Stagnation properties of an aircraft

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