# Chapter6-Turning Moment Diagram and Flywheel¶

## Ex1-pg175¶

In :
##CHAPTER 6 ILLUSRTATION 1 PAGE NO 175
##TITLE:Turning Moment Diagram and Flywheel

k=1.##         radius of gyration of flywheel in m
m=2000.##       mass of the flywheel in kg
T=1000.##      torque of the engine in Nm
w1=0.##        speedin the begining
t=10.##        time duration
##==============================
I=m*k**2.##         mass moment of inertia in kg-m**2
a=T/I##           angular acceleration of flywheel in rad/s**2
w2=w1+a*t##       angular speed after time t in rad/s
K=I*w2**2/2.##     kinetic energy of flywheel in Nm
##==============================
print'%s %.1f %s %.1f %s '%('Angular acceleration of the flywheel=',a,' rad/s**2'' Kinetic energy of flywheel= ',K,' N-m')

Angular acceleration of the flywheel= 0.5  rad/s**2 Kinetic energy of flywheel=  25000.0  N-m


## Ex2-pg176¶

In :
##CHAPTER 6 ILLUSRTATION 2 PAGE NO 176
##TITLE:Turning Moment Diagram and Flywheel
import math
pi=3.141
N1=225.##              maximum speed of flywheel in rpm
k=.5##                radius of gyration of flywheel in m
n=720.##               no of holes punched per hour
E1=15000.##            energy required by flywheel in Nm
N2=200.##              mimimum speedof flywheel in rpm
t=2.##                 time taking for punching a hole
##==========================
P=E1*n/3600.##              power required by motor per sec in watts
E2=P*t##                   energy supplied by motor to punch a hole in N-m
E=E1-E2##                  maximum fluctuation of energy in N-m
N=(N1+N2)/2.##              mean speed of the flywheel in rpm
m=E/(pi**2./900.*k**2.*N*(N1-N2))
print'%s %.1f %s %.1f %s'%('Power of the motor= ',P,' watts''Mass of the flywheel required= ',m,' kg')

Power of the motor=  3000.0  wattsMass of the flywheel required=  618.2  kg


## Ex3-pg176¶

In :
##CHAPTER 6 ILLUSRTATION 3 PAGE NO 176
##TITLE:Turning Moment Diagram and Flywheel
import math
pi=3.141
d=38.##              diameter of hole in cm
t=32.##              thickness of hole in cm
e1=7.##                energy required to punch one square mm
V=25.##                mean speed of the flywheel in m/s
S=100.##                stroke of the punch in cm
T=10.##                time required to punch a hole in s
Cs=.03##                coefficient of fluctuation of speed
##===================
A=pi*d*t##                sheared area in mm**2
E1=e1*A##                 energy required to punch entire area in Nm
P=E1/T##                 power of motor required in watts
T1=T/(2.*S)*t##           time required to punch a hole in 32 mm thick plate
E2=P*T1##               energy supplied by motor in T1 seconds
E=E1-E2##                maximum fluctuation of energy in Nm
m=E/(V**2.*Cs)##           mass of the flywheel required
print'%s %.1f %s'%('Mass of the flywheel required= ',m,' kg')


Mass of the flywheel required=  1197.8  kg


## Ex4-pg177¶

In :
##CHAPTER 6 ILLUSRTATION 4 PAGE NO 177
##TITLE:Turning Moment Diagram and Flywheel
##figure 6.4
import math
##===================
pi=3.141
N=480.##             speed of the engine in rpm
k=.6##            radius of gyration in m
Cs=.03##           coefficient of fluctuaion of speed
Ts=6000.##          turning moment scale in Nm per one cm
C=30.##             crank angle scale in degrees per cm
a=[0.5,-1.22,.9,-1.38,.83,-.7,1.07]##      areas between the output torque and mean resistance line in sq.cm
##======================
A=Ts*C*pi/180.##            1 cm**2 of turning moment diagram in Nm
E1=a##                max energy at B refer figure
E2=a+a+a+a
E=(E1-E2)*A##            fluctuation of energy in Nm
m=E/(k**2.*w**2*Cs)##        mass of the flywheel in kg
print'%s %.1f %s'%('Mass of the flywheel= ',m,' kg')

Mass of the flywheel=  195.8  kg


## Ex5-pg178¶

In :
##CHAPTER 6 ILLUSRTATION 5 PAGE NO 178
##TITLE:Turning Moment Diagram and Flywheel
##==============
pi=3.141
P=500.*10**3.##        power of the motor in N
k=.6##            radius of gyration in m
Cs=.03##          coefficient of fluctuation of spped
OA=750.##           REFER FIGURE
OF=6.*pi##          REFER FIGURE
AG=pi## REFER FIGURE
BG=3000.-750.## REFER FIGURE
GH=2.*pi## REFER FIGURE
CH=3000.-750.## REFER FIGURE
HD=pi## REFER FIGURE
LM=2.*pi## REFER FIGURE
T=OA*OF+1./2.*AG*BG+BG*GH+1./2.*CH*HD##    Torque required for one complete cycle in Nm
Tmean=T/(6.*pi)##                 mean torque in Nm
w=P/Tmean##                    angular velocity required in rad/s
BL=3000.-1875.## refer figure
KL=BL*AG/BG##   From similar trangles
CM=3000.-1875.## refer figure
MN=CM*HD/CH##from similar triangles
E=1./2.*KL*BL+BL*LM+1./2.*CM*MN##         Maximum fluctuaion of energy in Nm
m=E*100./(k**2*w**2.*Cs)##   mass of flywheel in kg
print'%s %.1f %s'%('Mass of the flywheel= ',m,' kg')

Mass of the flywheel=  1150.3  kg


## Ex6-pg179¶

In :
##CHAPTER 6 ILLUSRTATION 6 PAGE NO 179
##TITLE:Turning Moment Diagram and Flywheel
import math
pi=3.141
PI=180.##in degrees
theta1=0.
theta2=PI
m=400.##      mass of the flywheel in kg
N=250.##      speed in rpm
k=.4##       radius of gyration in m
n=2.*250./60000.##          no of working strokes per minute
W=1000.*pi-150.*math.cos((2*theta2)/57.3)-250.*math.sin((2*theta2)/57.3)-(1000.*theta1-150.*math.cos((2*theta1)/57.3)-250.*math.sin((2*theta1)/57.3))##     workdone per stroke in Nm
P=W*n##        power in KW
Tmean=W/pi##         mean torque in Nm
twotheta=math.atan((500/300)/57.3)##       angle at which T-Tmean becomes zero
THETA1=twotheta/2.
THETA2=(180.+twotheta)/2.
E=-150.*math.cos((2.*THETA2)/57.3)-250.*math.sin((2.*THETA2)/57.3)-(-150*math.cos((2.*THETA1)/57.3)-250.*math.sin((2*THETA1)/57.3))##    FLUCTUATION OF ENERGY IN Nm
Cs1=E*100./(k**2.*w**2.*m)##   fluctuation range
Cs=Cs1/2.##         tatal percentage of fluctuation of speed
Theta=60.
T1=300.*math.sin((2*Theta)/57.3)-500.*math.cos((2*Theta)/57.3)##        Accelerating torque in Nm(T-Tmean)
print'%s %.1f %s  %.3f %s  %.3f %s '%('Power delivered=',P,' kw''Total percentage of fluctuation speed=',Cs,' ''Angular acceleration=',alpha,'rad/s**2')
#in book ans is given wrong

Power delivered= 26.2  kwTotal percentage of fluctuation speed=  0.342  Angular acceleration=  7.965 rad/s**2


## Ex7-pg181¶

In :
##CHAPTER 6 ILLUSRTATION 7 PAGE NO 181
##TITLE:Turning Moment Diagram and Flywheel

pi=3.141
m=200.##      mass of the flywheel in kg
k=.5##       radius of gyration in m
N1=360.##      upper limit of speed in rpm
N2=240.##       lower limit of speed in rpm
##==========
I=m*k**2.##        mass moment of inertia in kg m**2
w1=2.*pi*N1/60.
w2=2.*pi*N2/60.
E=1./2.*I*(w1**2.-w2**2.)##    fluctuation of energy in Nm
Pmin=E/(4.*1000.)##       power in kw
Eex=Pmin*12.*1000.##  Energy expended in performing each operation in N-m
print'%s %.1f %s %.1f %s '%('Mimimum power required= ',Pmin,' kw' ' Energy expended in performing each operation= ',Eex,' N-m')

Mimimum power required=  4.9  kw Energy expended in performing each operation=  59195.3  N-m


## Ex8-pg182¶

In :
##CHAPTER 6 ILLUSRTATION 8 PAGE NO 182
##TITLE:Turning Moment Diagram and Flywheel
import math
pi=3.141
b=8.##    width of the strip in cm
t=2.##    thickness of the strip in cm
w=1.2*10**3.##          work required per square cm cut
N1=200.##                maximum speed of the flywheel in rpm
k=.80##                 radius of gyration in m
N2=(1.-.15)*N1##         minimum speed of the flywheel in rpm
T=3.##                   time required to punch a hole
##=======================
A=b*t##           area cut of each stroke in cm**2
W=w*A##            work required to cut a strip in Nm
w1=2.*pi*N1/60.##        speed before cut in rpm
w2=2.*pi*N2/60.##        speed after cut in rpm
m=2.*W/(k**2.*(w1**2.-w2**2.))##       mass of the flywheel required in kg
Ta=m*k**2.*a##             torque required in Nm
print'%s %.1f %s %.1f %s '%('Mass of the flywheel= ',m,' kg'' Amount of Torque required=',Ta,'Nm')

Mass of the flywheel=  493.1  kg Amount of Torque required= 330.4 Nm


## Ex9-pg182¶

In :
##CHAPTER 6 ILLUSRTATION 9 PAGE NO 182
##TITLE:Turning Moment Diagram and Flywheel

pi=3.141
P=5.*10**3.##            power delivered by motor in watts
N1=360.##               speed of the flywheel in rpm
I=60.##                mass moment of inertia in kg m**2
E1=7500.##             energy required by pressing machine for 1 second in Nm
##========================
Ehr=P*60.*60.##      energy sipplied per hour in Nm
n=Ehr/E1
E=E1-P##           total fluctuation of energy in Nm
w1=2.*pi*N1/60.##     angular speed before pressing in rpm
w2=((2.*pi*N1/60.)**2.-(2.*E/I))**.5##       angular speed after pressing in rpm
N2=w2*60./(2.*pi)
R=N1-N2##              reduction in speed in rpm
print'%s %.1f %s %.1f %s '%('No of pressings that can be made per hour= ',n,' Reduction in speed after the pressing is over= ',R,' rpm ')

No of pressings that can be made per hour=  2400.0  Reduction in speed after the pressing is over=  10.7  rpm


## Ex10-pg183¶

In :
##CHAPTER 6 ILLUSRTATION 10 PAGE NO 183
##TITLE:Turning Moment Diagram and Flywheel
import math
pi=3.141
Cs=.02##     coefficient of fluctuation of speed
N=200.##      speed of the engine in rpm

theta1=math.acos(0/57.3)
theta2=math.asin((-6000/16000)/57.3)
theta2=180.-theta2
##===============================================
##largest area,representing fluctuation of energy lies between theta1 and theta2
E=6000.*math.sin(theta2/57.3)-8000./2.*math.cos((2*theta2)/57.3)-(6000.*math.sin((theta1)/57.3)-8000./2.*math.cos((2*theta1)/57.3))##      total fluctuation of energy in Nm
Theta=180##    angle with which cycle will be repeated in degrees
Theta1=0
Tmean=1/pi*((15000*pi+(-8000*math.cos((2*Theta)/57.3))/2.)-((15000*Theta1+(-8000*math.cos((2*Theta1)/57.3))/2.)))##     mean torque of engine in Nm
P=2*pi*N*Tmean/60000.##      power of the engine in kw
w=2*pi*N/60.##           angular speed of the engine in rad/s
I=E/(w**2.*Cs)##          mass moment of inertia of flywheel in kg-m**2
print'%s %.1f %s %.1f %s '%('Power of the engine= ',P,' kw'' minimum mass moment of inertia of flywheel=',-I,' kg-m**2'' E value calculated in the textbook is wrong. Its value is -15,124. In textbook it is given as -1370.28')

Power of the engine=  314.1  kw minimum mass moment of inertia of flywheel= 19.5  kg-m**2 E value calculated in the textbook is wrong. Its value is -15,124. In textbook it is given as -1370.28