Chapter 5 : Newtonian Fluids

Example 5.2 page no. 42

In [1]:
from __future__ import division
print "Example 5.2 page no 42\n\n"
#To calculate the force to maintain movement of left plate 
#velocity of moving plate is equal to the velocity of the plate and velocity of the gas at the surface of the stationary plate is zero
k=1.66#kinamatic viscosity of gas
rho=0.08#density of gas
d=0.0833#distance between plate
v1=300#velocity of left plate
v2=0#velocity of stationary plate
g_c=4.17*10**(8)#gravitational constant
print "given \n kinamatic viscosity =%2f ft**2/hr\n rho=%2f lb/ft**3\n d=%4f ft\n v1=%0.2f ft/hr\n v2=%0.3e ft/hr\n gc=%0.3e (ft*lb/hr)/lbf*hr"%(k,rho,d,v1,v2,g_c)#
tou_xy=-k*rho*((v2-v1)/(g_c*d))#the frce necessary to mantain the movement of the left plate
print "\n force tou_xy=%0.3e  lbf/ft**2"%(tou_xy)#
Example 5.2 page no 42

 kinamatic viscosity =1.660000 ft**2/hr
 rho=0.080000 lb/ft**3
 d=0.083300 ft
 v1=300.00 ft/hr
 v2=0.000e+00 ft/hr
 gc=4.170e+08 (ft*lb/hr)/lbf*hr

 force tou_xy=1.147e-06  lbf/ft**2

Example 5.3 Page no. 45

In [2]:
from __future__ import division
from numpy import pi
print "Example  5.3 page no. 45\n\n"
D=0.25#diameter of fixed inner cylinder of viscometer
L=0.5#height of fixed inner cylinder of viscometer
T=15.3#measured torque
print "Given :\n diameter =%.2f ft\n height =%0.3f ft\n Torque=%0.3f ft.lbf"%(D,L,T)#
print "\n force =%0.3f lbf"%(F)#
#the shear stress(force parallel to the surface) using equation 5.11
print "\n shear stress tou=%0.3f psf"%( tou)#
Example  5.3 page no. 45

Given :
 diameter =0.25 ft
 height =0.500 ft
 Torque=15.300 ft.lbf

 force =122.400 lbf

 shear stress tou=311.689 psf

Example 5.4 page no. 45

In [3]:
from __future__ import division
print "Example 5.4 page no. 45\n\n"
#refer to example no 5.3
#determine dynamic viscosity and kinematic viscosity
omega=26.2#angular rotation speed 
D=0.25#diameter of fixed inner cylinder of viscometer
print "\n omega=%0.3f rad/s\n diameter D =%0.3f ft\n linear velocity =%2f ft/s"%(omega,D,v)#
d=0.001#clearance betwween two cylinder of visometer
vel_gradient =v/(d/12)#velocity gradient
gc=32.14#gravitational constant
print "\n clearance d=%5f ft\n vel. gradient=%0.3f 1/s\n gravitational constant gc=%3f ft/s*S"%(d,vel_gradient,gc)#
tou=311.7#shear stress tou
print "\n tou=%0.3f psf\n meu=%0.3f lb/ft*s"%(tou,meu)#
rho=60.528#density of oil
neu=meu/rho#kinamatic viscosity
print "\n kinematic viscosity=%5f (ft*ft)/s"%(neu)#
Example 5.4 page no. 45

 omega=26.200 rad/s
 diameter D =0.250 ft
 linear velocity =3.275000 ft/s

 clearance d=0.001000 ft
 vel. gradient=39300.000 1/s
 gravitational constant gc=32.140000 ft/s*S

 tou=311.700 psf
 meu=0.255 lb/ft*s

 kinematic viscosity=0.004211 (ft*ft)/s