# Chapter5-Inertia Force Analysis in Machines¶

## Ex1-pg160¶

In [1]:
##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
##===================

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,'')

Maximum velocity of the piston =  7.0  m/s
teeta 74.96


## Ex2-pg161¶

In [2]:
##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
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')

Velocity of Piston =  4.4  m/s Acceleration of piston = 83.5  m/s**2 position of crank from inner dead centre position for zero acceleration of piston= 79.3  degrees


## Ex3-pg161¶

In [3]:
##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
##=================================
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')

Piston effort =  60447.0  NTurning moment on the crank shaft =  12604.2  N-m


## Ex4-pg162¶

In [4]:
##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')

Net force on piston =  2639.3  N Resultant load on gudgeon pin =  2652.9  N Thrust on cylinder walls =  269.1  N speed at which other things remining same,the gudgeon pin load would be reversed in directionm=  2528.4  rpm


## Ex5-pg163¶

In [5]:
##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')

Net load on the gudgeon pin=  4241.2  NThrust in the connecting rod=  4278.9  N Reaction between the cylinder and piston=  566.8  N The engine speed at which the above values become zero=  3158.0  rpm


## Ex6-pg165¶

In [6]:
##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')

Torque exerted on the crank shaft=  4233.8  N-m Thrust on the crank shaft bearing=  16321.0 NAcceleration of the flywheel=   122.2  rad/s**2


## Ex7-pg166¶

In [7]:
##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
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')

Effective turning moment on the crank shaft=  2366.2  N-m


## Ex8-pg166¶

In [8]:
##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
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')

Effective turning moment on the crank shaft=  8850.3  N-m