Understanding Mach Number and Gas Compressibility

In fluid dynamics, the behavior of a fluid, whether it's considered compressible or incompressible, depends significantly on its speed relative to the speed of sound in that medium. For gases, which are inherently compressible, this distinction is particularly important.

The Mach number ($M$) is a dimensionless quantity that represents the ratio of the speed of an object or fluid flow ($V$) to the speed of sound ($a$) in the surrounding medium. Mathematically, it is expressed as:

$$M = \frac{V}{a}$$

When a gas flows at speeds significantly lower than the speed of sound, the density changes due to pressure variations are very small. In such cases, the flow can be approximated as incompressible, meaning its density remains nearly constant. This simplification greatly simplifies the analysis of fluid flow problems.

Condition for Incompressible Flow in Gases

The condition under which a gas flow can be treated as incompressible depends on the Mach number. When the Mach number is low, the effects of compressibility are negligible for many practical purposes. There isn't one single, strict Mach number value that universally defines the boundary between compressible and incompressible flow, as it often depends on the desired accuracy of the analysis.

However, a commonly accepted threshold in many engineering applications is that gas flow can be considered incompressible when the Mach number is less than approximately 0.3. Below this value, density variations are typically less than about 5%. For more stringent requirements or in certain contexts, a lower threshold might be used.

Analyzing the Given Options

Let's evaluate the provided options based on the relationship between Mach number and compressibility:

1. equal to 1.0: This corresponds to sonic flow. At $M = 1.0$, compressibility effects are very strong, and the flow is highly compressible.
2. equal to 0.5: At $M = 0.5$, compressibility effects are significant and cannot be ignored in most accurate analyses. This is considered a compressible flow regime.
3. more than 0.3: Any Mach number greater than 0.3 indicates that compressibility effects are present and generally need to be accounted for. This is the compressible flow regime.
4. less than 0.2: A Mach number less than 0.2 is well below the common threshold of 0.3. While 0.3 is often cited, a lower value like 0.2 represents an even stricter condition where compressibility effects are even smaller and the incompressible assumption is highly valid. Based on the options provided, this is the condition that best represents the incompressible flow assumption.

Considering the options, the condition where gases are considered incompressible is when the Mach number is a sufficiently low value. Among the given choices, "less than 0.2" represents the lowest Mach number range, where compressibility effects are minimal and the incompressible assumption is most justified.

Mach Number Regimes

Fluid flow regimes are often categorized based on the Mach number:

Revision Table - Mach Number and Compressibility

Additional Information - Why is M < 0.3 (or M < 0.2) Important?

The reason $M < 0.3$ is often used as a criterion for incompressible flow is rooted in the relationship between density change and Mach number. For isentropic flow of an ideal gas, the ratio of density ($\rho$) at a certain velocity to the stagnation density ($\rho_0$) can be related to the Mach number ($M$) by:

$$\frac{\rho}{\rho_0} = \left(1 + \frac{\gamma-1}{2} M^2\right)^{-\frac{1}{\gamma-1}}$$

where $\gamma$ is the ratio of specific heats (approximately 1.4 for air). The percentage change in density compared to stagnation density is $1 - \frac{\rho}{\rho_0}$.

Let's calculate the density ratio for $M = 0.2$ and $M = 0.3$ (using $\gamma = 1.4$ for air):

• For $M = 0.2$:
  $\frac{\rho}{\rho_0} = \left(1 + \frac{1.4-1}{2} (0.2)^2\right)^{-\frac{1}{1.4-1}} = \left(1 + 0.2 \times 0.04\right)^{-\frac{1}{0.4}} = (1.008)^{-2.5} \approx 0.9802$
  The density change is $1 - 0.9802 = 0.0198$, or about 1.98%.
• For $M = 0.3$:
  $\frac{\rho}{\rho_0} = \left(1 + \frac{1.4-1}{2} (0.3)^2\right)^{-\frac{1}{1.4-1}} = \left(1 + 0.2 \times 0.09\right)^{-\frac{1}{0.4}} = (1.018)^{-2.5} \approx 0.9564$
  The density change is $1 - 0.9564 = 0.0436$, or about 4.36%.

As you can see, for $M < 0.2$, the density change is very small (less than 2%). For $M < 0.3$, it's less than about 4.4%. These small changes are often considered negligible for engineering calculations where less than 5% error in density is acceptable, allowing the incompressible flow assumption to be used.