Question

A light plane flies at 720 km/hr in standard air at an altitude of 1000 m. Determine the stagnation pressure at the leading edge of the wing. Take, P_air = 9.0 × 10⁴ N/m², ρ = 1.1 kg/m³.
1. 90 kPa
2. 119 kPa
3. 112 kPa
4. 100 kPa

Answer 1

Correct Answer - Option 3 : 112 kPa

Concept:

Applying Bernoulli’s equation

\(\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + h = \frac{{{P_{stagnation}}}}{{\rho g}} + \frac{{V_{stag}^2}}{{2g}} + h\)

At stagnation point, V = 0

\( \frac{{{P_{stagnation}}}}{{\rho g}} =\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} \)

Calculation:

Given:

V = 720 km/hr = 720 × (1000/3600) = 200 m/sec, Pair = 9.0 × 104 N/m2, ρ = 1.1 kg/m3.

\(\frac{{{P_{stagnation}}}}{{\rho g}} = \frac{{9\; × \;{{10}^4}}}{{1.1\; × \;9.81}} + \frac{{{{\left( {200} \right)}^2}}}{{2\; × \;9.81}}\)

\(\frac{{{P_{stagnation}}}}{{\rho g}} = 10379.01\)

P_stagnation = 1.1 × 9.81 × 10379.01

P_stag = 112000 Pa

P_stagnation = 112 kPa