How Is Stagnation Pressure Calculated for an Airplane Wing at Altitude?

[4growler]

Homework Statement

When an airplane is flying 200mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. What suction pressure is developed on the wing at that point? What is the pressure at the leading edge ( a stagnation point) of the wing?
The pressure of air at this altitude is 12.228 lb/in^2 and the density of air at this altitude is .00238 slugs/ft^3

Homework Equations

p(stag) = p(static) + 1/2 rho (vel) ^2

The Attempt at a Solution

I have tried different routes and got wrong answers. The answers are -76.0 lb/ft^2 and 88.0 lb/ft^2. I guess I don't know what the question is asking for, I thought it was looking for the difference in the pressure between the plane and the wing...please help

 

[KataKoniK]

I would first confirm that the given values for air velocity, altitude, air pressure, and air density are accurate and appropriate for the problem. I would also clarify any additional information needed, such as the shape and size of the wing, and the angle of attack of the airplane.

Based on the given information, I would approach the problem by using the Bernoulli's equation, which states that the sum of pressure, kinetic energy, and potential energy per unit volume is constant along a streamline in an inviscid flow. In this case, the streamline would be the airflow around the wing.

Using the given values, I would first calculate the air velocity at the stagnation point, where the air velocity is zero, and then use that value to calculate the pressure at the leading edge of the wing. I would then use the given air velocity at the point on the wing to calculate the pressure at that point.

Using the Bernoulli's equation, I would set up the following equations:

p(stag) + 1/2 rho (vel(stag))^2 = p(static) + 1/2 rho (vel)^2
p(leading edge) + 1/2 rho (vel(leading edge))^2 = 12.228 lb/in^2 + 1/2 (0.00238 slugs/ft^3) (200 mph)^2
p(wing point) + 1/2 rho (vel(wing point))^2 = 12.228 lb/in^2 + 1/2 (0.00238 slugs/ft^3) (273 mph)^2

Solving these equations, I get the following values:

p(stag) = 12.228 lb/in^2
p(leading edge) = 12.228 lb/in^2 + 0.023 lb/in^2 = 12.251 lb/in^2
p(wing point) = 12.228 lb/in^2 + 0.116 lb/in^2 = 12.344 lb/in^2

Therefore, the suction pressure at the point on the wing is 12.344 lb/in^2 - 12.228 lb/in^2 = 0.116 lb/in^2, and the pressure at the leading edge is 12.251 lb/in^2.

One possible reason for getting incorrect answers could be using incorrect units or not taking into account the

 

[Mene]

I can provide some clarification on the concept of stagnation pressure and how it relates to the given scenario. Stagnation pressure is the pressure that a fluid would attain if it were brought to rest isentropically (without any change in entropy) from its actual state to a state of zero velocity. In other words, it is the maximum pressure that can be achieved by a fluid in motion.

In the case of an airplane flying at a constant speed and altitude, the fluid (air) flowing over the wing is in a steady state, meaning there is no change in velocity or pressure at any point on the wing. Therefore, the stagnation pressure at any point on the wing is equal to the static pressure at that point, as there is no change in velocity to cause a difference in pressure.

Using the given information, we can calculate the stagnation pressure at the point on the wing where the air velocity is 273 mph relative to the airplane. This can be done using the equation p(stag) = p(static) + 1/2 rho (vel)^2, where p(static) is the static pressure, rho is the density of air, and vel is the relative velocity.

Substituting the given values, we get:

p(stag) = 12.228 lb/in^2 + 1/2 (0.00238 slugs/ft^3) (273 mph)^2

= 12.228 lb/in^2 + 7.092 lb/in^2

= 19.32 lb/in^2

Therefore, the stagnation pressure at this point on the wing is 19.32 lb/in^2.

As for the pressure at the leading edge of the wing (stagnation point), it is important to note that this point is where the air is brought to a complete stop, i.e. its velocity is reduced to zero. This means that the stagnation pressure at this point is equal to the total pressure of the fluid, which is given by the equation p(total) = p(static) + 1/2 rho (vel)^2.

Substituting the given values, we get:

p(total) = 12.228 lb/in^2 + 1/2 (0.00238 slugs/ft^3) (200 mph)^2

= 12.228 lb/in^2 + 2.38 lb/in^2

= 14