EJERCICIO ECUACIÓN BREGUET (2017)

Ejercicio Inglés
Universidad Universidad Politécnica de Cataluña (UPC)
Grado Ingeniería de Aeronavegación - 3º curso
Asignatura Operacions Aèries
Año del apunte 2017
Páginas 3
Fecha de subida 27/08/2017
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Breguet’s equation exercise Air Operations Anna Reig 5GT31 November 11th, 2016 The aim of this exercise is to draw the Payload-Range diagram of a specific aircraft. This leads us to find the three characteristic points of this diagram which corresponds to the Maximum Payload, the Maximum Take-Off Weight and the Maximum Fuel Weight using Breguet’s equation.
The studied aircraft has the following given data.
Weight data Other data MTOW=70t M=0.8 at FL310 OEW=39t TAS=469kts MFW=18t E=13 MZFW=57t FF=2600kg/h MLW=61t T=42000N Breguet’s equation: 𝑊𝑖 𝑅 = 𝑘 · ln ( ) [𝑘𝑚] 𝑊𝑓 where: 𝑘= 𝑉·𝐸 𝑔·𝐶𝑗 [𝑘𝑚], 𝐶 = 𝐹𝐹 𝑇 [ 𝑘𝑔 𝑁·𝑆 ] First, let’s compute the specific fuel consumption and the constant k.
2600𝑘𝑔 1ℎ · 3600𝑠 𝑘𝑔 ℎ 𝐶= = 1.72 · 10−5 [ ] 42000𝑁 𝑁·𝑆 𝑘= 0.5144𝑚 1𝑘𝑚 · 3 · 13 𝑠 10 𝑚 = 18587 [𝑘𝑚] 9.81𝑚/𝑠 2 · 1.72 · 10−5 469𝑘𝑡𝑠 · Now, we can determine the range of each of the three points using the Breguet’s equation.
For the first characteristic point A, the MPL is carried, that is PLA=MZFWOEW=57-39=18t and consequently: 𝑀𝑇𝑂𝑊 70 𝑅 = 18587 · ln ( ) = 18587 · ln ( ) = 3819 [𝑘𝑚] 𝑀𝑍𝐹𝑊 57 For the second characteristic point B, the MTOW is used, so the payload can be computed as: PLB=MTOW-OEW-MFW=70-39-18=13t and consequently: 𝑀𝑇𝑂𝑊 70 𝑅 = 18587 · ln ( ) = 18587 · ln ( ) = 5525 [𝑘𝑚] 𝑀𝑇𝑂𝑊 − 𝑀𝐹𝑊 52 For the third characteristic point C, the MFW is used and no payload is carried, so PLC =0t, and consequently: 𝑀𝐹𝑊 + 𝑂𝐸𝑊 18 + 39 𝑅 = 18587 · ln ( ) = 18587 · ln ( ) = 7054 [𝑘𝑚] 𝑂𝐸𝑊 39 Summarizing the obtained results: Payload (t) Range (km) 18 3819 13 5525 0 7054 ...

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