Oct 14, 2012, 06:25 PM Registered User United States, UT, Salt Lake City Joined Oct 2007 8,216 Posts Apparantly -it seems necessary to ferret out the obvious Being a simple soul- I just look for energy being used - that resolves the "work " question. A hovering model is not magic - it uses power via a prop moving air - It matters not a whit to the battery if the model is moving or not -- Static thrust readings - are a very important part of propeller design- I use the readings to determine maximum possible power absorbed by a particular prop/motor . In the air -it will always use less - theorize -proove me wrong.
Oct 14, 2012, 09:20 PM
Sink stinks
United States, GA, Atlanta
Joined Apr 2005
4,628 Posts
Quote:
 Originally Posted by TangoKilo Gentlemen, this is becoming a bit silly. What on earth is “parasite power” or “profile power”?
Just terminology. Parasite power is power to overcome parasite drag (of the fuselage, etc.), induced power is the power required by the prop to accelerate the air, and profile power is the power due to profile drag of the propeller blades.

Quote:
 Originally Posted by TangoKilo Similarly, the propulsion system of a model moving at maximum speed is also producing thrust. The thrust is that required to equal the drag at that speed and is a factor of the output power.
True, and that's why the maximum speed of the ducted fan model must be less than its efflux velocity. Because if the speed is equal to the efflux velocity, no thrust is being produced. If the model were going as fast as the efflux velocity, drag would quickly slow it down until thrust equaling the drag was produced.
Oct 15, 2012, 12:00 AM
Registered User
Germany, BW, Stuttgart
Joined Mar 2012
1,029 Posts
Quote:
 Originally Posted by TangoKilo Incidently, shaft power is torque times angular velocity, not RPM.
Last I checked RPM was a measure of angular velocity
 Oct 15, 2012, 02:25 AM Registered User Germany, BW, Stuttgart Joined Mar 2012 1,029 Posts An airplane in level flight, whether propelled by a ducted fan, propeller, jet, or rocket, requires a certain amount of power to maintain constant speed. That power is equal to the airplane’s true airspeed times the drag acting on the airplane: Power Required = Drag*TAS. An airplane moving at a true airspeed of 10 m/s with a drag force of 10 Newtons acting on it will require 100 Watts of power to maintain constant speed in level flight. No propulsion system, regardless of its efficiency, will be able to keep the airplane level at 10 m/s unless its energy source is capable of providing at least 100 Watts. No propulsion system is 100% efficient, which means that the energy source must provide more than 100 Watts to keep the airplane at constant speed. Suppose we have a ducted fan driven by an electric motor, and we determine the motor is drawing 10 Amps at 12 Volts to keep the airplane in level flight at 10 m/s. The net propulsive efficiency of the ducted fan is: 100 Watts delivered to the airplane / 120 Watts consumed by the propulsion system = 83% efficiency. If the airplane weighs 10 Newtons and slows to a hover, the ducted fan will still need to provide 10 Newtons of thrust to keep the airplane level, but the Power Required (Drag*TAS) goes to zero. This doesn’t mean no power is required, it means no power would be required of a 100% efficient propulsion system. Since no (real) propulsion system is 100% efficient, the electric motor will still need to draw power from the battery to keep the airplane level. Suppose it draws 80 Watts in a hover: 0 Watts delivered to the airplane / 80 Watts consumed by the propulsion system = 0% efficiency. The distinction between induced power and profile power has to do with breaking down efficiency losses of the propulsion system into the underlying mechanisms. (induced losses due to kinetic energy supplied by the propulsion system to the air, and profile losses due to the profile drag acting on the blades).
Oct 15, 2012, 04:13 AM
Registered User
Joined Oct 2012
45 Posts
Quote:
 Originally Posted by ShoeDLG Last I checked RPM was a measure of angular velocity
It is, but radians/second is used to calculate shaft power.
 Oct 15, 2012, 04:20 AM Registered User Joined Oct 2012 45 Posts [QUOTE=Montag DP;23002637] Because if the speed is equal to the efflux velocity, no thrust is being produced. Mon, you need to show some maths or something to back that up. I, and many others, will totally disagree with that statement.
Oct 15, 2012, 05:18 AM
Registered User
Joined Oct 2012
45 Posts
Quote:
 Originally Posted by ShoeDLG An airplane moving at a true airspeed of 10 m/s with a drag force of 10 Newtons acting on it will require 100 Watts of power to maintain constant speed in level flight. No propulsion system, regardless of its efficiency, will be able to keep the airplane level at 10 m/s unless its energy source is capable of providing at least 100 Watts. No propulsion system is 100% efficient, which means that the energy source must provide more than 100 Watts to keep the airplane at constant speed. Suppose we have a ducted fan driven by an electric motor, and we determine the motor is drawing 10 Amps at 12 Volts to keep the airplane in level flight at 10 m/s. The net propulsive efficiency of the ducted fan is: 100 Watts delivered to the airplane / 120 Watts consumed by the propulsion system = 83% efficiency. No argument against that except to say that, in your example, the propulsion system must provide 100 Watts worth of moving air to keep the airplane at constant speed. If the airplane weighs 10 Newtons and slows to a hover, the ducted fan will still need to provide 10 Newtons of thrust to keep the airplane level, but the Power Required (Drag*TAS) goes to zero. This doesn’t mean no power is required, it means no power would be required of a 100% efficient propulsion system. Since no (real) propulsion system is 100% efficient, the electric motor will still need to draw power from the battery to keep the airplane level. Suppose it draws 80 Watts in a hover: 0 Watts delivered to the airplane / 80 Watts consumed by the propulsion system = 0% efficiency.
You are changing the frame of reference again. Your model airoplane contains a propulsion unit which moves air. The power output is calculated from the amount of air that is being moved. It doesn't matter if the frame of reference is moving in space.

If your system is drawing 80 Watts in a hover and is (for example) 75% efficient then there are 60 Watts worth of air being moved. That is a power output.

Richard Hanson has the right idea, quote,
"- it uses power via a prop moving air - It matters not a whit to the battery if the model is moving or not -".

You can not say that because the model is not moving there is no power output.
Last edited by TangoKilo; Oct 15, 2012 at 05:23 AM.
 Oct 15, 2012, 06:06 AM Registered User Germany, BW, Stuttgart Joined Mar 2012 1,029 Posts By referring to the airplane’s True Airspeed, the frame of reference remains the same in level flight or a steady hover: that being the frame of reference where the undisturbed air is at rest. You’re defining the efficiency of the propulsive system as: Power supplied to increase the kinetic energy of the air / Power supplied to the propulsion system That’s a way to define efficiency, but that’s not the standard definition for steady flight which is: Power supplied to keep the airplane moving (Drag*TAS) / Power supplied to the propulsion system One issue with your definition of efficiency can be illustrated with the following example: suppose you have two different systems that increase the kinetic energy to the air at the same rate (call it 60 Watts) and draw the same power (call it 80 Watts). It is entirely possible to have two such systems that, despite having the same kinetic energy transfer rate and power draw, produce different amounts of thrust. If you used these propulsion systems to power two otherwise identical airplanes in level flight, the airplanes would reach different steady state airspeeds (due to the difference in thrust). The airplane that reached a faster speed would be getting more power from the propulsion system (Drag1*TAS1) than the airplane that reached the slower airspeed (Drag2*TAS2)*. It makes sense to consider the propulsion system of the faster airplane more efficient because it is supplying more power to the airplane for the same power consumption. Your definition of propulsive efficiency would say that the efficiencies are the same. *assuming they are flying faster than the minimum drag airspeed.
 Oct 15, 2012, 06:19 AM Registered User Joined Oct 2004 3,103 Posts If the airflow through the tailpipe was slower than the airstream surrounding the plane, the plane would essentially suck air in through the tailpipe. How would that produce thrust? Extend for all possible combinations and integrate.
Oct 15, 2012, 08:29 AM
Sink stinks
United States, GA, Atlanta
Joined Apr 2005
4,628 Posts
Quote:
 Originally Posted by TangoKilo Mon, you need to show some maths or something to back that up. I, and many others, will totally disagree with that statement.
Mark Drela already showed you the math. It comes from a very basic application of Newton's 1st and 2nd laws to a control volume surrounding the propeller. If you refuse to accept that analysis, why should I waste my time trying to convince you?
Oct 15, 2012, 08:34 AM
Registered User
United States, UT, Salt Lake City
Joined Oct 2007
8,216 Posts
Quote:
 Originally Posted by ShoeDLG By referring to the airplane’s True Airspeed, the frame of reference remains the same in level flight or a steady hover: that being the frame of reference where the undisturbed air is at rest. You’re defining the efficiency of the propulsive system as: Power supplied to increase the kinetic energy of the air / Power supplied to the propulsion system That’s a way to define efficiency, but that’s not the standard definition for steady flight which is: Power supplied to keep the airplane moving (Drag*TAS) / Power supplied to the propulsion system One issue with your definition of efficiency can be illustrated with the following example: suppose you have two different systems that increase the kinetic energy to the air at the same rate (call it 60 Watts) and draw the same power (call it 80 Watts). It is entirely possible to have two such systems that, despite having the same kinetic energy transfer rate and power draw, produce different amounts of thrust. If you used these propulsion systems to power two otherwise identical airplanes in level flight, the airplanes would reach different steady state airspeeds (due to the difference in thrust). The airplane that reached a faster speed would be getting more power from the propulsion system (Drag1*TAS1) than the airplane that reached the slower airspeed (Drag2*TAS2)*. It makes sense to consider the propulsion system of the faster airplane more efficient because it is supplying more power to the airplane for the same power consumption. Your definition of propulsive efficiency would say that the efficiencies are the same. *assuming they are flying faster than the minimum drag airspeed.
Apparantly we all see questions a bit differently -
Little Mary asked "Mama, where did I come from?"
Mama gave her a condensed birds n bees reply.
Mary, being a bit puzzled said " "my friend Goldie said she came from Cleveland.".
Oct 15, 2012, 09:08 AM
Registered User
Germany, BW, Stuttgart
Joined Mar 2012
1,029 Posts
Quote:
 Originally Posted by richard hanson Apparantly we all see questions a bit differently - Little Mary asked "Mama, where did I come from?" Mama gave her a condensed birds n bees reply. Mary, being a bit puzzled said " "my friend Goldie said she came from Cleveland.".
And that's OK to an extent. But realize that your definition of efficiency has some "features" that are hard to reconcile.

Perhaps a better example...

Suppose you have an airplane powered by two ducted fans... one on each wing. Imagine you "cant" both engines inward by 45 degrees. I submit that the airplane will not go nearly as fast for the same power consumption, but that the rate of kinetic energy transfer to the air will be largely unchanged by canting the engines.

According to the standard definition of the net efficiency of the propulsion system (the ratio of power delivered to move the airplane through the air to the power supplied to the propulsion system), canting the engines inward reduces the net propulsive efficiency.

According to your definition of the efficiency of the propulsion system (the ratio of the rate of kinetic energy transfer to the air to the power supplied to the propulsion system), canting the engines inward has essentially no effect on the propulsive efficiency. I think it's pretty clear that's not a desireable feature of your definition.

I'm trying to picture the sales pitch:
Customer: "So airplane X uses twice the fuel to get from point A to point B, but has the same efficiency as airplane Y?"
Salesman: "Yeah, you've really got to see how this baby moves the air around."
Last edited by ShoeDLG; Oct 15, 2012 at 10:06 AM.
 Oct 15, 2012, 09:22 AM Registered User United States, UT, Salt Lake City Joined Oct 2007 8,216 Posts Again- two different questions Q1- is the propulsive system operating correcty Q2 was the propulsive system correctly applied . If we really want to nit pick- a pair of canted "jets" is desireable if one wants to use them to correct yaw issues. But that is yet another question. In doing far too many aerobatic designs and resolving issues of coupling etc.. I tend to compartmentalize each issue. THEN fix as req'd.
 Oct 15, 2012, 09:32 AM Registered User Germany, BW, Stuttgart Joined Mar 2012 1,029 Posts The canted jets example is just a construction to illustrate that you can have the same rate of kinetic energy addition to the air, and the same power consumption, but different net thrust. I think most people will be able to generalize this example, and recognize that it highlights a serious issue with your definition of efficiency.
Oct 15, 2012, 10:04 AM
Registered User
United States, UT, Salt Lake City
Joined Oct 2007
8,216 Posts
Quote:
 Originally Posted by ShoeDLG The canted jets example is just a construction to illustrate that you can have the same rate of kinetic energy addition to the air, and the same power consumption, but different net thrust. I think most people will be able to generalize this example, and recognize that it highlights a serious issue with your definition of efficiency.
I understand your viewpoint- I just look at the question differently.