Introduction to Compressible Flow

Mach number

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

where $V$ is the flow velocity and $a$ is the local speed of sound. It measures the importance of compressibility: the ratio of inertia force to elastic force.

Speed of sound

Consider a small pressure wave moving at speed $a$ into still fluid of density $\rho$ and pressure $p$. In a frame fixed to the wave, the fluid approaches at speed $a$ and leaves at $a - dV$, with properties $\rho + d\rho$ and $p + dp$.

   still fluid         wave        disturbed fluid
   p, rho, V = 0  -->  |  a  |    p+dp, rho+drho, dV

Continuity: $\rho a = (\rho + d\rho)(a - dV)$. Neglecting the product $d\rho\,dV$:

$$dV = a\frac{d\rho}{\rho}$$

Momentum: $p - (p + dp) = \rho a\,[(a - dV) - a]$, so $dp = \rho a\,dV$.

Eliminating $dV$:

$$a^2 = \frac{dp}{d\rho} \quad \Rightarrow \quad a = \sqrt{\frac{dp}{d\rho}} = \sqrt{\frac{E_v}{\rho}}$$

where $E_v$ is the bulk modulus. The small wave is nearly reversible and adiabatic, so the process is isentropic: $p/\rho^k =$ constant, which gives $dp/d\rho = kp/\rho$. With $p = \rho RT$:

$$a = \sqrt{kRT}$$

For air, $k = 1.4$, $R = 287\ \text{J/kg K}$, so $a = 20.05\sqrt{T}$ m/s (343 m/s at 20 °C).

Classification of flow

Incompressible treatment

For $Ma < 0.3$ the density change in the flow is less than about 5%, so the flow can be treated as incompressible. Above this value compressibility effects must be included.

Given data

$V = 250\ \text{m/s}$, $T = 233\ \text{K}$, $p = 26.5\ \text{kPa}$, $k = 1.4$, $R = 287\ \text{J/kg K}$.

(a) Speed of sound and Mach number

$$\begin{aligned} a &= \sqrt{kRT} = \sqrt{1.4 \times 287 \times 233} = 306.0\ \text{m/s} \\ Ma &= \frac{V}{a} = \frac{250}{306.0} = 0.817 \end{aligned}$$

The flow is subsonic, but $Ma > 0.3$, so compressibility must be considered.

(b) Stagnation properties (isentropic)

$$\frac{T_0}{T} = 1 + \frac{k - 1}{2}Ma^2 = 1 + 0.2(0.817)^2 = 1.1335$$ $$T_0 = 233 \times 1.1335 = 264.1\ \text{K}$$ $$\frac{p_0}{p} = \left(\frac{T_0}{T}\right)^{k/(k-1)} = (1.1335)^{3.5} = 1.5507$$ $$p_0 = 26.5 \times 1.5507 = 41.09\ \text{kPa}$$

(c) Incompressible estimate

Density: $\rho = \dfrac{p}{RT} = \dfrac{26\,500}{287 \times 233} = 0.3963\ \text{kg/m}^3$

$$p_0 - p \approx \tfrac12\rho V^2 = 0.5 \times 0.3963 \times 250^2 = 12.38\ \text{kPa}$$

The compressible (correct) rise is $41.09 - 26.5 = 14.59\ \text{kPa}$.

$$\text{Error} = \frac{14.59 - 12.38}{14.59} = 15.1\%$$

The incompressible formula under-predicts the stagnation pressure rise by about 15% at this Mach number.

Answer: $a = 306\ \text{m/s}$, $Ma = 0.817$, $T_0 = 264.1\ \text{K}$, $p_0 = 41.1\ \text{kPa}$; the incompressible formula gives $12.4\ \text{kPa}$ rise, about $15\%$ low