# Chapter 8 : Behaviour of Ideal and Viscous Fluids¶

### Example 8.2.1 page no : 190¶

In [1]:
#initialisation of variables

v= 5.*10**-6 			#gmsec/m**2
g= 32.2 			#ft/sec**2
g1= 981. 			#gm/cm**2

#CALCULATIONS
v1= v*2.2*30.5**2/1000.
v2= v1*g
v3= v*g1*100

#RESULTS
print  'viscosity = %.2e lbf sec/ft**2 '%(v1)
print  'viscosity = %.2e lb/ft sec '%(v2)
print  'viscosity = %.3f centi-poise '%(v3)

viscosity = 1.02e-05 lbf sec/ft**2
viscosity = 3.29e-04 lb/ft sec
viscosity = 0.490 centi-poise


### Example 8.3.1 page no : 192¶

In [2]:
#initialisation of variables

v= 3.732*10**-7 			#slug/ft sec
y= 0.

#CALCULATIONS
vbyy= 40000.*(1.-50*y)
q= v*vbyy

#RESULTS
print  ' viscous shear stress= %.4f lbf/ft**2 '%(q)

 viscous shear stress= 0.0149 lbf/ft**2


### Example 8.3.2 page no : 193¶

In [5]:
#initialisation of variables
import math

T= 2.95 			#lbf ft
y= 0.025 			#in
d= 3. 			    #in
w= 450. 			#r.p.m
g= 32.2 			#ft/sec**2
R = 1.5

#CALCULATIONS
A = math.pi*d*(2./3.)
u = (T*y*144*60*32.2)/(A*R**2*w*2*math.pi)

#RESULTS
print  ' coefficient of viscocity of oil= %.3f lb/ft sec '%(u)

 coefficient of viscocity of oil= 0.513 lb/ft sec


### Example 8.3.3 page no : 194¶

In [4]:
#initialisation of variables
import math

v= 0.02 			#lb/ft sec
L= 5. 			#in
D= 2.5 			#in
M= 26. 			#lbf in
w= 1200. 			#rev/min
g= 32.2 			#ft/sec**2

#CALCULATIONS
C= math.pi*v*w*2*math.pi*D**3*L/(2*M*g*60*144.)

#RESULTS
print  ' coefficient= %.4f in '%(C)

 coefficient= 0.0026 in


### Example 8.3.4 page no : 195¶

In [5]:
#initialisation of variables
import math

g= 32.2 			#ft/sec**2
l= 2.54 			#cm
g= 32.2 			#ft/sec**2
v= 3.22 			#centi-poise
f= 0.01
p= 1.74 			#lbf/in**2
w= 100. 			#rev

#CALCULATIONS
V= v*l/(453.6*g*12)
R= f*p*60/(math.pi*2*math.pi*w*V)

#RESULTS
print ' relevant ratio of diameter to clearance= %.1f '%(R)

 relevant ratio of diameter to clearance= 11.3


### Example 8.4.1 page no : 196¶

In [6]:
#initialisation of variables
g= 981. 			#cm/sec**2
d= 0.1 			#mm
v= 35. 			#centi-stokes
d1= 10. 			#mm
d2= 1. 			#mm

#CALCULATIONS
u= g*d**2.*100/(18*v*d1**2)
ub= (d2/d)**2*u

#RESULTS
print  ' rate for diameter 0.1 mm= %.4f cm/sec'%(u)
print  '  rate for diameter 1 mm= %.2f cm/sec'%(ub)

 rate for diameter 0.1 mm= 0.0156 cm/sec
rate for diameter 1 mm= 1.56 cm/sec


### Example 8.5.1 page no : 201¶

In [2]:
#initialisation of variables
import math
a= 0.25 			#ft
v= 1.2 			#poises
u= 10. 			#ft/sec
g= 32.2 			#ft/sec**2
s= 0.9
d= 6. 			#in

#CALCULATIONS
q= -2.*u*v*30.5/(a*454*g)
Q= math.pi*u*(d/24)**2/2
R= 2.5*30.5**2/(v)

#RESULTS
print  ' quantity flow = %.2f ft**3/sec'%(q)
print  '  shear stress in the oil = %.2f lbf/ft**2'%(Q)
print  '  Reynolds number = %.f '%(R)
print 'Which is less than 2300 .Hence flow is laminar.'
# Note : Answer in book is wrong for R. Please check it manually.

 quantity flow = -0.20 ft**3/sec
shear stress in the oil = 0.98 lbf/ft**2
Reynolds number = 1938
Which is less than 2300 .Hence flow is laminar.


### Example 8.5.2 page no : 202¶

In [5]:
#initialisation of variables
import math

s= 0.9          #relative density
v= 5. 			#ft/sec
l= 10. 			#ft
di= 0.5 			#in
n= 100.
u= 0.002 			#lbfsec/ft**2
w= 62.3 			#lbf/ft**3
g= 32.2 			#ft/sec**2

#CALCULATIONS
dp= 8.*u*v*l/(di/2)**2
hf= dp*144./(s*w)
hk= v**2/(2.*g)
ht=hf+hk
P= s*w*n*v*math.pi*ht*di**2/(144.*4*550)

#RESULTS
print  ' horse-power required = %.1f h.p'%(P)

 horse-power required = 2.3 h.p


### Example 8.6.2 pageno : 206¶

In [6]:
#initialisation of variables
import math

W= 50. 			#tonf
u= 0.1 			#lb/ft sec
d= 8. 			#in
g= 32.2 			#ft/sec**2
r= 0.01

#CALCULATIONS
P = (4*W**2)/(3*math.pi*u) * ((1./800)**3)*(12./8)*(g/550)*2240**2
Q= 4*W*2240*g*12*(r/d)**3/(3*math.pi*u*(d/12))

#RESULTS
print ' power required = %.2f'%P
print  ' rate = %.3f in/sec'%(Q)

 power required = 9.13
rate = 0.538 in/sec