Chapter 3:Fluid Flow Kinematics

Example 3.1 Page No74

In [1]:
import math

#variable initialisation

D=0.1                     #diameter in cm

L=2                       #length in m

V1=2.0                    #inlet discharge in m/s

V2=1.2                    #outlet dischargein m/s


#calculation

A=(math.pi/4)*(D*D)      #area calculation

Q1= round(V1*A,5)        #inlet discharge

Q2=round(V2*A,7)         #outlet discharge

Q_e=Q1-Q2                #discharge emitted through walls of the porous pipe

print "Answer given in the book for discharge emitted is wrong.It should come as"

print "Q_e=",Q_e,"m^3/s"

A_e=math.pi*D*L         #surface area of emission 

V_e=Q_e/A_e             #Velocity of emission

print "V_e=",round(V_e,2),"m/s"

Answer given in the book for discharge emitted is wrong.It should come as
Q_e= 0.0062852 m^3/s
V_e= 0.01 m/s

Example 3.3 Page No74

In [3]:
from __future__ import division

import math

#variable initialisation

Q=0.08                         #Discharge in m^3/s

D_a=12                         #Diameter of inlet in cm

V_r=25                         #Width in mm

D_i=30                         #Diameter of impeller in cm of impeller


#Calculation

A=round(math.pi*(D_i/100)*(25/1000)*(0.95),5)#Area calculation

R_v=Q/A                        #radial velocity at the edge of impeller

print "R_v=",round(R_v,3),"m/s"

#V_A*A_A=0.08 #V_a is the axial Velocity,A_z is the area of inlet

D=D_a/100                      #converting diameter of inlet into m

V_a=0.08/((math.pi/4)*(D*D))   #calculating axial velocity

print "V_a=",round(V_a,3),"m/s"

R_v= 3.575 m/s
V_a= 7.074 m/s

Example 3.10 Page No78

In [4]:
import math

#variable initialisation

#points of A

x=2

y=-3

z=1

t=2

#V=(10*t+x*y)i+(-y*z-10*t)j+(-y*z|+z*z/2)k given velocity equation  1

#V=4xi+(-4*y+3*t)j given velocit equation 2

#Calculation

#V is in the form ui+vj+zk

U_a=(10*t)+(x*y)

V_a=-(y*z)-(10*t)

W_a=-(y*z)+((z*z)/2)

V_A=math.sqrt((U_a**2)+(V_a**2)+(W_a**2))    #calculating Magnitude

print "(i) V_A=",round(V_A,1),"units"

#Second Equation,

U_a=4*x

V_a=(-4*y)+(3*t)

W_a=0

V_A=math.sqrt((U_a**2)+(V_a**2)+(W_a**2))   #Calculating magnitude

print "(ii) V_A=",round(V_A,1),"units"

(i) V_A= 22.3 units
(ii) V_A= 19.7 units

Example 3.11 Page No78

In [5]:
#variable initialisation

t=3                       #time in s from commencement of flow

x=0.5                     #distance from inlet into the nozzle in m

L=0.8 #in m

#Calculation

c=1-(x/(2*L))             #calculating local acceleration

c=c**2

lo=c*2

print "(i)Local acceleration doe_v/doe_t=",round(lo,3),"m/s^2"

c=(1-(x/(2*L)))**3        #calculating convective acceleration Vdoe_V/doe_x

b=-((4*t*t)/L)*c

print "(ii)Convective acceleration=",round(b,3),"m/s^2"

T=lo+b                    #calculating total acceleration

print "(iii)Total acceleration=",round(T,2),"m/s^2"

(i)Local acceleration doe_v/doe_t= 0.945 m/s^2
(ii)Convective acceleration= -14.623 m/s^2
(iii)Total acceleration= -13.68 m/s^2

Example 3.14 Page No80

In [6]:
from __future__ import division

import math

from sympy import symbols,diff,simplify,cancel

#Variable Initialisation

x, y,z=symbols("x y z")

u=4*x*y+y**2                                  #Given equations

v=6*x*y+3*x

#Calculation

#For steady,incompressible flow,the equation (du/dx)+(dv/dy)=0 must be satisfied.

#Let us take,a=(du/dx,b=du/dy)

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

c=a+b

print "(du/dx)+(du/dy)=",c,"not equal to 0.Hence flow is not possible"

u=2*x**2+y**2                                #Given equations 

v=-4*x*y

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

c=a+b

print "(du/dx)+(du/dy)=",c,".Hence flow is possible"

u=-(x/(x**2+y**2))                           #Given equations

v=-(y/(x**2+y**2))

a=diff(u,x,1)                                #differentiating to get (du/dx)

s=simplify(a)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

q=simplify(b)                                #simplifying
   
c=cancel(s+q)

print "(du/dx)+(du/dy)=",c,"Hence flow is possible"

(du/dx)+(du/dy)= 6*x + 4*y not equal to 0.Hence flow is not possible
(du/dx)+(du/dy)= 0 .Hence flow is possible
(du/dx)+(du/dy)= 0 Hence flow is possible

Example 3.15 Page No80

In [8]:
import math

from sympy import symbols,diff,ln,simplify,cancel,sin

#Variable Initialisation

x, y,z,c,A=symbols("x y z c A")

u=c*x                                        #Given equations

v=-c*y

#Calculation

#For steady,incompressible flow,the equation (du/dx)+(dv/dy)=0 must be satisfied.

#Let us take,a=(du/dx,b=du/dy)

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

c=a+b

print "(du/dx)+(du/dy)=",c,"Hence,the continuity equation is satisfied"

u=-c*c/y                                     #Given equations 

v=c*ln(x*y)

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

c=a+b

print "(du/dx)+(du/dy)=",c,".Hence,the continuity equation is satisfied"

u=A*sin(x*y)                                 #Given equations

v=-A*sin(x*y)

a=diff(u,x,1)                                #differentiating to get (du/dx)

s=simplify(a)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

q=simplify(b)                                #simplifying
   
c=cancel(s+q)

print "(du/dx)+(du/dy)=",c,",not equal to 0.Hence,the continuity equation is not satisfied"

u=x+y                                        #Given equations 

v=x-y

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

c=a+b

print "(du/dx)+(du/dy)=",c,".Hence,the continuity equation is satisfied"

u=2*x**2+z*y                                 #Given equations 

v=-2*x*y+3*y**3+3*z*y

w=-(3/2)*z**2-2*x*y-6*y*z

a=diff(u,x,1)                                #differentiating to get (du/dx)

b=diff(v,y,1)                                #differentiating to get (dv/dy)

s=diff(w,z,1)

c=a+b+s

print "(du/dx)+(du/dy)=",c,",not equal to 0.Hence,the continuity equation is not satisfied"

(du/dx)+(du/dy)= 0 Hence,the continuity equation is satisfied
(du/dx)+(du/dy)= 0 .Hence,the continuity equation is satisfied
(du/dx)+(du/dy)= -A*x*cos(x*y) + A*y*cos(x*y) ,not equal to 0.Hence,the continuity equation is not satisfied
(du/dx)+(du/dy)= 0 .Hence,the continuity equation is satisfied
(du/dx)+(du/dy)= 2*x + 9*y**2 - 6*y ,not equal to 0.Hence,the continuity equation is not satisfied

Example 3.16 Page No80

In [10]:
from sympy import integrate,diff,symbols

#Variable Initialisation

x, y, A,e,L=symbols("x y A e L")

u=A*(x**2+y**2)

du=-diff(u,x,1)                        #differentiating u with respect to x

a=integrate((du), y)                   #integrating with respect to y

print "(a)v=",a,"+f(x)"

u=A*(e**x)

du=-diff(u,x,1)                        #differentiating u with respect to x

a=integrate((du),x)                    #integrating with respect to x

print "(b)v=",a,"y+f(x)"

u=-A*ln(x/L)

du=-diff(u,x,1)                        #differentiating u with respect to x

a=integrate((du), y)                   #integrating with respect to y

print "(c)v=",a,"+f(x)"

v=-A*x*y

dv=-diff(v,y,1)                        #differentiating u with respect to y

a=-integrate((dv), x)                  #integrating with respect to x

print "(d)u=",a,"+f(y)"

(a)v= -2*A*x*y +f(x)
(b)v= -A*e**x y+f(x)
(c)v= A*y/x +f(x)
(d)u= -A*x**2/2 +f(y)

Example 3.17 Page No81

In [13]:
from __future__ import division

import math

from sympy import diff,symbols

#Variable Initialisation

x, y,z=symbols("x y z")

#Calculation

#let us take a=(dv/dx),b=(du/dy),c=(dw/dy),d=(dv/dz),e=du/dz,f=(dw/dx)

u=x*y**3*z

v=-y**2*z**2

w=y*z**2-(y**3*z**2)/2

a=diff(v,x,1)                                         #finding (dv/dx)

b=diff(u,y,1)                                         #finding (du/dy)

c=diff(w,y,1)                                         #finding (dw/dy)

d=diff(v,z,1)                                         #finding (dv/dz)

e=diff(u,z,1)                                         #finding (du/dz)

f=diff(w,x,1)                                         #finding (dw/dx)
    
omega_z=(1/2)*((a-b))                                 #calculating omega_z

print "(i)omega_z=",omega_z

omega_x=(1/2)*((c-d))                                 #calculating omega_x

print "omega_x=",omega_x

omega_y=(1/2)*(e-f)                                   #calculating omega_y

print "omega_y=",omega_y

#second values given 

u=3*x*y

v=(3/2)*x**2-(3/2)*y**2

a=diff(v,x,1)                                          #finding (dv/dx)

b=diff(u,y,1)                                          #finding (du/dy)

omega_z=(1/2)*((a-b))                                  #calculating omega_z

print "(ii)omega_z=",omega_z,"As the flow is two-dimensional in the x-y plane omega_z=omega_y=0"

#third values given 

u=y**2

v=-3*x

a=diff(v,x,1)                                          #finding (dv/dx)

b=diff(u,y,1)                                          #finding (du/dy)

omega_z=(1/2)*((a-b))                                  #calculating omega_z

print "(iii)omega_z=",omega_z,"As the flow is two-dimensional in x-y plane omega_z=omega_y=0"

(i)omega_z= -1.5*x*y**2*z
omega_x= -0.75*y**2*z**2 + 1.0*y**2*z + 0.5*z**2
omega_y= 0.5*x*y**3
(ii)omega_z= 0 As the flow is two-dimensional in the x-y plane omega_z=omega_y=0
(iii)omega_z= -1.0*y - 1.5 As the flow is two-dimensional in x-y plane omega_z=omega_y=0

Example 3.19 Page No81

In [14]:
import math

from sympy import symbols,diff,integrate

#Variable Initialisation

x,y,z,a=symbols("x y z a")

#Calculation

u=2*x*y

v=a**2+x**2-y**2

#let us take c=(du/dx),b=dv/dy

c=diff(u,x,1)                                         #finding (dv/dx)

b=diff(v,y,1)                                         #finding (du/dy)

#The continuity equation for steady,incompressible flow is satisfied.Hence the flow is possible.The stream, function psi is related to u and v as,

psi=integrate((u),y)                                  #integrating with respect to y

print "psi",psi,"+f(x)"

d_psi=diff(psi,x,1)                                   #finding (dpsi/dx)

#take f'(x) as ff

ff=-(a**2)+(x**2)                                     # assigning f'(x)

f_x=integrate(ff, x)                                  #integrating to get f(x)

print "psi=",psi,f_x,"+constant"                      #replacing f(x)

psi x*y**2 +f(x)
psi= x*y**2 -a**2*x + x**3/3 +constant

Example 3.24 Page No83

In [15]:
from sympy import symbols,diff,integrate

#Variable Initialisation

x,y,z,U_in,a,r=symbols("x y z U_in a r")

#Calculation

n=3*x*y                                   #given value

u=diff(n,x,1)                             #d_psi/dy

psi=integrate(u, y)  #integrating to find psi    

v=diff(n,y,1)                             #d_psi/dx

ff=-v                                     #assinging to f'(x)

f_x=integrate(ff, x)                      #integrating to get f(x)

psi=psi-(-f_x)                            #substituting in psi

print "(i)psi=",psi,"+c,where c is a constant"

#second equation

n=4*(x**2-y**2)                           #given value

u=diff(n,x,1)                             #d_psi/dy

psi=integrate(u, y)                       #integrating to find psi    

v=diff(n,y,1)                             #d_psi/dx

ff=-v                                     #assinging to f'(x)

f_x=diff(ff,x)                            #differentiating to get f(x)

psi=psi-(-f_x)                            #substituting in psi

print "(i)psi=",psi,"+c"


#Third value

n=x+y+3                                   #given value

u=diff(n,x,1)                             #d_psi/dy

psi=integrate(u, y)  #integrating to find psi    

v=diff(n,y,1)                             #d_psi/dx

ff=-v                                     #assinging to f'(x)

f_x=integrate(ff, x)                      #integrating to get f(x)

psi=psi-(-f_x)                            #substituting in psi

print "(i)psi=",psi,"+c,where c is a constant"

(i)psi= -3*x**2/2 + 3*y**2/2 +c,where c is a constant
(i)psi= 8*x*y +c
(i)psi= -x + y +c,where c is a constant

Example 3.25 Page No84

In [16]:
from sympy import symbols,diff

#Variable Initialisation

x, y,z,c,A,m=symbols("x y z c A m")

#Calculation

#A valid potential function satisfies the Laplace equation.(d**2u/dx**2)+(d**2v/dy**2)=0.

phi=A*x*y                                            #Given equations

a=diff(phi,x,2)                                      #differentiating to get (d**2u/dx**2)

b=diff(phi,y,2)                                      #Differentiating to get (d**2v/dy**2)

c=a+b

print "(d**2u/dx**2)+(d**2u/dy**2)=",c,"Hence,phi=",phi,"is a valid potential function"

phi=m*ln(x)                                          #Given equations

a=diff(phi,x,2)                                      #differentiating to get (d**2u/dx**2)

b=diff(phi,y,2)                                      #Differentiating to get (d**2v/dy**2)

c=a+b

print "(d**2u/dx**2)+(d**2u/dy**2)=",c,"Hence,phi=",phi,"is not a valid potential function"

phi=A*(x**2-y**2)                                    #Given equations

a=diff(phi,x,2)                                      #differentiating to get (d**2u/dx**2)

b=diff(phi,y,2)                                      #Differentiating to get (d**2v/dy**2)

c=a+b

print "(d**2u/dx**2)+(d**2u/dy**2)=",c,"Hence,phi=",phi,"is a valid potential function"

phi=A*cos(x)                                         #Given equations

a=diff(phi,x,2)                                      #differentiating to get (d**2u/dx**2)

b=diff(phi,y,2)                                      #Differentiating to get (d**2v/dy**2)

c=a+b

print "(d**2u/dx**2)+(d**2u/dy**2)=",c,"Hence,phi=",phi,"is not a valid potential function"

(d**2u/dx**2)+(d**2u/dy**2)= 0 Hence,phi= A*x*y is a valid potential function
(d**2u/dx**2)+(d**2u/dy**2)= -m/x**2 Hence,phi= m*log(x) is not a valid potential function
(d**2u/dx**2)+(d**2u/dy**2)= 0 Hence,phi= A*(x**2 - y**2) is a valid potential function
(d**2u/dx**2)+(d**2u/dy**2)= -A*cos(x) Hence,phi= A*cos(x) is not a valid potential function

Example 3.26 Page No85

In [17]:
from sympy import symbols,diff,integrate

#Variable Initialisation

x,y,z,A=symbols("x y z A")

#Calculation

psi=A*((x**2)-(y**2))                                           #given

u=diff(psi,y,1)                                                 #differentiating to find dn/dx

n=integrate(u, x)                                               #integrating to find n

v=-diff(psi,x,1)                                                #differentiating to find dn/dy

#v=diff(n,y,1)=-2*A*x+f'(y),By comparing those two

#f'(y)=0

print "n=",n,"+constant"

#second equation is taken

psi=y**3-(3*x**2*y)                                             #given

u=diff(psi,y,1)                                                 #differentiating to find dn/dx

n=integrate(u, x)                                               #integrating to find n

v=-diff(psi,x,1)                                                #differentiating to find dn/dy

#v=diff(n,y,1)=-2*A*x+f'(y),By comparing those two

#f'(y)=0

print "n=",n,"+constant"

n= -2*A*x*y +constant
n= -x**3 + 3*x*y**2 +constant