Problem 7.1
a) By definition:
u=
∂Ψ
∂Ψ
, v= −
∂y
∂x
Then
∂Ψ
∂Ψ
=y
= x,
∂y
∂x
After integrating, we get the expression of the stream function:
Ψ= xy + const
Please see DGSJ, Example 7.1, if more details about integration are required.
Assuming the constant to be zero, we have
Ψ =xy
Streamlines:
y
x
b) The expression for the stream function is
Ψ= xy + y= y ( x + 1)
Streamlines:
y
x
c) The expression for the stream function is
Ψ= xy + y= y ( x + 1)
∂Ψ
∂Φ ∂Ψ
∂Φ
,
= y= −
= x +1 =
∂y
∂x
∂x
∂y
After integrating the expressions above, for the velocity potential we have
x2
y2
Φ=
+x−
+ const
2
2
Please see DGSJ, Example 7.1, if more details about integration are required.
The constant can be assumed to be zero.
Problem 7.3
By definition
u=
And
u=
∂Ψ
∂Ψ
, v= −
∂y
∂x
∂Φ
∂Φ
, v=
∂y
∂x
Then
∂Ψ ∂Φ
= = 3V ( x 2 − y 2 ) .
∂y
∂x
After integrating, we have:
3Vy 3
=
Ψ 3Vx y −
+ f1 ( x )
3
2
On the other hand:
∂Ψ
∂Φ
=
−
=
6Vxy .
∂x
∂y
After integrating, we have:
=
Ψ 3Vx 2 y + f 2 ( y )
Both expressions for the stream function are compatible when
=
Ψ 3Vx 2 y − Vy 3=
+ C Vy ( 3 x 2 − y 2 ) + C
Please see DGSJ, Example 7.1, if more details about integration are required.
Assuming the constant to be zero, we have
=
Ψ Vy ( 3 x 2 − y 2 )
Consider streamlines corresponding to Ψ =0 .
=
Ψ Vy ( 3 x 2 =
− y2 ) 0
The solutions to this equation are:
y=0
y = ± 3x
The expressions above are equations for straight lines passing through the origin of coordinate
system. The first line is the horizontal line, coincides with the x-axis. The inclinations of the other two
lines are:
± tan −1 3 =
±60°
θ 2,3 =
Velocity components are:
=
u
Velocity magnitude is:
v=
∂Ψ
= 3V ( x 2 − y 2 ) ,
∂y
∂Ψ
v=
−
=
−6Vxy
∂x
u 2 + v 2 = 3V x 4 − 2 x 2 y 2 + y 4 + 4 x 2 y 2 = 3V ( x 2 + y 2 )
Direction:
−2 xy
−1 v
=
=
θ v tan
tan −1 2
u
x − y2
Note that for y = 3 x we have
2
v
−1 −2 3 x
tan = tan 2
= tan −1 3= 60°
θ=
v
2
u
x − 3x
−1
Problem 7.4
C
y
A
0
θ1
θ2
1B
D x
=
Ψ
=
Ψ
q1q1 q2q 2
+
2π
2π
q1 = 30
m2s-1
q2 = 20
m2s-1
30θθ
20 2
1
+
2π
2π
tan θ1 =
tan θ 2 =
=
Ψ
m2s-1
y
x
y
x −1
30
 y  20
 y 
arctan   +
arctan 

2π
 x  2π
 x −1 
m2s-1
∂Ψ 15
1
1 10
1
1
=
+
vx =
m 2 s −1 =
2
2
∂y π
 y x π
 y  x −1
1+  
1+ 

x
 x −1 
x −1
15 x
10
2 −1
=
+
m
s
2
2
2
π x +y
π ( x − 1) + y 2
2015/2016
Page 1 of 2
∂Ψ 15
1
y 10
1
y
vy =
m 2 s −1 =
−
=
+
2
2
2
2
∂x π
π
 y x
 y  ( x − 1)
1+  
1+ 

x
 x −1 
15 y
10
y
2 −1
m
s
=
+
2
2
2
π x +y
π ( x − 1) + y 2
At point (-1, 0):
 15 ( −1) 10 ( −2 ) 
20
−1
vx =
+
m
⋅
s
=
−
m ⋅ s −1


π 4 
π
π 1
=
v y 0 m ⋅ s −1
At point (1,1):
15
m ⋅ s −1
2π
35
 15 10 
vy = 
m ⋅ s −1
+  m ⋅ s −1 =
2π
 2π π 
=
vx
Dynamic pressure:
=
pdyn
1 2 1
=
ρv
ρ ( vx2 + v y2 )
2
2
At point (-1, 0):
1 2 2 400
⋅ 2 Nm −2 ≈ 40.53 Nm −2
pdyn =
ρv =
2
2 p
At point (1,1):
1 2 2 25
ρv =
⋅ 2 ( 9 + 49 ) Nm −2 ≈ 36.73 Nm −2
pdyn =
2
2 4p
2015/2016
Page 2 of 2