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Kona64
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Discussion Starter · #1 ·
Could some one check my maths please, & useful info for others
( That if the charge unit drinks say 20kw of electrons your car battery does not go up by that same 20kw)

I've just put 2hours of home 7kw PodPoint Solo in to a Kona 64.
The car dashboard shows it went up from 45% SOC to 63% ---- so 18% lift up.
The PodPoint app says 14.8kw was used over these 2 hours.

But that electricity is 1.55 times the amount the battery took in - right ?
as 18% of a 64kw battery is 11.5, but the apps says 14.8kw used
14.8 is is 23% of a 64kw battery, so that 23% = 1.55 x the 18% the dashboard SOC has increased
Am a bit surprised, I knew there'd be overhead of kw absorbed in charging the vehicle --- but 1.55 times ?!?!
And, I'd just driven the vehicle 35miles so it was not "a cold engine"

If I had plugged in at 10% SOC, and put the battery to "full", ie 90% increase in SOC
in theory that's 90% of 64kw total capacity ( 57.6 )
but with the1.55 factor I would need to feed it 1.55 x 90% of 64kw ==== 89kw !
So given my home electricity is 14p a unit, that's £12.50 or so to fill up from 10-100% instead of a raw £8 or so ?
Still a lot cheaper than £1.25 a litre of unleaded granted

Final sum
So if my 2hr charge cost me according to the app £2.08
( according to app, 14.05p per kWh unit price,times 14.8kw absorbed by the PodPoint)
which is 18% lift in battery, then given the charge screens in the in car display suggest 100% is 303miles ( Kona64 !)
Then the 18% is 54.5 miles on range added for the £2.08 == 3.81p a mile cost if all paid charging at home


For those wondering on the economics of an EV ---- unleaded £1.25 litre, 45mpg say, = 12.6p mile.
 

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Could some one check my maths please, & useful info for others
( That if the charge unit drinks say 20kw of electrons your car battery does not go up by that same 20kw)

I've just put 2hours of home 7kw PodPoint Solo in to a Kona 64.
The car dashboard shows it went up from 45% SOC to 63% ---- so 18% lift up.
The PodPoint app says 14.8kw was used over these 2 hours.

But that electricity is 1.55 times the amount the battery took in - right ?
as 18% of a 64kw battery is 11.5, but the apps says 14.8kw used
14.8 is is 23% of a 64kw battery, so that 23% = 1.55 x the 18% the dashboard SOC has increased
Am a bit surprised, I knew there'd be overhead of kw absorbed in charging the vehicle --- but 1.55 times ?!?!
And, I'd just driven the vehicle 35miles so it was not "a cold engine"

If I had plugged in at 10% SOC, and put the battery to "full", ie 90% increase in SOC
in theory that's 90% of 64kw total capacity ( 57.6 )
but with the1.55 factor I would need to feed it 1.55 x 90% of 64kw ==== 89kw !
So given my home electricity is 14p a unit, that's £12.50 or so to fill up from 10-100% instead of a raw £8 or so ?
Still a lot cheaper than £1.25 a litre of unleaded granted

Final sum
So if my 2hr charge cost me according to the app £2.08
( according to app, 14.05p per kWh unit price,times 14.8kw absorbed by the PodPoint)
which is 18% lift in battery, then given the charge screens in the in car display suggest 100% is 303miles ( Kona64 !)
Then the 18% is 54.5 miles on range added for the £2.08 == 3.81p a mile cost if all paid charging at home


For those wondering on the economics of an EV ---- unleaded £1.25 litre, 45mpg say, = 12.6p mile.
Your equations need to start with the right units, eg Coulomb's and Joules.
 

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Audi eTron 55
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To help understand this you need to be clear on the difference between charging power (kW) and energy consumed (kWh). Plenty online, YouTube videos etc depending on your learning style.

The first big assumption you are making is that your 7kW chargepoint is filling the car at that rate - the second issue is that small charges will always give you a big uncertainty in the measurements. To see what the true position is you would need to measure the actual energy delivered over a large SoC range (e.g. 10-90%). I've done this, albeit with a different vehicle, and find a charging efficiency of between 90 and 95%. Some loss is inevitable, primarily due to heat lost in the car's on-board charger. Some vehicles (e.g. Zoe) have much lower efficiency due to their charging system. Lower charge rates will always be less efficient (i.e. losses will be relatively higher at 3kW than 7kW).
 

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Kona64
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Discussion Starter · #6 ·
The first big assumption you are making is that your 7kW chargepoint is filling the car at that rate - the second issue is that small charges will always give you a big uncertainty in the measurements. To see what the true position is you would need to measure the actual energy delivered over a large SoC range (e.g. 10-90%). I've done this, albeit with a different vehicle, and find a charging efficiency of between 90 and 95%. Some loss is inevitable, primarily due to heat lost in the car's on-board charger. Some vehicles (e.g. Zoe) have much lower efficiency due to their charging system. Lower charge rates will always be less efficient (i.e. losses will be relatively higher at 3kW than 7kW).
Thanks - the car dash shows drinking at 7.1, but I will give a long charge a blast so see about a 10-90% rise and efficiency
And, you answer another question I had , I was wondering about 3+11kw speeds. (so another reason for upping a home charge point to 3phase 11kw would also be that you'd get more efficient charge).
 

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Puma GTE > Suzuki Jimny > BMW Z4 > Cupra Leon ST
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Another assumption that you have is that the SOC is representing a direct % of the advertised battery size.

In reality the OEM generally leave a buffer on the top and bottom part of the battery, so an 100% SOC might in reality be only 95% of the battery and by 0% SOC the battery mighty still have 7% of charge in it.
 

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A longer charge will help get a more realistic figure and I do think shorter charges are less efficient for whatever reason. I charged from 86 to 100% once and it used 13 kWh, i.e. a "loss" of 31% of the used electricity. For more normal charges, the average loss has been 9% for me.
 

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Puma GTE > Suzuki Jimny > BMW Z4 > Cupra Leon ST
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A longer charge will help get a more realistic figure and I do think shorter charges are less efficient for whatever reason. I charged from 86 to 100% once and it used 13 kWh, i.e. a "loss" of 31% of the used electricity. For more normal charges, the average loss has been 9% for me.
For shorter charges a higher porcentage of the energy will be used for battery conditioning leadibg to a lower chargibg efficiency.
 

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For shorter charges a higher porcentage of the energy will be used for battery conditioning leadibg to a lower chargibg efficiency.
Makes sense.(y)

Strangely, I consistently saw less loss in December & January compared to September to November when it was warmer. I'd have expected it to be the other way around! For example, 24% to 100% SOC in late November had 12% loss, whereas 27% to 100% SOC in early January had 5% loss.
 

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Puma GTE > Suzuki Jimny > BMW Z4 > Cupra Leon ST
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Makes sense.(y)

Strangely, I consistently saw less loss in December & January compared to September to November when it was warmer. I'd have expected it to be the other way around! For example, 24% to 100% SOC in late November had 12% loss, whereas 27% to 100% SOC in early January had 5% loss.
How are you calculating your "loss"?
 

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An important point about these packs is that percentage is actually Ah (which is actually annoying).
As the voltage drops off the energy stored for each percent also drops off.

More details on the post by @Richard Wood
1st Motorway trip

"1. There's a lot more energy in the top end of the battery than the bottom - as the voltage drops (405V down to 320V-ish) - the percentage relates to amp-hours (180Ah) not kilowatt-hours (64kWh). Only 40-50% = 6.4kWh - top 10% = about 7.2kWh, bottom 10% maybe 5.9kWh. "
 

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Measurements I did on my ID.3 showed no noticeable difference between charging it at 6A, 10A or 32A. I was getting between 90 & 93% efficiency. Energy in measured by domestic-type grid meter dedicated to my EVSE, energy in battery based on reported SOC & 58 kWh capacity.
While you'ld expect 6A to be less efficient than 10A, and 10A less than 32A, the thermal losses should be reduced at lower currents as they're I*I*R. Maybe that compensates for the car having to be active for longer to get the same SOC increase, etc.
 

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Ioniq 38kwh 2020
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Note that if it is cold then some of that loss could be the battery heater. I'm not sure if it's the same on the kona, but on the ioniq if your charge at >3.7kW and the temperature is below about 20, the battery heater comes on and draws 2.2kW and heats at about 0.5°C per minute.
Charge below 3.7kW and no battery heater.
So slower charging can actually be more efficient in many cases (and as @HandyAndy has found) there isn't much evidence that slower charging per se is any less efficient.
Indeed it would make sense that any perceived additional losses in keeping the charging systems active for longer could easily be offset by the reduced heat and resistance in the battery and cabling etc.
Also small charging amounts will be less accurate, so your starting Soc may have been 44.5% and your ending Soc may have been 63.49% , so it's actually a nearly 20% charge. That is roughly 1.5kWh, which makes your numbers look better already.
 
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Could some one check my maths please, & useful info for others
( That if the charge unit drinks say 20kw of electrons your car battery does not go up by that same 20kw)

I've just put 2hours of home 7kw PodPoint Solo in to a Kona 64.
The car dashboard shows it went up from 45% SOC to 63% ---- so 18% lift up.
The PodPoint app says 14.8kw was used over these 2 hours.

But that electricity is 1.55 times the amount the battery took in - right ?
as 18% of a 64kw battery is 11.5, but the apps says 14.8kw used
14.8 is is 23% of a 64kw battery, so that 23% = 1.55 x the 18% the dashboard SOC has increased
Am a bit surprised, I knew there'd be overhead of kw absorbed in charging the vehicle --- but 1.55 times ?!?!
And, I'd just driven the vehicle 35miles so it was not "a cold engine"

If I had plugged in at 10% SOC, and put the battery to "full", ie 90% increase in SOC
in theory that's 90% of 64kw total capacity ( 57.6 )
but with the1.55 factor I would need to feed it 1.55 x 90% of 64kw ==== 89kw !
So given my home electricity is 14p a unit, that's £12.50 or so to fill up from 10-100% instead of a raw £8 or so ?
Still a lot cheaper than £1.25 a litre of unleaded granted

Final sum
So if my 2hr charge cost me according to the app £2.08
( according to app, 14.05p per kWh unit price,times 14.8kw absorbed by the PodPoint)
which is 18% lift in battery, then given the charge screens in the in car display suggest 100% is 303miles ( Kona64 !)
Then the 18% is 54.5 miles on range added for the £2.08 == 3.81p a mile cost if all paid charging at home


For those wondering on the economics of an EV ---- unleaded £1.25 litre, 45mpg say, = 12.6p mile.
First, energy meter showing kWh. So meter gave 14.8 kWh and car absorbing 11.5 kWh. You lost 3.3 kWh and this is 3.3/14.8= 22.3% and not 55% you mentioned. During longer charging this % will be lower
 

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First, energy meter showing kWh. So meter gave 14.8 kWh and car absorbing 11.5 kWh. You lost 3.3 kWh and this is 3.3/14.8= 22.3% and not 55% you mentioned. During longer charging this % will be lower
In other forums this has been discussed as well. Charging los is between 15% and 25%, depending mainly on the charging level. Charging at 2.4 kW (granny charger so AC) results in higher loss than charging at a public charger (22 kW?) and charging at a fast charger (DC) will result in the lowest loss.
 

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In other forums this has been discussed as well. Charging los is between 15% and 25%, depending mainly on the charging level. Charging at 2.4 kW (granny charger so AC) results in higher loss than charging at a public charger (22 kW?) and charging at a fast charger (DC) will result in the lowest loss.
in my garage I have option how fast I like to charge. Can adjust 6 A (1.5 kW), 8 A (2 kW), 10 A (2.5 kW), ...., up to 16 A (4 kW). tested all options and my observation is: faster charging -> more energy loosing
 
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