# Chapter12-Balancing of reciprocating of masses¶

## Ex1-pg310¶

In :
##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
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')

Magnitude of balance mass required=  90.0 Residual unbalanced forces=  3263.971  N


## Ex2-pg310¶

In :
##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
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
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')

speed= 86.476  swaying couple= 21.812  kNm


## Ex3-pg313¶

In :
##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
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")

Hammer blow= 32975.566  in Ntractive effort=  29018.117  in Nswaying couple=  10156.341  in Nm
The answer is a bit different due to rounding off error in textbook


## Ex4-pg314¶

In :
##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
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')

maximum unbalanced primary couple= 45.788   k Nm


## Ex5-pg315¶

In :
##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
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")

Maximum unbalanced secondary force= -29560.284  N in anticlockwise direction thats why - sign
The answer is a bit different due to rounding off error in textbook


## Ex6-pg316¶

In :
##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
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")

Hammer blow per wheel=  39182.3  N
The answer is a bit different due to rounding off error in textbook


## Ex7-pg318¶

In :
##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
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
##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')

portion of reciprocating mass to be balanced= 0.664  variation in tractive effort= 36980.420  N maximum swaying couple= 12943.147  N-m


## Ex8-pg320¶

In :
##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
##====================

unbalanced primary couple= 675 N-munbalanced secondary couple= 479.663  N-m