Question 4.7: If a stream function exists for the velocity field of Exampl...

If a stream function exists for the velocity field of Example 4.5

u = a(x^2 - y^2)               v = -2axy                 w = 0

find it, plot it, and interpret it.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

• Assumptions: Incompressible, two-dimensional flow.

• Approach: Use the definition of stream function derivatives, Eqs. (4.85), to find ψ(x, y).

u = \frac{\partial \psi}{\partial y}                v=-\frac{\partial \psi}{\partial x}           or               V=i\frac{\partial \psi}{\partial y}-j\frac{\partial \psi}{\partial x}             (4.85)

• Solution step 1: Note that this velocity distribution was also examined in Example 4.3. It satisfies continuity, Eq. (4.83), but let’s check that; otherwise ψ will not exist:

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0                   (4.83)

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=\frac{\partial}{\partial x}[a(x^2-y^2)]+\frac{\partial}{\partial y}(-2axy)=2ax+(-2ax)≡0            checks

Thus we are certain that a stream function exists.

• Solution step 2: To find ψ, write out Eqs. (4.85) and integrate:

u=\frac{\partial \psi}{\partial y}=ax^2-ay^2                       (1)

v=-\frac{\partial \psi}{\partial x}=-2axy                              (2)

and work from either one toward the other. Integrate (1) partially

\psi=ax^2y-\frac{ay^3}{3}+f(x)                       (3)

Differentiate (3) with respect to x and compare with (2)

\frac{\partial \psi}{\partial x}=2axy+f^{\prime}(x)=2axy                            (4)

Therefore f^{\prime}(x) = 0, or f = constant. The complete stream function is thus found:

\psi=a\left(x^2y-\frac{y^3}{3}\right)+C                              Ans. (5)

To plot this, set C = 0 for convenience and plot the function

3x^2y-y^3=\frac{3\psi}{a}                   (6)

for constant values of ψ. The result is shown in Fig. E4.7a to be six 60° wedges of circulating motion, each with identical flow patterns except for the arrows. Once the streamlines are labeled, the flow directions follow from the sign convention of Fig. 4.9. How can the flow be interpreted? Since there is slip along all streamlines, no streamline can truly represent a solid surface in a viscous flow. However, the flow could represent the impingement of three incoming streams at 60, 180, and 300°. This would be a rather unrealistic yet exact solution to the Navier-Stokes equations, as we showed in Example 4.5.

By allowing the flow to slip as a frictionless approximation, we could let any given streamline be a body shape. Some examples are shown in Fig. E4.7b.

capture-20220123-215558
capture-20220123-215610
capture-20220123-215629

Related Answered Questions