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Computing Energy consumption by voltage


EricGhost
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Hi everybody,

I have recently moved from a KS16B 640 Wh (on which I did my first 3000Km)  to a KS18-L mainly for safety reasons: stability due to 18" and also higher torque due to a more powerful engine 2000W.
So far I'm happy with my choiche, the only concern is about the battery consumption which I know can change a lot due to a lot of factor (Esaj states from 10-25Wh/Km) but still something is missing to me.
I read many posts of high range (my range of interest is max 40Km so not a main issue for me) with this wheel which are not consistent with my experience.
I'm a 90Kg for 1.9mt tall and drive on a flat terrain mainly and the range is quite far from what I read but more consistent with the experience of my previous wheel keeping into account the Higher Wh, but also the higher weigth and the bigger wheel!
So my worry is that either something is not working properly on my wheel or it is just some KS sponsorship over the forum ?!?

I made some investigation on Li-Ion 18650 or similar discharge capacity there are many curves as below,  but majority agrees the one more consistent with my case is the one where we have:
Battery Cell Charge Max Voltage        Vbc    4.20    [V]
Battery Cell Discharge Min Voltage    Vbd    3.00    [V]
Battery Cell Nominal Voltage            Vbn    3.60    [V]
18650chargeDischarge-web.jpg.4351021fba7b86061ba507642968d00f.jpg

According to what I read and maybe understood the battery packeg of my wheel can be defined by
Battery Pack Charge Max Voltage        Vpc    84.00    [V]
Number of cells in Battery Pack        Nc    20.00            as     Nc=Vpc/Vbc

therefore we have
Pack Discharge Min Voltage            Vpd    60.00    [V]        as     Vpd=Vbd*Nc
Pack Nominal Voltage                Vpn    72.00    [V]        as     Vpn=Vbn*Nc

Always according to what I read and maybe understood then I have the manufacturer Battery Pack Capacity as Epc=1036 [Wh], which allows to compute the discharge nominal capacity in [Ah]:
Pack Discharge Capacity Nominal            Ipd    14.39    [Ah]        as     Ipd=Epc/Vpn

The above data allow the construction of an approssimated model for the Li-Ion discharge capacity, the figure is a rectangle trapezoid which area is the battery pack capacity:
Epc = 0.5*Ipd*(Vpc+Vpd)    Epc(KS18L)=0.5*14.39*(84+60)=1036.08[Wh] 

DischargeCapacity.JPG.72b9463d5fab23d6b8eb95b2deb90c04.JPG


The area can be seen as a section of a fuel tank which is filled when you have 84[V] and it is depleting while the battery voltage "Vp" drops, look at the yellow arrow as a floating sensor
thanks to the linear approximation of the discharging curve it is possible to compute the discharging current as function of the battery voltage Ip(Vp) 
Ip = Ipd * (Vpc-Vp)/(Vpc-Vpd)
so if the voltage from 84[V] drops to a Vp=76.3[V] then Ip=4.6 [Ah] and the used energy is Wh=0.5*Ip*(Vpc+Vp)    Wh(KS18L)=0.5*4.6*(84+76.3)=368.69
as we have Ip(Vp) then we can also compute Wh(Vp) which is not linear:

Wh = (Vpc^2 - Vp^2)*Ipd*0.5/(Vpc-Vpd)    see figure below
WhConsumption.JPG.14f8e2d63b2f036291db60578195b926.JPG


Wh(KS18L)=(84^2-Vp^2)*14.39*0.5/(84-60)=0.3*(7056-Vp^2) 

if Vp=76.3 then Wh=0.3*(7056-5821.7)=370.3 [Wh] (some rounding here in excel everythings matches)

the above means I can compute the depletion of my battery by reading the voltage and therefore knowing also the residual energy left in my fuel tank.

Then I filled at a full 84V my wheel and start testing the above stuff by  wheeling one of my GPSies run of about 11.2Km forward and back for a total of 22.5Km confirmed by GPsies and OruxMap,.
I point out this because the KS app tells another story a story of 27Km, so here it is still the old story of the KS app with the 20% discrepancy from real Km and KSapp Km.
Of course as a consequece all performance are improved of 20% as with the same energy the app gives you a bonus of 20% more range, unfortunately it is virtual and not real.
But let's go back to the numbers, after the run at an average of 20 Km/h I had a voltage left of 76.3V, and it is quite difficult to believe at the 100Km range, at least for me 55km  is more realistic.

Maybe a light chinese of 50Kg can do it as it 90/50=1.8 *55km =99km

Eric2work.pdf

Energy Consumption by Voltage drop    
                    [V]           [Wh] 
                    76.3        370.0 consumption
                    
Residual                            [Wh]     665.99
Residual                            [%]        64.3%
Real range                        [Km]    22.5
App range                         [Km]    27.0
App error Real/app         [%]        83.2%
Consumption                  [Wh/Km]    16.4

Range Forecast No margin              [Km]    63.0
Wh safety margin                              [ %]        10%
Range Forecast with Wh margin    [Km]    56.7
 

 

 

Edited by EricGhost
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17 minutes ago, EricGhost said:

...

18650chargeDischarge-web.jpg.4351021fba7b86061ba507642968d00f.jpg

...

Always according to what I read and maybe understood then I have the manufacturer Battery Pack Capacity as Epc=1036 [Wh],

This 1036 Wh are computed as follows:

4 packs with 20 cells with 3,7V nominal voltage and each cell having a capacity of 3,5Ah -> 4*20*3,7*3,5=1036

17 minutes ago, EricGhost said:

which allows to compute the discharge nominal capacity in [Ah]:

Pack Discharge Capacity Nominal            Ipd    14.39    [Ah]        as     Ipd=Epc/Vpn

But as you see in the above example until 3.0V the cell only delivers ~2300 mAh at 2C. At 1C it would deliver about 2800mAh...

The full capacity is not even available at an constant load of 0,2C and discharging until 2.5V

17 minutes ago, EricGhost said:

The above data allow the construction of an approssimated model for the Li-Ion discharge capacity, the figure is a rectangle trapezoid which area is the battery pack capacity:
Epc = 0.5*Ipd*(Vpc+Vpd)    Epc(KS18L)=0.5*14.39*(84+60)=1036.08[Wh] 

DischargeCapacity.JPG.72b9463d5fab23d6b8eb95b2deb90c04.JPG

So this formula/graph has to be adapted to the load (amperage taken...), as the capacity of the cells gets lower the higher the load is...

A very nice idea to get some rough assumption - but as you see from the above graph there is an uncertainty of capacity for the cells, depending on load between 0,2C und 2C of ~1000mAh! So a difference of (3300-2300)/3400=29%....

PS.: I hope you did not consider this somewhere in the post... ;) ... just scanned quickly over it...:ph34r:

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Like in life there are peaks and valleys but overall it is the average that matters. You do not drive your wheel at 2C or 0.2C all the time, and if you check at 1C things are not so far away from my math, plus I considered a 10% margin (maybe 5% will be better)for the gray area 3500/3000.

Your statement means even a more restricted range

Edited by EricGhost
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  • 3 weeks later...

@Chriull I investigated more on the modeling of the discharge and I come out with a model taking into account discharging currents from 1A up to 10A which means 4A to 40A on the package

dischargeEB2.thumb.JPG.4ce61c74ad36550bb851b652112d5541.JPG

so now the model graph "discharge package can work from the blue line 4A current -> I=4A,Vpc(4A)=84V, Ipd(4A)=14 Ah,  to the red one 40A current -> Vpc(40A)=69V, Ipd(40A)=13.1 Ah , I can evaluate the lost of energy in the battery pack due to the current usage now, while using a current "I" at a voltage V(I) battery I use a power W(I)=I*V(I) but for the battery is like they are still working at the higher voltage V(4A) so you have W(4A)=I*V(4A) > W(I) and you consume more energy from you battery. Knowing the working current "I" and voltage "V" it is possible to compute the V(4A) and therefore to know the real drain of energy from your batteries which is the W(4A) and not the W(I) like in my first model. Of course for currents near 4A there is a small difference  or if you are near the 65V. Some simulations below

dischargeEB.thumb.JPG.c0ae8144962d6eeed5c97a8f554727c4.JPG

I    10    [A]
V(I)    75    [V]
Vc(I)    80.50    [V]
K(I)    -1.50    [V]/[Ah]
V(4A)    77.84    [V]
dt    3600    
Wh(I)    750    
Wh(4A)    778.4    
Wh(4A)/Wh(I)    103.79%    

I    20    [A]
V(I)    75    [V]
Vc(I)    75.50    [V]
K(I)    -1.20    [V]/[Ah]
V(4A)    83.3    [V]
dt    3600    
Wh(I)    1500    
Wh(4A)    1666    
Wh(4A)/Wh(I)    111.07%    
 

Edited by EricGhost
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