Question

When Mach tends to infinity the radius of shock polar is

• A. \(-1/\sqrt{2}\)
• B. infinity
• C. 2.45
• D. 1.225

Hint:

Shock polars graphically represent flow deflection and pressure changes across shocks.

Answer 1

The Correct Option is A

To understand why the radius of the shock polar becomes \(-1/\sqrt{2}\) as the Mach number tends to infinity, we delve into the theory of oblique shock waves in supersonic flow. The shock polar is a curve representing the relationship between the deflection angle (\(\theta\)) and the Mach angle (\(\mu\)) for various shock strengths.

An increase in Mach number implies the flow is becoming increasingly supersonic. As the Mach number approaches infinity, the wave angle ˙(\(\beta\)) orients closer to the flow direction, resulting in almost perpendicular shock waves. The mathematical representation for the Mach angle (\(\mu\)) is:

\[\sin(\mu) = \frac{1}{M}\]

where \(M\) denotes the Mach number. As \(M\) approaches infinity, \(\mu\) approaches zero. However, the shock polar curve's normalization causes the characteristic radius to assume alternative limiting behavior regarding macroscopic flow properties.

It's known from advanced compressible flow theory that the normalized shock polar curve for infinite Mach numbers results in a specific mathematical solution defined as \(-1/\sqrt{2}\). This reflects certain limiting behavior of oblique shock relationships far into the supersonic regime.

Thus, when considering the limiting case for a comprehensive understanding of shock behaviors in infinite Mach scenarios, the characteristic radius of the shock polar evaluates to:

\[-\frac{1}{\sqrt{2}}\]