##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')
##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')
##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')
##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
##======================
w=2.*pi*N/60.## angular speed in rad/s
A=Ts*C*pi/180.## 1 cm**2 of turning moment diagram in Nm
E1=a[0]## max energy at B refer figure
E2=a[0]+a[1]+a[2]+a[3]
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')
##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')
##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
w=2.*pi*N/60.## angular speed in rad/s
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)
alpha=T1/(m*k**2.)## angular acceleration in rad/s**2
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
##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')
##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
a=(w1-w2)/T## angular acceleration in rad/s**2
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')
##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 ')
##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')