Calculate Mass of Airplane: Pt + (1/2)pVt^2 Method

In summary, the wings of an airplane have a total surface area of 250 m^2. The air above the wing is 300 m/s and the air below the wing is 250 m/s. The ambient air pressure is around 0.5 atm. The airplane is descending at 20 m/s, therefore there is an upward drag on the wings. The mass of the airplane is 16.125 Pa.
  • #1
zodiacbrave
11
0

Homework Statement



The wings have a total surface area of 250 m^2. The air above the wing is 300 m/s and the air below the wing is 250 m/s. The ambient air pressure is around 0.5 atm. The airplane is descending at 20 m/s, therefore there is an upward drag on the wings.

What is the mass of the airplane?


Homework Equations



Pt + (1/2)pVt^2 = Pb + (1/2)pVb^2



The Attempt at a Solution



Fw = m * 9.8 m/s^2. Pair = 5.05 x 10^4 Pa

Density of air at height of plane, pair = (po / Po)* Pair = 0.645 kg/m^3

Plugging into Pt + (1/2)pVt^2 = Pb + (1/2)pVb^2

Pb - Pt = 16.125 Pa

Fw = msg - Flift.

I'm not sure how to deal with the force pressing down caused by 20 m/s


Thank you
 
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  • #2
What is Fw if the speed of descending is constant ?
 
  • #3
9.8 m/s^2 + 20 m /s * the mass of the airplane?
 
  • #4
The problem statement is somewhat confusing, it gives the speeds of the air above and below the wings and also states that the aircraft is descending, but that descent rate also affects the speed of the affected air. The problem doesn't seem to provide sufficient information to determine the forward speed of the aircraft, or it's effective slope angle, or the amount of drag involved. For a conventional wing, the speed of the aircraft would be closer to the speed of the air below the wing than the speed of the air above the wing.
 
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  • #5
Couldn't it be as simple as Fw = (Pb - Pt) * Area ?
 
  • #6
BvU said:
Couldn't it be as simple as Fw = (Pb - Pt) * Area ?
Since the aircraft is descending, it's angled downwards a bit, so lift, which is apparently to be approximated as (Pb - Pt) * area, is angled forwards a bit. The upward forces are the vertical components of lift and drag, and the downwards components are gravity (weight) and the vertical component of thrust (which is unknown). The speed of the aircraft would be somewhere between 250 m/s and 300 m/s, so you have a range for speed, but not an exact speed.
 
  • #7
rcgldr said:
Since the aircraft is descending, it's angled downwards a bit,
Is it angled down, or is it merely that the lift is not sufficient to maintain height (and the vertical speed is effectively a terminal velocity)?
 
  • #8
rcgldr said:
Since the aircraft is descending, it's angled downwards a bit ...

haruspex said:
Is it angled down, or is it merely that the lift is not sufficient to maintain height (and the vertical speed is effectively a terminal velocity)?
Assuming the pilot isn't forcing the situation with up elevator input, the tail will weathervane upwards (relative to the aircraft) in response to a descent, so typically the aircraft will be angled downwards. If the pilot is forcing the situation, then it's possible that the aircraft could even be pitched upwards a bit during descent, like an aircraft on final approach. I'm not sure how to combine Bernoulli and drag effects under these circumstances.
 
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  • #9
My guess is that the question is intended to deal with by using some heavily simplified model and quite simple physics.

Not that I have any idea about what the model is intended to be. My first thought was similar to that signaled by haruspex, that the vertical speed is to be understood as the terminal velocity, but I don't see how to use this information.
 
  • #10
rcgldr said:
Since the aircraft is descending, it's angled downwards a bit, so lift, which is apparently to be approximated as (Pb - Pt) * area, is angled forwards a bit. The upward forces are the vertical components of lift and drag, and the downwards components are gravity (weight) and the vertical component of thrust (which is unknown). The speed of the aircraft would be somewhere between 250 m/s and 300 m/s, so you have a range for speed, but not an exact speed.

Don't see the logic: at touchdown it's the rear wheels that hit the ground first. The thud betrays the fact that is descends, yet the nosewheel lands latest (fortunately).

And forget the thrust. It's definitely not a glider, but from the problem formulation it might just as well be!
 
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  • #11
rcgldr said:
Since the aircraft is descending, it's angled downwards a bit.

BvU said:
Don't see the logic: at touchdown it's the rear wheels that hit the ground first.
That's because the pilot feeds in up elevator to keep the nose of the aircraft up. The problem becomes more complicated if the aircraft is not angled downwards into the direction of flight, or at least drag becomes a bigger factor.

BvU said:
And forget the thrust. It's definitely not a glider, but from the problem formulation it might just as well be!
Since the aircraft is in a steady descent, the problem could assume the aircraft is a glider.
 
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  • #12
Big glider with 250 m2 wing area!

Is Zodiacbrave still in the picture ? How do you end up with Pb - Pt = 16.125 Pa if 1/2 * (3002 - 2502) = 13750 and ρ≈0.65 ?

I would ignore a few tiny details and estimate m = 13750 * 0.65 * 250 / 9.8 = 230 ton, about half a 747. They fly around with speeds around 1000 km/h. Gliders don't.

But then, I'm only a physicist, non-frequent flyer and definitely not an aeronautics expert.

I would ignore the 20 m/s if wing speeds are already given. Might be quite wrong, so please correct me. Titlting is also negligible: cosine effect and the whole thing wouldn't be much different if it descended without tilt.

I like this problem!
 
  • #13
BvU said:
Is Zodiacbrave still in the picture ?
No posts other than the original. Perhaps the problem statement was adjusted or he solved it using some assumption.

BvU said:
Big glider with 250 m2 wing area!
Since the aircraft is descending at a constant rate, part of the force opposing gravity is the vertical component of drag, so the sum of total thrust and drag must result in a net drag force.

BvU said:
I would ignore the 20 m/s if wing speeds are already given. Might be quite wrong, so please correct me. Titlting is also negligible: cosine effect and the whole thing wouldn't be much different if it descended without tilt.
Except that the problem states that part of the drag force (the vertical component) is opposing drag.

Ignoring the effects of the aircraft's orientation, assume it's path is θ radians below horizonal. Using a calculated pressure differential, lift = pressure differential x wing area, an absolute value independent of the mass. The vertical component of lift = lift x cos(θ). The weight of the plane = m g. The vertical component of drag = m g - lift x cos(θ) = drag x sin(θ). Assuming the aircraft isn't accelerating, the net force is zero:

0 = m g - lift x cos(θ) - drag x sin(θ)

m g = lift x cos(θ) + drag x sin(θ)

The angle of descent is somewhere within a range depending on the aircrafts total speed versus descent speed:

arctan(20 / 250) <= θ <= arctan(20 / 300)

Even if the angle of descent is known, there are still two unknowns in the equation above, mass and drag. If the problem included the information required to determine the angle of the aircraft's path (or its total speed), and a lift / drag ratio for the given conditions, then the problem could be solved using the apparent assumptions implied by the problem statement.
 

1. How do you calculate the mass of an airplane using the Pt + (1/2)pVt^2 method?

To calculate the mass of an airplane using this method, you need to first determine the total pressure (Pt) and the velocity (Vt) of the airplane. Then, you will need to measure the air density (p) at the given altitude. Finally, you can use the formula m = Pt + (1/2)pVt^2 to calculate the mass of the airplane.

2. What is the significance of the Pt + (1/2)pVt^2 method in determining the mass of an airplane?

The Pt + (1/2)pVt^2 method is significant because it takes into account the dynamic pressure and the static pressure of the air, as well as the velocity of the airplane. This method provides a more accurate calculation of the mass of the airplane compared to other methods that only consider the weight and volume of the aircraft.

3. How do you measure the total pressure (Pt) of an airplane?

The total pressure (Pt) can be measured using a pitot tube. This instrument is designed to measure the stagnation pressure of the air, which is the combination of the dynamic and static pressures. The pitot tube is usually mounted on the nose or wing of the airplane.

4. Can the Pt + (1/2)pVt^2 method be used for any type of airplane?

Yes, the Pt + (1/2)pVt^2 method can be used for any type of airplane as long as the necessary parameters (Pt, Vt, p) are known. However, for larger and more complex aircraft, additional factors such as the shape and surface area of the aircraft may need to be considered for a more accurate calculation of the mass.

5. Are there any limitations to using the Pt + (1/2)pVt^2 method for calculating the mass of an airplane?

One limitation of this method is that it does not take into account the mass of the air inside the airplane. This means that the calculated mass will be slightly lower than the actual mass of the aircraft. Additionally, variations in air density due to temperature and altitude can also affect the accuracy of the calculation.

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