Chapter 3,Laminar Flow and Lubrication

Example 3.2,Page 63

In [1]:
#variable decleration
Re =2000.0;
d =0.008; #m

#calculation
L1 =0.058* Re*d;

#result
print"The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)=",round(L1,3);

The furtherest distance the fluid can flow into the 8 mm inside diameter pipe (m)= 0.928

Example 3.4,Page 67

In [2]:
from math import pi

#variable decleration
del_p =90*10**3; # N/m^2
d =0.126; # m
R =0.126/2; # m
u =1.2;
L =60; # m
Rho =1260;


#calculation
Q= pi * del_p * R**4 / (8*u*L);
Re =4* Rho *Q/(u*pi*d);

#result
print "The glycerol delivery rate is (m^3/s) ",round(Q,4);
print"The Reynolds number is ",round(Re,3);
print"As Re is below 2000 , so, laminar flow"

The glycerol delivery rate is (m^3/s)  0.0077
The Reynolds number is  82.047
As Re is below 2000 , so, laminar flow

Example 3.5,Page 69

In [6]:
from math import pi

#variable decleration
u =0.015; #Ns/m^2
Q =0.004/60; #m^3/ s
dp =100;
rho =1100.0; #kg /m^3


#calculation
R =(8* u*Q/( pi *dp)) **(0.25) ;
Re =(4* rho *Q/( pi*u *(2* R)));

#result
print"Diameter of the pipe(m) =",round(R,3);
print"Reynolds number =",round(Re,2)

Diameter of the pipe(m) = 0.013
Reynolds number = 246.38

Example 3.6,Page 71

In [5]:
from math import pi

#variable decleration
mu =0.03; #Ns/m^2
Q =10.0**( -7) ; #m^3/ s
u=(7.0);

#calculation
dp= 8*mu*Q*u/3/pi/0.005**4;

#result
print" pressure difference (N/m^2)=",round(dp,2)

 pressure difference (N/m^2)= 28.52

Example 3.8,Page 75

In [9]:
from math import pi

#variable decleration
u =0.1; #Ns/m^2
d =0.1; #m
R =0.05; # m
Rho =900; #kg /m^3
v_max =2; # m/ s
v= v_max /2; # m/ s

#calculation
Tw = 2*u* v_max /R;
del_p =4* u* v_max /R **2;

#result
print "At the pipe wa l l ( r =R) , shear stress (N/m^2)= ",Tw;
print" pressure drop per metre lengthh of pipe is (N/m^2)",del_p;

At the pipe wa l l ( r =R) , shear stress (N/m^2)=  8.0
 pressure drop per metre lengthh of pipe is (N/m^2) 320.0

Example 3.9,Page 77

In [7]:
#variable decleration
import math
u =0.032; #Ns/m^2
Re =2000.0; #maximum v a l u e
Rho =854.0;
del_p =150.0; #N/m^2

#calculation
d =(32* u **2* Re /( Rho* del_p ))**(0.33)  ;

print "The maximum inside diameter is found to be (m) ",round(d,3);

The maximum inside diameter is found to be (m)  0.082

Example 3.10,Page 79

In [4]:
from math import pi
#variable decleration
rho =1000; #kg /m^3
u =0.1; #Ns/m^2
g =9.81; #m/ s ^2
L =10; #m
H =2; #m
Q =14.0/3600; #m^3/ s
d =0.05; #m

#calculation
dp=rho*g*(L+H) - (128* Q*u*L/pi /d**4) ;

#result
print" Pressure drop across the valve (N/m^2)=",round(dp,2)

 Pressure drop across the valve (N/m^2)= 92368.39

Example 3.12,Page 83

In [3]:
#variable decleration
from math import pi
import math
Q =3*10**( -6) ; #m^3/ s
u =0.001; #Ns/m^2
W =1;
rho =1000; #kg /m^3
g =9.81; #m/ s ^2
d =1.016*10**( -4) ; #m
d1 =1.25*10**( -4) ; # m


#calculation
theta = math.asin (3*Q*u/W/rho/g/d **3) ;
u1=W*rho *g* math.sin ( theta )*( d1 **3) /(3* Q);

#result
print"Exact angle of inclination (radians) =",round(theta,3)
print" Viscosity of the second liquid (Ns/m^2)=",round(u1,4)

Exact angle of inclination (radians) = 1.065
 Viscosity of the second liquid (Ns/m^2)= 0.0019

Example 3.17,Page 93

In [34]:
#variable decleration
u =1.5; #Ns/m^2
v =0.5; # m/ s
H =0.02/2; # m

#calculation
t=-u*3*v/H;

#result
print "The shear stress (N/m^2)=",t
print"acting on opp. direction of flow" 

The shear stress (N/m^2)= -225.0
acting on opp. direction of flow

Example 3.18,Page 95

In [18]:
from math import pi
from scipy import integrate

#variable decleration
N =600.0/60; #r e v o l u t i o n s pe r s e c
r =0.025; #m
t =400; # N/m^2
l =0.002; # m
time=2.0/1000; #s

#calculation
w =2* pi*N;
u=t*l/w/r;
def integrand (R,w,u,time):
    return 2*pi*u*w/time*R**3

#a=lambda r:2*pi*0.509*62.8/2*r**3*1000;
T=integrate.quad(integrand, 0,0.025,args=(w,u,time));

#result
print" Viscosity,(Ns/m^2)=",round(u,3)
print"Torque(Nm) =",round(T[0],4)

 Viscosity,(Ns/m^2)= 0.509
Torque(Nm) = 0.0098

Example 3.19,Page 97

In [19]:
from math import pi

#variable decleration
u =0.153; #Ns/m^2
r =0.05; # m
N =30; # rps
t =2.0/10**5; # s
L =0.2; #m


#calculation
tau =u *(2* pi *N*r/t);
F= tau *2* pi *r*L;
T=F*r;
w =2* pi*N;
P=T*w;

#result
print "The torque on the bearing is found to be (Nm)= ",round(T,3);
print " and the power required to overcome the resistance is (W)=",round(P,3);

The torque on the bearing is found to be (Nm)=  226.507
 and the power required to overcome the resistance is (W)= 42695.643

Example 3.20,Page 99

In [20]:
#variable decleration
t =0.0005;#s
P =22; 
r =300.0/60; 
R_1 =0.1; 
R_2 =0.0625; 
pi=3.14;

#calculation
w =2* pi*r;
u =2* t*P/( pi *w **2*(( R_1 )**4 -( R_2) **4) );

#result
print"The viscosity of the oil is (Ns/m^2)= ",round(u,4);

The viscosity of the oil is (Ns/m^2)=  0.0839
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