Question

An aircraft is flying at an altitude of 4500 m above sea level, where the ambient pressure, temperature, and density are 57 kPa, 259 K, and 0.777 kg/m\(^3\), respectively. The speed of the aircraft \( V \) is 230 m/s. Gas constant \( R = 287 \, {J/kg/K} \), and specific heat ratio \( \gamma = 1.4 \). If the stagnation pressure is \( p_0 \), and static pressure is \( p \), the value of \[ \frac{p_0 - p}{\frac{1}{2} \rho V^2} \] is \_\_\_\_\_\_\_\_ (rounded off to two decimal places).

Hint:

For calculating the stagnation pressure and the related quantities in compressible flow, remember to apply the Bernoulli equation in the form \( \frac{p_0}{p} = \left( 1 + \frac{\gamma - 1}{2} \cdot \frac{V^2}{R T} \right)^{\frac{\gamma}{\gamma - 1}} \).

Answer 1

Step 1: Calculate the Mach Number
First, compute the speed of sound (\(a\)): \[ a = \sqrt{\gamma R T} = \sqrt{1.4 \times 287 \times 259} \approx 322.5 \, {m/s} \] Next, the Mach number (\(M\)): \[ M = \frac{V}{a} = \frac{230}{322.5} \approx 0.713 \] Step 2: Compute Stagnation Pressure (\(p_0\))
Using the isentropic relation: \[ \frac{p_0}{p} = \left(1 + \frac{\gamma - 1}{2} M^2\right)^{\frac{\gamma}{\gamma - 1}} \] Substitute \(\gamma = 1.4\) and \(M = 0.713\): \[ \frac{p_0}{p} = \left(1 + \frac{0.4}{2} \times 0.713^2\right)^{3.5} = \left(1 + 0.2 \times 0.508\right)^{3.5} \approx 1.396 \] Thus: \[ p_0 = 1.396 \times p = 1.396 \times 57 \, {kPa} \approx 79.57 \, {kPa} \] Step 3: Compute \(p_0 - p\)
\[ p_0 - p = 79.57 - 57 = 22.57 \, {kPa} = 22570 \, {Pa} \] Step 4: Dynamic Pressure Term
\[ \frac{1}{2} p V^2 = \frac{1}{2} \times 0.777 \times (230)^2 \approx 20550 \, {Pa} \] Step 5: Evaluate the Expression
\[ \frac{p_0 - p}{\frac{1}{2} p V^2} = \frac{22570}{20550} \approx 1.098 \approx \boxed{1.10} \]