##CHAPTER 5 ILLUSRTATION 1 PAGE NO 160
##TITLE:Inertia Force Analysis in Machines
import math
pi=3.141
r=.3## radius of crank in m
l=1.## length of connecting rod in m
N=200.## speed of the engine in rpm
n=l/r
##===================
w=2.*pi*N/60.## angular speed in rad/s
teeta=math.acos((-n+((n**2)+4*2*1)**.5)/(2*2))*57.3## angle of inclination of crank in degrees
Vp=w*r*(math.sin(teeta/57.3)+(math.sin((2*teeta)/57.3)/n))## maximum velocity of the piston in m/s
print'%s %.1f %s'%('Maximum velocity of the piston = ',Vp,' m/s')
print'%s %.2f %s'%('teeta',teeta,'')
##CHAPTER 5 ILLUSRTATION 2 PAGE NO 161
##TITLE:Inertia Force Analysis in Machines
import math
PI=3.141
r=.3## length of crank in metres
l=1.5## length of connecting rod in metres
N=180.## speed of rotation in rpm
teeta=40.## angle of inclination of crank in degrees
##============================
n=l/r
w=2.*PI*N/60## angular speed in rad/s
Vp=w*r*(math.sin(teeta/57.3)+math.sin((2.*teeta/57.3)/(2.*n)))## velocity of piston in m/s
fp=w**2.*r*(math.cos(teeta/57.3)+math.cos(2.*teeta/57.3)/(2.*n))## acceleration of piston in m/s**2
costeeta1=(-n+(n**2.+4.*2.*1.)**.5)/4.
teeta1=math.acos(costeeta1)*(57.3)## position of crank from inner dead centre position for zero acceleration of piston
##===========================
print'%s %.1f %s %.1f %s %.1f %s'%('Velocity of Piston = ',Vp,' m/s'' Acceleration of piston =',fp,' m/s**2'' position of crank from inner dead centre position for zero acceleration of piston=',teeta1,' degrees')
##CHAPTER 5 ILLUSRTATION 3 PAGE NO 161
##TITLE:Inertia Force Analysis in Machines
import math
pi=3.141
D=.3## Diameter of steam engine in m
L=.5## length of stroke in m
r=L/2.
mR=100.## equivalent of mass of reciprocating parts in kg
N=200.## speed of engine in rpm
teeta=45## angle of inclination of crank in degrees
p1=1.*10**6## gas pressure in N/m**2
p2=35.*10**3## back pressure in N/m**2
n=4.## ratio of crank radius to the length of stroke
##=================================
w=2.*pi*N/60## angular speed in rad/s
Fl=pi/4.*D**2.*(p1-p2)## Net load on piston in N
Fi=mR*w**2*r*(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/(2*n))## inertia force due to reciprocating parts
Fp=Fl-Fi## Piston effort
T=Fp*r*(math.sin(teeta/57.3)+(math.sin((2*teeta)/57.3))/(2.*(n**2-(math.sin(teeta/57.3))**2)**.5))
print'%s %.1f %s %.1f %s '%('Piston effort = ',Fp,' N' 'Turning moment on the crank shaft = ',T,' N-m')
##CHAPTER 5 ILLUSRTATION 4 PAGE NO 162
##TITLE:Inertia Force Analysis in Machines
import math
pi=3.141
D=.10## Diameter of petrol engine in m
L=.12## Stroke length in m
l=.25## length of connecting in m
r=L/2.
mR=1.2## mass of piston in kg
N=1800.## speed in rpm
teeta=25.## angle of inclination of crank in degrees
p=680.*10**3## gas pressure in N/m**2
n=l/r
g=9.81## acceleration due to gravity
##=======================================
w=2.*pi*N/60.## angular speed in rpm
Fl=pi/4.*D**2.*p## force due to gas pressure in N
Fi=mR*w**2.*r*(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/(n))## inertia force due to reciprocating parts in N
Fp=Fl-Fi+mR*g## net force on piston in N
Fq=n*Fp/((n**2-(math.sin(teeta/57.3))**2.)**.5)## resultant load on gudgeon pin in N
Fn=Fp*math.sin(teeta/57.3)/((n**2-(math.sin(teeta/57.3))**2.)**.5)## thrust on cylinder walls in N
fi=Fl+mR*g## inertia force of the reciprocating parts before the gudgeon pin load is reversed in N
w1=(fi/mR/r/(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/(n)))**.5
N1=60.*w1/(2.*pi)
print'%s %.1f %s %.1f %s %.1f %s %.1f %s '%('Net force on piston = ',Fp,' N'' Resultant load on gudgeon pin = ',Fq,' N'' Thrust on cylinder walls = ',Fn,' N'' speed at which other things remining same,the gudgeon pin load would be reversed in directionm= ',N1,' rpm')
##CHAPTER 5 ILLUSRTATION 5 PAGE NO 163
##TITLE:Inertia Force Analysis in Machines
##Figure 5.3
import math
pi=3.141
N=1800.## speed of the petrol engine in rpm
r=.06## radius of crank in m
l=.240## length of connecting rod in m
D=.1## diameter of the piston in m
mR=1## mass of piston in kg
p=.8*10**6## gas pressure in N/m**2
x=.012## distance moved by piston in m
##===============================================
w=2.*pi*N/60.## angular velocity of the engine in rad/s
n=l/r
Fl=pi/4.*D**2.*p## load on the piston in N
teeta=32.## by mearument from the figure 5.3
Fi=mR*w**2.*r*(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/n)## inertia force due to reciprocating parts in N
Fp=Fl-Fi## net load on the gudgeon pin in N
Fq=n*Fp/((n**2.-(math.sin(teeta/57.3))**2.)**.5)## thrust in the connecting rod in N
Fn=Fp*math.sin(teeta/57.3)/((n**2-(math.sin(teeta/57.3))**2)**.5)## reaction between the piston and cylinder in N
w1=(Fl/mR/r/(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/(n)))**.5
N1=60.*w1/(2.*pi)##
print'%s %.1f %s %.1f %s %.1f %s %.1f %s'%('Net load on the gudgeon pin= ',Fp,' N''Thrust in the connecting rod= ',Fq,' N'' Reaction between the cylinder and piston= ',Fn,' N'' The engine speed at which the above values become zero= ',N1,' rpm')
##CHAPTER 5 ILLUSRTATION 6 PAGE NO 165
##TITLE:Inertia Force Analysis in Machines
import math
pi=3.141
D=.25## diameter of horizontal steam engine in m
N=180.## speed of the engine in rpm
d=.05## diameter of piston in m
P=36000.## power of the engine in watts
n=3.## ration of length of connecting rod to the crank radius
p1=5.8*10**5## pressure on cover end side in N/m**2
p2=0.5*10**5## pressure on crank end side in N/m**2
teeta=40.## angle of inclination of crank in degrees
m=45.## mass of flywheel in kg
k=.65## radius of gyration in m
##==============================
Fl=(pi/4.*D**2.*p1)-(pi/4.*(D**2.-d**2.)*p2)## load on the piston in N
ph=(math.sin(teeta/57.3)/n)
phi=math.asin(ph)*57.3## angle of inclination of the connecting rod to the line of stroke in degrees
r=1.6*D/2.
T=Fl*math.sin((teeta+phi)/57.3)/math.cos(phi/57.3)*r## torque exerted on crank shaft in N-m
Fb=Fl*math.cos((teeta+phi)/57.3)/math.cos(phi/57.3)## thrust on the crank shaft bearing in N
TR=P*60./(2.*pi*N)## steady resisting torque in N-m
Ts=T-TR## surplus torque available in N-m
a=Ts/(m*k**2)## acceleration of the flywheel in rad/s**2
print'%s %.1f %s %.1f %s %.1f %s '%('Torque exerted on the crank shaft= ',T,' N-m'' Thrust on the crank shaft bearing= ',Fb,'N''Acceleration of the flywheel= ',a,' rad/s**2')
##CHAPTER 5 ILLUSRTATION 7 PAGE NO 166
##TITLE:Inertia Force Analysis in Machines
import math
pi=3.141
D=.25## diameter of vertical cylinder of steam engine in m
L=.45## stroke length in m
r=L/2.
n=4.
N=360.## speed of the engine in rpm
teeta=45.## angle of inclination of crank in degrees
p=1050000.## net pressure in N/m**2
mR=180.## mass of reciprocating parts in kg
g=9.81## acceleration due to gravity
##========================
Fl=p*pi*D**2./4.## force on piston due to steam pressure in N
w=2.*pi*N/60.## angular speed in rad/s
Fi=mR*w**2.*r*(math.cos(teeta/57.3)+math.cos((2*teeta)/57.3)/(n))## inertia force due to reciprocating parts in N
Fp=Fl-Fi+mR*g## piston effort in N
phi=math.asin((math.sin(teeta/57.3)/n))*57.3## angle of inclination of the connecting rod to the line of stroke in degrees
T=Fp*math.sin((teeta+phi)/57.3)/math.cos(phi/57.3)*r## torque exerted on crank shaft in N-m
print'%s %.1f %s'%('Effective turning moment on the crank shaft= ',T,' N-m')
##CHAPTER 5 ILLUSRTATION 8 PAGE NO 166
##TITLE:Inertia Force Analysis in Machines
##figure 5.4
import math
pi=3.141
D=.25## diameter of vertical cylinder of diesel engine in m
L=.40## stroke length in m
r=L/2.
n=4.
N=300.## speed of the engine in rpm
teeta=60.## angle of inclination of crank in degrees
mR=200.## mass of reciprocating parts in kg
g=9.81## acceleration due to gravity
l=.8## length of connecting rod in m
c=14.## compression ratio=v1/v2
p1=.1*10**6.## suction pressure in n/m**2
i=1.35## index of the law of expansion and compression
##==============================================================
Vs=pi/4.*D**2.*L## swept volume in m**3
w=2.*pi*N/60.## angular speed in rad/s
Vc=Vs/(c-1.)
V3=Vc+Vs/10.## volume at the end of injection of fuel in m**3
p2=p1*c**i## final pressure in N/m**2
p3=p2## from figure
x=r*((1.-math.cos(teeta/57.3)+(math.sin(teeta/57.3))**2/(2.*n)))## the displacement of the piston when the crank makes an angle 60 degrees with T.D.C
Va=Vc+pi*D**2.*x/4.
pa=p3*(V3/Va)**i
p=pa-p1## difference of pressues on 2 sides of piston in N/m**2
Fl=p*pi*D**2./4.## net load on piston in N
Fi=mR*w**2.*r*(math.cos(teeta/57.3)+math.cos(2.*teeta/57.3)/(n))## inertia force due to reciprocating parts in N
Fp=Fl-Fi+mR*g## piston effort in N
phi=math.asin((math.sin(teeta/57.3)/n))*57.3## angle of inclination of the connecting rod to the line of stroke in degrees
T=Fp*math.sin((teeta+phi)/57.3)/math.cos(phi/57.3)*r## torque exerted on crank shaft in N-m
print'%s %.1f %s'%('Effective turning moment on the crank shaft= ',T,' N-m')