Question

A two–dimensional potential–flow solution for flow past an airfoil has the streamline pattern shown. Which additional condition is required to satisfy the Kutta condition? \begin{center} \includegraphics[width=0.5\textwidth]{03.jpeg} \end{center}

• A. Addition of a source of strength \(Q>0\)
• B. Addition of a source of strength \(Q<0\)
• C. Addition of a circulation of strength \(\Gamma>0\) (counter–clockwise)
• D. Addition of a circulation of strength \(\Gamma<0\) (clockwise)

Hint:

In inviscid 2-D airfoil theory, "Kutta condition \(\Rightarrow\) choose \(\Gamma\) so the trailing-edge velocity is finite." The sign of \(\Gamma\) sets the lift direction (CCW \(\Rightarrow\) upward for left-to-right flow).

Answer 1

The Correct Option is C

Step 1: What the Kutta condition demands.
For a sharp–edged airfoil in inviscid potential flow, the Kutta condition requires that the flow leaves the trailing edge smoothly, i.e. the velocity there is finite and the rear stagnation point sits at the trailing edge.

Step 2: How to enforce it in potential flow.
Potential–flow models are built by superposing elementary solutions. Uniform flow (plus a doublet) alone around a lifting shape produces infinite speed at the trailing edge. To make the rear stagnation point coincide with the trailing edge and keep the velocity finite, we must add a circulation \(\Gamma\) about the airfoil.

Step 3: Direction (sign) of circulation.
With the conventional left–to–right free stream and the depicted streamline pattern (higher speed over the upper surface producing upward lift), the necessary circulation is counter–clockwise, i.e. \(\Gamma>0\). This superposes a velocity that augments the upper–surface speed and reduces the lower–surface speed, moving the stagnation point to the trailing edge, thereby satisfying Kutta.

Step 4: Eliminate other options.
Sources/sinks (A,B) change mass flux and cannot by themselves regularize the trailing–edge singularity. A clockwise circulation (D) would shift the stagnation point the wrong way for the shown pattern.

Final Answer:
\[ \boxed{\text{Add a counter–clockwise circulation of strength }\Gamma>0.} \]