Chapter 12: Dynamics of Machines. Turning Moment. The Flywheel¶

Example 2, Page 414¶

In [2]:
#Variable declaration
ne=31
na=25
nb=90
nc=83
Ta=10 #lbft

#Calculations
#Ne-Nf/(Nc-Nf)=-83/31
k=114./83#k=Nc/Nf As Ne = 0, on simplification we get Nc/Nf= 114/83
j=-90./25#j=Na/Nb
#Nc=Nb, Thus Na/Nc=-90/25
#Na/Nf=(Na/Nc)*(Nc/Nf) ie Na/Nf=k*j
#Tf*Nf=Ta*Na
Tf=Ta*k*j

#Result
print "Torque exerted on driven shaft = %.1f lb.ft"%Tf

Torque exerted on driven shaft = -49.4 lb.ft


Example 3, Page 415¶

In [4]:
import math

#Variable declaration
D=9.#in
stroke=24.#in
d=2#in
l=60.#in
CP=l
N=120
theta=40#degrees
x=theta*math.pi/180
P1=160#lb/in^2
P2=32#lb/in^2

#Calculations
OC=stroke/2
F=math.pi*(D/2)**2*P1-math.pi*(D/2)**2*P2+math.pi*(d/2)**2*P2
#Ft*Vc=F*Vp; Where Vc and Vp are velocities of crank and pin respectively
#Vp/Vc=IP/IC=OM/OC - From similar triangles  ; fig 274
n=CP/OC
OM=OC*(math.sin(x) + (math.sin(2*x)/(2*n)))#from 3.11
T=F*OM/12#torque exerted on crankshaft
Torque=math.floor(T)

#Result
print "The torque exerted on crankshaft= F*OM = %.f lb ft"%Torque

The torque exerted on crankshaft= F*OM = 6110 lb ft


Example 4, Page 420¶

In [5]:
#Variable declaration
AB=12.5#in
IB=10.15#in
IA=10.75#in
IX=2.92#in
IY=5.5#in
w=3#lb
Fi=5#lb
Fa1=9#lb

#Calculations
Fb1=(Fa1*IA-w*IY-Fi*IX)/IB
#From the polygon of forces
Fa2=7.66#lb
Fb2=3.0#lb
Fa=(Fa1**2+Fa2**2)**(1./2)
Fb=(Fb1**2+Fb2**2)**(1./2)

#Results
print "The total force applied to the link AB at the pin A = Fa = %.2f lb\nThe total force applied to the link AB" \
"at the pin B = Fb = %.2f lb"%(Fa,Fb)

The total force applied to the link AB at the pin A = Fa = 11.82 lb
The total force applied to the link ABat the pin B = Fb = 7.13 lb


Example 5, Page 424¶

In [9]:
import math

#Variable declaration
CP=60.#in
l=CP/12
a=41.
cg=19.
g=32.2#ft/s^2
m1=580.#lb
Mr=500.#lb
n=5.#from example 12.3
x=40*math.pi/180
N=120.
r=1.#ft
k=25.

#Calculations
w=N*math.pi/30
Rm=m1+(cg/CP)*Mr
fp=w**2*r*(math.cos(x)+math.cos(2*x)/n)
Fp=-Rm*fp/g
OM=0.7413#ft -from example 12.3
Tp=Fp*OM#from 12.6
L=a+k**2/a#length for simple equivalent pendulum
L1=L/12
Tc=-Mr*(a/12)*(l-L1)*w**2*math.sin(2*x*math.pi/180)/(g*2*n**2)#from 12.10
Tw=-Mr*a*math.cos(x*math.pi/180)/(n*12)
T=Tp+Tc+Tw

#Results
print "Tp= %.f lbft\nTc = %.1f lbft\nTw = %.1f lbft\nTotal torque exerted on the crankshaft due to the inertia of "\
"the moving parts = Tp+Tc+tw = %.1f lbft"%(Tp,Tc,Tw,T)

Tp= -2149 lbft
Tc = -1.3 lbft
Tw = -341.6 lbft
Total torque exerted on the crankshaft due to the inertia of the moving parts = Tp+Tc+tw = -2492.3 lbft


Example 6, Page 428¶

In [11]:
import math

#Variable declaration
AB=2.5#in
BC=7.#in
CD=4.5#in
ED=2.3#from figure
N=180
w=N*math.pi/30
m=3.#lb
g=32.2#ft/s**2
QT=1.35#inches from figure

#Calculations
alpha=w**2*(QT/CD)
Torque=m*(k/12)**2*alpha/g
Torque1=Torque*12
Fc1=Tc/CD
M=5#lb
gA=1.8#in
fg=w**2*(gA/12)
F=M*fg/g
OaG=5.6#in
Kg=2.9#in
GZ=Kg**2/OaG
#scaled from figure
IB=9#in
IC=5.8#in
IX=2.49#in
IY=1.93#in
Fb1=(Fc1*IC+F*IX+M*IY)/IB
Tor=Fb1*AB
#from force polygon
Fc2=1#lb
Fb2=15.2#lb
Fb=(Fb1**2+Fb2**2)**(1./2)
Fc=(Fc1**2+Fc2**2)**(1./2)

#Results
print "The torque which must be exerted on AB in order to overcome the inertia of the links = Fb1*AB = %.1f lb.in"\
"\nThe total force applied to the link BC \nAt pin C = %.2f lb\nAt pin B = %.1f lb"%(Tor,Fc,Fb)

The torque which must be exerted on AB in order to overcome the inertia of the links = Fb1*AB = 14.5 lb.in
The total force applied to the link BC
At pin C = 3.92 lb
At pin B = 16.3 lb


Example 7, Page 441¶

In [12]:
import math

#Variable declaration
N=210#rpm
w=N*math.pi/30
F=50

#Calculations
p1=F*120/(N*2)#N*p=F*120
p2=math.floor(p1)#no of poles must be a whole number ; P2=P/2
p=2*p2
N1=F*120/p
n=3#no of impulse per second
Ks=n/(6*p)#equation 12.13

#Results
print "Ks = %.4f\n\nActual speed = %.1f rpm\nNumber of poles = %.f"%(Ks,N1,p)

Ks = 0.0179

Actual speed = 214.3 rpm
Number of poles = 28


Example 8, Page 443¶

In [13]:
import math

#Variable declaration
N=120#rpm
k=3.5#ft
Ef=2500#ft lb
Ks=.01
g=32.2#ft/s^2

#Calculations
w=math.pi*N/30#angular velocity
W=g*Ef/(w**2*k**2*Ks*2240)#Weight of flying wheel

#Result
print "Weight of flying wheel, W = %.2f tons"%W

Weight of flying wheel, W = 1.86 tons


Example 9, Page 443¶

In [16]:
import math

#Variable declaration
N=270#rpm
ihp=35.8
k=2.25#ft
g=32.2#ft/s^2
ke=1.93#from table on p 440

#Calculations
E=ihp*33000/N
Ef=ke*E
w=math.pi*N/30
W=1000#lb
MOI=2*W*k**2#moment of inertia of both wheel
ks=Ef*g/(MOI*w**2)#formula for ks
p=ks/2

#Results
print "The fluctuation speed is therefore %.1f%% or %.1f%% on either side of the mean speed"%(ks*100,p*100)

The fluctuation speed is therefore 3.4% or 1.7% on either side of the mean speed


Example 10, Page 444¶

In [17]:
import math

#Variable declaration
ihp=25.
N=300.#rpm
Ks=2./100#given
u=2.3#work done by gases during expansion is u(2.3) times that during compression

#Calculations
E=ihp*33000/N#indicated work done per revolution
E1=E*2#indicated work done per cycle
We=E1/(1-1./u)#work done by gases during expansion
AB=We*2./math.pi#the maximum torque from fig 290
AC=E/(2*math.pi)#mean turning moment
CB=AB-AC#maximum excess turning moment
Ef=(CB/AB)**2*We#fluctuation of energy
Ke=Ef/E
w=math.pi*N/30#angular speed
g=32.2#ft/s^2
moi=g*Ef/(w**2*Ks)#moment of inertia

#Result
print "Moment of inertia of the flywheel = %.f lb ft^2"%moi

#answer is not EXACT due to the approximations in calculations done by the author of the book

Moment of inertia of the flywheel = 13710 lb ft^2


Example 11, Page 445¶

In [18]:
import math

#Variable declaration
N=100#rpm
ke=1.93#As per given figure
l=15#1 inch of fig = 15 ton ft
x=40#degrees; 1 inch = 40 degree
I=150#ton ft^2
g = 32.2

#Calculations
w=math.pi*N/30#angular speed
E=l*x*math.pi/180#energy
Ef=E*ke#fluctuation energy
Ks=Ef*g/(w**2*I)#from equation 12.14
p=Ks*100/2#dummy variables
q=p*2#dummy variables

#Results
print "The total fluctuation of speed is %.2f percent and the variation in speed is %.2f percent on either side of "\
"\n the mean speed"%(q,p)

The total fluctuation of speed is 3.96 percent and the variation in speed is 1.98 percent on either side of
the mean speed