ME 6135: Advanced Aerodynamics
Dr. A.B.M. Toufique Hasan
Professor
Department of Mechanical Engineering
Bangladesh University of Engineering & Technology (BUET), Dhaka
Lecture-05
29/10/2024
toufiquehasan.buet.ac.bd
toufiquehasan@me.buet.ac.bd
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
1
Continuity Equation for steady two-dimensional flows (in differential form):
∇ ⋅ (𝜌𝑉) = 0
𝜕(𝜌𝑢) 𝜕(𝜌𝑣)
+
=0
𝜕𝑥
𝜕𝑦
for compressible flow
∇⋅𝑉 =0
u v
+
=0
x y
for incompressible flow
Condition of irrotationality in case of two-dimensional flows: (curl of velocity =0)


curl V = 0 (  V = 0)
© Dr. A.B.M. Toufique Hasan (BUET)
v u
−
=0
x y
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
2
Stream function, ψ
Stream function, ψ (x, y, t) is a single function by which the two entities of velocity components u (x, y, t)
and v (x, y, t) of a two-dimensional incompressible flow can be defined. Consider continuity equation-
u v
+ =0
x y
Define stream function by the following definition (for incompressible flow) -

u
and
y

v −
x
1 

vr 
and v  −
(r ,  coordinate )
r 
r
This definition automatically satisfy the continuity equation as-
u v          2
 2
 +  −
+ = 
−
=0
=
x y x  y  y  x  xy
yx
Thus, stream function is a single function which satisfy the first governing equation in fluid dynamics i.e.
the continuity equation.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
3
Stream function, ψ
Continuity equation for two-dimensional compressible flow -
𝜕(𝜌𝑢) 𝜕(𝜌𝑣)
+
=0
𝜕𝑥
𝜕𝑦
Define stream function by the following definition (for compressible flow) -
1 𝜕𝜓
𝑢≡
𝜌 𝜕𝑦
and
1 𝜕𝜓
𝑣≡−
𝜌 𝜕𝑥
1 1 𝜕𝜓
1 𝜕𝜓
𝑣𝑟 ≡
and 𝑣𝜃 ≡ −
(𝑟, 𝜃coordinate)
𝜌 𝑟 𝜕𝜃
𝜌 𝜕𝑟
This definition automatically satisfy the continuity equation as𝜕(𝜌𝑢) 𝜕(𝜌𝑣)
𝜕
1 𝜕𝜓
𝜕
1 𝜕𝜓
𝜕2𝜓
𝜕2𝜓
+
=
𝜌×
+
−𝜌 ×
=
−
=0
𝜕𝑥
𝜕𝑦
𝜕𝑥
𝜌 𝜕𝑦
𝜕𝑦
𝜌 𝜕𝑥
𝜕𝑥𝜕𝑦 𝜕𝑦𝜕𝑥
Thus, stream function is a single function which satisfy the first governing equation in fluid dynamics i.e.
the continuity equation.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
4
Velocity Potential, φ
Velocity Potential, Φ (x, y, t) is another function by which the two entities of velocity components u (x, y, t)
and v (x, y, t) of a two-dimensional irrotational incompressible flow can be defined.
Consider the condition of irrotationality
v u
−
=0
x y
Define velocity potential by the following definition-

u
and
x

1 
vr 
and v 
(r ,  coordinate )
r
r 

v
y
This definition automatically satisfy the condition of irrotationality asv u          2
 2
−
=   −   =
−
=0
x y x  y  y  x  xy
yx
Thus, velocity potential is a function which satisfy the condition of irrotationality .
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
5
Relation between φ and ψ
It can be seen that-
v
 dy 
=
 
u
 dx  =constant
u
 dy 
=
−
 
v
 dx  =constant
streamline
equipotential line
Streamlines and equipotential lines are mutually perpendicular.
Velocity potential
streamline
Flow net
© Dr. A.B.M. Toufique Hasan (BUET)
Flow net
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
6
Laplace Equation
Consider 2D irrotational, incompressible flow: the velocity components can be defined in terms of both
the stream function and velocity potential-
u=


; v=−
y
x
u=


; v=
x
y
Now use the expression of stream function in the condition of irrotationality:
v u
−
=0
x y

       
 = 0
−
 − 
x  x  y  y 
−


 2
x
2
 2
x
2
−
+
 2
y
2
 2
y
2
=0
=0
 2 = 0
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
7
Laplace Equation
Similarly, use the expression of velocity potential in the continuity equation:
u v
+ =0
x y



       
  +   = 0
x  x  y  y 
 2
x
2
+
 2
y
2
=0
 2 = 0
 2 = 0
 2 = 0
Ψ and Φ both satisfy the Laplace equation
The equations of stream function and velocity potential are in the forms of Laplace’s equation- an
equation that arise in many areas of physical sciences and engineering.
The functions ψ and Φ that satisfy the Laplace’s equation represents a possible two-dimensional,
incompressible, inviscid, irrotational flow field i.e. the Potential flow.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
8
Problem
Consider a flow field given by
 = 3( x 2 − y 2 )
Show that the flow is irrotational. Determine the velocity potential for this flow.
Solution:


=
(3 x 2 − 3 y 2 ) = −6 y
y y


y − velocity component, v = −
= − (3 x 2 − 3 y 2 ) = −6 x
x
x
x − velocity component, u =
angular velocity,  z =
 z =
 z =
1  v u 
 −

2  x y 
1   (6 x )  (6 y ) 


−
2  x
y 
1
( 6 − 6) = 0
2
Another approach:
If the flow is irrotational ; Laplace equation ,  2 = 0
So, the flow is irrotational.
© Dr. A.B.M. Toufique Hasan (BUET)
 2  2
 = 2 + 2
x
y
2
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
9
Since the flow is irrotational, there must exist a velocity potential for this flow.
Again, from the definition of velocity potential,

x − velocity component, u =
= −6 y
x
  =  − 6 y dx + f ( y )
; f ( y ) is an arbitrary function of y
  = −6 xy + f ( y )

y − velocity component, v =
= −6 x
y

(− 6 xy + f ( y ) ) = −6 x

y
Φ from earlier
df ( y )
 −6 x +
= −6 x
dy
df ( y )

=0
 f = constant
dy
  = −6 xy + constant
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
10
Since  and  are used to determine the velocity components by differentiation,
the constant is of no concern; it is usually set to zero. Hence
 = −6 xy
 = 3( x 2 − y 2 )
 d = 6 xdx − 6 ydy = 0 at = C

 = −6 xy
 d = −6 xdy − 6 ydx = 0 at  = C


dy
x
=
dx  =C y
dy
y
=−
dx  =C
x
dy
dy
x
y

=  − = −1
dx  =C dx  =C y
x
Therefore lines of constant ϕ are orthogonal to lines of constant ψ.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
11
Problem
The velocity in a flow field is given by
3


y
2
2 ˆ
2 ˆ
V = ( x y − xy ) i +  − xy  j
 3

(a)Does a stream function exist? If a stream function exists, what is it?
(b)Does a potential function exist? If a potential function exists, what is it?
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
12
Problem
A velocity field is proposed to be

V=
10 y ˆ
10 x ˆ
i− 2
j
2
2
2
x +y
x +y
(a) Is this a possible incompressible flow?
(b) If so, find the pressure gradient with z-axis vertical. Use ρ =1.23 kg/m3 and consider the fluid is
frictionless.
Solution
(a) The differential continuity equation is to be checked to determine if the velocity field is possible or not.
u=

10 x
x2 + y2

v=−
Now,
u
− 20 xy
= 2
x ( x + y 2 ) 2
10 y
x2 + y2
v
20 xy
= 2
y ( x + y 2 ) 2
u v
− 20 xy
20 xy
+
= 2
+ 2
2 2
x y ( x + y ) ( x + y 2 ) 2
=0
Since the given velocity field satisfies the continuity equation, thus this field represents a possible incompressible flow.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
13
Problem
(b) Consider Euler equation (for frictionless fluid)
Du
p
= f x −
Dt
x
 u
u
u
u 
p
   + u
+ v + w  = f x −
x
y
z 
x
 t
x − momentum : 
; steady 2D flow, z-axis vertical
 u
u 
p
   u
+ v  = −
y 
x
 x
p
123 x

= 2
x ( x + y 2 ) 2
y − momentum : 
Dv
p
= f y −
Dt
y
 v
v
v
v 
p
   + u + v + w  = f y −
x
y
z 
y
 t
; steady 2D flow, z-axis vertical
 v
v 
p
   u + v  = −
y 
y
 x
p
123 y

= 2
y ( x + y 2 ) 2
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
14
Problem
Dw
p
= f z −
Dt
z
p
 0 =  (− g ) −
z
p

= (1.23)(−9.81) = −12.07
z
z − momentum : 
; steady 2D flow, z-axis vertical, fz = −g = − 9.81m/s2
So, the pressure gradient:
p =
p =
p ˆ p ˆ p ˆ
i+
j+ k
x
y
z
123
( xiˆ + yˆj ) − 12.07 kˆ
2
2 2
(x + y )
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
15
Simple problems to be solved1. Determination of stream function and velocity potential
2. Confirmation of possible potential flow etc.
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
16
Coordinate systems
To apply the governing equations, a coordinate
system: (x,y,z) or (r,θ,z) is to be chosen that best
fits the geometry of the flow problem to be solved.
The velocity field can be expressed by :

(Cartesian (x,y,z))
V = uiˆ + vˆj + wkˆ

(Cylindrical (r,θ,z))
V = vr iˆr + v iˆ + v z iˆz
y
x = r cos 
(r, θ)
P (x, y)
y = r sin 
 y
x
 = tan −1  
r
θ
x
(0,0)
2D coordinate system
The “vector” or “del” operator has the following two
forms depending on coordinate system:



 = iˆ + ˆj + kˆ
x
y
z
(Cartesian (x,y,z))
 ˆ 1  ˆ 
 = iˆr
+ i
+ iz
r
r 
z
(Cylindrical (r,θ,z))
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
17
Continuity equation for steady inviscid
incompressible flows:

(divergence free velocity field)
 V = 0
y
x = r cos 
(r, θ)
P (x, y)
y = r sin 
 y
x
 = tan −1  
r
For 2D flows:
(
)
  
 
  V =  iˆ + ˆj   uiˆ + vˆj = 0
y 
 x
u v

+
=0
x y
θ
(Cartesian (x,y))
(
x
(0,0)
2D coordinate system
)
  
1   ˆ
  V =  iˆr
+ iˆ
  vr ir + v iˆ = 0
r  
 r
1   (rvr ) v 
 
+
=0
r  r
 

 (rvr ) v
+
=0
r

© Dr. A.B.M. Toufique Hasan (BUET)
(Cylindrical (r,θ))
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
18
Stream function,ψ in (r,θ)
For 2D flows in polar coordinate (r, θ) continuity equation:
 (rvr ) v

+
=0
r

y

and

v  −
r
y = r sin 
 y
x
 = tan −1  
r
Define stream function by the following definition-
1 
vr 
r 
x = r cos 
(r, θ)
P (x, y)
θ
x
(0,0)
2D coordinate system
 (rvr )   1        
= r
= 
=
r
r  r   r    r
2
v    
 2
=
and 
−
 =−
   r 
r
Now
 (rvr ) v
 2
 2
+
=
−
=0
r

r r
© Dr. A.B.M. Toufique Hasan (BUET)
which satisfy the continuity equation.
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
19
Condition of irrotationality for steady inviscid flows:
y

 V = 0
x = r cos 
(r, θ)
P (x, y)
(Curl of velocity field is zero)
y = r sin 
 y
x
 = tan −1  
r
For 2D flows:
(
)
  
 
  V =  iˆ + ˆj   uiˆ + vˆj = 0
y 
 x
v u
 −
=0
x y
θ
(Cartesian (x,y))
(
x
(0,0)
2D coordinate system
)
  
1  
  V =  iˆr
+ iˆ
  vr iˆr + v iˆ = 0
r  
 r
v 1 
v 
  +  v − r  = 0
r r 
 

v v 1 vr
+ −
=0
r
r r 
© Dr. A.B.M. Toufique Hasan (BUET)
(Cylindrical (r,θ))
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
20
Potential function,φ in (r,θ)
For 2D flows in polar coordinate (r, θ) the condition of
irrotationality:
v
v 1 vr
  +  −
=0
r
r r 
y

and
1 
v 
r 
y = r sin 
 y
x
 = tan −1  
r
Define potential function by the following definition-

vr 
r
x = r cos 
(r, θ)
P (x, y)
θ
x
(0,0)
2D coordinate system
vr      
=
 =
   r  r
2
and
v   1  
v 1  2
 1 1  2
1 1  2
= 
+
= −(v r ) 2 +
=− +
 =−
2
r r  r  
 r
r r
r
r r
r r r
Now
v v 1 vr  v 1  2  v 1   2 
 + − 
 = 0
+ −
=  − +
r
r r   r r r  r r  r 
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
which satisfy the condition of
irrotationality
ME 6135: Advanced Aerodynamics
21
Problem
A velocity field is given in polar coordinates for a perfect fluid flow as:
 2 
vr =  − 1 and v = ( − 2r )
 r

Find the stream function for this flow.
Solution:
1 
vr =
r 



 =
3
3
 2 
=  − 1
 r

(
= 2 −r
)
− r + f ( r )

= ( − 2r ) (given )
r

  3
 −  − r + f ( r ) 
= ( − 2r )
r  3

v = −
df (r )
= ( − 2r )
dr
df (r )
−
= − 2r
dr
 f (r ) = r 2 + constant
 −
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
 =
3
3
− r + r 2 + constant
ME 6135: Advanced Aerodynamics
22
Problem
A physically possible irrotational flow is:

V = (2 x + 1)iˆ − (2 y ) ˆj
Find the velocity potential function for this flow.
Solution:
  = x 2 + x − y 2 + constant
© Dr. A.B.M. Toufique Hasan (BUET)
M.Sc. Eng. (April 2024)
ME 6135: Advanced Aerodynamics
23