Lecture P4: Connecting Aircraft Performance and Engine Requirements

I covered all of Chapter IV of the notes. The most important concepts are 1) given aircraft aerodynamic and weight information you can calculate thrust required for steady level flight, 2) flight at max range requires flying at a condition that minimizes drag (for a given weight--i.e. that maximizes L/D), flight at max endurance requires flying at a condition that minimizes power required (drag times velocity), and 3) Power available minus power required = time rate of change of potential plus kinetic energy. Requirements for T/W are set by maneuverability requirements. These can be found knowing aircraft weight and aerodynamic information. Note that there are typically three kinds of calculations requested for manueverability: a) steady climbing flight (so no acceleration term), b) constant altitude acceleration (no term for change in potential energy), and c) steady, constant altitude turning flight (no terms for acceleration or change in potential energy, but drag increases in response to increased lift required to overcome centrifugal acceleration).

We did two PRS questions (PRS #1, PRS#2). The first was designed to test your understanding of the conditions under which endurance is maximized (the goal for your dragonflies). The second was designed to give you practice with a manueverability calculation.

Next lecture we will discuss rocket performance. Please read Chapter V and VI of the notes.

Responses to 'Muddiest Part of the Lecture Cards'

(8 respondents)

1) How can you measure overall efficiency if propulsive efficiency is zero? (1 student) If the propulsive efficiency is zero, then the overall efficiency is zero.

2) Why are the energy components equal to zero for steady turning flight? Isn't there acceleration? (1 student) You will get some practice with this on your next homework assignment. For steady turning flight,there is no change in kinetic energy or potential energy, but drag increases in response to increased lift required to overcome centrifugal acceleration. Therefore, the power required increases.

2) I don't understand why power available is equal to TV and power required is equal to DV. I thought that TV was power consumed?? (1 student) TV is the thrust times the velocity, so it is the power provided to the vehicle.

3) No mud (3 students). Good.

3) Thanks for the donuts (2 students). You are welcome!