The stream function for a two dimensional flow is given by ψ=8xy. Find the velocity potential function.
Potential Flow Analysis
Problem Statement
The stream function for a two dimensional flow is given by:
1. Determine the Velocity Components
For a stream function ψ, the velocity components are related by:
u = -∂ψ/∂y
v = ∂ψ/∂x
Given ψ = 8xy:
u = -∂ψ/∂y = -8x
v = ∂ψ/∂x = 8y
2. Relationship with Velocity Potential
For a velocity potential φ, the velocity components are related by:
u = ∂φ/∂x
v = ∂φ/∂y
Comparing with our velocity components:
∂φ/∂x = -8x (a)
∂φ/∂y = 8y (b)
3. Finding the Velocity Potential
Step 3.1: Integrate equation (a) with respect to x:
φ = ∫(-8x)dx = -4x² + f(y)
Where f(y) is a function that depends only on y.
Step 3.2: Differentiate this expression with respect to y:
∂φ/∂y = f'(y)
Step 3.3: Using equation (b):
f'(y) = 8y
Step 3.4: Integrate to find f(y):
f(y) = ∫(8y)dy = 4y² + C
Where C is a constant.
Step 3.5: Substitute back to get the complete velocity potential:
φ = -4x² + 4y² + C
Since we’re only interested in the form of the potential function (not its absolute value), we can set C = 0:
Verification
Let’s verify our solution by checking if the velocity components derived from φ match those from ψ:
u = ∂φ/∂x = -8x
v = ∂φ/∂y = 8y
From ψ = 8xy:
u = -∂ψ/∂y = -8x
v = ∂ψ/∂x = 8y
Physical Interpretation
This potential flow has the following characteristics:
- The flow is irrotational (∇×V = 0) as expected for a potential flow
- The streamlines form hyperbolas (since ψ = 8xy = constant defines hyperbolas)
- The equipotential lines (φ = constant) form orthogonal hyperbolas to the streamlines
- The flow pattern resembles a saddle point at the origin, with flow approaching along the y-axis and departing along the x-axis
