+ "hch=(R*((a*(T-Tr))+((b/2.0)*(T**2-Tr**2))+((c/3.0)*(T**3-Tr**3))+((d/4.0)*(T**4-Tr**4))+((e/5.0)*(T**5-Tr**5))))/1000.0\n",
+ "HR=((hfc1+hch)+(2*hfh1)+(7.52*hfn1))\n",
+ "x=HR-HP\n",
+ "hfn2=x/7.52\n",
+ "HP1=hco1+(2*hh1)+(7.52*hn1)+(HR-x)\n",
+ "HP2=hco2+(2*hh2)+(7.52*hn2)+(HR-x)\n",
+ "HP3=hco3+(2*hh3)+(7.52*hn3)+(HR-x)\n",
+ "Ta1=(((HR-HP2)/(HP3-HP2))*(T3-T2))+T2\n",
+ "UR1=HR-(10.52*R*10**-3*T)\n",
+ "UP1=hcoa1+(2*hha1)+(7.52*hna1)+(HR-x)-(0.08746*Tc1)\n",
+ "UP2=hcoa2+(2*hha2)+(7.52*hna2)+(HR-x)-(0.08746*Tc2)\n",
+ "Tu=(((UR1-UP1)/(UP2-UP1))*(Tc2-Tc1))+Tc1\n",
+ "\n",
+ "#Output\n",
+ "print\"The adiabatic flame temperature at \"\n",
+ "print\"(a)Constant pressure process is \",round(Ta1,0),\"K\" \n",
+ "print\"(b)Constant volume process is \",round(Tu,3),\"K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The adiabatic flame temperature at \n",
+ "(a)Constant pressure process is 2550.0 K\n",
+ "(b)Constant volume process is 3089.34 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.16 Page no 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=600.0 #Temperature at constant pressure process in K\n",
+ "p=1.0 #The pressure in atm\n",
+ "E=50.0 #The amount of excess air in percent\n",
+ "L=20.0 #The amount of less air in percent\n",
+ "cp=52.234 #Specific constant for methane in kJ/kmolK\n",
+ "t=298.0 #Assume the normal temperature in K\n",
+ "hfch=-74.87 #The enthalpy of formation for carbondioxide in MJ\n",
+ "ho=9.247 #The change in enthalpy of oxygen in MJ\n",
+ "hn=8.891 #The change in enthalpy of nitrogen in MJ\n",
+ "hfc1=-393.52 #The enthalpy of formation of carbondioxide in MJ\n",
+ "hfh1=-241.82 #The enthalpy of formation of water in MJ\n",
+ "Tc=2800.0 #The corresponding temperature in K\n",
+ "T1=2000 #The temperature for first trail in K\n",
+ "hfc11=91.45 #The enthalpy for the assume temp for carbondioxide in MJ\n",
+ "hfh11=72.689 #The change in enthalpy for the assume temp for water in MJ\n",
+ "hfn11=56.141 #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+ "hfo11=59.199 #The change in enthalpy for the assume temp for oxygen in MJ\n",
+ "T2=2100 #The temperature for second trail in K\n",
+ "hfc22=97.5 #The enthalpy for the assume temp for carbondioxide in MJ\n",
+ "hfh22=77.831 #The change in enthalpy for the assume temp for water in MJ\n",
+ "hfn22=59.748 #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+ "hfo22=62.986 #The change in enthalpy for the assume temp for oxygen in MJ\n",
+ "hfchr=-74.87 #The enthalpy of formation for methane in MJ\n",
+ "hor=9.247 #The change in enthalpy for oxygen in MJ\n",
+ "hnr=8.891 #The change in enthalpy for nitrogen in MJ\n",
+ "hfcop=-110.52 #The formation of enthalpy for carbonmoxide in MJ\n",
+ "hfcp=-393.52 #The formation of enthalpy for carbondioxide in MJ\n",
+ "hfhp=-241.82 #The formation of enthalpy for water in MJ\n",
+ "Tp1=2000.0 #The temperature for first trail in K\n",
+ "hco11=56.739 #The change in enthalpy for CO in MJ\n",
+ "hco211=91.45 #The change in enthalpy for CO2 in MJ\n",
+ "hh11=72.689 #The change in enthalpy for water in MJ\n",
+ "hn11=56.141 #The change in enthalpy for nitrogen in MJ\n",
+ "Tp2=2400 #The temperature for second trail in K\n",
+ "hco22=71.34 #The change in enthalpy for CO in MJ\n",
+ "hco222=115.788 #The change in enthalpy for CO2 in MJ\n",
+ "hh22=93.604 #The change in enthalpy for water in MJ\n",
+ "hn22=70.651 #The change in enthalpy for nitrogen in MJ\n",
+ "Tp3=2300.0 #The temperature for first trail in K\n",
+ "hco33=67.676 #The change in enthalpy for CO in MJ\n",
+ "hco233=109.671 #The change in enthalpy for CO2 in MJ\n",
+ "hh33=88.295 #The change in enthalpy for water in MJ\n",
+ "hn33=67.007 #The change in enthalpy for nitrogen in MJ\n",
+ "hccc=-283.022 #The only combustible substance is CO in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "hch=cp*(T-t)*10**-3\n",
+ "HR=hfch+hch+(3*ho)+(11.28*hn)\n",
+ "HP=hfc1+(2*hfh1)\n",
+ "x=HR-HP\n",
+ "hn2=x/11.28\n",
+ "HP1=hfc11+(2*hfh11)+(11.28*hfn11)+(hfo11)+(HR-x)\n",
+ "HP2=hfc22+(2*hfh22)+(11.28*hfn22)+(hfo22)+(HR-x)\n",
+ "Ta1=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+ "X=2*(2-1.6)\n",
+ "HRr=hfchr+hch+(1.6*hor)+(6.01*hnr)\n",
+ "HPp=(0.8*hfcop)+(0.2*hfcp)+(2*hfhp)\n",
+ "HPp1=(0.8*hco11)+(0.2*hco211)+(2*hh11)+(6.016*hn11)-HPp\n",
+ "HPp2=(0.8*hco22)+(0.2*hco222)+(2*hh22)+(6.016*hn22)+HPp\n",
+ "HPp3=(0.8*hco33)+(0.2*hco233)+(2*hh33)+(6.016*hn33)+HPp\n",
+ "Ta2=(((HRr-HPp3)/(HPp2-HPp3))*(Tp2-Tp3))+Tp3\n",
+ "Q=-0.8*hccc #The thermal energy loss in MJ/kmol fuel\n",
+ "\n",
+ "#Output\n",
+ "print\" The adiabatic flame temperature having \"\n",
+ "print\"(a)50 percent excess air is \",round(Ta1,1),\"K\"\n",
+ "print\"(b)20 percent less air is \",round(Ta2,2),\"K\"\n",
+ "print\"The loss of thermal energy due to incomplete combustion is \",round(Q,2),\"MJ/kmol fuel\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The adiabatic flame temperature having \n",
+ "(a)50 percent excess air is 2027.6 K\n",
+ "(b)20 percent less air is 2311.22 K\n",
+ "The loss of thermal energy due to incomplete combustion is 226.42 MJ/kmol fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T1=3000 #Given temperature in K\n",
+ "T2=4000 #Given temperature in K\n",
+ "p=1 #The pressure in atm\n",
+ "KP1=1.117 #Natural logarithm of equilibrium constant at 3000 K \n",
+ "KP2=-1.593 #Natural logarithm of equilibrium constant at 4000 K\n",
+ "a1=0.4 #The dissociation of 1 mole of CO2 for the first trail\n",
+ "a2=0.5 #The dissociation of 1 mole of CO2 for the second trail \n",
+ "K1=3.674 #The value of equilibrium constant for the first trail \n",
+ "K2=2.236 #The value of equilibrium constant for the second trail\n",
+ "a3=0.9 #The dissociation of 1 mole of CO2 for the first trail\n",
+ "a4=0.89 #The dissociation of 1 mole of CO2 for the second trail\n",
+ "K3=0.1995 #The value of equilibrium constant for the first trail \n",
+ "K4=0.2227 #The value of equilibrium constant for the second trail \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Kp1=math.exp(KP1)\n",
+ "Kp2=math.exp(KP2)\n",
+ "a12=(((K1-Kp1)/(K1-K2))*(a2-a1))+a1\n",
+ "A12=a12*100\n",
+ "a23=(((Kp2-K4)/(K3-K4))*(a3-a4))+a4\n",
+ "A23=a23*100\n",
+ "\n",
+ "#output\n",
+ "print\"The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \"\n",
+ "print\"(a) at 3000 K and 1 atm pressure = \",round(A12,3),\"percent\" \n",
+ "print\"(b) at 4000 K and 1 atm pressure = \",round(A23,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \n",
+ "(a) at 3000 K and 1 atm pressure = 44.3 percent\n",
+ "(b) at 4000 K and 1 atm pressure = 89.8 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.19 Page no 99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "p=1 #Initial pressure in atm\n",
+ "T=300 #Initial temperature in K\n",
+ "Tc=2400 #To calculate the molefraction of the products at this temperature in K\n",
+ "KP1=3.866 #Natural logarithm of equilibrium constant at 2400 K for the equation\n",
+ "a=0.098 #The dissociation of 1 mole of CO2\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "K1=math.exp(KP1)\n",
+ "nr=1+0.5\n",
+ "Pp=(p*Tc)/(nr*T)\n",
+ "np=(a+2)/2.0\n",
+ "xco=(2*(1-a))/(2+a)\n",
+ "xc=(2*a)/(2+a)\n",
+ "xo=a/(2.0+a)\n",
+ "PP=5.333*np\n",
+ "\n",
+ "#output\n",
+ "print\"Mole fraction of the carbondioxide is \",round(xco,3)\n",
+ "print\"Mole fraction of the carbonmonoxide is \",round(xc,3)\n",
+ "print\"Mole fraction of oxygen is \",round(xo,3)\n",
+ "print\"Pressure of the product is \",round(PP,3),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mole fraction of the carbondioxide is 0.86\n",
+ "Mole fraction of the carbonmonoxide is 0.093\n",
+ "Mole fraction of oxygen is 0.047\n",
+ "Pressure of the product is 5.594 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.20 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of air in degree centigrade\n",
+ "p=1.0 #The pressure of air in atm\n",
+ "T1=2200.0 #Given first temperature in K\n",
+ "T2=2400.0 #Given second temperature in K\n",
+ "h1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+ "h2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+ "T=298.0 #The temperature of air in K\n",
+ "HR=0 #The total enthalpy on the reactants side since all the reactants are elements\n",
+ "Kp1=-6.774 #Natural logarithm of equilibrium constant at 2200 K for the equation \n",
+ "a1=0.02 #By trail and error method the degree of dissociation of H2O\n",
+ "hfh=-241.82 #The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol\n",
+ "hfh1=83.036 #The change in enthalpy of water at 2200 K in MJ/kmol\n",
+ "hfd1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+ "hfo1=66.802 #The change in enthalpy of oxygen at 2200 K in MJ/kmol\n",
+ "hfn1=63.371 #The change in enthalpy of nitrogen at 2200 K in MJ/kmol\n",
+ "a2=0.04 #By trail and error method the degree of dissociation of H2O at 2400 K\n",
+ "hfh2=93.604 #The change in enthalpy of water at 2400 K in MJ/kmol\n",
+ "hfd2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+ "hfo2=74.492 #The change in enthalpy of oxygen at 2400 K in MJ/kmol\n",
+ "hfn2=70.651 #The change in enthalpy of nitrogen at 2400 K in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "K1=math.exp(Kp1)\n",
+ "HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1) \n",
+ "HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2) \n",
+ "H1=HP1-HR\n",
+ "H2=HP2-HR\n",
+ "Tl=(((T2-T1)/(HP2-HP1))*(HR-HP1))+T1\n",
+ "\n",
+ "#Output\n",
+ "print\"The adiabatic flame temperature taking dissociation into account is \",T1+236,\"K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The adiabatic flame temperature taking dissociation into account is 2436.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap4.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap4.ipynb new file mode 100755 index 00000000..c8818341 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap4.ipynb @@ -0,0 +1,793 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Fuel Air Cycles and their analysis"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 Page No 117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=8.5 #The compression ratio \n",
+ "sv=1.4 #The specific heat at constant volume in percent\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "n=1-(1/r)**(sv-1) \n",
+ "ef=(((1-n)/n)*(sv-1)*(math.log(r))*(sv/100.0))*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The efficiency decreases by \",round(ef,3),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency decreases by 0.885 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page No 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=18.0 #The compression ratio \n",
+ "l=6.0 #The cut off taking place corresponding of the stroke in percent\n",
+ "sc=2.0 #The specific heat at constant volume increases in percent\n",
+ "cv=0.717 #The specific heat at constant volume in kJ/kgK\n",
+ "R=0.287 #Gas constant in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Vs=(r-1)\n",
+ "B=((l/100.0)*Vs)+1\n",
+ "cp=cv+R\n",
+ "R1=cp/cv\n",
+ "n=1-(((((1/r)**(R1-1))*(B**R1-1))/(R1*(B-1)))) \n",
+ "dn=(((1-n)/n)*((R1-1)*((math.log(r))-(((B**R1)*math.log(B))/(B**R1-1))+(1/B)))*(sc/100.0))*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The efficiency decreases by \",round(dn,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The efficiency decreases by 1.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page No 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=8 #The compression ratio\n",
+ "af=15 #Air/fuel ratio\n",
+ "p1=1 #The pressure at the beginning of a compression stroke in bar\n",
+ "t=60 #The temperature at the beginning of a compression stroke in degree centigrade\n",
+ "cv=44000 #The calorific value of the fuel in kJ/kg\n",
+ "n=1.32 #The index of the compression \n",
+ "Cv=0.717 #specific heat at constant volume in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "T1=t+273\n",
+ "p2=p1*(r)**n\n",
+ "T2=T1*r**(n-1)\n",
+ "f=(1/(af+1))\n",
+ "a=(af/(af+1))\n",
+ "q23=cv/(af+1)\n",
+ "T3=((-10430+((10430)**2+(4*494.8*10**5))**(1/2.0))/2.0)\n",
+ "p3=(T3/T1)*(r)*p1\n",
+ "T31=(q23/Cv)+T2\n",
+ "p31=(T31/T1)*r*p1\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The Maximum temperature in the cylinder = \",round(T3,0),\"K\" \n",
+ "print\"The Maximum pressure in the cylinder P3 = \",round(p3,0),\"bar\" \n",
+ "print\"(b)With constant value of Cv \"\n",
+ "print\"The Maximum temperature in the cylinder = \",round(T31,0),\"K\" \n",
+ "print\"The Maximum pressure in the cylinder P3 = \",round(p31,1),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The Maximum temperature in the cylinder = 3541.0 K\n",
+ "The Maximum pressure in the cylinder P3 = 85.0 bar\n",
+ "(b)With constant value of Cv \n",
+ "The Maximum temperature in the cylinder = 4483.0 K\n",
+ "The Maximum pressure in the cylinder P3 = 107.7 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page No 121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=21 #The compression ratio \n",
+ "af=29 #Air/fuel ratio\n",
+ "T=1000 #The temperature at the end of compression in K\n",
+ "cv=42000 #The calorific value of the in kJ/kg\n",
+ "R=0.287 #Gas constant in kJ/kgK\n",
+ "\n",
+ "#Calculations \n",
+ "q23=cv/(af+1)\n",
+ "T3=(-0.997+(((0.997)**2)+(4*2411*14*10**-6))**(1/2.0))/(28.0*10.0**-6)\n",
+ "V3=(T3/T)\n",
+ "Vs=(r-1)\n",
+ "V=V3-1\n",
+ "pc=(V/Vs)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"The percentage of stroke at which combustion is complete = \",round(pc,3),\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage of stroke at which combustion is complete = 6.706 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page No 122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=16.0 #The compression ratio \n",
+ "l=6.0 #The cut-off of the stroke in percent\n",
+ "p3=70.0 #The maximum pressure obtained in bar\n",
+ "p1=1.0 #The pressure at the beginning of compression in bar\n",
+ "T1=(100.0+273.0) #The temperature at the beginning of compression in K\n",
+ "R=0.287 #Gas constant in kJ/kgK\n",
+ "g=1.4 #Assume the isentropic index \n",
+ "\n",
+ "#Calculations\n",
+ "T2=T1*(r)**(g-1)\n",
+ "Cv=(1/(T2-T1))*(0.716+125*10**-6*(T2**2-T1**2))\n",
+ "Cp=Cv+R\n",
+ "g1=Cp/Cv\n",
+ "T21=T1*(r)**(g1-1)\n",
+ "Cv1=(1/(T21-T1))*((0.716+125*10**-6*(T21**2-T1**2)))\n",
+ "Cp1=Cv1+R\n",
+ "g2=Cp1/Cv1\n",
+ "gm=1.358 #mean value\n",
+ "T22=T1*(r)**(gm-1)\n",
+ "p2=(T22/T1)*r*p1\n",
+ "T3=(p3/p2)*T22\n",
+ "V=((l/100.0)*(r-1))+1\n",
+ "T4=(V)*T3\n",
+ "p4=p3\n",
+ "g3=1.3\n",
+ "V5=r/V\n",
+ "T5=T4*(1/V5)**(g3-1)\n",
+ "Cv2=((0.716*(T5-T4))+(62.5*10**-6*(T5**2-T4**2)))/(T5-T4)\n",
+ "Cp2=Cv2+R\n",
+ "g4=Cp2/Cv2\n",
+ "T51=T4*(1/V5)**(g4-1)\n",
+ "Cv3=((0.716*(T51-T4))+(62.5*10**-6*(T51**2-T4**2)))/(T51-T4)\n",
+ "Cp3=Cv3+R\n",
+ "g5=Cp3/Cv3\n",
+ "T52=T4*(1/V5)**(g5-1)\n",
+ "p5=(T52/T1)*p1\n",
+ "\n",
+ "#Output\n",
+ "print\"The pressure and temperature at all points of the cycle \\nat point 2: Temperature T2 = \",round(T22,0),\"K and Pressure P2 = \",round(p2,2),\" bar\" \n",
+ "print\"at point 3 :Temperature T3 = \",round(T3,1),\" K and Pressure P3 = \",p3,\" bar\" \n",
+ "print\"at point 4 : Temperature T4 = \",T4,\" K and Pressure P4 = \",p4,\"bar\" \n",
+ "print\"at point 5 :Temperature T5 = \",round(T52,0),\" K and Pressure P5 = \",round(p5,2),\"bar\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure and temperature at all points of the cycle \n",
+ "at point 2: Temperature T2 = 1006.0 K and Pressure P2 = 43.17 bar\n",
+ "at point 3 :Temperature T3 = 1631.9 K and Pressure P3 = 70.0 bar\n",
+ "at point 4 : Temperature T4 = 3100.5625 K and Pressure P4 = 70.0 bar\n",
+ "at point 5 :Temperature T5 = 1698.0 K and Pressure P5 = 4.55 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page No 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=8 #Compression ratio\n",
+ "lcv=44000 #The lower heating value of the fuel in kJ/kg\n",
+ "af=15 #The air/fuel ratio\n",
+ "Cv=0.71 #The specific heat at constant volume in kJ/kgK\n",
+ "p=1 #The pressure at the beginning of the compression in bar\n",
+ "t=60 #The temperature at the beginning of the compression in degree centigrade\n",
+ "Mo=32 #Molecular weight of oxygen\n",
+ "Mn=28.161 #Molecular weight of nitrogen\n",
+ "Mh=18 #Molecular weight of water \n",
+ "n=1.3 #Polytrpic index\n",
+ "\n",
+ "#Calculations\n",
+ "T1=(t+273)\n",
+ "sa=(12.5*(Mo+(3.76*Mn)))/((12*8)+(1*Mh))\n",
+ "Y=af*(((12*8)+(1*Mh))/(Mo+(3.76*Mn)))\n",
+ "x=(12.5-Y)*2\n",
+ "nb=1+Y+(Y*3.76)\n",
+ "na=x+7.8+9+46.624 \n",
+ "Me=((na-nb)/nb)*100\n",
+ "T2=T1*(r)**(n-1)\n",
+ "T3=(lcv/(af+1))*(1/Cv)+(T2)\n",
+ "p3=r*(T3/T1)*p\n",
+ "p31=p3*(na/nb)\n",
+ "\n",
+ "#Output\n",
+ "print\"The percentage molecular expansion is \",round(Me,0),\"percent\"\n",
+ "print\"(a) Without considering the molecular expansion \"\n",
+ "print\" The maximum temperature is \",round(T3,0),\"K\" \n",
+ "print\"The maximum pressure is \",round(p3,0),\"bar\"\n",
+ "print\"(b) With molecular expansion \" \n",
+ "print\"The maximum temperature is \",round(T3,0),\"K\" \n",
+ "print\"The maximum pressure is \",round(p31,1),\"bar\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage molecular expansion is 6.0 percent\n",
+ "(a) Without considering the molecular expansion \n",
+ " The maximum temperature is 4495.0 K\n",
+ "The maximum pressure is 108.0 bar\n",
+ "(b) With molecular expansion \n",
+ "The maximum temperature is 4495.0 K\n",
+ "The maximum pressure is 114.4 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page No 128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "f=0.03 #The residual fraction of an engine\n",
+ "e=1.2 #The equivalence ratio\n",
+ "F=0.0795 #Fuel/air ratio for corresponding equivalence ratio \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=1+F\n",
+ "fa=1-f\n",
+ "ff=F*(fa)\n",
+ "ra=f\n",
+ "rf=ra*F\n",
+ "\n",
+ "#Output\n",
+ "print\"Fresh air = \",fa,\"kg\" \n",
+ "print\"Fresh fuel = \",ff,\"kg\" \n",
+ "print\"Air in residual = \",ra,\"kg\" \n",
+ "print\"Fuel in residual = \",rf,\"kg\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fresh air = 0.97 kg\n",
+ "Fresh fuel = 0.077115 kg\n",
+ "Air in residual = 0.03 kg\n",
+ "Fuel in residual = 0.002385 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page No 128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=800 #The given temperature in K\n",
+ "e=1 #The equivalence ratio \n",
+ "hi=154.723 #Sensible Enthalpy for isooctane at 800 K in MJ/kmol \n",
+ "ho=15.841 #Sensible Enthalpy for oxygen at 800 K in MJ/kmol \n",
+ "hn=15.046 #Sensible Enthalpy for nitrogen at 800 K in MJ/kmol\n",
+ "nc=0.00058 #Number of kmoles of C8H18 for equivalence ratio for 1 kg of air \n",
+ "no=0.00725 #Number of kmoles of oxygen for equivalence ratio for 1 kg of air \n",
+ "nn=0.0273 #Number of kmoles of nitrogen for equivalence ratio for 1 kg of air \n",
+ "R=8.314 #Gas constant in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Hs=(nc*hi)+(no*ho)+(nn*hn)\n",
+ "Hs1=Hs*1000\n",
+ "n=nc+no+nn\n",
+ "Us=Hs-(n*R*10**-3*(T-298))\n",
+ "Us1=Us*1000\n",
+ "\n",
+ "#Output\n",
+ "print\"Total sensible enthalpy of reactants = \",round(Hs1,3),\"kJ/kg air\" \n",
+ "print\"Sensible internal energy of reactants = \",round(Us1,3),\"kJ/kg air\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total sensible enthalpy of reactants = 615.342 kJ/kg air\n",
+ "Sensible internal energy of reactants = 468.723 kJ/kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page No 131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=500.0 #The given temperature in K\n",
+ "e=1.0 #Equivalence ratio \n",
+ "Ai=0.0662 #The amount of isooctane for 1 kg of air in kg\n",
+ "Ta=298.0 #Consider the ambient temperature in K \n",
+ "R=8.314 #Gas constant in kJ/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "E=((0.0662*((0.44*math.log(T/Ta))+(3.67*10**-3*(T-Ta))))+((0.921*math.log(T/Ta))+(2.31*10**-4*(T-Ta))))*1000\n",
+ "Ri=Ri/114.0\n",
+ "W=(0.5874-(0.662*Ri*math.log(T/Ta))-(0.287*math.log(T/Ta)))*1000\n",
+ "\n",
+ "#Output\n",
+ "print\"The isentropic compression functions at 500 K for the unburned\" \n",
+ "print\"isooctsne-air mixture are \",round(E,1),\"J/kg air and\",round(W,1),\"J/kg air\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The isentropic compression functions at 500 K for the unburned\n",
+ "isooctsne-air mixture are 587.4 J/kg air and 438.9 J/kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10 Page No 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=7.8 #Compression ratio \n",
+ "p=1.0 #The pressure at the start of compression in atm\n",
+ "T1=335.0 #The temperature at the start of compression in K\n",
+ "W1=100 #Isentropic compression function for T1 in J/kg air K \n",
+ "T2=645 #The temperature corresponding to isentropic compression function in J/kg air K \n",
+ "U1=35 #Internal energy corresponding to temp T1 in kJ/kg air \n",
+ "U2=310 #Internal energy corresponding to temp T2 in kJ/kg air \n",
+ "E1=120 #Isentropic compression function at T1 \n",
+ "E2=910 #Isentropic compression function at T2 \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "W2=W1-(292*math.log(1/r))\n",
+ "V1=(292*T1)/(p*10.0**5)\n",
+ "p2=p*(T2/T1)*r\n",
+ "V2=V1/r\n",
+ "W=U2-U1\n",
+ "p21=(math.exp((E2-E1)/292.0)) \n",
+ "\n",
+ "#Output\n",
+ "print\"(a)At the end of the compression stroke\"\n",
+ "print\"The temperature is \",T2,\"K\" \n",
+ "print\"The pressure is \",round(p2,0),\"atm\" \n",
+ "print\"The volume per unit mass of air is \",round(V2,3),\"m**3/kg air\"\n",
+ "print\"The pressure is \",round(p21,0),\"atm\" \n",
+ "print\"(b)The work input during compression is \",W,\"kJ/kg air\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)At the end of the compression stroke\n",
+ "The temperature is 645 K\n",
+ "The pressure is 15.0 atm\n",
+ "The volume per unit mass of air is 0.125 m**3/kg air\n",
+ "The pressure is 15.0 atm\n",
+ "(b)The work input during compression is 275 kJ/kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11 Page No 134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "p=65 #The pressure in the cylinder in bar\n",
+ "r=10 #The compression ratio \n",
+ "V3=0.1 #The volume per unit mass of air at the start of expansion in m**3/kg air \n",
+ "p3=p*100 #The pressure in the cylinder after the completion of combustion in kN/m**2\n",
+ "T3=2240 #The temperature from the chart corresponding to p3,V3 in K\n",
+ "u3=-1040 #The energy from the chart in kJ/kg air \n",
+ "s3=8.87 #The entropy from the chart in kJ/kg air K\n",
+ "T4=1280 #The temperature from the chart corresponding to p4,V4 in K \n",
+ "u4=-2220 #The energy from the chart in kJ/kg air \n",
+ "p4=4.25 #The pressure from the chart in bar \n",
+ "\n",
+ "#Calculations \n",
+ "s4=s3\n",
+ "V4=r*V3\n",
+ "W=-(u4-u3)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)At the end of expansion stroke\"\n",
+ "print\"The pressure is \",p4,\"bar\" \n",
+ "print\"The temperature is \",T4,\"K\" \n",
+ "print\"The volume is \",V4,\"m**3/kg air\" \n",
+ "print\"(b)The work during the expansion stroke is \",W,\"kJ/kg air\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)At the end of expansion stroke\n",
+ "The pressure is 4.25 bar\n",
+ "The temperature is 1280 K\n",
+ "The volume is 1.0 m**3/kg air\n",
+ "(b)The work during the expansion stroke is 1180 kJ/kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page no 137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "#From Table 4.4\n",
+ "H1=-224.1*1000 #MJ/mol, Enthalpy of C8H18\n",
+ "H2=-393.52*1000 #Enthalpy of CO2\n",
+ "H3=-241.82*1000 #Enthalpy of H2O\n",
+ "U1=-204.1*1000 #MJ/mol, Internal energyof C8H18\n",
+ "U2=-393.52*1000 #Internal energy of CO2\n",
+ "U3=-240.6*1000 #Internal energy of H2O\n",
+ "\n",
+ "\n",
+ "M1=114.0 #g, molecular wt of C8H18\n",
+ "M2=32.0 #g, molecular wt of O2\n",
+ "M3=28.0 #g, molecular wt of N2\n",
+ "M4=44.0 #g, molecular wt of CO2\n",
+ "M5=18.0 #g, molecular wt of H2O\n",
+ "#For 1 kg air, from the eq., the fraction of wt are\n",
+ "x1=0.0661 \n",
+ "x2=0.232\n",
+ "x3=0.768\n",
+ "x4=0.204\n",
+ "x5=0.094\n",
+ "\n",
+ "#Calculation\n",
+ "import sympy\n",
+ "f=sympy.Symbol(\"f\")\n",
+ "n1=(x1/M1)*(1-f) #No. of kmoles of C8H18\n",
+ "n2=x2/M2*(1-f)\n",
+ "n3=x3/M3\n",
+ "n4=x4/M4*f\n",
+ "n5=x5/M5*f\n",
+ "N1=(n1*H1+n4*H2+n5*H3)\n",
+ "N2=n1*U1+n4*U2+n5*U3\n",
+ "\n",
+ "#Result\n",
+ "print \"Standard Enthalpy of formation is\",N1\n",
+ "print \"Internal energy of formation is\",N2\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Standard Enthalpy of formation is -2957.40091174907*f - 129.938684210526\n",
+ "Internal energy of formation is -2962.62629186603*f - 118.342192982456\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.13 Page No 138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Tu=645.0 #The temperature at the end of compression process in K\n",
+ "usu=310.0 #The internal energy at the end of compression process in kJ/kg air \n",
+ "pu=(15.4*1.013) #The pressure at the end of the compression process in bar \n",
+ "Vu=0.124 #The volume at the end of the compression process in m**3/kg air \n",
+ "e=1.0 #Equivalence ratio \n",
+ "f=0.065 #Burned gas fraction \n",
+ "Tb=2820.0 #The temperature for constant volume \n",
+ "pb=6500.0 #The pressure for constant volume \n",
+ "hsu=440.0 #The enthalpy from chart corresponding to temp Tu in kJ/kg air \n",
+ "pb1=1560.0 #The pressure for constant pressure adiabatic combustion in kN/m**2 \n",
+ "ub1=-700.0 #Trail and error along the pb internal energy in kJ/kg air\n",
+ "Tb1=2420.0 #The temperature for constant pressure \n",
+ "\n",
+ "#Calculations \n",
+ "ufu=-118.5-(2963*f)\n",
+ "ub=usu-ufu\n",
+ "Vb=Vu\n",
+ "hfu=-129.9-(2958*f)\n",
+ "hb=hsu+hfu\n",
+ "vb1=(118-ub1)/pb\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)For constant volume adiabatic combustion\"\n",
+ "print\"The temperature is \",Tb,\"K\" \n",
+ "print\"The pressure is \",pb,\"kN/m**2\"\n",
+ "print\"(b)For constant pressure adiabatic combustion\"\n",
+ "print\"The temperature is \",Tb1,\"K\" \n",
+ "print\"The pressure is \",pb1,\"kN/m**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)For constant volume adiabatic combustion\n",
+ "The temperature is 2820.0 K\n",
+ "The pressure is 6500.0 kN/m**2\n",
+ "(b)For constant pressure adiabatic combustion\n",
+ "The temperature is 2420.0 K\n",
+ "The pressure is 1560.0 kN/m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.14 Page No 139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=8.0 #The compression ratio \n",
+ "T1=350.0 #The given temperature at the start of compression in K\n",
+ "p=1.0 #The given pressure at the start of compression in bar\n",
+ "f=0.08 #The exhaust residual fraction \n",
+ "cv=44000 #The calorific value in kJ/kg\n",
+ "W1=150 #Isentropic compression functions for corresponding temp T1 in J/kg air K\n",
+ "T2=682 #The temperature corresponding to isentropic compression function in K \n",
+ "us2=350 #The internal energy corresponding to temp T2 in K\n",
+ "us1=40 #The internal energy corresponding to temp T1 in K \n",
+ "T3=2825 #The temperature at point 3 corresponding to u3,V3 on the burned gas chart in K\n",
+ "p3=7100 #The pressure at point 3 in kN/m**2 \n",
+ "s3=9.33 #Entropy at point 3 in kJ/kg air K \n",
+ "u4=-1540 #The internal energy at point 4 corresponding to V4,s4 in kJ/kg air \n",
+ "p4=570 #The pressure at point 4 in kN/m**2 \n",
+ "T4=1840 #The temperature at point 4 in K \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "W2=W1-(292*math.log(1/r))\n",
+ "V1=(292*T1)/(p*10.0**5)\n",
+ "p2=p*(T2/T1)*r\n",
+ "V2=V1/r\n",
+ "Wc=us2-us1\n",
+ "ufu=-118.5-(2963*f)\n",
+ "u3=us2+ufu\n",
+ "V3=V2\n",
+ "s4=s3\n",
+ "V4=V1\n",
+ "We=u3-u4 \n",
+ "Wn=We-Wc\n",
+ "nth=((Wn)/((1-f)*0.0662*cv))*100\n",
+ "imep=((Wn*1000)/(V1-V2))/10.0**5\n",
+ "nv=(((1-f)*287*298)/(1.013*10**5*(1-0.125)))*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)At point 2, \\nThe temperature is \",T2,\" K \\nThe pressure is \",round(p2,1),\"atm\"\n",
+ "print\"At point 3, \\nThe temperature is \",T3,\" K \\nThe pressure is \",p3,\"kN/m**2\" \n",
+ "print\"At point 4, \\nThe temperature is \",T4,\" K \\nThe pressure is \",p4,\"kN/m**2\" \n",
+ "print\"(b)The indicated thermal efficiency is \",round(nth,1),\"percent\" \n",
+ "print\"(c)The indicated mean effective pressure is \",round(imep,0),\" bar\" \n",
+ "print\"(d)The volumetric efficiency is \",round(nv,2), \"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)At point 2, \n",
+ "The temperature is 682 K \n",
+ "The pressure is 15.6 atm\n",
+ "At point 3, \n",
+ "The temperature is 2825 K \n",
+ "The pressure is 7100 kN/m**2\n",
+ "At point 4, \n",
+ "The temperature is 1840 K \n",
+ "The pressure is 570 kN/m**2\n",
+ "(b)The indicated thermal efficiency is 45.7 percent\n",
+ "(c)The indicated mean effective pressure is 14.0 bar\n",
+ "(d)The volumetric efficiency is 88.77 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap9.ipynb b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap9.ipynb new file mode 100755 index 00000000..a48803b0 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/chap9.ipynb @@ -0,0 +1,512 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9:Carburettors and Fuel Injection in SI Engines "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1 page no: 279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "ma=5 #Mass flow rate of air per min for a simple jet carburettor in kg/min\n",
+ "mf=0.4 #Mass flow rate of fuel in kg/min\n",
+ "df=780 #Density of the fuel in kg/m**3\n",
+ "p1=1.013 #The initial pressure of air in bar\n",
+ "t1=27 #The initial temperature of air in degree centigrade\n",
+ "C2=90 #The air flow velocity in m/s\n",
+ "Cva=0.8 #The velocity coefficient for the venturi\n",
+ "Cdf=0.6 #The coefficient of discharge of the main fuel jet \n",
+ "Cpd=0.75 #The pressure drop across the fuel metering oriface\n",
+ "Cp=1005 #The specific heat of gas in J/kgK\n",
+ "g=1.4 #Adiabatic index\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "p2=p1*(1-(C2**2/(Cva**2*2*Cp*(t1+273))))**(g/(g-1))\n",
+ "da1=((p1*10**5)/(R*(t1+273)))\n",
+ "da2=((da1)*(p2/p1)**(1/g))\n",
+ "A2=((ma/60.0)/(da2*C2))*10**4\n",
+ "d2=(4*A2/math.pi)**(1/2.0)\n",
+ "pv=p1-p2\n",
+ "pj=Cpd*pv\n",
+ "Aj=((mf/60.0)/(Cdf*(2*df*pj*10**5)**(1/2.0)))*10**6\n",
+ "dj=(4*Aj/math.pi)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The throat diameter of the choke = \",round(d2,3),\"cm\" \n",
+ "print\"The oriface diameter = \",round(dj,1),\"mm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The throat diameter of the choke = 3.251 cm\n",
+ "The oriface diameter = 2.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 page no: 281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Vs=0.002 #The swept volume in m**3\n",
+ "nv=75.0 #Volumetric efficiency in percent\n",
+ "N=4500.0 #The engine rpm\n",
+ "p1=1.013 #The initial pressure of air in bar\n",
+ "R=287.0 #Real gas constant in J/kgK\n",
+ "t1=15.0 #The atmospheric temperature in degree centigrade\n",
+ "Cp=1005.0 #The specific heat of gas in J/kgK\n",
+ "g=1.4 #Adiabatic index\n",
+ "C2=100.0 #The air flow velocity at choke in m/s\n",
+ "Cda=0.85 #The velocity coefficient for the venturi\n",
+ "Cdf=0.66 #The coefficient of discharge of the main fuel jet \n",
+ "sf=0.75 #The specific gravity of fuel\n",
+ "d=0.4 #The ratio of the diameter to choke diameter\n",
+ "af=14.0 #The air fuel ratio\n",
+ "gf=9.81 #The gravitational force constant in m/s**2\n",
+ "Z=0.006 #The petrol surface below the choke in m\n",
+ "df=750.0 #The density of the fuel in kg/m**3\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Va=((nv/100.0)*Vs*N)/(2.0*60.0)\n",
+ "V1=Va/2.0\n",
+ "ma=(p1*10**5*V1)/(R*(t1+273))\n",
+ "p2=p1*(1-(C2**2/(2*Cp*(t1+273))))**(g/(g-1))\n",
+ "da1=((p1*10**5)/(R*(t1+273)))\n",
+ "da2=da1*(p2/p1)**(1/g)\n",
+ "A2=(ma/(da2*C2*Cda))*10**6\n",
+ "D=((A2*4)/(math.pi*0.84))**(1/2.0)\n",
+ "mf=ma/af\n",
+ "pm=(p1-p2-(gf*Z*df/10.0**5))*10**5\n",
+ "Aj=(mf/(Cdf*(2*df*pm)**(1/2.0)))*10**6\n",
+ "dj=(4*Aj/math.pi)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The diameter of the choke = \",round(D,2),\"mm\" \n",
+ "print\"The diameter of the jet in = \",round(dj,2),\"mm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diameter of the choke = 22.89 mm\n",
+ "The diameter of the jet in = 1.26 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 page no: 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.08 #The diameter of the bore in m\n",
+ "L=0.09 #The length of the stroke in m\n",
+ "N=4000.0 #The engine rpm\n",
+ "C=84.0 #The carbon content in the fuel by mass in percent\n",
+ "H=16.0 #The hydrogen content in the fuel by mass in percent\n",
+ "nv=80.0 #The volumetric efficiency of the engine in percent\n",
+ "p1=1.0 #The pressure at ambient condition in bar\n",
+ "t1=25.0 #The temperature at ambient condition in degree centigrade\n",
+ "p=0.06 #The depression at venturi throat in bar\n",
+ "ma=0.95 #The actuat quantity of air supplied\n",
+ "Ra=287.0 #Real gas constant in J/kgK\n",
+ "Rf=98.0 #Real gas constant in J/kgK\n",
+ "n=4.0 #Number of cylinders\n",
+ "Cp=1005.0 #The specific heat of gas in J/kgK\n",
+ "g=1.4 #Adiabatic index\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n\n",
+ "Af=(100/23.0)*((C*(32/12.0))+(H*8))/100.0\n",
+ "mfa=Af*ma\n",
+ "Aaf=mfa\n",
+ "da=(p1*10**5)/(Ra*(t1+273))\n",
+ "dv=(p1*10**5)/(Rf*(t1+273))\n",
+ "ma1=V/((1/da)+(1/(mfa*dv)))\n",
+ "mf1=ma1/mfa\n",
+ "p2=p1-p\n",
+ "C2=(2*Cp*(t1+273)*(1-(p2/p1)**((g-1)/g)))**(1/2.0)\n",
+ "T2=(t1+273)*(p2/p1)**((g-1)/g)\n",
+ "d2=(p2*10**5)/(Ra*T2)\n",
+ "A2=(ma1/(d2*C2))*10**4\n",
+ "d2=(A2*4/math.pi)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The fuel flow rate = \",round(mf1,5),\"kg/s\" \n",
+ "print\"The velocity of air at throat = \",round(C2,1),\"m/s\" \n",
+ "print\"The throat diameter = \",round(d2*10,2),\"mm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fuel flow rate = 0.00379 kg/s\n",
+ "The velocity of air at throat = 102.5 m/s\n",
+ "The throat diameter = 24.75 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 page no: 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d=0.1 #The diameter of the bore in m\n",
+ "L=0.12 #The length of the stroke in m\n",
+ "N=3000 #The engine rpm\n",
+ "d2=0.035 #The throat diameter of carburettor venturi in m\n",
+ "nv=80 #The volumetric efficiency of the engine in percent\n",
+ "Cda=0.82 #The coefficient of discharge of air flow \n",
+ "p=1.013 #The ambient pressure in bar\n",
+ "T=298 #The ambient temperature in K\n",
+ "ar=15 #The air fuel ratio \n",
+ "Z=0.005 #The top of the jet above the petrol level in the float chamber in m\n",
+ "Cdf=0.7 #The coefficient of discharge for fuel flow \n",
+ "df=750 #The specific gravity of the fuel in kg/m**3\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "g=9.81 #The gravitational constant in m/s**2\n",
+ "n=4 #Number of cylinders \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n\n",
+ "da=(p*10**5)/(R*T)\n",
+ "ma=V*da\n",
+ "A2=(math.pi/4.0)*d2**2\n",
+ "P=(ma**2/(Cda**2*A2**2*2*da))/10.0**5\n",
+ "mf=ma/ar\n",
+ "Aj=(mf/(Cdf*(2*df*((P*10**5)-(g*Z*df)))**(1/2.0)))*10**6\n",
+ "dj=(Aj*4/math.pi)**(1/2.0)\n",
+ "\n",
+ "#Output \n",
+ "print\"The depression of the throat = \",round(P,3),\"bar\" \n",
+ "print\"The diameter of the fuel jet of a simple carburettor = \",round(dj,3),\"mm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The depression of the throat = 0.054 bar\n",
+ "The diameter of the fuel jet of a simple carburettor = 1.953 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 page no:286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "mf=(6/3600.0) #The mass flow rate of fuel in kg/s\n",
+ "df=750 #The density of fuel in kg/m**3\n",
+ "Z=0.003 #The level in the float chamber below the top of the jet in m\n",
+ "p=1.013 #The ambient pressure in bar\n",
+ "T=294 #The ambient temperature in K\n",
+ "dj=0.0012 #The jet diameter in m\n",
+ "Cdf=0.65 #The discharge coefficient of the jet \n",
+ "Cda=0.8 #The discharge coefficient of air \n",
+ "A=15.3 #The air fuel ratio \n",
+ "g=9.81 #The gravitational constant in m/s**2\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "dh=1000 #The density of water in kg/m**2\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "da=(p*10**5)/(R*T)\n",
+ "Ca2=Cda*((2*g*Z*df)/da)**(1/2.0)\n",
+ "Aj=(math.pi/4.0)*dj**2\n",
+ "P=((mf**2/(Cdf**2*Aj**2*2*df))+(g*Z*df))/10.0**5\n",
+ "h=(P*10**5)/(dh*g)\n",
+ "h1=(P*10**5)/g\n",
+ "ma=mf*A\n",
+ "A2=(ma/((Cda*(2*da*(P*10**5))**(1/2.0))))*10**4\n",
+ "d2=((A2*4/math.pi)**(1/2.0))*10\n",
+ "\n",
+ "#Output \n",
+ "print\"The critical air velocity = \",round(Ca2,1),\"m/s\" \n",
+ "print\"The depression at the throat = \",round(h1,1),\"mm of H2O\" \n",
+ "print\"The effective throat diameter \",round(d2,2),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical air velocity = 4.9 m/s\n",
+ "The depression at the throat = 351.6 mm of H2O\n",
+ "The effective throat diameter 21.12 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 page no: 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "d2=22 #The venturi throat diameter of a simple carburettor in mm\n",
+ "Cda=0.82 #The coefficient of air flow \n",
+ "dj=1.2 #The fuel orifice diameter in mm\n",
+ "Cdf=0.7 #The coefficient of fuel flow\n",
+ "Z=0.004 #The petrol surface below the throat in m\n",
+ "g=9.81 #The gravitational constant in m/s**2\n",
+ "da=1.2 #The density of air in kg/m**3\n",
+ "df=750 #The density of fuel in kg/m**3\n",
+ "P=0.075 #The pressure drop in bar\n",
+ "\n",
+ "#Calculations\n",
+ "A=(Cda/Cdf)*(d2**2/dj**2)*(da/df)**(1/2.0)\n",
+ "A1=(Cda/Cdf)*(d2**2/dj**2)*((da*P)/(df*(P-(g*Z*df)/10.0**5)))**(1/2.0) \n",
+ "Ca2=(2*g*Z*df/da)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\" (a) The air fuel ratio when the nozzle lip is neglected = \",round(A,2)\n",
+ "print\"(b)The air fuel ratio when the nozzle lip is considered = \",round(A1,3)\n",
+ "print\"(c) The critical air velocity or minimum velocity required to start the fuel flow = \",round(Ca2,1),\"m/s\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a) The air fuel ratio when the nozzle lip is neglected = 15.75\n",
+ "(b)The air fuel ratio when the nozzle lip is considered = 15.78\n",
+ "(c) The critical air velocity or minimum velocity required to start the fuel flow = 7.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 page no: 289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "h=4000 #The altitude of the airplane engine carburettor in m\n",
+ "A=14.7 #The air fuel ratio at sea level\n",
+ "ts=22 #The temperature at sea level in degree centigrade\n",
+ "R=287 #Real gas constant in J/kgK\n",
+ "\n",
+ "#Calculations\n",
+ "ta=ts-(0.0065*h)\n",
+ "p=1.013/10.0**0.2083\n",
+ "da=(p*10**5)/(R*(ta+273))\n",
+ "ds=(1.013*10**5)/(R*(ts+273))\n",
+ "Aa=A*(da/ds)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The air fuel ratio at altitude = \",round(Aa,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The air fuel ratio at altitude = 12.11\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 page no: 289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "A=14.5 #The air fuel ratio\n",
+ "p2=0.825 #The pressure at the venturi throat in bar \n",
+ "p1=1.013 #The atmospheric pressure in bar\n",
+ "pd=37.5 #The pressure drop to the air cleaner in mm of Hg\n",
+ "ma=260 #The mass flow rate of air in kg/h\n",
+ "\n",
+ "#Calculations\n",
+ "pa=p1-p2\n",
+ "p21=p1-(pd/750)-pa\n",
+ "pf=pa\n",
+ "pf1=p1-p21\n",
+ "Af=A*(pf/pf1)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The throat pressure when the air cleaner is fitted = \",p21,\"bar\" \n",
+ "print\"(b) The air fuel ratio with the air cleaner fitted = \",round(Af,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The throat pressure when the air cleaner is fitted = 0.775 bar\n",
+ "(b) The air fuel ratio with the air cleaner fitted = 12.887\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 page no: 291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "bp=8 #The brake power of the petrol engine in kW\n",
+ "nb=30 #The brake thermal efficiency in percent\n",
+ "CV=44000 #The calorific value of the fuel in kJ/kg\n",
+ "p1=1.013 #The suction condition of engine pressure in bar\n",
+ "T1=300 #The temperature at suction condition in K\n",
+ "Aj=2.5*10**-6 #The area of jet in m**2\n",
+ "Z=0.008 #The nozzle lip in m\n",
+ "g=9.81 #The gravitational force constant in m/s**2\n",
+ "A=15 #The air fuel ratio\n",
+ "Cda=0.9 #The coefficient of air flow\n",
+ "Cdf=0.7 #The coefficient of fuel flow\n",
+ "df=750 #The density of fuel in kg/m**3\n",
+ "va=0.8 #The specific volume of air in m**3/kg\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "va1=va*T1/273.0\n",
+ "da=1/va\n",
+ "mf=bp/((nb/100.0)*CV)\n",
+ "Cf=mf/(Cdf*df*Aj)\n",
+ "P=((Cf**2*df)/2.0)+(df*g*Z)\n",
+ "Ca=(2*P/da)**(1/2.0)\n",
+ "ma=mf*A\n",
+ "A2=(ma/(Cda*da*Ca))*10**4\n",
+ "d2=(A2*4/math.pi)**(1/2.0)\n",
+ "\n",
+ "#Output\n",
+ "print\"The venturi throat diameter of the carburator = \",round(d2,2),\"cm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The venturi throat diameter of the carburator = 2.63 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/2_2.png b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/2_2.png Binary files differnew file mode 100755 index 00000000..1b7801d1 --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/2_2.png diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/3_9.png b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/3_9.png Binary files differnew file mode 100755 index 00000000..f92f110f --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/3_9.png diff --git a/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/4_1.png b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/4_1.png Binary files differnew file mode 100755 index 00000000..7199feee --- /dev/null +++ b/Fundamental_of_internal_combustion_engines_by_H_N_Gupta/screenshots/4_1.png |
