An airplane flies at an altitude of 40,000 ft (temperature= -67, pressure=2.72 psia) with a speed of 400 mph. Neglecting frictional effects, calculate the critical velocity of the air relative to the aircraft and the maximum possible velocity of the air relative to the aircraft.
The correct answer and explanation is :
To determine the critical velocity and maximum possible velocity of air relative to the aircraft, we use compressible flow equations.
Step 1: Given Data
- Altitude = 40,000 ft
- Temperature (T = -67^\circ F = 460 – 67 = 393^\circ R)
- Pressure (P = 2.72) psia
- Aircraft speed (V = 400) mph
Step 2: Critical Velocity
The critical velocity is the velocity at which the flow reaches Mach 1 (sonic conditions) in an isentropic flow.
The formula for critical velocity is:
[
V^* = a^* = \sqrt{\frac{2 \gamma}{\gamma + 1} R T}
]
where:
- ( \gamma = 1.4 ) (ratio of specific heats for air)
- ( R = 1716 ) ft·lb/(slug·°R) (gas constant for air)
- ( T = 393^\circ R )
[
V^* = \sqrt{\frac{2(1.4)}{(1.4+1)} \times 1716 \times 393}
]
[
V^* = \sqrt{\frac{2.8}{2.4} \times 1716 \times 393}
]
[
V^* = \sqrt{1.167 \times 673428} = \sqrt{785930.7} \approx 887.6 \text{ ft/s}
]
Converting to mph:
[
V^* = \frac{887.6 \times 3600}{5280} \approx 605.4 \text{ mph}
]
Step 3: Maximum Possible Velocity (Velocity at M=∞)
The maximum velocity occurs when the flow is completely expanded to vacuum conditions ((P_0 \rightarrow 0)), given by:
[
V_{\max} = \sqrt{\frac{2 \gamma}{\gamma – 1} R T_0}
]
where ( T_0 ) is the stagnation temperature:
[
T_0 = T \left(1 + \frac{\gamma – 1}{2} M^2 \right)
]
First, calculate Mach number:
[
a = \sqrt{\gamma R T} = \sqrt{1.4 \times 1716 \times 393} = \sqrt{944799.6} \approx 972 \text{ ft/s}
]
[
M = \frac{400 \times 1.467}{972} \approx 0.603
]
Now, stagnation temperature:
[
T_0 = 393 \left(1 + \frac{0.4}{2} \times 0.603^2 \right)
]
[
T_0 = 393 \times (1 + 0.0724) = 393 \times 1.0724 \approx 421.4^\circ R
]
Now, calculating ( V_{\max} ):
[
V_{\max} = \sqrt{\frac{2(1.4)}{(1.4-1)} \times 1716 \times 421.4}
]
[
V_{\max} = \sqrt{\frac{2.8}{0.4} \times 1716 \times 421.4}
]
[
V_{\max} = \sqrt{11.9 \times 723722.4} = \sqrt{8612287} \approx 2935.6 \text{ ft/s}
]
Converting to mph:
[
V_{\max} = \frac{2935.6 \times 3600}{5280} \approx 2002 \text{ mph}
]
Final Answers
- Critical Velocity = 605.4 mph
- Maximum Velocity = 2002 mph
Explanation (300 words)
The critical velocity represents the speed at which the local airflow becomes sonic (Mach 1). This velocity depends on temperature and is derived from the speed of sound in compressible flow. At 40,000 ft, the temperature is -67°F, leading to a critical velocity of approximately 605.4 mph. This means that if the local flow speed reaches this value, it will become sonic.
The maximum possible velocity represents the upper limit of speed if the airflow expands ideally to vacuum conditions. It is derived from the energy equation, considering isentropic flow conditions. This velocity is significantly higher, calculated to be around 2002 mph. This represents the theoretical maximum speed that the air can attain relative to the aircraft in an idealized scenario.
These calculations are crucial in aerospace engineering, as they help determine shock wave behavior and aircraft performance at high altitudes. For supersonic and hypersonic aircraft, understanding these limits ensures proper aerodynamic design and stability.