Ch:17 Sliding contact bearings

exa 17-1 - Page 482

In [1]:
from __future__ import division
from math import pi, sqrt
Ta=22#
u=7/10**9#
nj=20#
r=25#
l=2*r#
Ao=30000#
Uo=15.3/10**3#
c=0.025#
#specific weight of the material is rho
rho=8.46*(10**-6)#
Cp=179.8#
Tf=Ta+(16*pi**3*u*nj**2*l*r**3/(Uo*Ao*c))#
# avg mean film temperature is Tav
Tav=(Tf-Ta)/2#
x= l*c*rho*pi*r*nj*Cp*10**3#
y=Ao*Tav*Uo#
delT=y/x#
print " Tav is %0.2f degC "%(Tav)#
print "\n delT is %0.1f degC "%(delT)#

 Tav is 47.29 degC 

 delT is 7.3 degC

exa 17-2 - Page 482

In [2]:
l=60#
d=60#
r=d/2#
ho=0.008#
c=0.04#
S=0.0446#
nj=1260/60#
W=6000#
p=W/(l*d)#
u=S*(c/r)**2*p/nj#
u=u*10**9#
print "u is %0.3f cP "%(u)#

u is 6.293 cP

exa 17-3 - Page 483

In [3]:
d=60#
r=30#
l=60#
c=0.8*10**-3*r#
ho=0.2*c#
W=21000/2#
p=W/(l*d)#
S=0.0446#
nj=1440/60#
u=S*(c/r)**2*p/nj#
u=u*10**9#
# since Q/(r*nj*l)=4.62
Q=4.62*r*c*nj*l#
Q=Q*60/10**6#
print " u is %0.3f cP "%(u)#
print "\n Q is %0.4f lpm "%(Q)#

 u is 3.469 cP 

 Q is 0.2874 lpm

exa 17-4 - Page 483

In [4]:
from numpy import mat,pi,sqrt
l=60#
d=60#
r=d/2#
W=3000#
p=W/(l*d)#
u=30*10**-9#
c=0.06#
nj=1440/60#
S=(r/c)**2*u*nj/p#
#For ratio l/d=1, values of different parameters are given in matrix A corresponding to S
A=mat([[0,0,0,0,0,0],[0,1, 0.264, 0.6, 5.79, 3.99],[0,1, 0.121, 0.4 ,3.22, 4.33]])#
#let ho/c=x
x=(A[1,3])-((A[1,3]-(A[2,3]))*((A[1,2])-S)/((A[1,2])-(A[2,2])))#
#let y= (r/c)*f=CFV
y=(A[1,4])-(A[1,4]-(A[2,4]))*((A[1,2])-S)/((A[1,2])-(A[2,2]))#
#let z=Q/(r*c*nj*l)=FV
z=(A[1,5])-((A[1,5]-(A[2,5]))*((A[1,2])-S)/((A[1,2])-(A[2,2])))#
f=y*c/r#
ho=x*c#
Q=z*r*c*nj*l#
Q=Q*60/10**6#
delT=8.3*p*y/z#
#let power lost in friction be Pf
Pf=2*pi*nj*f*W*r/10**6#
print " f is %0.5f  "%(f)#
print "\n ho is %0.3f mm "%(ho)#
print "\n Q is %0.3f lpm "%(Q)#
print "\n delT is %0.1f degC "%(delT)#
print "\n Pf is %0.4f KW "%(Pf)#

 f is 0.00985  

 ho is 0.032 mm 

 Q is 0.638 lpm 

 delT is 8.3 degC 

 Pf is 0.1337 KW

exa 17-5 - Page 484

In [5]:
from math import sqrt, pi
W=22000#
nj=960/60#
p=2.4#
u=20*10**-9#
d=sqrt(W/p)#
d=96#
r=d/2#
l=d#
S=0.0446#
pact=W/(l*d)#
#x=r/c#
x=sqrt(S*pact/(u*nj))#
c=r/x#
ho=0.2*c#
Q=r*c*nj*l*4.62#
Q=Q*60/10**6#
print " d is %0.0f mm "%(d)#
print "\n l is %0.0f mm "%(l)#
print "\n ho is %0.4f mm "%(ho)#
print "\n Q is %0.3f lpm "%(Q)#
#The difference in answer to Q is due to rounding -off the value of c.

 d is 96 mm 

 l is 96 mm 

 ho is 0.0166 mm 

 Q is 1.701 lpm

exa 17-6 - Page 485

In [6]:
from math import log,pi
W=400*10**3#
Ro=200#
Ri=160#
ho=0.1#
t=150#
# specific gravity is rho
rho=0.86#
pi=2*W*log(Ro/Ri)/(pi*(Ro**2-Ri**2))#
zk=(0.22*t)-(180/t)#
z=rho*zk#
u=z/(10**9)#
Q=pi*pi*ho**3/(6*u*log(Ro/Ri))#
Q=Q*60/10**6#
print " pi is %0.3f MPa "%(pi)#
print "\n Q is %0.2f lpm "%(Q)#
  
#The difference in answer to Q is due to rounding -off of values.

 pi is 3.946 MPa 

 Q is 25.52 lpm

exa 17-7 - Page 486

In [7]:
from math import log, pi
#let number of pads be n
n=4#
W=100*10**3#
Ro=125#
Ri=50#
t=200#
ho=0.15#
pi=2*W*log(Ro/Ri)/(pi*(Ro**2-Ri**2))#
zk=(0.22*t)-(180/t)#
# specific gravity is rho
rho=0.86#
z=rho*zk#
u=z/(10**9)#
Q=pi*pi*ho**3/(6*u*log(Ro/Ri))#
Q=Q*60/10**6#
print "pi is %0.2f MPa "%(pi)#
print "\nQ is %0.3f lpm "%(Q)#

pi is 4.44 MPa 

Q is 19.629 lpm