Confusion with units for Power Required Formula

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Hi all,

I'm having trouble with the power required formula for steady level flight.

Here's the formula:

Preq = 4/3*W*sqrt(W/S*2/ρ*sqrt(3*CD0/(π*ε*A)))

I'm using this for a small indoor RC fixed wing. The values I'm using are as follows:
Weight, W = 0.3924N
Wing Area, S = 0.073 m^2
Density, ρ = 1.226 kg/m^3
Zero lift drag, CD0 = 0.03
Span efficiency, ε = 0.9
Aspect Ratio, A = 15

These values are giving me a power required of 0.051 W, which is almost exactly 1000 times less than what I know the electric motor uses, since it holds SLF at half throttle and the max power rating for the motor is 100W.

Where am I going wrong? I know I've gotten my units mixed up but I can't work out where. Help!

:/
 

Answers and Replies

  • #2
AlephZero
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Your weight of 0.392N corresponds to a mass of about 40 grams or 1.5 ounces.

An electric motor producing 100W with a mass as small as that (plus the mass of the battery to power it) doesn't seem very plausible to me.
 
  • #3
jack action
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I agree with AlephZero. Also, there is something to say about your formula and calculations. When you plug in the numbers into the formula given, it doesn't return 0.051 W. But it does with this one:

Pmin = (1/2 * (4/3)¾ + 1 / (1/2 * (4/3)¼) ) * sqrt(W³/ρ/S * sqrt(CD0/(π*ε*A)³))

This represents the minimum power required (Preq) for SLF. It happens only at a certain velocity Vminimum power. At any other velocity, it will take more power. You can find the formulae for both here and combine them to obtain the previous equation.
 
  • #4
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I agree with AlephZero. Also, there is something to say about your formula and calculations. When you plug in the numbers into the formula given, it doesn't return 0.051 W. But it does with this one:

Pmin = (1/2 * (4/3)¾ + 1 / (1/2 * (4/3)¼) ) * sqrt(W³/ρ/S * sqrt(CD0/(π*ε*A)³))

This represents the minimum power required (Preq) for SLF. It happens only at a certain velocity Vminimum power. At any other velocity, it will take more power. You can find the formulae for both here and combine them to obtain the previous equation.
There was a typo in my formula, sorry. It's supposed to be:
Preq,min = 4/3*W*√(W/S*2/ρ*√(3*CD0/(π*ε*A)3)), which is equivalent to what you are using.

The reason I'm confused is that a power drain of 0.051W would take 70.8 hours to deplete the battery I'm using (500mAh or 13 000 J), whereas 51W depletes it in 4.24 minutes, which is about the flight time of this aircraft, give or take.

Here's the motor I'm using: http://www.hobbyking.com/hobbyking/store/uh_viewItem.asp?idProduct=17591 [Broken]
Max Current= 14A, Max Voltage = 7V, therefore max power consumption is 98W ~ 100W
It produces about 0.4N thrust with the GWS prop i'm using.
 
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  • #5
jack action
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You're missing the point. According to the hobby site and the wing dimensions you gave, your plane mass is easily between 2 and 3 kg, which represents a weight between 19.62 and 29.43 N (W = mg).

Putting that in the equations gives:

For W = 19.62 N, Pmin = 18.05 W @ V = 15 m/s;

For W = 29.43 N, Pmin = 33.17 W @ V = 18.3 m/s.

At any other speed (slower or faster), the power consumption will increase.
 
  • #6
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You're missing the point. According to the hobby site and the wing dimensions you gave, your plane mass is easily between 2 and 3 kg, which represents a weight between 19.62 and 29.43 N (W = mg).

Putting that in the equations gives:

For W = 19.62 N, Pmin = 18.05 W @ V = 15 m/s;

For W = 29.43 N, Pmin = 33.17 W @ V = 18.3 m/s.

At any other speed (slower or faster), the power consumption will increase.
Sorry, your explanation isn't helping. My plane weighs 40 grams, and depletes a 500mAh battery in 5 minutes during SLF.
 
  • #7
jack action
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You will have to describe your airplane in more details because something doesn't add up.

Your motor weight is 23 g. A 500 mAh battery, 7.4 V (it is the voltage you use for your calculations) weight around 30 g. That is already 53 g. Even using a 450 mAh, 3.7 V battery, the weight is still 11 g, which combined to your motor gives a total of 34 g.

With the wing area (0.073 m²) and aspect ratio (15) given, your wingspan should be 1046 mm.

The only airplanes I found that are close to your specs are these one:

Micro F4 jet: 100 g, 7.4 V-500 mAh battery, 540 mm wingspan;
Micro 3D Piaget: 50 g, 100 mAh to 200 mAh battery, 418 mm wingspan;
Freeflight P-26: 41 g, rubber band powered, wingspan 466 mm.

All of these airplanes seem to be about half the size of yours.

Another point about the motor specs. The maximum efficiency is 78% @ about 7.5A, which give an output of 43 W @ 7.4V. The efficiency will be lower at 14 A, so you will never obtain an output of 98 W.

Finally, with the specs you gave, the speed for minimum power should be 2.11 m/s, which is slightly faster than a person walking. How do you evaluate your speed?
 
  • #8
AlephZero
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The maximum efficiency is 78% @ about 7.5A, which give an output of 43 W @ 7.4V.
If you are discharging a 500mAh battery in 5 minutes, that is an average current of 6A.

So at least we have SOME numbers that make sense.
 
  • #9
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You will have to describe your airplane in more details because something doesn't add up.

Your motor weight is 23 g. A 500 mAh battery, 7.4 V (it is the voltage you use for your calculations) weight around 30 g. That is already 53 g. Even using a 450 mAh, 3.7 V battery, the weight is still 11 g, which combined to your motor gives a total of 34 g.

With the wing area (0.073 m²) and aspect ratio (15) given, your wingspan should be 1046 mm.

Finally, with the specs you gave, the speed for minimum power should be 2.11 m/s, which is slightly faster than a person walking. How do you evaluate your speed?
I'll give you all of the specs for the plane in question. It's scratch-built, so you won't find it online anywhere.

I've been using approximations up until now, but no approximation should knock my power required out by 100x...

Mass: 23g(motor) + 36g (battery) + 10g(airframe) + 20g(All other electronics) = 89g (more than twice what I quoted)
Wing Area: 0.09m^2
Aspect Ratio: 4 (much much lower than what I quoted)
Span efficiency: ~0.7 (assumed)
CD0: ~0.03 (assumed)
density: 1.226 kg/m^3 (Sea Level Conditions)
Battery capacity: 500 mAh (~13 000 J)
Average flight time: 4-5 mins
Average loiter speed: 2-3 m/s (d/t measured at an indoor basketball court)

With these values, I get Preq of 0.4966 W, which is still too small!
 
  • #10
jack action
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I don' t think 0.5 W is too small. If you use the equation for the power required at a speed of 2.5 m/s, you get 0.65 W (because with the new data, the speed for Pmin is now 4.2 m/s).

You said in an earlier post that your propeller produced 0.4 N of thrust. Combined with a speed of 2.5 m/s, it gives a power of 1 W for your airplane. So we are still in that region, far from the 50 W you are expecting.

Is your battery OK? Rechargeable batteries can have problems if they are not cared for properly (example). From Wikipedia about LiPo batteries:
During discharge on load, the load has to be removed as soon as the voltage drops below approximately 3.0 V[citation needed] per cell (used in a series combination), or else the battery[citation needed]will subsequently no longer accept a full charge and may experience problems holding voltage under load. Li-poly batteries can be protected by circuitry that prevents over-charge and deep-discharge.
If it is NiCd battery, it can have a memory effect.
 
  • #11
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I don' t think 0.5 W is too small. If you use the equation for the power required at a speed of 2.5 m/s, you get 0.65 W (because with the new data, the speed for Pmin is now 4.2 m/s).

You said in an earlier post that your propeller produced 0.4 N of thrust. Combined with a speed of 2.5 m/s, it gives a power of 1 W for your airplane. So we are still in that region, far from the 50 W you are expecting.
The problem is this. My battery has 500mAh capacity, which, at a voltage of 7.4V, give a capacity of 0.5*7.4 = 3.7 Watt-hours. If my motor only draws 0.5W, then the battery should last for 7.4 hours! Something must be wrong here. Am I calculating my battery capacity wrong?
 

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