Basics of Fluid Mechanics, 2014a

10.3. POTENTIAL FLOW FUNCTIONS INVENTORY 365
value Γ=−2 √ 2 − √ 3 πU 0 a can be evaluated as
−Γ
{ }} {
2 √ 2 − √ √ √
3 πU 0 a 2 − 3
sin θ =
=
(10.183)
4 πU 0 a
2
The solution for equation (theta, θ) (10.183) is 15 ◦ or π/12 and 165 ◦ or 11 π/12. For
various stagnation points can be found in similar way.
The rest of the points of the stagnation stream lines are found from the equation
(10.179). For the previous example with specific value of the ratio, Γ as
sin θ =
√
2 −
√
3 a
2 r
( a
) 2
ln
r
( a
) 2
(10.184)
1 −
r
There is a special point where the two points are merging 0 and π.
For all other points stream function can be calculated from equation (10.175) can
be written as
ψ
U 0 a = r ( ( a
) ) 2
a sin θ Γ
1 − +
r 2 πU 0 a ln r a
(10.185)
or in a previous dimensionless form plus multiply by r as
( ( ) ) 2
r ψ
1
sin θ = Γr
r2 1 − +
ln r (10.186)
r 2 πU 0 a sin θ
After some rearrangement of moving the left hand side to right and denoting Γ=
Γ
along with the previous definition of ψ =2n equation (10.186) becomes
4 πU 0 a
0=r 2 − r ψ Γ r ln r
− 1+2 (10.187)
sin θ sin θ
Note the sign in front the last term with the Γ is changed because the ratio in the
logarithm is reversed.
The stagnation line occur when n =0hence equation (10.187) satisfied for all
r =1regardless to value of the θ. However, these are not the only solutions. To obtain
the solution equation (stagnation line) (10.187) is rearranged as
( ) 2 Γ r ln r
θ = sin −1 1 − r 2 (10.188)
Equation (10.187) has three roots (sometime only one) in the most zone and
parameters. One roots is in the vicinity of zero. The second roots is around the one