In [1]:

```
import math
from numpy import linalg
# Variables:
m = 6.5*1000 #kg
k = 1.8 #m
deltaE = 56.*1000 #N-m
N = 120. #rpm
#Solution:
#Calculating the maximum and minimum speeds
#We know that fluctuation of energy deltaE = math.pi**2/900*m*k**2*N*(N1-N2) or N1-N2 = (deltaE/(math.pi**2/900*m*k**2*N)) .....(i)
#Also mean speed N = (N1+N2)/2 or N1+N2 = 2*N .....(ii)
A = [[1, -1],[ 1, 1]]
B = [deltaE/(math.pi**2/900*m*k**2*N), 2*N]
V = linalg.solve(A,B)
N1 = round(V[0]) #rpm
N2 = round(V[1]) #rpm
#Results:
print " Maximum speed N1 = %d rpm."%(N1)
print " Minimum speed N2 = %d rpm."%(N2)
```

In [1]:

```
import math
# Variables:
k = 1. #m
m = 2500. #kg
T = 1500. #N-m
#Solution:
#Angular acceleration of the flywheel:
#Calculating the mass moment of inertia of the flywheel
I = m*k**2 #kg-m**2
#Calculating the angular acceleration of the flywheel
alpha = T/I #rad/s**2
#Kinetic energy of the flywheel:
omega1 = 0 #Angular speed at rest
#Calculating the angular speed after 10 seconds
omega2 = omega1+alpha*10 #rad/s
#Calculating the kinetic energy of the flywheel
KE = 1./2*I*(omega2)**2/1000 #Kinetic energy of the flywheel kN-m
#Results:
print " Angular acceleration of the flywheel alpha = %.1f rad/s**2."%(alpha)
print " Kinetic energy of the flywheel = %.1f kN-m."%(KE)
```

In [4]:

```
import math
# Variables:
P = 300.*1000 #W
N = 90. #rpm
CE = 0.1
k = 2. #m
#Solution:
#Calculating the mean angular speed
omega = 2*math.pi*N/60 #rad/s
#Calculating the coefficient of fluctuation of speed
CS = 1./100
#Calculating the work done per cycle
WD = P*60/N #Work done per cycle N-m
#Calculating the maximum fluctuation of energy
deltaE = WD*CE #N-m
#Calculating the mass of the flywheel
m = deltaE/(k**2*omega**2*CS) #kg
#Results:
print " Mass of the flywheel, m = %d kg."%(m)
```

In [5]:

```
import math
# Variables:
m = 36. #kg
k = 150./1000 #m
N = 1800. #rpm
#Solution:
#Refer Fig. 16.6
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Calculating the value of 1 mm**2 on the turning moment diagram
c = 5*math.pi/180 #Value of 1 mm**2 on turning miment diagram N-m
#Calculating the maximum fluctuation of energy
#From the turning moment diagram maximum energy = E+295 and minimum energy = E-690
deltaE = (285-(-690))*c #N-m
#Calculating the coefficient of fluctuation of energy
CS = deltaE/(m*k**2*omega**2)*100 #%
#Results:
print " Coefficient of fluctuation of speed CS = %.1f %%."%(CS)
```

In [6]:

```
import math
# Variables:
N = 600. #rpm
R = 0.5 #m
#Solution:
#Refer Fig. 16.7
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Calculating the coefficient of fluctuation of speed
CS = 3./100
#Calculating the value of 1 mm**2 on turning moment diagram
c = 600*math.pi/60 #Value of 1 mm**2 on turning moment diagram N-m
#Calculating the maximum fluctuation of energy
#From the turning moment diagram maximum fluctuation = E+52 and minimum fluctuation = E-120
deltaE = (52.-(-120))*c #N-m
#Calculating the mass of the flywheel
m = deltaE/(R**2*omega**2*CS) #kg
#Results:
print " Mass of the flywheel m = %d kg."%(m)
```

In [10]:

```
import math
# Variables:
N = 250. #rpm
m = 500. #kg
k = 600./1000 #m
#Solution:
#Refer Fig. 16.8
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Calculating the torque required for one complete cycle
T = (6*math.pi*750)+(1./2*math.pi*(3000-750))+(2*math.pi*(3000-750))+(1./2*math.pi*(3000-750)) #N-m
#Calculating the mean torque
Tmean = T/(6*math.pi) #N-m
#Calculating the power required to drive the machine
P = Tmean*omega/1000 #kW
#Coefficient of fluctuation of speed:
#Calculating the value of LM
LM = math.pi*((3000.-1875)/(3000-750.))
#Calculating the value of NP
NP = math.pi*((3000.-1875)/(3000-750))
#Calculating the value of BM
BM = 3000-1875. #N-m CN = BM
#Calculating the value of MN
MN = 2*math.pi
#Calculating the maximum fluctuation of energy
deltaE = (1./2*LM*BM)+(MN*BM)+(1./2*NP*BM) #N-m
#Calculating the coefficient of fluctuation of speed
CS = deltaE/(m*k**2*omega**2)
#Results:
print " Power required to drive the machine P = %.3f kW."%(P)
print " Coefficient of speed CS = %.3f."%(CS)
```

In [3]:

```
import math
# Variables:
N = 100. #rpm
k = 1.75 #m
#Solution:
#Refer Fig. 16.9
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Calculating the coefficient of fluctuation of speed
CS = 1.5/100
#Coefficient of fluctuation of energy:
AB = 2000.
LM = 1500. #N-m
#Calculating the work done per cycle
WD = (1./2*math.pi*AB)+(1./2*math.pi*LM) #Work done per cycle N-m
#Calculating the mean resisting torque
Tmean = WD/(2*math.pi) #N-m
#Calculating the value of CD
CD = math.pi/2000*(2000-875) #rad
#Calculating the maximum fluctuation of energy
deltaE = 1./2*CD*(2000-875) #N-m
#Calculating the coefficient of fluctuation of energy
Ce = deltaE/WD*100 #%
#Calculating the mass of the flywheel
m = deltaE/(k**2*omega**2*CS) #kg
#Crank angles for minimum and maximum speeds:
#Calculating the value of CE
CE = (2000.-875)/2000*(4*math.pi/9) #rad
#Calculating the crank angle for minimum speed
thetaC = ((4.*math.pi/9)-CE)*180/math.pi #degrees
#Calculating the value of ED
ED = (2000.-875)/2000*(math.pi-(4*math.pi/9)) #rad
#Calculating the crank angle for maximum speed
thetaD = ((4.*math.pi/9)+ED)*180/math.pi #degrees
#Results:
print " Coefficient of fluctuation of energy CE = %d %%."%(Ce)
print " Mass of the flywheel, m = %.1f kg."%(m)
print " Crank angle from IDC for the minimum speed, thetaC = %d degrees."%(thetaC)
print " Crank angle from IDC for the maximum speed, thetaD = %d degrees."%(thetaD)
```

In [12]:

```
import math
# Variables:
N = 600. #rpm
Tmax = 90. #N-m
m = 12. #kg
k = 80./1000 #m
#Solution:
#Refer Fig. 16.10
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Power developed:
#Calculating the work done per cycle
WD = 3*1./2*math.pi*90 #Work done per cycle N-m
#Calculating the mean torque
Tmean = WD/(2*math.pi) #N-m\
#Calculating the power developed
P = Tmean*omega/1000 #Power developed kW
#Coefficient of fluctuation of speed:
#Calculating the maximum fluctuation of energy
#From the torque-crank angle diagram maximum energy = E+5.89 and minimum energy = E-5.89
deltaE = 5.89-(-5.89) #N-m
#Calculating the coefficient of fluctuation of speed
CS = round(deltaE/(m*k**2*omega**2)*100) #%
#Calculating the coefficient of fluctuation of energy
CE = deltaE/WD*100 #%
#Calculating the maximum angular acceleration of the flywheel
alpha = (Tmax-Tmean)/(m*k**2) #rad/s**2
#Results:
print " Power developed = %.2f kW."%(P)
print " Coefficient of fluctuation of speed CS = %d %%."%(CS)
print " Coefficient of fluctuation of energy CE = %.2f %%."%(CE)
print " Maximum angular acceleration of the flywheel alpha = %d rad/s**2."%(alpha)
```

In [11]:

```
import math
# Variables:
P = 20.*1000 #W
N = 300. #rpm
#Solution:
#Refer Fig. 16.11
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #ra/s
#Calculating the coefficient of fluctuation of speed
CS = 4./100
#Calculating the number of working strokes per cycle for a four stroke engine
n = N/2
#Calculating the work done per cycle
WD = P*60/n #Work done per cycle N-m
#Calculating the work done during expansion cycle
WE = WD*3./2 #N-m
#Calculating the maximum turning moment
Tmax = WE*2/math.pi #N-m
#Calculating the mean turning moment
Tmean = WD/(4*math.pi) #N-m
#Calculating the excess turning moment
Texcess = Tmax-Tmean #N-m
#Calculating the value of DE
DE = Texcess/Tmax*math.pi #rad
#Calculating the maximum fluctuation of energy
deltaE = (1./2*DE*Texcess) #N-m
#Calculating the moment of inertia of the flywheel
I = deltaE/(omega**2*CS) #kg-m**2
#Results:
print " Moment of inertia of the flywheel I = %.1f kg-m**2."%(I)
```

In [11]:

```
import math
# Variables:
a1 = 0.45*10**-3
a2 = 1.7*10**-3
a3 = 6.8*10**-3
a4 = 0.65*10**-3 #m**2
N1 = 202.
N2 = 198. #rpm
R = 1.2 #m
#Solution:
#Refer Fig. 16.12
#Calculating the net area
a = a3-(a1+a2+a4) #Net area m**2
#Calculating the energy scale constant
c = 3*10**6 #Energy scale constant N-m
#Calculating the net work done per cycle
WD = a*c #Net work done per cycle N-m
#Calculating the mean torque
Tmean = round(WD/(4*math.pi)) #N-m
#Calculating the value of FG
FG = Tmean #N-m
#Calculating the work done during expansion stroke
WDe = a3*c #Work done during expansion stroke N-m
#Calculating the value of AG
AG = WDe/(1./2*math.pi) #N-m
#Calculating the excess torque
Texcess = round(AG-FG,-1) #N-m
#Calculating the value of AF
AF = Texcess #N-m
#Calculating the value of DE
DE = round(AF/AG*math.pi,1) #rad
#Calculating the maximum fluctuation of energy
deltaE = 1./2*DE*AF #N-m
#Mass of the rim of a flywheel:
#Calculating the mean speed of the flywheel
N = (N1+N2)/2 #rpm
#Calculating the mass of the rim of a flywheel
m = deltaE/(math.pi**2/900*R**2*N*(N1-N2)) #kg
#Results:
print " Mass of the rim of the flywheel m = %.f kg."%(m)
```

In [22]:

```
import math
from scipy.integrate import quad
# variables
w = math.pi*2*180./60 # rad/s
T = 180 # rpm
Cs = 0.01 # speed
# Calculations
work_done_r = 20000 * 2 # pi N-m
Tmean = work_done_r/2 # N-m
power = round(Tmean * w,-3)/1000
deltaE = 11078
energy = deltaE/round((w**2*Cs),2)
excess = 9500*math.sin(math.radians(90)) - 5700*math.cos(math.radians(90))
alpha = excess/energy
# results
print "power developed by the engine : %.f kW"%power
print "maximum fluctuation of energy : %.f kg-m**2"%energy
print "Alpha a = %.3f rad/s**2"%alpha
```

In [24]:

```
import math
from scipy.integrate import quad
# Variables:
m = 500. #kg
k = 0.4 #m
N = 150. #rpm
#Solution:
#Refer Fig. 16.14
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Fluctuation of energy:
#Equating the change in torque to zero and calculating the value of theta
thetaA = math.sin(math.radians(0))
thetaC = math.sin(math.radians(0))+180
thetaE = math.sin(math.radians(0))+360 #degrees
thetaB = 65.4
thetaD = 294.6
#Calculating the maximum fluctuation of energy
def f4(theta):
return (5000+600*math.sin(2*theta))-(5000+500*math.sin(theta))
deltaE = round( quad(f4 ,thetaC*math.pi/180,thetaD*math.pi/180)[0])
#Calculating the total percentage fluctuation of speed
CS = deltaE/(m*k**2*omega**2)*100 #%
#Maximum and minimum angular acceleration of the flywheel and the corresponding shaft positions:
#Calculating the maximum or minimum values of theta
#Differentiating (600*math.sin(2*theta))-500*math.sin(theta) = 0 with respect to theta and equating to zero
#we get 12*2*(math.cos(theta))**2-5*math.cos(theta)-12 = 0
a = 12.*2
b = -5.
c = -12.
costheta1 = (-b+math.sqrt(b**2-4*a*c))/(2*a)
costheta2 = (-b-math.sqrt(b**2-4*a*c))/(2*a)
theta1 = math.degrees(math.acos(costheta1))
theta2 = math.degrees(math.acos(costheta2)) #degrees
#Calculating the maximum torque
Tmax = 600*math.sin(math.radians(2*theta1))-500*math.sin(math.radians(theta1)) #N-m
#Calculating the minimum torque
Tmin = 600*math.sin(math.radians(2*theta2))-500*math.sin(math.radians(theta2)) #N-m
#Calculating the maximum acceleration
alphamax = Tmax/(m*k**2) #rad/s**2
#Calculating the minimum acceleration
alphamin = abs(Tmin)/(m*k**2) #rad/s**2
#Results:
print " Fluctuation of energy deltaE = %d N-m."%(deltaE)
print " Total percentage fluctuation of speed CS = %.1f %%."%(CS)
print " Shaft position corresponding to maximum and minimum accelerations\
theta = %d degrees and %.1f degrees."%(theta1,theta2)
print " Maximum acceleration, alphamax = %.2f rad/s**2."%(alphamax)
print " Minimum acceleration alphamin = %.1f rad/s**2."%(alphamin)
```

In [20]:

```
import math
from scipy.integrate import quad
# Variables:
I = 1000. #kg-m**2
N = 300. #rpm
#Solution:
#Refer Fig. 16.15 and Fig. 16.16
#Calculating the angular speed of the crank
omega = 2*math.pi*N/60 #rad/s
#Power of the engine:
#Calculating the work done per revolution
def f0(theta):
return 5000+1500*math.sin(3*theta)
WD = quad(f0,0,2*math.pi)[0]
#Calculating the mean resisting torque
Tmean = WD/(2*math.pi) #N-m
#Calculating the power of the engine
P = Tmean*omega/1000 #kW
#Maximum fluctuation of the speed of the flywheel when resisting torque is consmath.tant:
#Calculating the value of theta
theta = (5000-5000)/1500
theta = 1./3*(math.sin(math.radians((theta)))+180) #degrees
#Calculating the maximum fluctuation of energy
def f1(theta):
return 5000+1500*math.sin(3*theta)-5000
deltaE = quad(f1,0,60*math.pi/180)[0]
#Calculating the maximum fluctuation of speed of the flywheel
CS1 = deltaE/(I*omega**2)*100 #%
#Maximum fluctuation of speed of the flywheel when resisting torque (5000+600*math.sin(theta)) N-m:
#Calculating the values of theta thetaB and thetaC
thetaB = math.sin(math.radians(math.sqrt((1./4*(3-600./1500))))) #degrees
thetaC = 180-thetaB #degrees
#Calculating the maximum fluctuation of energy
def f2(theta):
return (5000+1500*math.sin(3*theta))-(5000+600*math.sin(theta))
deltaE = round( quad(f2,thetaB*math.pi/180,thetaC*math.pi/180)[0])
#Calculating the maximum fluctuation of speed of the flywheel
CS2 = abs(deltaE)/(I*omega**2)*100 #%
#Results:
print " Power of the engine P = %.1f kW."%(P)
print " Maximum fluctuation of the speed of the flywheel when resisting torque\
is constant, CS = %.1f %%."%(CS1)
print " Maximum fluctuation of speed of the flywheel when resisting torque \
5000+600*sintheta N-m CS = %.3f %%."%(CS2)
```

In [21]:

```
import math
# Variables:
N = 800. #rpm
stroke = 300. #mm
sigma = 7.*10**6 #N/m**2
rho = 7200. #kg/m**3
#Solution:
#Refer Fig. 16.18
#Calculating the angular speed of the engine
omega = 2*math.pi*N/60 #rad/s
#Calculating the coefficient of fluctuation of speed
CS = 4./100
#Diameter of the flywheel rim:
#Calculating the peripheral velocity of the flywheel rim
v = math.sqrt(sigma/rho) #m/s
#Calculating the diameter of the flywheel rim
D = v*60/(math.pi*N) #m
#Cross-section of the flywheel rim:
#Calculating the value of 1 mm**2 on the turning moment diagram
c = 500.*math.pi/30 #Value of 1 mm**2 on the turning moment diagram N-m
#Calculating the maximum fluctuation of energy
deltaE = round((420.-(-30))*c) #N-m
#Calculating the mass of the flywheel rim
m = deltaE/(v**2*CS) #kg
#Calculating the thickness of the flywheel rim
t = math.sqrt(m/(math.pi*D*5*rho))*1000 #mm
#Calculating the width of the flywheel rim
b = 5*t #mm
#Results:
print " Diameter of the flywheel rim D = %.3f m."%(D)
print " Thickness of the flywheel rim t = %d mm."%(t)
print " Width of the flywheel rim b = %d mm."%(b)
```

In [30]:

```
import math
# Variables:
P = 150.*1000 #W
N = 80. #rpm
CE = 0.1
D = 2.
R = D/2. #m
rho = 7200. #kg/m**3
#Solution:
#Calculating the angular speed of the engine
omega = round(2*math.pi*N/60,1) #rad/s
#Calculating the coefficient of fluctuation of speed
CS = 4./100
#Mass of the flywheel rim:
#Calculating the work done per cycle
WD = P*60/N #Work done per cycle N-m
#Calculating the maximum fluctuation of energy
deltaE = WD*CE #N-m
#Calculating the mass moment of inertia of the flywheel
I = deltaE/(omega**2*CS) #kg-m**2
#Calculating the mass moment of inertia of the flywheel rim
Irim = 0.95*I #kg-m**2
#Calculating the mass of the flywheel rim
k = R #Radius of gyration m
m = Irim/k**2 #kg
#Calculating the cross-sectional area of the flywheel rim
A = m/(2*math.pi*R*rho) #m**2
#Resilts:
print " Mass of the flywheel rim m = %.f kg."%(m)
print " Cross-sectional area of the flywheel rim A = %.3f m**2."%(A)
```

In [23]:

```
import math
# Variables:
N = 600. #rpm
rho = 7250. #kg/m**3
sigma = 6.*10**6 #N/m**2
#Solution:
#Refer Fig. 16.19
#Calculating the angular speed of the engine
omega = 2.*math.pi*N/60 #rad/s
#Calculating the total fluctuation of speed
CS = 2./100
#Moment of inertia of the flywheel:
#Calculating the value of 1 mm**2 of turning moment diagram
c = 250.*math.pi/60 #Value of 1 mm**2 of turning moment diagram N-m
#Calculating the maximum fluctuation of energy
deltaE = round((162.-(-35))*c) #N-m
#Calculating the moment of inertia of the flywheel
I = deltaE/(omega**2*CS) #kg-m**2
#Dimensions of the flywheel rim:
#Calculating the peripheral velocity of the flywheel
v = math.sqrt(sigma/rho) #m/s
#Calculating the mean diameter of the flywheel
D = v*60/(math.pi*N) #m
#Calculating the maximum fluctuation of energy of the flywheel rim
deltaErim = 0.92*deltaE #N-m
#Calculating the mass of the flywheel rim
m = deltaErim/(v**2*CS) #kg
#Calculating the thickness of the flywheel rim
t = math.sqrt(m/(math.pi*D*2*rho))*1000 #mm
#Calculating the breadth of the flywheel rim
b = 2*t #mm
#Results:
print " Moment of inertia of the flywheel I = %.1f kg-m**2."%(I)
print " Thickness of the flywheel rim t = %.1f mm."%(t)
print " Breadth of the flywheel rim b = %.1f mm."%(b)
```

In [19]:

```
import math
# Variables:
a1 = 5.*10**-5 #m**2
a2 = 21.*10**-5 #m**2
a3 = 85.*10**-5 #m**2
a4 = 8.*10**-5 #m**2
N2 = 98.
N1 = 102. #rpm
rho = 8150. #kg/m**3
sigma = 7.5*10**6 #N/m**2
#Solution:
#Refer Fig. 16.20
#Calculating the net area
a = a3-(a1+a2+a4) #Net area m**2
#Calculating the value of 1 m**2 on the turning moment diagram in terms of work
c = 14*10**6 #Value of 1 m**2 on the turning moment diagram N-m
#Calculating the net work done per cycle
WD = a*c #Net work done per cycle N-m
#Calculating the mean torque on the flywheel
Tmean = round(WD/(4*math.pi)) #N-m
FG = Tmean #N-m
#Calculating the work done during expansion stroke
WDe = int(a3*c) #Work done during expansion stroke N-m
#Calculating the value of AG
AG = int(WDe/(1./2*math.pi)) #N-m
#Calculating the excess torque
Texcess = AG-FG #Excess torque N-m
AF = Texcess #N-m
#Calculating the value of DE
DE = round(AF/AG*math.pi,1) #rad
#Calculating the maximum fluctuation of energy
deltaE = 1./2*DE*AF #N-m
#Moment of inertia of the flywheel:
#Calculating the mean speed during the cycle
N = (N1+N2)/2 #rpm
#Calculating the corresponding angular mean speed
omega = 2*math.pi*N/60 #rad/s
#Calculating the coefficient of fluctuation of speed
CS = (N1-N2)/N
#Calculating the moment of inertia of the flywheel
I = deltaE/(omega**2*CS) #kg-m**2
#Size of flywheel:
#Calculating the peripheral velocity of the flywheel
v = math.sqrt(sigma/rho) #m/s
#Calculating the mean diameter of the flywheel
D = v*60/(math.pi*N) #m
#Calculating the mass of the flywheel rim
m = deltaE/(v**2*CS) #kg
#Calculating the thickness of the flywheel rim
t = math.sqrt(m/(math.pi*D*4*rho))*1000 #mm
#Calculating the width of the flywheel rim
b = 4*t #mm
#Results:
print " Moment of inertia of the flywheel I = %.f kg-m**2."%(I)
print " Thickness of the flywheel rim t = %.1f mm."%(t)
print " Width of the flywheel rim b = %.1f mm."%(b)
# rounding off error.
```

In [27]:

```
import math
# Variables:
P = 50.*1000 #W
N = 150. #rpm
n = 75.
sigma = 4.*10**6 #N/m**2
rho = 7200. #kg/m**3
#Solution:
#Refer Fig. 16.21
#Calculating the angular speed of the engine
omega = 2*math.pi*N/60 #rad/s
#Calculating the mean torque transmitted by the flywheel
Tmean = P/omega #N-m
FG = Tmean #N-m
#Calculating the work done per cycle
WD = Tmean*4*math.pi #Work done per cycle N-m
#Calculating the work done during power stroke
WDp = 1.4*WD #Work done during power stroke N-m
#Calculating the maximum torque transmitted by the flywheel
Tmax = WDp/(1./2*math.pi) #N-m
BF = Tmax #N-m
#Calculating the excess torque
Texcess = Tmax-Tmean #N-m
BG = Texcess #N-m
#Calculating the value of DE
DE = BG/BF*math.pi #N-m
#Calculating the maximum fluctuation of energy
deltaE = 1./2*DE*BG #N-m
#Mean diameter of the flywheel:
#Calculating the peripheral velocity of the flywheel
v = math.sqrt(sigma/rho) #m/s
#Calculating the mean diameter of the flywheel
D = v*60./(math.pi*N) #m
#Cross-sectional dimensions of the rim:
#Calculating the coefficient of fluctuation of speed
CS = 1./100
#Calculating the total energy of the flywheel
E = deltaE/(2*CS) #N-m
#Calculating the energy of the rim
Erim = 15./16*E #N-m
#Calculating the mass of the flywheel rim
m = Erim/(1./2*v**2) #kg
#Calculating the thickness of the rim
t = round(math.sqrt(m/(math.pi*D*4*rho))*1000) #mm
#Calculating the width of the rim
b = 4*t #mm
#Results:
print " Mean diameter of the flywheel D = %d m."%(D)
print " Thickness of the flywheel rim t = %d mm."%(t)
print " Width of the flywheel rim b = %d mm."%(b)
```

In [28]:

```
import math
# Variables:
N1 = 225.
N2 = 200. #rpm
k = 0.5 #m
E1 = 15.*1000 #N-m
HolePunched = 720. #per hour
#Solution:
#Power of the motor:
#Calculating the total energy required per second
E = E1*HolePunched/3600 #N-m/s
#Calculating the power of the motor
P = E/1000 #kW
#Minimum mass of the flywheel:
#Calculating the energy supplied by the motor in 2 seconds
E2 = E*2 #N-m
#Calculating the energy supplied by the flywheel during punching
deltaE = E1-E2 #N-m
#Calculating the mean speed of the flywheel
N = (N1+N2)/2 #rpm
#Calculating the minimum mass of the flywheel
m = round(deltaE*900/(math.pi**2*k**2*N*(N1-N2))) #kg
#Results:
print " Power of the motor P = %d kW."%(P)
print " Minimum mass of the flywheel m = %d kg."%(m)
```

In [29]:

```
import math
# Variables:
d = 38. #mm
t = 32. #mm
s = 100. #mm
E1 = 7. #N-m/mm**2 of sheared area
v = 25. #m/s
#Solution:
#Power of the motor required:
#Calculating the sheared area
A = round(math.pi*d*t) #mm**2
#Calculating the total energy required per hole
E1 = E1*A #N-m
#Calculating the energy required for punching work per second
E = E1/10 #Energy required for punching work per second N-m/s
#Calculating the power of the motor required
P = E/1000 #Power of the motor required kW
#Mass of the flywheel required:
#Calculating the time required to punch a hole in a 32 mm thick plate
t32 = 10/(2*s)*t #Time required to punch a hole in 32 mm thick plate seconds
#Calculating the energy supplied by the motor in t32 seconds
E2 = E*t32 #N-m
#Calculating the energy to be supplied by the flywheel during punching
deltaE = E1-E2 #N-m
#Calculating the coefficient of fluctuation of speed
CS = 3/100.
#Calculating the mass of the flywheel required
m = round(deltaE/(v**2*CS)) #kg
#Results:
print " Power of the motor required P = %.3f kW."%(P)
print " Mass of the flywheel required m = %d kg."%(m)
```

In [30]:

```
import math
# Variables:
P = 3. #kW
m = 150. #kg
k = 0.6 #m
N1 = 300. #rpm
#Solution:
#Calculating the angular speed of the flywheel before riveting
omega1 = 2*math.pi*N1/60 #rad/s
#Speed of the flywheel immediately after riveting:
#Calculating the energy supplied by the motor
E2 = P*1000 #N-m/s
#Calculating the energy absorbed during one riveting operation which takes 1 second
E1 = 10000 #N-m
#Calculating the energy to be supplied by the flywheel for each riveting operation per second
deltaE = E1-E2 #N-m
#Calculating the angular speed of the flywheel immediately after riveting
omega2 = math.sqrt(omega1**2-(2*deltaE/(m*k**2))) #rad/s
#Calculating the corresponding speed in rpm
N2 = omega2*60/(2*math.pi) #rpm
#Calculating the number of rivets that can be closed per minute
n = E2/E1*60 #Number of rivets that can be closed per minute
#Results:
print " Speed of the flywheel immediately after riveting N2 = %.1f rpm."%(N2)
print " Number of rivets that can be closed per minute = %d rivets."%(n)
```

In [31]:

```
import math
# Variables:
d = 40. #mm
t = 15. #mm
NoofHoles = 30. #per minute
EnergyRequired = 6. #N-m/mm**2
Time = 1./10 #seconds
N1 = 160.
N2 = 140. #rpm
k = 1. #m
#Solution:
#Calculating the sheared area per hole
A = round(math.pi*d*t) #Sheared area per hole mm**2
#Calculating the energy required to punch a hole
E1 = EnergyRequired*A #N-m
#Calculating the energy required for punching work per second
E = E1*NoofHoles/60 #Energy required for punching work per second N-m/s
#Calculating the energy supplied by the motor during the time of punching
E2 = E*Time #N-m
#Calculating the energy to be supplied by the flywheel during punching a hole
deltaE = E1-E2 #N-m
#Calculating the mean speed of the flywheel
N = (N1+N2)/2 #rpm
#Calculating the mass of the flywheel required
m = round(deltaE*900/(math.pi**2*k**2*N*(N1-N2))) #kg
#Results:
print " Mass of the flywheel required m = %d kg."%(m)
```

In [32]:

```
import math
# Variables:
n = 25.
d1 = 25./1000 #m
t1 = 18./1000 #m
D = 1.4
R = D/2 #m
touu = 300.*10**6 #N/m**2
etam = 95./100
CS = 0.1
sigma = 6.*10**6 #N/m**2
rho = 7250. #kg/m**3
#Solution:
#Power needed for the driving motor:
#Calculating the area of the plate sheared
AS = math.pi*d1*t1 #m**2
#Calculating the maximum shearing force required for punching
FS = AS*touu #N
#Calculating the energy required per stroke
E = 1./2*FS*t1 #Energy required per stroke N-m
#Calculating the energy required per minute
E1 = E*n #Energy required per minute N-m
#Calculating the power required for the driving motor
P = E1/(60*etam)/1000 #Energy required for the driving motor kW
#Dimensions for the rim cross-section:
#Calculating the maximum fluctuation of energy
deltaE = 9./10*E #N-m
#Calculating the maximum fluctuation of energy provided by the rim
deltaErim = 0.95*deltaE #N-m
#Calculating the mean speed of the flywheel
N = 9.*25 #rpm
#Calculating the mean angular speed
omega = 2*math.pi*N/60 #rad/s
#Calculating the mass of the flywheel
m = round(deltaErim/(R**2*omega**2*CS)) #kg
#Calculating the thickness of rim
t = math.sqrt(m/(math.pi*D*2*rho))*1000 #mm
#Calculating the width of rim
b = 2*t #mm
#Results:
print " Power needed for the driving motor = %.3f kW."%(P)
print " Thickness of the flywheel rim t = %d mm."%(t)
print " Width of the flywheel rim b = %d mm."%(b)
#Answers vary due to rounding-off errors
```