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1. (30 points). Consider an aircraft climbing steadily in the Earth's atmo- sphere with a constant s ...


1. (30 points). Consider an aircraft climbing steadily in the Earth's atmo- sphere with a constant speed of 1100 km/hr. As the aircraft goes to higher altitudes, the temperature of the air changes approximately linearly with elevation at a rate of -6.6 °C/km. Assume that the temperature at sea level is 25 °C. (a) Compute the Mach number of the aircraft when it reaches an altitude of 3000 m. (b) Compute the Mach number of the aircraft when it reaches an altitude of 15000 m. (c) How does the Mach number of the aircraft vary with elevation? At what altitude does the flow around it become sonic?





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We are given the following information:

  • The speed of the aircraft v=1100 km/hv = 1100 \, \text{km/h}v=1100km/h (which we will need to convert into m/s).
  • The rate of temperature change with altitude is given by −6.6 °C/km-6.6 \, \text{°C/km}−6.6°C/km.
  • The temperature at sea level is T0=25 °CT_0 = 25 \, \text{°C}T0?=25°C.
  • The speed of sound ccc in air varies with temperature according to the formula: c=γRTc = \sqrt{\gamma R T}c=γRT? where:
    • γ=1.4\gamma = 1.4γ=1.4 (ratio of specific heats for air),
    • R=287 J/kg\cdotpKR = 287 \, \text{J/kg·K}R=287J/kg\cdotpK (specific gas constant for air),
    • TTT is the temperature in Kelvin.

Let's first convert the speed of the aircraft into m/s:

v=1100 km/h=1100×10003600=305.56 m/sv = 1100 \, \text{km/h} = \frac{1100 \times 1000}{3600} = 305.56 \, \text{m/s}v=1100km/h=36001100×1000?=305.56m/s

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