# Chapter 16 : Turning Moment Diagrams and Flywheel¶

## Example 16.1 Page No : 573¶

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)

 Maximum speed N1  =  121 rpm.
Minimum speed N2  =  119 rpm.


## Example 16.2 Page No : 573¶

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
#Kinetic energy of the flywheel:
omega1 = 0 			#Angular speed at rest
#Calculating the angular speed after 10 seconds
#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)

 Angular acceleration of the flywheel alpha  =  0.6 rad/s**2.
Kinetic energy of the flywheel  =  45.0 kN-m.


## Example 16.3 Page No : 574¶

In [4]:
import math

# Variables:
P = 300.*1000 	#W
N = 90. 		#rpm
CE = 0.1
k = 2. 			#m

#Solution:
#Calculating the mean angular speed
#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)

 Mass of the flywheel, m  =  5628 kg.


## Example 16.4 Page No : 574¶

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
#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)

 Coefficient of fluctuation of speed CS  =  0.3 %.


## Example 16.5 Page No : 575¶

In [6]:
import math

# Variables:
N = 600. 			#rpm
R = 0.5 			#m

#Solution:
#Refer Fig. 16.7
#Calculating the angular speed of the crank
#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)

 Mass of the flywheel m  =  182 kg.


## Example 16.6 Page No : 584¶

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
#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)

 Power required to drive the machine P  =  49.087 kW.
Coefficient of speed CS  =  0.072.


## Example 16.7 Page No : 578¶

In [3]:
import math

# Variables:
N = 100. 			#rpm
k = 1.75 			#m

#Solution:
#Refer Fig. 16.9
#Calculating the angular speed of the crank
#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
#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
#Calculating the crank angle for minimum speed
thetaC = ((4.*math.pi/9)-CE)*180/math.pi 			#degrees
#Calculating the value of ED
#Calculating the crank angle for maximum speed

#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)

 Coefficient of fluctuation of energy CE  =  18 %.
Mass of the flywheel, m  =  197.3 kg.
Crank angle from IDC for the minimum speed, thetaC  =  35 degrees.
Crank angle from IDC for the maximum speed, thetaD  =  136 degrees.


## Example 16.8 Page No : 580¶

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
#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

#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)

 Power developed  =  4.24 kW.
Coefficient of fluctuation of speed CS  =  4 %.
Coefficient of fluctuation of energy CE  =  2.78 %.
Maximum angular acceleration of the flywheel alpha  =  292 rad/s**2.


## Example 16.9 Page No : 582¶

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
#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)

 Moment of inertia of the flywheel I  =  255.4 kg-m**2.


## Example 16.10 Page No : 584¶

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
#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)

 Mass of the rim of the flywheel m  =  1381 kg.


## Example 16.11 page no : 585¶

In [22]:
import math

# variables
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)
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

power developed by the engine : 377 kW
maximum fluctuation of energy : 3121 kg-m**2


## Example 16.12 Page No : 587¶

In [24]:
import math

# Variables:
m = 500. 			#kg
k = 0.4 			#m
N = 150. 			#rpm

#Solution:
#Refer Fig. 16.14
#Calculating the angular speed of the crank
#Fluctuation of energy:
#Equating the change in torque to zero and calculating the value of theta
thetaB = 65.4

#Calculating the maximum fluctuation of energy
def f4(theta):
return (5000+600*math.sin(2*theta))-(5000+500*math.sin(theta))

#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
#Calculating the minimum torque
#Calculating the maximum acceleration
#Calculating the minimum acceleration

#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)

 Fluctuation of energy deltaE  =  1204 N-m.
Total percentage fluctuation of speed CS  =  6.1 %.
Shaft position corresponding to maximum and minimum accelerations theta  =  35 degrees and 127.6 degrees.
Maximum acceleration, alphamax  =  3.46 rad/s**2.
Minimum acceleration alphamin  =  12.2 rad/s**2.


## Example 16.13 Page No : 589¶

In [20]:
import math

# 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
#Power of the engine:
#Calculating the work done per revolution
def f0(theta):
return 5000+1500*math.sin(3*theta)

#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
#Calculating the maximum fluctuation of energy
def f1(theta):
return 5000+1500*math.sin(3*theta)-5000

#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
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))

#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)

 Power of the engine P  =  157.1 kW.
Maximum fluctuation of the speed of the flywheel when resisting torque is constant, CS  =  0.1 %.
Maximum fluctuation of speed of the flywheel when resisting torque  5000+600*sintheta N-m CS  =  0.020 %.


## Example 16.14 Page No : 592¶

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
#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)

 Diameter of the flywheel rim D  =  0.744 m.
Thickness of the flywheel rim t  =  84 mm.
Width of the flywheel rim b  =  424 mm.


## Example 16.15 Page No : 594¶

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
#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)

 Mass of the flywheel rim m  =  3787 kg.
Cross-sectional area of the flywheel rim A  =  0.084 m**2.


## Example 16.16 Page No : 595¶

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
#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)

 Moment of inertia of the flywheel I  =  32.7 kg-m**2.
Thickness of the flywheel rim t  =  58.6 mm.
Breadth of the flywheel rim b  =  117.2 mm.


## Example 16.17 Page No : 596¶

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
#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
#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.

 Moment of inertia of the flywheel I  =  2316 kg-m**2.
Thickness of the flywheel rim t  =  21.6 mm.
Width of the flywheel rim b  =  86.3 mm.


## Example 16.18 Page No : 599¶

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
#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)

 Mean diameter of the flywheel D  =  3 m.
Thickness of the flywheel rim t  =  170 mm.
Width of the flywheel rim b  =  680 mm.


## Example 16.19 Page No : 603¶

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)

 Power of the motor P  =  3 kW.
Minimum mass of the flywheel m  =  618 kg.


## Example 16.20 Page No : 603¶

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)

 Power of the motor required P  =  2.674 kW.
Mass of the flywheel required m  =  1198 kg.


## Example 16.21 Page No : 604¶

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
#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
#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)

 Speed of the flywheel immediately after riveting N2  =  257.6 rpm.
Number of rivets that can be closed per minute  =  18 rivets.


## Example 16.22 Page No : 605¶

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)

 Mass of the flywheel required m  =  327 kg.


## Example 16.23 Page No : 606¶

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
#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

 Power needed for the driving motor  =  1.674 kW.
Thickness of the flywheel rim t  =  43 mm.
Width of the flywheel rim b  =  86 mm.