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Patrick

Tue Jun 10 2014, 05:08PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Uspring wrote ...

The second one is wrong.

im trying to rewreite it in terms of only vwlocity

Ash Small

Tue Jun 10 2014, 05:59PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Patrick wrote ...

Uspring wrote ...

The second one is wrong.

im trying to rewreite it in terms of only vwlocity

If the starting velocity is zero, then it can be written F=mv/t, where v is the velocity of the accelerated air.

Velocity is metres per second, acceleration is metres per second per second, which (I think) is metres per second squared, not velocity squared.

F=mass times metres per second squared, I think. (it's over 30 years since I did this stuff)

Patrick

Tue Jun 10 2014, 06:09PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Ash Small wrote ...

Patrick wrote ...

Uspring wrote ...

The second one is wrong.

im trying to rewreite it in terms of only vwlocity

dam it, that's what I did wrong... ( m/s^2 ) = a

Uspring

Tue Jun 10 2014, 06:10PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

@Bigbad:
Your usage of the symbol m is unconventional. The way you are using it, it denotes a mass rate or mass flow with units of e.g. grams/second. I think, calling this dm/dt is ok.

@Patrick:
Following Bigbads reasoning

dm/dt = rho * A * v, which makes

F = rho * A * vÂ²

@Ash: Even if there would be no vortices, there is a big advantage to large propellers. Thrust is given by the equation above, power consumption by:

P = 1/2 * rho * A * vÂ³

Smaller props need higher v for the same thrust, which makes them less efficient since power consumption rises faster with v than thrust does. Roughly battery lifetime would be about proportional to prop diameter.

Patrick

Tue Jun 10 2014, 06:16PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Uspring wrote ...

Smaller props need higher v for the same thrust, which makes them less efficient since power consumption rises faster with v than thrust does. Roughly battery lifetime would be about proportional to prop diameter.

USping, this was the relationship i was searching for, i want to make or modify props for "high mass induction, low velocity change" for max static thrust at lower power than a conventional prop would allow.

Uspring

Tue Jun 10 2014, 06:51PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

"high mass induction, low velocity change" looks to me like "big props".
Anyway, one can read off

, that the effective propeller area is the cross section of the tail of the slip stream, which is just half as large as the propeller area. This can be avoided by a ducted prop, which keeps the slip stream cylindrical.

Patrick

Tue Jun 10 2014, 07:08PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Uspring, i have read that PDF many times.

and im trying to break recirculation currents with ducts as seen above.

But getting the forces to not wobble the duct apart is hard, much less randomly guessing what the slipstream shape is since i dont have a FEA fluid program.

Ash Small

Tue Jun 10 2014, 07:31PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Uspring wrote ...

@Ash: Even if there would be no vortices, there is a big advantage to large propellers. Thrust is given by the equation above, power consumption by:

P = 1/2 * rho * A * vÂ³

Smaller props need higher v for the same thrust, which makes them less efficient since power consumption rises faster with v than thrust does. Roughly battery lifetime would be about proportional to prop diameter.

Correct me if I'm mistaken, but the same equation is used to calculate the 'drag' of aerodynamic vehicles, but in this case is multiplied by the 'drag coefficient' (for most cars, around 0.3). If (and I'm making an assumption here), the 'drag coefficient' here is 1, then doesn't this equation relate to the 'peripheral vortex losses' (drag) that I was referring to?

EDIT: I'm not sure I have all the terminology correct here, but what I mean is, it's the column of air being accelerated here (comparable to the vehicle), and it's the 'drag' of this column of air that results in the vast majority of losses (comparable to the drag losses of the vehicle), at least, that's what I always understood.

Uspring

Tue Jun 10 2014, 08:29PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

The physics for drag and propellers are similar. Both are based on acceleration of air masses.

The above equation is based on the kinetic energy of the air. P would be the minimum power which needs to be put into the propeller to accelerate the air. This equation assumes a 100% efficient prop.

Ash Small

Tue Jun 10 2014, 08:43PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Uspring wrote ...

This equation assumes a 100% efficient prop.

That's what I assumed. I edited the previous post as you replied, not sure if you noticed.

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