Chapter2-Transmission of Motion and Power by Belts and Pulleys

Ex1-pg57

In [1]:
##CHAPTER 2 ILLUSRTATION 1 PAGE NO 57
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##===========================================================================================
##INPUT DATA
Na=300.;##driving shaft running speed in rpm
Nb=400.;##driven shaft running speed in rpm
Da=60.;##diameter of driving shaft in mm
t=.8;##belt thickness in mm
s=.05;##slip in percentage(5%)
##==========================================================================================
##calculation
Db=(Da*Na)/Nb;##finding out the diameter of driven shaft without considering the thickness of belt
Db1=(((Da+t)*Na)/Nb)-t##/considering the thickness
Db2=(1.-s)*(Da+t)*(Na/Nb)-t##considering slip also
##=========================================================================================
##output
print'%s %.1f %s'%('the value of Db is',Db,' cm')
print'%s %.1f %s'%('the value of Db1 is',Db1,' cm')
print'%s %.1f %s'%('the value of Db2 is',Db2,' cm')

the value of Db is 45.0  cm
the value of Db1 is 44.8  cm
the value of Db2 is 42.5  cm

Ex2-pg57

In [2]:
##CHAPTER 2,ILLUSRTATION 2 PAGE NO 57
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS

##====================================================================================
##input
n1=1200##rpm of motor shaft
d1=40##diameter of motor pulley in cm
d2=70##diameter of 1st pulley on the shaft in cm
s=.03##percentage slip(3%)
d3=45##diameter of 2nd pulley
d4=65##diameter of the pulley on the counnter shaft
##=========================================================================================
##calculation
n2=n1*d1*(1-s)/d2##rpm of driven shaft
n3=n2##both the pulleys are mounted on the same shaft
n4=n3*(1-s)*d3/d4##rpm of counter shaft

##output
print'%s %.1f %s %.1f %s '%('the speed of driven shaft is',n2,' rpm''the speed of counter shaft is ',n4,' rpm')

the speed of driven shaft is 665.1  rpmthe speed of counter shaft is  446.7  rpm

Ex3-pg58

In [3]:
##CHAPTER 2 ILLUSTRATION 3 PAGE NO:58
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##==============================================================================
##input
d1=30.##diameter of 1st shaft in cm
d2=50.##diameter 2nd shaft in cm
pi=3.141
c=500.##centre distance between the shafts in cm
##==============================================================================
##calculation
L1=((d1+d2)*pi/2.)+(2.*c)+((d1+d2)**2.)/(4.*c)##lenth of cross belt
L2=((d1+d2)*pi/2.)+(2.*c)+((d1-d2)**2.)/(4.*c)##lenth of open belt
r=L1-L2##remedy
##==============================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s '%('length of cross belt is ',L1,'cm '' length of open belt is ',L2,'cm''the length of the belt to be shortened is ',r,' cm')

length of cross belt is  1128.8 cm  length of open belt is  1125.8 cmthe length of the belt to be shortened is  3.0  cm

Ex4-pg59

In [4]:
##CHAPTER 2,ILLUSTRATION 4 PAGE 59
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##====================================================================================
##INPUT
D1=.5##            DIAMETER OF 1ST SHAFT IN m
D2=.25##           DIAMETER OF 2nd SHAFT IN m
C=2.##              CENTRE DISTANCE IN m
N1=220.##           SPEED OF 1st SHAFT
T1=1250.##          TENSION ON TIGHT SIDE IN N
U=.25##            COEFFICIENT OF FRICTION
PI=3.141
e=2.71
##====================================================================================
##CALCULATION
L=(D1+D2)*PI/2.+((D1+D2)**2./(4.*C))+2.*C
F=(D1+D2)/(2.*C)
ALPHA=math.asin(F/57.3)
THETA=(180.+(2.*ALPHA))*PI/180.##   ANGLE OF CONTACT IN radians
T2=T1/(e**(U*THETA))##            TENSION ON SLACK SIDE IN N
V=PI*D1*N1/60.##                   VELOCITY IN m/s
P=(T1-T2)*V/1000.##                POWER IN kW
##====================================================================================
##OUTPUT
print'%s %.1f %s'%('LENGTH OF BELT REQUIRED =',L,' m')
print'%s %.1f %s'%('ANGLE OF CONTACT =',THETA,' radians')
print'%s %.1f %s'%('POWER CAN BE TRANSMITTED=',P,' kW')

LENGTH OF BELT REQUIRED = 5.2  m
ANGLE OF CONTACT = 3.1  radians
POWER CAN BE TRANSMITTED= 3.9  kW

Ex5-pg59

In [5]:
##CHAPTER 2,ILLUSTRATION 5 PAGE 5
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##=====================================================================================================
##input
n1=100.## of driving shaft
n2=240.##speed of driven shaft
p=11000.##power to be transmitted in watts
c=250.##centre distance in cm
d2=60.##diameter in cm
b=11.5*10**-2##width of belt in metres
t=1.2*10**-2##thickness in metres
u=.25##co-efficient of friction 
pi=3.141
e=2.71
##===================================================================================================
##calculation for open bely drive
d1=n2*d2/n1
f=(d1-d2)/(2.*c)##sin(alpha) for open bely drive
##angle of arc of contact for open belt drive is,theta=180-2*alpha
alpha=math.asin(f)*57.3
teta=(180.-(2*alpha))*3.147/180.##pi/180 is used to convert into radians
x=(e**(u*teta))##finding out the value of t1/t2
v=pi*d2*10.*n2/60.##finding out the value of t1-t2
y=p*1000./(v)
t1=(y*x)/(x-1.)
Fb=t1/(t*b)/1000.
##=======================================================================================================
##calculation for cross belt drive bely drive
F=(d1+d2)/(2.*c)##for cross belt drive bely drive
ALPHA=math.asin(F)*57.3
THETA=(180.+(2.*ALPHA))*pi/180.##pi/180 is used to convert into radians
X=(e**(u*THETA))##finding out the value of t1/t2
V=pi*d2*10.*n2/60.##finding out the value of t1-t2
Y=p*1000./(V)
T1=(Y*X)/(X-1.)
Fb2=T1/(t*b)/1000.
##========================================================================================================
##output
print('for a open belt drive:')
print'%s %.1f %s %.1f %s'%('the tension in  belt is ',t1,'N'     'stress induced is ',Fb,' kN/m**2')
print('for a cross belt drive:')
print'%s %.1f %s %.1f %s '%('the tension in belt is ',T1,'N'     'stress induced is ',Fb2,' kN/m**2')

for a open belt drive:
the tension in  belt is  2898.4 Nstress induced is  2100.3  kN/m**2
for a cross belt drive:
the tension in belt is  2318.8 Nstress induced is  1680.3  kN/m**2

Ex6-pg61

In [6]:
##CHAPTER 2,ILLUSTRATION 6 PAGE 61
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##========================================================================================
##INPUT
D1=80.##DIAMETER OF SHAFT IN cm
N1=160.##SPEED OF 1ST SHAFT IN rpm
N2=320.##SPEED OF 2ND SHAFT IN rpm
C=250.##CENTRE DISTANCE IN CM
U=.3##COEFFICIENT OF FRICTION
P=4.##POWER IN KILO WATTS
e=2.71
PI=3.141
f=110.##STRESS PER cm WIDTH OF BELT
##========================================================================================
##CALCULATION
V=PI*D1*math.pow(10,-2)*N1/60.##VELOCITY IN m/s
Y=P*1000./V##Y=T1-T2
D2=D1*(N1/N2)##DIAMETER OF DRIVEN SHAFT
F=(D1-D2)/(2.*C)
ALPHA=math.asin(F/57.3)
THETA=(180.-(2.*ALPHA))*PI/180.##ANGLE OF CONTACT IN radians
X=e**(U*THETA)##VALUE OF T1/T2
T1=X*Y/(X-1.)
b=T1/f##WIDTH OF THE BELT REQUIRED 
##=======================================================================================
##OUTPUT
print'%s %.1f %s'%('THE WIDTH OF THE BELT IS ',b,' cm')
#apporximate ans is correct 

THE WIDTH OF THE BELT IS  8.9  cm

Ex7-pg62

In [7]:
##CHAPTER 2 ILLUSRTATION 7 PAGE NO 62
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS

##===========================================================================================
##INPUT DATA
m=1000.##                             MASS OF THE CASTING IN kg
PI=3.141
THETA=2.75*2*PI##                    ANGLE OF CONTACT IN radians
D=.26##                               DIAMETER OF DRUM IN m
N=24.##                                SPEED OF THE DRUM IN rpm
U=.25##                               COEFFICIENT OF FRICTION
e=2.71
T1=9810##                             TENSION ON TIGHTSIDE IN N
##=============================================================================================
##CALCULATION
T2=T1/(e**(U*THETA))##                 tension on slack side of belt in N
W=m*9.81##                            WEIGHT OF CASTING IN N
R=D/2.##                               RADIUS OF DRUM IN m
P=2*PI*N*W*R/60000.##                  POWER REQUIRED IN kW
P2=(T1-T2)*PI*D*N/60000.##                  POWER SUPPLIED BY DRUM IN kW
##============================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s %.1f %s '%('FORCE REQUIRED BY MAN=',T2,' N'and 'POWER REQUIRED TO RAISE CASTING=',P,' kW' 'POWER SUPPLIED BY DRUM=',P2,' kW')

FORCE REQUIRED BY MAN= 132.4 POWER REQUIRED TO RAISE CASTING= 3.2  kWPOWER SUPPLIED BY DRUM= 3.2  kW

Ex8-pg62

In [9]:
##CHAPTER 2,ILLUSTRATION 8 PAGE 62
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##INPUT
t=9.##THICKNESS IN mm
b=250.##WIDTH IN mm
D=90.##DIAMETER OF PULLEY IN cm
N=336.##SPEED IN rpm
PI=3.141
U=.35##COEFFICIENT FRICTION
e=2.71
THETA=120.*PI/180.
Fb=2.##STRESS IN MPa
d=1000.##DENSITY IN KG/M**3

##CALCULATION
M=b*10**-3.*t*10**-3.*d##MASS IN KG
V=PI*D*10**-2.*N/60.##VELOCITY IN m/s
Tc=M*V**2##CENTRIFUGAL TENSION
Tmax=b*t*Fb##MAX TENSION IN N
T1=Tmax-Tc
T2=T1/(e**(U*THETA))
P=(T1-T2)*V/1000.

##OUTPUT
print'%s %.1f %s'%('THE TENSION ON TIGHT SIDE OF THE BELT IS',T1,' N')
print'%s %.1f %s'%('THE TENSION ON SLACK SIDE OF THE BELT IS ',T2,' N')
print'%s %.1f %s'%('CENTRIFUGAL TENSION =',Tc,'N')
print'%s %.1f %s'%('THE POWER CAPACITY OF BELT IS ',P,' KW')

THE TENSION ON TIGHT SIDE OF THE BELT IS 3936.1  N
THE TENSION ON SLACK SIDE OF THE BELT IS  1895.6  N
CENTRIFUGAL TENSION = 563.9 N
THE POWER CAPACITY OF BELT IS  32.3  KW

Ex9-pg63

In [10]:
##CHAPTER 2,ILLUSTRATION 9 PAGE 63
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##INPUT
P=35000.##POWER TO BE TRANSMITTED IN WATTS
D=1.5##EFFECTIVE DIAMETER OF PULLEY IN METRES
N=300.##SPEED IN rpm
e=2.71
U=.3##COEFFICIENT OF FRICTION
PI=3.141
THETA=(11/24.)*360.*PI/180.##ANGLE OF CONTACT
density=1.1##density of belt material in Mg/m**3
L=1.##in metre
t=9.5##THICKNESS OF BELT IN mm
Fb=2.5##PERMISSIBLE WORK STRESS IN N/mm**2

##CALCULATION
V=PI*D*N/60.##VELOCITY IN m/s
X=P/V##X=T1-T2
Y=e**(U*THETA)##Y=T1/T2
T1=X*Y/(Y-1)
Mb=t*density*L/10**3.##value of m/b
Tc=Mb*V**2.##centrifugal tension/b
Tmaxb=t*Fb##max tension/b
b=T1/(Tmaxb-Tc)##thickness in mm
##output
print'%s %.1f %s'%('TENSION IN TIGHT SIDE OF THE BELT =',T1,' N')
print'%s %.1f %s'%('THICKNESS OF THE BELT IS =',b,' mm')

TENSION IN TIGHT SIDE OF THE BELT = 2573.5  N
THICKNESS OF THE BELT IS = 143.4  mm

Ex10-pg64

In [11]:
##CHAPTER 2,ILLUSTRATION 10 PAGE 64
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##INPUT
t=5.##THICKNESS OF BELT IN m
PI=3.141
U=.3
e=2.71
THETA=155.*PI/180.##ANGLE OF CONTACT IN radians
V=30.##VELOCITY IN m/s
density=1.##in m/cm**3
L=1.##LENGTH

##calculation
Xb=80.##           (T1-T2)=80b;so let (T1-T2)/b=Xb
Y=e**(U*THETA)##    LET Y=T1/T2
Zb=80.*Y/(Y-1.)##   LET  T1/b=Zb;BY SOLVING THE ABOVE 2 EQUATIONS WE WILL GET THIS EXPRESSION
Mb=t*L*density*10**-2.## m/b in N
Tcb=Mb*V**2.##          centrifugal tension/b
Tmaxb=Zb+Tcb##         MAX TENSION/b
Fb=Tmaxb/t##STRESS INDUCED IN TIGHT BELT

##OUTPUT
print'%s %.1f %s'%('THE STRESS DEVELOPED ON THE TIGHT SIDE OF BELT=',Fb,' N/cm**2')

THE STRESS DEVELOPED ON THE TIGHT SIDE OF BELT= 37.8  N/cm**2

Ex11-pg65

In [12]:
##CHAPTER 2,ILLUSTRATION 11 PAGE 65
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##INPUT
C=4.5##      CENTRE DISTANCE IN metres
D1=1.35##    DIAMETER OF LARGER PULLEY IN metres
D2=.9##      DIAMETER OF SMALLER PULLEY IN metres
To=2100.##    INITIAL TENSION IN newtons
b=12.##       WIDTH OF BELT IN cm
t=12.##       THICKNESS OF BELT IN mm
d=1.##        DENSITY IN gm/cm**3
U=.3##       COEFFICIENT OF FRICTION
L=1.##        length in metres
PI=3.141
e=2.71

##CALCULATION
M=b*t*d*L*10**-2.##               mass of belt per metre length in KG
V=(To/3./M)**.5##                 VELOCITY OF FOR MAX POWER TO BE TRANSMITTED IN m/s
Tc=M*V**2.##                      CENTRIFUGAL TENSION IN newtons
##                              LET (T1+T2)=X
X=2.*To-2.*Tc   ##                THE VALUE OF (T1+T2)
F=(D1-D2)/(2.*C)
ALPHA=math.asin(F/57.3)
THETA=(180.-(2.*ALPHA))*PI/180.##   ANGLE OF CONTACT IN radians
##                               LET T1/T2=Y
Y=e**(U*THETA)##                  THE VALUE OF T1/T2
T1=X*Y/(Y+1.)##                   BY SOLVING X AND Y WE WILL GET THIS EQN
T2=X-T1
P=(T1-T2)*V/1000.##                 MAX POWER TRANSMITTED IN kilowatts
N1=V*60./(PI*D1)##                   SPEED OF LARGER PULLEY IN rpm
N2=V*60./(PI*D2)##                   SPEED OF SMALLER PULLEY IN rpm
##OUTPUT
print'%s %.1f %s'%(' MAX POWER TO BE TRANSMITTED =',P,' KW')
print'%s %.1f %s'%(' SPEED OF THE LARGER PULLEY =',N1,' rpm')
print'%s %.1f %s'%(' SPEED OF THE SMALLER PULLEY =',N2,' rpm')

 MAX POWER TO BE TRANSMITTED = 27.0  KW
 SPEED OF THE LARGER PULLEY = 312.0  rpm
 SPEED OF THE SMALLER PULLEY = 468.0  rpm

Ex12-pg66

In [14]:
##CHAPTER 2,ILLUSTRATION 12 PAGE 66
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##============================================================================================================================
##INPUT
PI=3.141
e=2.71
D1=1.20##               DIAMETER OF DRIVING SHAFT IN m
D2=.50##                DIAMETER OF DRIVEN SHAFT IN m
C=4.##                   CENTRE DISTANCE BETWEEN THE SHAFTS IN m
M=.9##                  MASS OF BELT PER METRE LENGTH IN kg
Tmax=2000##             MAX TENSION IN N
U=.3##                  COEFFICIENT OF FRICTION
N1=200.##                SPEED  OF DRIVING SHAFT IN rpm
N2=450.##                SPEED OF DRIVEN SHAFT IN rpm
##==============================================================================================================================
##CALCULATION
V=PI*D1*N1/60.##         VELOCITY OF BELT IN m/s
Tc=M*V**2.##              CENTRIFUGAL TENSION IN N
T1=Tmax-Tc##            TENSION ON TIGHTSIDE IN N
F=(D1-D2)/(2.*C)
ALPHA=math.asin(F/57.3)
THETA=(180.-(2.*ALPHA))*PI/180.##   ANGLE OF CONTACT IN radians
T2=T1/(e**(U*THETA))##            TENSION ON SLACK SIDE IN N
TL=(T1-T2)*D1/2.##                TORQUE ON THE SHAFT OF LARGER PULLEY IN N-m
TS=(T1-T2)*D2/2.##                TORQUE ON THE SHAFT OF SMALLER PULLEY IN N-m
P=(T1-T2)*V/1000.##               POWER TRANSMITTED IN kW
Pi=2.*PI*N1*TL/60000.##            INPUT POWER
Po=2.*PI*N2*TS/60000.##            OUTPUT POWER
Pl=Pi-Po##                       POWER LOST DUE TO FRICTION IN kW
n=Po/Pi*100.##                    EFFICIENCY OF DRIVE IN %
##==================================================================================================================================
##OUTPUT
print'%s %.1f %s'%('TORQUE ON LARGER SHAFT =',TL,'N-m')
print'%s %.1f %s'%('TORQUE ON SMALLER SHAFT =',TS,' N-m')
print'%s %.1f %s'%('POWER TRANSMITTED =',P,' kW')
print'%s %.1f %s'%('POWER LOST DUE TO FRICTION =',Pl,' kW')
print'%s %.1f %s'%('EFFICIENCY OF DRINE =',n,' percentage')

TORQUE ON LARGER SHAFT = 679.0 N-m
TORQUE ON SMALLER SHAFT = 282.9  N-m
POWER TRANSMITTED = 14.2  kW
POWER LOST DUE TO FRICTION = 0.9  kW
EFFICIENCY OF DRINE = 93.8  percentage

Ex13-pg67

In [15]:
##CHAPTER 2,ILLUSTRATION 13 PAGE 67
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##============================================================================================================================
##INPUT
PI=3.141
e=2.71
P=90##                      POWER OF A COMPRESSOR IN kW
N2=250.##                   SPEED OF DRIVEN SHAFT IN rpm
N1=750.##                   SPEED OF DRIVER SHAFT IN rpm
D2=1.##                     DIAMETER OF DRIVEN SHAFT IN m
C=1.75##                    CENTRE DISTANCE IN m
V=1600./60.##                 VELOCITY IN m/s
a=375.##                     CROSECTIONAL AREA IN mm**2
density=1000.##              BELT DENSITY IN kg/m**3
L=1##                       length to be considered
Fb=2.5##                    STRESSS INDUCED IN MPa
beeta=35./2.##                THE GROOVE ANGLE OF PULLEY
U=.25##                     COEFFICIENT OF FRICTION
##=================================================================================================================================
##CALCULATION
D1=N2*D2/N1##               DIAMETER OF DRIVING SHAFT IN m
m=a*density*10**-6.*L##       MASS OF THE BELT IN kg
Tmax=a*Fb##                 MAX TENSION IN N
Tc=m*V**2.##                  CENTRIFUGAL TENSION IN N
T1=Tmax-Tc##                TENSION ON TIGHTSIDE OF BELT IN N
F=(D2-D1)/(2.*C)
ALPHA=math.asin(F/57.3)
THETA=(180.-(2.*ALPHA))*PI/180.##   ANGLE OF CONTACT IN radians
T2=T1/(e**(U*THETA/math.sin(beeta/57.3)))##TENSION ON SLACKSIDE IN N
P2=(T1-T2)*V/1000.##              POWER TRANSMITTED PER BELT kW
N=P/P2##                          NO OF V-BELTS
N3=N+1.
##======================================================================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('NO OF BELTS REQUIRED TO TRANSMIT POWER=',N,' APPROXIMATELY=',N3,'')

NO OF BELTS REQUIRED TO TRANSMIT POWER= 5.4  APPROXIMATELY= 6.4

Ex14-pg68

In [16]:
##CHAPTER 2,ILLUSTRATION 14 PAGE 68
##TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS
import math
##============================================================================================================================
##INPUT
PI=3.141
e=2.71
P=75.##             POWER IN kW
D=1.5##            DIAMETER OF PULLEY IN m
U=.3##             COEFFICIENT OF FRICTION
beeta=45./2.##       GROOVE ANGLE
THETA=160.*PI/180.## ANGLE OF CONTACT IN radians
m=.6##             MASS OF BELT IN kg/m
Tmax=800.##         MAX TENSION IN N
N=200.##            SPEED OF SHAFT IN rpm
##=============================================================================================================================
##calculation
V=PI*D*N/60.##               VELOCITY OF ROPE IN m/s
Tc=m*V**2.##                  CENTRIFUGAL TENSION IN N
T1=Tmax-Tc##                     TENSION ON TIGHT SIDE IN N
T2=T1/(e**(U*THETA/math.sin(beeta/57.3)))##TENSION ON SLACKSIDE IN N
P2=(T1-T2)*V/1000.##              POWER TRANSMITTED PER BELT kW
No=P/P2##                          NO OF V-BELTS
N3=No+1.##                           ROUNDING OFF
To=(T1+T2+Tc*2.)/2.##                 INITIAL TENSION
##================================================================================================================================
##OUTPUT
print'%s %.1f %s %.1f %s '%('NO OF BELTS REQUIRED TO TRANSMIT POWER=',No,'' 'APPROXIMATELY=',N3,'')
print'%s %.1f %s'%('INITIAL ROPE TENSION=',To,' N')

NO OF BELTS REQUIRED TO TRANSMIT POWER= 8.3 APPROXIMATELY= 9.3  
INITIAL ROPE TENSION= 510.8  N