##CHAPTER 12 ILLUSRTATION 1 PAGE NO 310
##TITLE:Balancing of reciprocating of masses
import math
#calculate the magnitude of balance mass required and residual balance error
pi=3.141
N=250.## speed of the reciprocating engine in rpm
s=18.## length of stroke in mm
mR=120.## mass of reciprocating parts in kg
m=70.## mass of revolving parts in kg
r=.09## radius of revolution of revolving parts in m
b=.15## distance at which balancing mass located in m
c=2./3.## portion of reciprocating mass balanced
teeta=30.## crank angle from inner dead centre in degrees
##===============================
B=r*(m+c*mR)/b## balance mass required in kg
w=2.*math.pi*N/60.## angular speed in rad/s
F=mR*w**2.*r*(((1.-c)**2.*(math.cos(teeta/57.3))**2.)+(c**2.*(math.sin(teeta/57.3))**2.))**.5## residual unbalanced forces in N
print'%s %.1f %s %.3f %s'%('Magnitude of balance mass required= ',B,'kg' and 'Residual unbalanced forces= ',F,' N')
##CHAPTER 12 ILLUSRTATION 2 PAGE NO 310
##TITLE:Balancing of reciprocating of masses
#calculate speed and swaying couples
pi=3.141
g=10.## acceleration due to gravity approximately in m/s**2
mR=240.## mass of reciprocating parts per cylinder in kg
m=300.## mass of rotating parts per cylinder in kg
a=1.8##distance between cylinder centres in m
c=.67## portion of reciprocating mass to be balanced
b=.60## radius of balance masses in m
r=24.## crank radius in cm
R=.8##radius of thread of wheels in m
M=40.
##=======================================
Ma=m+c*mR## total mass to be balanced in kg
mD=211.9## mass of wheel D from figure in kg
mC=211.9##..... mass of wheel C from figure in kg
theta=171.## angular position of balancing mass C in degrees
Br=c*mR/Ma*mC## balancing mass for reciprocating parts in kg
w=(M*g**3./Br/b)**.5## angular speed in rad/s
v=w*R*3600./1000.## speed in km/h
S=a*(1.-c)*mR*w**2*r/2.**.5/100./1000.## swaying couple in kNm
print'%s %.3f %s %.3f %s'%('speed=',v,' kmph'and ' swaying couple=',S,' kNm')
##CHAPTER 12 ILLUSRTATION 3 PAGE NO 313
##TITLE:Balancing of reciprocating of masses
#calculate hammer blow and tractive effort and swaying couple
import math
pi=3.141
g=10.## acceleration due to gravity approximately in m/s**2
a=.70##distance between cylinder centres in m
r=60.## crank radius in cm
m=130.##mass of rotating parts per cylinder in kg
mR=210.## mass of reciprocating parts per cylinder in kg
c=.67## portion of reciprocating mass to be balanced
N=300.##e2engine speed in rpm
b=.64## radius of balance masses in m
##============================
Ma=m+c*mR## total mass to be balanced in kg
mA=100.44## mass of wheel A from figure in kg
Br=c*mR/Ma*mA## balancing mass for reciprocating parts in kg
H=Br*(2.*math.pi*N/60.)**2*b## hammer blow in N
w=(2.*math.pi*N/60.)## angular speed
T=2**.5*(1.-c)*mR*w**2.*r/2./100.##tractive effort in N
S=a*(1.-c)*mR*w**2.*r/2./2.**.5/100.## swaying couple in Nm
print'%s %.3f %s %.3f %s %.3f %s'%('Hammer blow=',H,' in N' 'tractive effort= ',T,' in N' 'swaying couple= ',S,' in Nm')
print '%s'%("The answer is a bit different due to rounding off error in textbook")
##CHAPTER 12 ILLUSRTATION 4 PAGE NO 314
##TITLE:Balancing of reciprocating of masses
import math
#calculate maximum unbalanced primary couples
pi=3.141
mR=900.## mass of reciprocating parts in kg
N=90.## speed of the engine in rpm
r=.45##crank radius in m
cP=.9*mR*(2.*math.pi*N/60.)**2.*r*2.**.5/1000.## maximum unbalanced primary couple in kNm
print'%s %.3f %s'%('maximum unbalanced primary couple=',cP,' k Nm')
##CHAPTER 12 ILLUSRTATION 5 PAGE NO 315
##TITLE:Balancing of reciprocating of masses
import math
#calculate maximum unbalanced secondary force and with reasons
pi=3.141
mRA=160.## mass of reciprocating cylinder A in kg
mRD=160.## mass of reciprocating cylinder D in kg
r=.05## stroke lenght in m
l=.2## connecting rod length in m
N=450.## engine speed in rpm
##===========================
theta2=78.69## crank angle between A & B cylinders in degrees
mRB=576.88## mass of cylinder B in kg
n=l/r## ratio between connecting rod length and stroke length
w=2.*math.pi*N/60.## angular speed in rad/s
F=mRB*2.*w**2.*r*math.cos((2.*theta2)/57.3)/n
print'%s %.3f %s'%('Maximum unbalanced secondary force=',F,' N in anticlockwise direction thats why - sign')
print '%s'%("The answer is a bit different due to rounding off error in textbook")
##CHAPTER 12 ILLUSRTATION 6 PAGE NO 316
##TITLE:Balancing of reciprocating of masses
import math
pi=3.141
rA=.25## stroke length of A piston in m
rB=.25## stroke length of B piston in m
rC=.25## stroke length C piston in m
N=300.## engine speed in rpm
mRL=280.## mass of reciprocating parts in inside cylinder kg
mRO=240.## mass of reciprocating parts in outside cylinder kg
c=.5## portion ofreciprocating masses to be balanced
b1=.5## radius at which masses to be balanced in m
##======================
mA=c*mRO## mass of the reciprocating parts to be balanced foreach outside cylinder in kg
mB=c*mRL## mass of the reciprocating parts to be balanced foreach inside cylinder in kg
B1=79.4## balancing mass for reciprocating parts in kg
w=2.*math.pi*N/60.## angular speed in rad/s
H=B1*w**2*b1## hammer blow per wheel in N
print'%s %.1f %s'%('Hammer blow per wheel= ',H,' N')
print '%s'%("The answer is a bit different due to rounding off error in textbook")
##CHAPTER 12 ILLUSRTATION 7 PAGE NO 318
##TITLE:Balancing of reciprocating of masses
import math
pi=3.141
mR=300.## reciprocating mass per cylinder in kg
r=.3## crank radius in m
D=1.7## driving wheel diameter in m
a=.7## distance between cylinder centre lines in m
H=40.## hammer blow in kN
v=90.## speed in kmph
##=======================================
R=D/2.## radius of driving wheel in m
w=90.*1000./3600./R## angular velocity in rad/s
##Br*b=69.625*c by mearument from diagram
c=H*1000./(w**2.)/69.625## portion of reciprocating mass to be balanced
T=2.**.5*(1-c)*mR*w**2.*r## variation in tractive effort in N
M=a*(1.-c)*mR*w**2.*r/2.**.5## maximum swaying couple in N-m
print'%s %.3f %s %.3f %s %.3f %s'%('portion of reciprocating mass to be balanced=',c,' ''variation in tractive effort=',T,' N'' maximum swaying couple=',M,' N-m')
##CHAPTER 12 ILLUSRTATION 8 PAGE NO 320
##TITLE:Balancing of reciprocating of masses
import math
pi=3.141
N=1800.## speed of the engine in rpm
r=6.## length of crank in cm
l=24.## length of connecting rod in cm
m=1.5## mass of reciprocating cylinder in kg
##====================
w=2.*math.pi*N/60.## angular speed in rad/s
UPC=.019*w**2.## unbalanced primary couple in N-m
n=l/r## ratio of length of crank to the connecting rod
USC=.054*w**2./n## unbalanced secondary couple in N-m
print'%s %.f %s %.3f %s '%('unbalanced primary couple=',UPC,'N-m' 'unbalanced secondary couple=',USC,' N-m')