An electric motor is pulling some amount of energy from the battery. The model climbs at speed which is much less than it would correspond to this battery power. About 2/3 of the energy dissipates on the way. Where it goes to?
The data and calculation in this article are based on the Rook – motor MVVS 3.5/960, propeller Ae12x6.5, 3S1600, wingspan 1900 mm, area 44 dm2, airfoil E 178, and AUW 1000 g, measured battery power 275 W, and measured rate of climb (RoC) 10 m/s.
I developed a kind of the Calculator. The motor parameters are calculated using the Drivecalc (www.drivecalc.de).
The propeller model is based on the NACA Report No. 14 by William Durand, 1918 (see NASA archive at http://ntrs.nasa.gov/search.jsp) which is the only source of information on propellers I have been able to find. The program uses the efficiency curves (for propellers Nos 3, 7, and 11) determined from this report.
Sorry, the diagrams herein are not translated but I will do it here. On the horizontal co-ordinate is the ratio of the speed of flight to the product of RPM and propeller diameter, on the vertical co-ordinate is propeller eficiency. There are three curves for various pitch/diameter ratio. The green line is an „envelope“ to the curves and represents a maximal achievable propeller efficiency (General efficiency curve).
The propeller absorbed power is also from the Drivecalc, the propeller unloading during the flight is derived from this NACA report.
The airframe performance is then calculated using known procedures (if anybody is interested in the Excel calculation sheet, please let me know).
So, the motor/propeller data from the Drivecalc :
For voltage 11.1 V, and aeronaut propellers, it returns following static (at v=0) information:
|Propeller||Current (A)||RPM (1/min)||Efficiency|
Based on this information the program then calculates the model rate of climb with different propellers as follows:
On the horizontal co-ordinate is the speed of flight, on the vertical co-ordinate is the RoC. As the model could not climb faster than it flies, the thin black line indicates just the vertical flight. Accordingly, the curves on the left from „vertical“ make no sense.
Please note that:
1. It may be a demanding task to keep a low powered sailplane in the optimum climbing angle/speed (red line).
2. Even if the model is capable of vertical climb, the optimum angle could be less (green line).
3. Should the model have even more powerful drive, the peak would lie on the left from vertical line, ie. the actual efficiency would be less than maximal. More on this point is below.
The efficiency is a ratio of the useful power or energy and the input power or energy. In our case the input power is (voltage x current), the useful power is (RoC x model weight x acceleration of gravity(9.81)).
On the horizontal co-ordinate is the flight speed, on the vertical co-ordinate is total efficiency.
So, about 2/3 of the battery power vanishes, just 1/3 is converted the the height.
Total efficiency = motor efficiency x propeller efficiency x airframe efficiency
The motor efficiency (and that of ESC) are given in the table above. Known and uninteresting …
The propeller efficiencies look as follows:
Again, the speed on the horizontal axis and the propeller efficiency on the vertical.
It is obvious that the propeller is much faster than the airframe. The efficiency peaks are located far to the right.
Total drive unit efficiency (motor+propeller).
As above, the drives are „faster“ than the airframe.
The airframe efficiency can be defined as follows: the net propeller power (battery power x motor efficiency x propeller efficiency) pulls the airframe up with some speed. If there were no drag and lift this speed would be higher than it actually is. The ratio of the actual RoC to the „no drag, no lift“ speed is the airframe efficiency.
Airframe efficiency, speed of flight on horizontal, airframe efficiency on vertical axis.
Conclusion: the drive efficiency raises with the increasing speed of flight; the airframe efficiency drops with the speed. This, the total efficiency must have a peak.
The total efficiency can be determined as the ratio: (rate of climb x weight x 9.81) to (voltage x current).
– motor efficiency 70%
– propeller efficiency 60%
– airframe efficiency 80%
– total 33.6%
How can be this value increased:
– either make the airframe faster,
– or make the drive slower (gearbox).
Every answer brings about a new bunch of questions: does the efficiency depends on the power level? From the available data it seems that the higher power means higher efficiency, but then why they do use so low powered models in the Autonomy contest?
Let’s assume that:
– the model is quite efficient thermal ship (1000 g, 2500 mm, 40 dm2);
– powers from 100 to 600 W;
– motor efficiency always set to 80%, RPM 4000, 8000 and 12000 (on diagrams indicated as 4, 8 a 12 – i.e. geared motor, outrunner, and inrunner); it is assumed that the motor with the required power, efficiency and RPM can always be selected;
– power absorbed by a propeller is a function of the RPM and the diameter; thus the propeller diameter can be found from the power and RPM;
– propeller efficiency is then defined by the general efficiency curve (see above).
Well, the lower RPM means higher efficiency, that is generally known. However, that there is an optimum power that seems to be suprising.
The reason for this optimum appears to be a good match between the drive and the airframe as mentioned above. Also please note a faintly visible change in the slope of the curves at about 250 W. This power is just about to suffice for the vertical flight.
The conclusion of this case study is that the optimal power should allow the TD model to go vertical but in fact not „very quickly“.
Well, in reality the Rook needs about 13. Thus, there is a room for the improvement.