Understanding propulsive efficiency

1. Aug 28, 2015

msat

Hello,

I've been trying to wrap my head around this topic and came across such a thread here: https://www.physicsforums.com/threads/propulsive-efficiency.643646/ but still dont get it. Why is it that that accelerating one unit of mass by two units less efficient than accelerating two units of mass by one unit? Are the energy requirements non-linear as compressibility effects of a gas come into play? Or is it less efficient to extract work from a compressed gas? Otherwise, from my understanding of Newton's laws of force and motion state there should be no difference. What am I missing?

Thanks,
Mark

2. Aug 28, 2015

Staff: Mentor

Welcome to PF!

The difference is actually pretty simple: look at the equations for momentum and kinetic energy:
p = mv
e = .5mv^2

Momentum change directly translates into force and energy change into power. But since energy is a square function of velocity, a higher velocity and lower mass flow means more energy input for the same thrust. This is why helicopters have such large rotors.

3. Aug 30, 2015

msat

Thanks, Russ

The equations are simple enough, but I'm still confused

I guess I don't grasp how force, momentum, and kinetic energy are all related. It would seem that p and e describe something similar, but alas they are different. A mass can have both momentum attributed to it as well as kinetic energy, but their values are different. And yet, if f = ma, even if you switch the two terms, the force stays the same, so according to that, for a given force, you can double the velocity if you halve the mass, and vice versa. Looking into it a bit more, it seems that the discrepancy is between force and work. I'm still trying to wrap my head around these. Is force instantaneous whereas work is not?

Thanks,
Mark

4. Sep 1, 2015

SCP

Hi msat.
Propulsive efficiency is a measure of how much energy is transmitted to the aircraft versus how much is left behind in the airstream. You're right about changing up the massflow and the velocity, so run with that concept and see where it leads:

Consider a high-massflow propulsion system producing thrust and a low-massflow propulsion system producing the same amount of thrust with exactly half of the massflow. To produce the same amount of thrust, the low-massflow system will need to eject the airflow at twice the velocity (just as you noted above). The kinetic energy left in the airstream is then halved by the reduced massflow, but quadrupled by the increased velocity -- for a net effect of doubling the wasted energy. This wasted energy is ultimately dissipated as heat in the atmosphere.

5. Sep 7, 2015

msat

This is the bit confusing to me, as while I understand the energy may be transmitted to the atmosphere above ambient, why would it matter, since as stated by Newton, every action has an equal and opposite reaction, thus work(?) was extracted in equal but opposite amounts. I know what you and Russ say is true, as can be observed by his example of a helicopter rotor (and I'd add any large airfoil such as a wing as an example an example of an efficient propulsion source). I just don't see how momentum and kinetic energy describe similar, but different things.

Besides the equation for kinetic energy, is there perhaps a more intuitive way to explain why energy increases with the square of the velocity? Basically, why does doubling the the velocity require 4x the energy?

Thanks,
Mark

6. Sep 8, 2015

SCP

The error you're making here is equating force and work. They aren't the same.

Think of propulsion as the airflow exerting a force on the aircraft and the aircraft exerting an equal and opposite force on the airflow. So far so good. But force is not work. In order to get work, we need to multiply the force by the distance travelled. The distance travelled in a given amount of time depends on speed.

Ok, so the aircraft and the airflow are moving at equal and opposite speeds, right? Nope. The aircraft is travelling at a speed that is equal and opposite to the freestream flow. But the portion of airflow being used to generate thrust is not the freestream - it's being influenced by the engine and/or prop. This portion of the flow is moving faster than the aircraft by virtue of the momentum imparted by the propulsion unit. The faster it's moving, the further it travels in a given time, and the more energy it takes away with it.

So it's possible to have equal and opposite forces without equal and opposite energy transfers (work).

Rather than thinking of energy and momentum as "similar, but different", think of them as "related, but different". They really aren't similar. One is a vector, the other is a scalar. One is related to force, the other to work (i.e. - energy transferred). One is linear, the other quadratic.

I can take a run at this. Start by sketching a graph of momentum (mv) versus speed (v): v on the horizontal axis, mv on the vertical. What you'll get is a straight line that passes through the origin (0,0). The kinetic energy is represented on this graph as the area under the curve (line) from v=0 to v=whatever speed you're interested in. Because the curve is a straight line through the origin, the area under it is the area of a triangle: A = 1/2 (base)(height). The base is v, and the height is mv, so KE = A = 1/2 (v)(mv), which reduces to KE = 1/2mv^2.

I'm not sure how much more intuitive I can make it with a forum post. The leap of logic here is accepting that the energy is in fact the area under the curve. This assumption/definition is justified if we dig a bit deeper into the definition and mathematics of work and energy.

You can also think of it this way: momentum is determined from the duration of a force application. Energy is determined from the distance covered during a force application. For a given change of momentum (force applied for a fixed time), the work done increases at higher speeds (given time --> larger distance).