Potential Flow Analysis
Problem Statement
If, for a two dimensional potential flow, the velocity potential is given by:
- The velocity at point (2, 3)
- The value of stream function ψ at point (2, 3)
1. Velocity Components
First, let’s expand the velocity potential:
φ = 4x(3y-4) = 12xy – 16x
u = ∂φ/∂x = 12y – 16
At point (2, 3): u = 12(3) – 16 = 36 – 16 = 20 m/s
v = ∂φ/∂y = 12x
At point (2, 3): v = 12(2) = 24 m/s
2. Resultant Velocity
R = √(u² + v²) = √(20² + 24²)
R = √(400 + 576) = √976 ≈ 31.2 m/s
3. Stream Function Determination
(a) From u = ∂φ/∂x = 12y – 16
(b) From v = ∂φ/∂y = 12x
(c) Integrating equation (a) with respect to y:
ψ = ∫(12y – 16)dy = 6y² – 16y + c(x)
where c(x) is a constant which is independent of y
(d) Differentiating equation (c) with respect to x:
∂ψ/∂x = c'(x)
From equation (b) and stream function properties:
∂ψ/∂x = v = 12x
Therefore:
c'(x) = 12x
Integrating c'(x):
c(x) = 6x² + C
where C is a constant
Substituting back into equation (c):
ψ = 6y² – 16y + 6x² + C
To find the constant C, we use the fact that at point (2, 3), ψ = -18:
ψ(2,3) = 6(3)² – 16(3) + 6(2)² + C = -18
6(9) – 16(3) + 6(4) + C = -18
54 – 48 + 24 + C = -18
30 + C = -18
C = -48
Therefore:
ψ = 6y² – 16y + 6x² – 48
At point (2, 3): ψ = -18
Flow Characteristics
Key features of this potential flow:
- The flow is irrotational as it is derived from a velocity potential
- The velocity field has constant y-component gradients (∂v/∂x = 12)
- The x-component varies linearly with y (∂u/∂y = 12)
- The continuity equation is satisfied: ∂u/∂x + ∂v/∂y = 0 + 0 = 0
- u = -∂ψ/∂y = -(12y – 16) = -12y + 16 (equals -36 + 16 = -20 at point (2, 3))
- v = ∂ψ/∂x = 12x (equals 24 at point (2, 3))
- Note: The negative sign for u is consistent with the relationship between velocity potential and stream function
