Answer and Explanation:
To obtain the potential function from the given current function, we can use the fact that the velocity components are related to the current function through partial derivatives.
The velocity components can be expressed as:
Vx = ∂ψ/∂y
Vy = -∂ψ/∂x
From the given current function ψ(x,y) = 4.33y + 2.5x m^2/s, we can calculate the velocity components as follows:
Vx = ∂ψ/∂y = 4.33 m^2/s
Vy = -∂ψ/∂x = -2.5 m^2/s
Now, to obtain the potential function ϕ(x,y), we need to integrate the velocity components with respect to the corresponding variables:
ϕ(x,y) = ∫Vx dx = ∫4.33 dx = 4.33x + C1(y)
ϕ(x,y) = ∫Vy dy = -∫2.5 dy = -2.5y + C2(x)
Here, C1(y) and C2(x) are integration constants that can depend on y and x, respectively.
Since the potential function ϕ(x,y) is not uniquely determined from the given current function, we cannot find an exact expression for ϕ(x,y) without additional information or boundary conditions.
To sketch the velocity lines for ϕ(x,y) = 0 and ϕ(x,y) = 1 m^2/s for x = 1, 2, 3, 4 m, we can plot the streamlines, which are the lines tangent to the velocity vector at each point.
The streamlines for ϕ(x,y) = 0 will follow the equation 4.33x + C1(y) = 0, and the streamlines for ϕ(x,y) = 1 m^2/s will follow the equation 4.33x + C1(y) = 1.
For each given value of x, we can solve these equations for y to obtain the corresponding streamline. By plotting these streamlines, we can sketch the velocity lines.
Note: Without knowing the specific values of the integration constants C1(y) and C2(x), we cannot determine the exact shape of the velocity lines.