Question

Consider steady, two-dimensional, incompressible flow over a non-porous flat plate as shown in the figure. For the control volume PQRS, the speed \(u_{\infty}\) at section PQ is uniform and the speed at section RS is given by \(u(y)=A_0\left(\frac{y}{h}\right)^n\), where \(n\) is a positive integer. The value of \(A_0\) for which the flow through section PS will vanish is:

• A. \( \frac{u_{\infty}}{n+1} \)
• B. \( u_{\infty}(n+1) \)
• C. \( \frac{u_{\infty}}{n-1} \)
• D. \( u_{\infty}(n-1) \)

Hint:

Whenever a side flow vanishes in a control volume, simply equate inflow and outflow. Velocity profiles usually reduce to integrals of the form \( \int z^n dz \).

Answer 1

The Correct Option is B

Step 1: Write inflow at PQ.
Velocity is uniform \(u_{\infty}\). Height is \(h\).
\[ Q_{PQ} = u_{\infty} h \]
Step 2: Write outflow at RS.
Velocity profile is \(u(y) = A_0\left(\frac{y}{h}\right)^n\).
\[ Q_{RS} = \int_0^h A_0 \left(\frac{y}{h}\right)^n dy \]
Let \( y = h z \). Then:
\[ Q_{RS} = A_0 h \int_0^1 z^n dz \]
\[ \int_0^1 z^n dz = \frac{1}{n+1} \]
Thus:
\[ Q_{RS} = \frac{A_0 h}{n+1} \]
Step 3: Apply continuity for zero flow across PS.
Flow through PS vanishes ⟹ inflow = outflow:
\[ Q_{PQ} = Q_{RS} \]
\[ u_{\infty} h = \frac{A_0 h}{n+1} \]
Cancel \(h\):
\[ u_{\infty} = \frac{A_0}{n+1} \]
\[ A_0 = u_{\infty}(n+1) \]
Step 4: Conclusion.
The required value is:
\[ A_0 = u_{\infty}(n+1) \]
Hence, the correct answer is option (B).