I. INTRODUCTION
Air crash has many investigations[1]. This paper studies that air crash would be caused by deference of pressures between outside and inside of passenger's cargo. Such deference of pressures breaks the window or door and thus the plane crashes.
In order to calculate pressure outside the passenger's cargo, we study three type equations: A "wind - temperature equation" ("w-t eq." in short), B "wind - pressure /density equation" ("w - P/peq." in short). C "jet - pressure /density equation" ("J - P/peq." in short).
In Section 2, we derive the "w - T eq". It states that the derivative of wind speed respect to time proportions to the derivative of temperature respect to track (space).
In Section 3, the solution of "w - T eq." is obtained by method of separating variables. And it has been checked by weather forecasting, by dimensional check, and by other model check.
In Section 4, the "w - P/peq." is derived by the combination of Boyles law and Charles' law.
In Section 5, the solution of "w - P/peq." can be obtained directly from the solution of "w - T eq." in Section 3. In Section 5.1, the air density is calculated. Where the traditional method and engineering tool bar considered the air density is the functions of temperature and highness, but no connected to wind speed. Here, the air density is connected with wind speed.
In section 6, set up the "J - P/peq." The absolute motion (the jet plane motion relative a referenced point on Earth) is equal to the linking motion (wind speed motion relative ) and the relative motion (jet speed motion relative wind motion).
In Section 7, the solution of "J - P/peq." is directly obtained from section 3. Two cases: 1, density of jet plane is a constant; 2, the density of the jet plane is variable.
In Section 8, as an example of calculation of pressure outside the passenger's cargo, MH370 has been used. Where it had been studied in cases of nonpowered flying[6].
In Section 10, a conclusion is made.
II. WIND-TEMPERATURE EQUATION OF A POINT IN AIR ("W-T EQ." IN SHORT)
The derivation of "w - T equation" follows [3].
According to combination of Boyles' law and Charles' law (B-C law, in short), we have:
Where = the air pressure, accuracy. is a stress tensor, with the same normal stress and zero shearing stress
compressive normal stress; shearing stress i, j, k are unit vectors in Cartesian co-ordinates. volume of a point with are the cross section areas of the element (point), respectively. c is a constant, T is the absolute temperature. (2-2) is the matrix form of
Let operator be an inner product on both sides of (2-1), then, we have
Where
Where = the applied force, = the displacement
By Newton's second law, we have
Where is the mass, is the wind speed, . Substituting (2-9) into (2-8), we have
(2-10) is called the "Wind -Temperature Equation of A Point (Mass) in Air", or ("W - T equation" in short). It states that is in proportion to , i.e., the derivative of wind speed respect to time proportions to the derivative of temperature respect to track (space).
III. SOLUTION OF "W - TEQ" BY METHOD OF SEPARATING VARIABLES
(2-10) is a vector PDE. It is hard to solve. Changing (2-10) to scalar function (its component form) is an easy way to solve. By (2-2) -- (2-4), we have:
Where . , , .
Looking at (3-3):
Where
Suppose that variables of and can be separated:
Where , , and are unknown functions and have continuous derivatives. We need sufficient relations to determine these unknowns.
The dimensions of and are the same (see the following red part), therefore, let (3-4) = (3-5), and separating variables, we have
Where the function of the left term is , while the function of the middle term is , if they are equal to each other, they must be a constant--- .
For simple, we choose two relations:
then, integrating both sides of (3-6) for and by (3-8), we have:
Where is a constant, determined by initial condition, i.e., , , , . Then, (3-9) gives:
Substituting (3-7) into (3-6), then, integrating both sides of (3-6) for , we have:
Where is an integral constant, determined by the boundary condition, i.e., . Therefore,
Substituting (3-14), (3-10), into (3-4), we have
Substituting (3-11), (3-13), into (3-5), we have
Similar to (3-16), for (3-1), (3-2), we have:
Expressed by vector form:
Dimensional check: (3-19)
Dimensional check is a tool often used to check the correctness of calculation. The dimensions of each term of an equality must be the same.
The dimensional check (red part) shows that the dimensions on both sides of the equality of (3-19) are the same. The dimensions in (3-19) is correct.
a) Checking the "w - T eq." by weather forecasting (China Weather Net), and motion equation of mushroom cloud.
Every winter, the weather forecast (China Weather Net) alerts people be wear that cold wave comes with strong wind accompanied with temperature sharp dropping. These description on temperature sharp dropping company with strong wind confirms that the difference of temperature between positions in track causes strong wind. Which agrees with (2-10) very well.
Moreover, a motion equation of mushroom cloud[3] derived by different model based on modifying of Navier-Stock equation, with result well agreed to (2-10). Which again confirms that (2-10) is believable.
IV. THE "WIND - PRESSURE/DENSITY EQUATION"(W-P/ρ EQ.IN SHORT)
Dividing both sides of B - C law(combination of Boyles law and Charles law) (2-1) by M (mass), we have
Where is the density of air. Substituting (4-1) into (2-10), we have:
(4-2) is called the "W-P/ρ equation". (4-2) shows that is in proportion to , i.e., the derivative of wind speed respect to time proportions to the derivative of respect to track (space).
V. SOLUTION OF "W - P/ρ EQ." (4-2)
The solution of scalar form of (4-2) can be obtained from (3-17), (3-18), and (3-16) by changing to , i.e.,
Kp'(s,t) = u_x(s,t) = \exp[k(x-t^2)], \(k = \rho_{air}^{-1}) \tag{5-1}Kp'(s,t) = u_y(s,t) = \exp[k(y-t^2)],\(k = \rho_{air}^{-1})\(5-2)Kp'(s,t) = u_z(s,t) = \exp[k(z-t^2)],\(k = \rho_{air}^{-1})\(5-3)The solution of vector form of (4-2) is:
The solution of vector form of (4-2) can be obtained from (3-19) by changing to , i.e.,
a) Calculation of
Traditionally, methods, tools for calculation of as functions of and , based on Boyles law and Charles law for static description, i.e., they have no connection with wind speed. However, our treatment of is different to that of traditional. It connects with wind speed by "w - p/ equation". (4-2), based on Boyles law, Charles' law and together with the Newton's second law. The calculation of , can be found in Appendix.
VI. SET UP THE "JET - PRESSURE/DENSITY EQUATION" ("J- P/ρ EQ." IN SHORT)
A jet plane flies in a atmospheric environment. In which the absolute motion (the jet plane motion relative a referenced point P on Earth) is equal to the linking motion (wind speed motion relative P) and the relative motion (jet speed motion relative wind motion).
Let the relative motion of the jet plane speed, density of air and pressure be , and . Then, set up the "J- Equation" similar to (4-2), we have:
Where Density of jet plane; is the mass of the jet plane. it varies with time t (jet fuel consuming); is the volume of the jet plane.
(6-2) shows that is in proportion to i.e., the derivative of jet plane speed respect to time , proportions to the derivative of pressure respect to track (space)of the jet plane.
VII. THE SOLUTION OF "J - P/ρ EQUATION" (6-2)
- If, then the solution of(6-2)is the same as(5-5), just by replacing,instead of,, respectively. That is:
The solutions of similar (5-1) - (5-4) are:
- If, is a known function. Wheredepends on flying status, e.g., flying at a constant speed, then
Where for , jet plsne mass of full fuel.; for
plane mass of empty fuel
The solution of (6-2) is the same as (7-1) with inside , i.e.,
Where is instead of by z.
Similar treatment can be used for the scalar form, e.g., (7-5).
VIII. EXAMPLE
MH370—a missing plane [5] is used as an example to show the calculation. Suppose that the plane was crashed due to pressures difference between inside and out side of passenger's cargo. Here, we calculate the pressure out side the cargo. The Boeing
747 cruise speed , cruise high , The masses of jet plane of full fuel and empty fuel are and , respectively.
From (81), we get t.
Calculation of
Substituting (8-2) into (5-3), we have:
Substituting , , and (7-6) into (7-8), we have:
Substituting (6-2) and (8-4) into relative motion and linking motion, We have:
If the moist air instead of the dray air, the density of moist air can be calculated by (A-3) with weight of 18 of water molecular.
Now, the pressure out side the passenger's cargo is calculated by (8-5), and the difference of pressure between inside and out side of passenger's cargo is known.
IX. APPENDIX
a) Density of Dry air
In the atmosphere around us, Nitrogen , Oxygen , and other gases. The N has a molecular weight of 14, so has a molecular weight of 28. Oxygen has a molecular weight of 16, so has molecular weight of 32. Given the mixture of gases of molecules weight is around 29. The total weight of air of is: w = .
Where gravity acceleration. Assumption: is independent with position , . Then, by (A-1), we have:
b) Density of moist air
Water , Hydrogen H is the lightest element and has a molecular weight of 1. So a water molecular weight is . Which shows that the water molecular weight is much lighter than the average weight of the molecular found in air.
The density of moist air can be calculated as the sum of two gases: dry air and water vapor in proportion with their partial proportion .
X. CONCLUSION
- The "w - T eq." is basic. The solution of "w - P/peq." and "J - P/peq." can be obtained directly from "w - T eq."
- The calculation of pressure outside passenger's cargo is used by (6-1) and (6-2).