Respuesta :
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:
[tex]((pressure head + velocity head + datum head))_{1}[/tex] = [tex]((pressure head + velocity head + datum head))_{2}[/tex]
[tex]\frac{P_{1} }{gρ} +\frac{V^{1} }{2g} + Z_{1} = \frac{P_{2} }{gρ} +\frac{V^{2} }{2g} + Z_{2}[/tex]
A both points (1) and (2) are on the same level;
- [tex]Z_{1} = Z_{2}[/tex]
- Velocity of air at point (2), [tex]V_{2}[/tex] = 0
from the properties of air table: Density of air at 10500 m is 0.3885 [tex]\frac{kg}{m^{3} }[/tex]
[tex]\frac{P_{1} }{gρ} +\frac{V^{1} }{2g} + Z_{1}[/tex] = [tex]\frac{P_{2} }{gρ}[/tex]
[tex]\frac{P_{2}-P_{1} }{2g} = \frac{V_{1}^{2}}{2g}[/tex]
[tex]\frac{P_{2} - P_{1}}{ρg} = [tex]\frac{1}{2} V_{1}^{2}ρ[/tex] [/tex]
[tex]\frac{P_{2} - P_{1}}= \frac{1}{2} x [tex]97.222^{2} x0.3885[/tex]
[tex]\frac{P_{2} - P_{1}} = 1836.07 [tex]pa[/tex]
At stagnation point, the gauge pressure = 1836.07 [tex]pa[/tex]
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.
The gauge pressure at the stagnation point on the nose of the plane is; 1836.07 Pa
What is the gauge pressure?
Since both points fall on the same streamline, we can apply bernoulli's equation to get;
P₁/g + V₁/2g + Z₁ = P₂/g + V₂/2g + Z₂
However, both points fall on the same streamline and so Z₁ = Z₂. Since this is a one-dimensional continuity equation, we have;
P₁ - P₂ = ¹/₂ρ(V₂² - V₁²)
The speed of air at point 2 is zero and so V₂ = 0 m/s.
Also, from properties of air table, the density of air at 10,500 m is ρ = 0.3885 kg/m³.
We are given V₁ = 350 km/h = 97.222 m/s
Thus;
P₁ - P₂ = ¹/₂*0.3885(0² - 97.222²)
Rearranging, we have;
P₂ - P₁ = 1836.07 Pa
If the speed is increased to 1050 km/h, it means that the flow will be considered as compressible because it is only at lower speed that they are considered to be incompressible.
Read more about Gauge pressure at; https://brainly.com/question/15358899
