Question

An airplane is flying at an altitude of 10,500 m. Determine the gage pressure at the stagnation point on the nose of the plane if the speed of the plane is 350 km/h. Can the flow be considered incompressible if the speed is increased to 1050 km/h

Answer 1

Answer:

gauge pressure at stagnation point is 1836.07  pa, the flow will be compressible at 1050 km/h.

Explanation:

The airplane is flying at a velocity of 350 km/hr. Then assume that air is also flowing at a speed of 350 km/hr  The point at which velocity of flow of fluid is zero at that point kinetic energy of the fluid is converted into pressure head, that pressure head is called stagnation pressure.

Applying Bernoulli's equation between points (1) and (2), we can get:

$((pressure head + velocity head + datum head))_{1}$ = $((pressure head + velocity head + datum head))_{2}$

$\frac{P_{1} }{gρ} +\frac{V^{1} }{2g} + Z_{1} = \frac{P_{2} }{gρ} +\frac{V^{2} }{2g} + Z_{2}$

A both points (1) and (2) are on the same level;

• $Z_{1} = Z_{2}$

• Velocity of air at point (2), $V_{2}$ = 0

from the properties of air table: Density of air at  10500 m is 0.3885 $\frac{kg}{m^{3} }$

$\frac{P_{1} }{gρ} +\frac{V^{1} }{2g} + Z_{1}$ = $\frac{P_{2} }{gρ}$

$\frac{P_{2}-P_{1} }{2g} = \frac{V_{1}^{2}}{2g}$

$\frac{P_{2} - P_{1}}{ρg} = [tex]\frac{1}{2} V_{1}^{2}ρ$ [/tex]

$\frac{P_{2} - P_{1}}= \frac{1}{2} x [tex]97.222^{2} x0.3885$

$\frac{P_{2} - P_{1}} = 1836.07 [tex]pa$

At stagnation point, the gauge pressure = 1836.07 $pa$

If the airplane speed is increased to 1050 km/h(this is about 3 time the initial speed), the flow will be considered compressible because at low speed gases are considered to be incompressible.