constant. Both potential and kinetic energy can change in value, but the total energy must remain the same. For example, when a ball is thrown upward, if the height of the thrower is the reference plane, its energy is all kinetic when it leaves the thrower’s hand. As it rises, PE is continually increasing, but KE is always decreasing by the same amount, so the sum remains constant. At the top of its travel, PE is at its maximum (the same amount as the KE it had when it left the thrower’s hand) and KE is zero. Energy cannot be created or destroyed, but can change in form.
EXAMPLE
An aircraft that weighs 15 000 lb is flying at 10 000 ft altitude at an airspeed of 210 kts. Calculate the potential energy, kinetic energy, and the total energy.
PE: PE = Wh → PE = 15000 lb × 10000 ft → PE = 1.5 × 108
KE:
Total Energy: TE = PE + KE → TE = 1.79 × 108
Application 1.2
Consider a general aviation airplane that weighs 3000 lb with a designated approach speed over the runway threshold of 65 kts., calculate the KE. Now, consider if that same airplane approaches the runway with an extra 10 kts. of speed due to poor planning, calculate the new KE.
Why does only a 10 kts. change in approach speed result in such a wide margin of KE? What are the consequences of this “extra” energy?
POWER
In our discussion of work and energy, we have not mentioned time. Power is defined as “the rate of doing work” or work/time. We know:
and
James Watt defined the term horsepower (HP) as 550 ft‐lb/s:
If the speed is measured in knots, Vk, and the force is the thrust, T, of a jet engine, then
EXAMPLE
An aircraft’s turbojet engine produces 8000 lb of thrust at 180 kts., what is the equivalent horsepower that engine is producing?
Equation 1.13 is very useful in comparing thrust‐producing aircraft (turbojets) with power‐producing aircraft (propeller aircraft and helicopters); a more detailed discussion will follow in future chapters.
Application 1.3
Consider the example calculation provided to solve for horsepower (HP).
Would the horsepower remain the same if the thrust remained 8000 lb but the aircraft slowed to a speed of 160 kts.? Why or why not? How can the equation be altered to solve for thrust (T) if an aircraft was maintaining a constant speed with a known HP?
FRICTION
If two surfaces are in contact with each other, then a force develops between them when an attempt is made to move them relative to each other. This force is called friction. Generally, we think of friction as something to be avoided because it wastes energy and causes parts to wear. In our discussion on drag, we will discuss the parasite drag on an airplane in flight and the thrust or power to overcome that force. Friction is not always our enemy; however, without it there would be no traction between an aircraft’s tires and the runway. Once an aircraft lands, lift is reduced and a portion of the weight contributes to frictional force. Depending on the aircraft type, aerodynamic braking, thrust reversers, and spoilers will be used to assist the brakes and shorten the landing, or rejected takeoff distance.
Figure 1.9 Coefficients of friction for airplane tires on a runway.
At the microscopic level, as in the surface of a wing, friction causes resistance and slows down the velocity of the air as it passes over it. The layer of air that is impacted by the friction of the wing, or any other surface of the aircraft, is referred to as the boundary layer.
Several factors are involved in determining friction effects on aircraft during takeoff and landing operations. Among these are runway surfacing material, condition of the runway, tire material and tread, and the amount of brake slippage. All of these variables determine a coefficient of friction μ (mu). The actual braking force, Fb, is the product of this coefficient μ (Greek symbol mu) and the normal force, N, between the tires and the runway (Eq. 1.14):
Figure 1.9 shows typical values of the coefficient of friction for various conditions. Note the value of μ for dry concrete is ~0.7 with ~10% wheel slip, while the μ on smooth, clear ice is ~0.2. This means that an airplane wheel rolling on smooth, clear ice will experience much lower friction (increased stopping distance) than a wheel rolling on dry concrete.
EXAMPLE
Calculate the braking force on dry concrete when the normal force (N) is 2000 lb.
SYMBOLS
| a | acceleration (ft/s2) |
|---|---|
| E | Energy (ft‐lb) |
| KE | Kinetic energy (ft‐lb) |
| PE | Potential energy (ft‐lb) |
| TE | Total energy (ft‐lb) |
| F | Force (lb) |
|
F
b
|
