T-Pain Is On A Boat: [Electric Boats] Re: Regeneration in an Electric Sailboat -- what speeds to use

Hi Jim,

Wow.

But read this excellent post from James L:

http://groups.yahoo.com/group/electricboats/message/11466

I think the issue discussed therein represents the primary "disconnect" between assumptions and reality in regen - the prop will only move in proportion to it's pitch and the boat speed.

-Keith

--- In electricboats@yahoogroups.com, "james_c_rock" <jcelrock@...> wrote:
>
> This is my first attempt at analyzing regeneration for a sailboat and if anyone finds errors or can improve upon it, Please do so.
>
> BRIEF BACKGROUND: Thermodynamics says energy is neither created nor destroyed, but only changes its form. It is surprising how much insight one can achieve by tracking energy through a system. In what follows efficiency is the coefficient between total energy to usable energy for propulsion or regen. Power is energy per unit time. For equal durations one gets equal energy from the same power level.
>
> BOUNDARY CONDITIONS: Assume an electric drive converts 60% of the energy going into the electric motor into useful propulsive work. This is a very aggressive assumption about a hypothetical electrical propulsion system, as I think I recall that the Highly Engineered Torqeedo Propulsion System is only ~ 50% efficient from electrical power in to propulsive power out, and most other systems do not achieve that.
>
> In this example, hypothetical Losses result from electrical efficiency of controller and motor (say 90%), efficiency of transmission and bearings (95%), and propeller efficiency under cruising conditions at 80% of hull speed (this assigns an unlikely 70% efficiency for a large diameter slow turning propeller in free stream conditions).
>
> QUANTITATIVE EXAMPLE:
> At 60% efficiency 2 kW electrical input power to the controller yields 1.2 kW propulsive power.
>
> Under regeneration conditions at the same speed, assume the system efficiencies remain the same. (In fact the efficiency of the motor generator is a function of rpm, so its efficiency is probably somewhat lower than the assumed 90%, and the propeller efficiency is probably somewhat lower than 60%). At the same speed, the fluid power available over the diameter of the propeller is 1.2 kW. Multiply that by the 60% efficiency to find power back to the battery bank is 0.72 kW, or 36% of the cruising power consumption. To a first order, given a propulsive efficiency of eff, propulsive power of Wpp and electrical power of Welec, regenerated power Wrgen to the battery is
> Wregen = Welec*eff^2.
>
> Regeneration at the same speed used for electric propulsion yields returned power of (eff)^2. At 60% system efficiency, 36% of the power used for propulsion is available for regeneration. This reality makes my plan to power up the front side of the wave and generate down the back side with a fixed speed over water a bit doubtful. Assume a fixed pitched propeller has rpm proportional to speed over water. Its rpm during regen will be 36% of its rpm during powering. Most generators have an efficiency curve that falls with rpm, so both the propeller and generator will be less efficient during regen than during powering.
>
> But what if I power up the wave at a slower speed than I slide down the wave. Is there a ratio of speeds where I can equalize the regen power and the propulsive power?
>
> The power density in a moving fluid depends on the fluid velocity and fluid density. For a propeller with constant diameter, the power associated with its travel through the water is proportional to the power density of the moving water in [W/m^2 ]. It is sufficient to perform calculation on the basis of fluid power density.
>
> Power density per unit area is half the product of density and the cube of velocity. For a simple explanation of this equation, you can go to the web,
> <http://www.seic.okstate.edu/owpi_old/about/library/lesson1_windenergycalc.pdf>
>
> Assume my boat can achieve hull speed relative to the water while surfing down the back side. The power density availble to the propeller is 1/2 (density) (Vhull)^3 [W/m^2]. Using our default system efficiency of 60%, the power density available to recharge the battery is
> Wregen = ( 0.6) (0.5) (density) (Vhull)^3 = 0.3 (density) (Vhull)^3 [W/m^2]
>
> Now find the speed to use while powering up the front side of the wave to achieve the same power density. The propulsive power density at boat speed relative to the water of k*Vhull (with k < 1), is
> Wpp = (1/2) (density) (k * Vhull)^3 [W/m^2].
> The electrical power necessary to achieve this is larger by a factor of 1/eff = 1/0.6 = 5/3 = 1.67.
> The electrical power needed to achive a speed of k*Vhull is
> Welec = 5/3 Wpp = (5/6) (density) (k * Vhull)^3 [W/m^2]
>
> Now, I can set the power consumed climbing the front side of the wave equal to the power returned to the battery surfing down the back side of the wave and solve for k, the fraction of hull speed to be used on the front side of the wave.
> Welec = Wpp ==>
> 5/3* 1/2 (density) (k * Vhull)^3 == 3/10 (density) (Vhull)^3
> After a bit of algebra, k = (3/5)^(2/3) = 0.592
>
> Note that if I power up the front side of a wave at ~ 60% of Vhull and slide down the back side of a wave at ~ Vhull, both measured relative to the water, the electrical power out of the battery during powering up equals the electrical power returned to the battery terminals while surfing down. If the durations of powering up and of sliding down are equal, the battery sees energy out during climb = energy in during slide.
>
> There are three further adjustments needed for a first order analysis.
>
> First, It is likely that the slide down is shorter than the climb up. Because energy = power * time, and the goal is to maintain the energy in the battery bank, it is necessary to further reduce the power during the climb in proportion the ratio of powering time to generation time. This means the speed over water needs to be further reduced during the climb phase.
>
> Second, because the Peukert effect causes losses internal to the battery that have not been considered in the analysis above, the speed over water while climbing needs to be reduced even further than mentioned above. This also changes the duration ratio above, so the calculation needs to be iterated.
>
> Third, the analysis assumed propeller efficiency and motor/generator efficiency were both constant. It is likely that each is lower during regen than during cruise for a cruise optimized propeller. The speed over water during climb needs to be further reduced. This reduces losses due to Peukert effect and increases the time ratio. More iteration is needed.
>
> SUMMARY:
> To summarize, assuming a clever system with eff% efficiency between electrical power in and propuslive power out, and the same efficiency between fluid power to the propeller and electrical power for regeneration, and with batterys unaffected by the Peukert effect, returns, per unit time, eff^2 of the energy during regen that is consumed during powering, if the regen and powering occur at the same speed. For a high quality 40% efficient system operated at constant speed, the power returned to the battery terminals during regen is 16% of the power delivered to the controller terminals for cruising.
>
> If one generates at hull speed, the cruising speed for the same power passing through the controller terminals is:
> eff^(2/3)
>
> EQUATION SUMMARY
> Given : Wpp, Welec, Wregen, Vhull, k and eff.
> Welec = Wpp/eff
> Wregen = eff (1/2) (density) (Vhull)^3 *(Propeller disc area)
> For Welec = Wregen, k = eff^(2/3) with cruise speed over water = k * Vhull, and regen speed over water = Vhull.
>
> SAMPLE CALCULATION FOR eff = 40%
> For a 40% efficient system, one maintains the same power passing through the controller terminals when generating at Vhull and powering at Vhull*(0.4 ^(2/3)) = 0.54 Vhull.
>
> CLOSING THOUGHTS
> The equality of efficiency for regen and propulsion is not achieved in practice because of the Peukert Effect (internal electrical losses in the battery bank) and propeller efficiency differences at different speeds. A variable pitch propeller can help somewhat to keep the propeller efficiency high at both speeds, and very expensive battery designs (not lead acid) can help to minimize battery related losses. For most of us, cruising at 30% to 40% Vhull and regenerating at Vhull would allow energy per unit time to be approximately equal during discharge and recharge. Alternatively, recharging for 3 or MORE times as long as we motor, would return similar usable energy to our batteries. This means that one should, as a starting point, set ones speed controller so that generation occurs 3 times more frequently than propulsion under varying forces of wind and waves.
>
> To move from the ratios in this email to actual power levels, one can use density = 1000 Kilo Gram / Meter^3, to get power density per square meter as a function of speed over water. Then multiply by your propeller area to find the power available from your speed over water.
>
> If there are electrical loads on your battery pack while sailing, or if your battery is largely discharged, you will want to use higher ratio of generation time to propulsive time, perhaps 10:1 when the wind is fresh. Empirical testing is the way to find out. Further calculations are unlikely to be sufficiently close to reality to be helpful.
>
> Finally, anchoring in tidal or river currents can allow regen while on the hook, even for non-sailboats. I have seen little discussion of this as an option for electric boats.
>
> HAVE FUN! Jim Rock
>