# How do I work out my car's Watt/Hours?



## KiwiEV (Jul 26, 2007)

I've seen people talk about their battery packs in Watt Hours (Wh) but never been sure how exactly I could work out the W/h in my pack.
My battery pack has a low A/h rating of 12 x 85A/h, however the voltage is reasonable at 155v when charged. 
What would be the formula to work out W/h using that information?
If someone can show the formula here (in simple English) then I'll add it to the _Formulas_ Wiki.


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## mattW (Sep 14, 2007)

Power = Voltage x Amps
Energy= Power x time
so 144 x 85=12.2kWh
and if you are going to be technical you can times it by the factor to go from 20C to 1C I think it is around 0.53 or something like that for LA.


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## KiwiEV (Jul 26, 2007)

Thanks Matt, I understand some of that, although I'm not sure how you got your answer.
When I use that formula I get this:
Power = Voltage x Amps (power = 144v x 515 Cold Cranking Amps: 72000)
Energy= Power x time (72000 x 60 seconds? = 4320000???)

Help!


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## rankhornjp (Nov 26, 2007)

KiwiEV said:


> Thanks Matt, I understand some of that, although I'm not sure how you got your answer.
> When I use that formula I get this:
> Power = Voltage x Amps (power = 144v x 515 Cold Cranking Amps: 72000)
> Energy= Power x time (72000 x 60 seconds? = 4320000???)
> ...


I think he is using Ah and you are using cranking amps, which are much higher.


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## mattW (Sep 14, 2007)

Yeah I sorta cheated because you already have amps x time in Ah so you can just multiply by the voltage to get V x I x t. Lets say you can use 200A at 144V for 45mins before your batteries go dead, the energy in the system is 100 x 144 x 0.75 (3/4 of an hour) =10.8kWh. Power is instantaneous whereas energy is measured over time.


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## KiwiEV (Jul 26, 2007)

Aaahhh, gotcha! Thanks Matt!


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## NobleTruths (Jan 14, 2008)

My question ties closely (I think) to yours. I am trying to work out the equivelant cents/mile of ICE vs EV. Since I am still in early planning, I have no definative components yet. I wish to use lithium batteries for sure, however. My electric company charges 12cents/Kwh. Can anyone help me with?


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## CPLTECH (Sep 14, 2007)

My on-board E-Meter can be set up on either Amp Hours (AH) or Kilowatt Hours (KWH), but I prefer the AH setting. My brain logic seems to accept that better while driving. Either way, it computes your usage similar to how the electric utility does: How many watts or amps did you use over a given time?

To answer the question, the utility charges by the equivalent thousand watts per hour (KWH). Ex: A 100 watt lamp uses 100 watt IF kept on for an hour. My E-Meter readout tells me AH & my battery pack happens to be120V. Amps times Volts equals Watts. In my case, a reading of 100AH x 120V = 12,000 watts or 12KWH. Always add 20% to the on-board figure to allow for lead acid battery losses & charging. All batts have losses, not sure about lith losses. Hope this helps.

I bought an old electric meter (one removed from a house) at a hamfest flea market. I use it to see how much of the utility bill is for the EV alone. The 94S10 used 1630KWH in 2700 miles = *.*6KWH per mile. Next question… How much is battery replacement going to cost per mile? Inquiring minds want to know.


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## mattW (Sep 14, 2007)

Just to clarify *watts* is an *instantaneous* measure of *power*, just like horsepower. A 100W light bulb uses 100W whenever it is on irrespective of how long it is on for. *Watt hours* or Kilowatt hours are measures of *energy* just like joules or calories which *is effected by time*. A 100W lightbulb left on for 10h will use 1kW of energy but at any time in that period it will atill be using 100W. Amps times Volts = power (in watts) but amps times volts times time = energy (in watt hours).

For an example your battery's AH rating is usually given at rate that will discharge it in 20 hours. So a 100Ah battery will put out 5A for twenty hours. The power of a 12V battery at 5A is 60W (Volts x Amps), the energy provided by the battery at 5A for twenty hours is 1.2kW (Volts x Amps x time). But you can skip that step and just say 100Ah (amps x time) by 12V = 1.2kW. 

Apparently 250Wh/mile is pretty average for an EV so 0.25kWh times 12c =3c per mile (theoretically). I hope that clears some stuff up!


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## Mr. Sharkey (Jul 26, 2007)

> Apparently 250Wh/mile is pretty average for an EV so 0.25kWh...


I suspect this figure is low, perhaps calculated only on the DC consumption. My EV, with aerodynamic enhancements, ultra low rolling resistance tires, and very effective regen consumes about 300 wH per mile DC, but 500 wH as measured on an accurate utility type AC kilowatt-hour meter. From my research while on the EVDL, I found that overall efficiency of the average EV was about .5 KwH per mile if battery and charging efficiencies were properly considered. CPLTECH's figure of .6 KwH/mile seems very believable to me.


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## NobleTruths (Jan 14, 2008)

Thanks, Matt and Sharkey

As a follow up question, does the fact of using an AC vs DC motor effect the "fuel economy"? And does using Lead acid vs Lithium batteries effect the "fuel economy"? I suspect "no" to the second question, but not sure about the first. Also, to the comment of "battery cost per mile" of Lead acid vs Lithium.......that, to me, is a moot point. My venture has little to do with "financial pay-off." I want a car with performance, and Lead batteries do not compete with Lithium to that end. As I see it, I can plunk $100K for a nice ICE, or I can build a unique EV for a little less and feel real good about it, lol.


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## John (Sep 11, 2007)

Generally DC cannot do regenerative braking and AC can so that affects your mileage especially under stop and go conditions and is responsible for the lions share of the difference in efficiency. Regens effectiveness depends very much on your configuration also. AC motors also don't have the brush drag of DC. The different charge recovery of different battery chemistries will certainly affect your mileage and particularly so in a performance oriented vehicle. Kokam quote charge recoveries in the mid 90's and higher for some quite high currents up to 3C.
For most vehicles the higher the performance the higher the energy consumption per distance. From what I can tell the component most responsible for this inverse performance/efficiency relationship in an EV is the battery. As the power demand increases on the battery the usable kWh declines. Flooded lead acids are very poor at this. Some LiIons are very good at it allowing for very good performance with a very modest efficiency penalty.


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## speleocaver (Nov 17, 2009)

13,175 kwh
155v x 85


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## few2many (Jun 23, 2009)

Heres a good question, I've tried to figure out, but I'm getting stuck on the efficiency of my ice.
I want to convert the mpg of my car, 31mpg, to the watt/hrs it would need if I were to convert to electric. Using the watts availible in a gallon of gas, I havent had much luck.


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## Tesseract (Sep 27, 2008)

few2many said:


> ...
> I want to convert the mpg of my car, 31mpg, to the watt/hrs it would need if I were to convert to electric...


The energy content of gasoline varies somewhat between summer and winter blends, and with ethanol content, but Wikipedia says is is 36.6kWh/gal or 9.7kWh/l on average.

Thus, the calculation is relatively straightforward: 36,600/31 = 1180Wh/mi.

EDIT: oops - I misunderstood the question. You wanted to figure out, essentially, how big of a battery pack you'll need which means, as subsequent comments have pointed out, that you need to multiply that Wh/mi. figure by the efficiency of the ICE, then divide by the efficiency of EV motor/controller which will replace it. In the end just multiplying by 0.25 to 0.30 seems reasonable, since the modern ICE is about 15-20% efficient.


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## natemartinsf (Jul 18, 2010)

Tesseract said:


> The energy content of gasoline varies somewhat between summer and winter blends, and with ethanol content, but Wikipedia says is is 36.6kWh/gal or 9.7kWh/l on average.
> 
> Thus, the calculation is relatively straightforward: 36,600/31 = 1180Wh/mi.



Of course, converting this to Wh/mi after conversion is complicated by the fact that you don't know the efficiency of your engine. 1180Wh of gas is needed to move the car down the street, but most of that is turning into heat.

If we assume that the engine is 25% efficient, then that comes out to 295Wh/mi.

But without knowing the actual efficiency, it's a complete guess.


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## dimitri (May 16, 2008)

natemartinsf said:


> Of course, converting this to Wh/mi after conversion is complicated by the fact that you don't know the efficiency of your engine. 1180Wh of gas is needed to move the car down the street, but most of that is turning into heat.
> 
> If we assume that the engine is 25% efficient, then that comes out to 295Wh/mi.
> 
> But without knowing the actual efficiency, it's a complete guess.


Actually, these numbers and this conversion method is pretty accurate. My ICE car was about 30mpg and now after conversion its about 350Wh/mi on average. I guesstimate electric drivetrain to be about 80%-85% efficient, compared to ICE's 25%-30%.


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## few2many (Jun 23, 2009)

Thanks all, that was what I was looking for. I had numbers relatively close, but messed up at guessing the efficiency of the ice.


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## michaeljayclark (Apr 3, 2008)

do the kilowatt hours of a battery pack change depending on if you have 144 vdc in 6 volt 200 Ah batteries or 12 vdc 200 Ah batteries?

youd have 24 6 vdc 200 Ah batteries versus 12 vdc 200 Ah batteries.

whats the difference?

I think a nice online calculator would work out well for us math dunces.


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## John (Sep 11, 2007)

michaeljayclark said:


> do the kilowatt hours of a battery pack change depending on if you have 144 vdc in 6 volt 200 Ah batteries or 12 vdc 200 Ah batteries?
> 
> youd have 24 6 vdc 200 Ah batteries versus 12 vdc 200 Ah batteries.
> 
> ...


Both packs would be the same size. The 200Ah 6V batteries would be about half the size of the 200 Ah 12V batteries. A lead acid cell is about 2V and each in this case would be 200Ah. The 6V batteries have 3 cells and the 12V batteries have 6. Each pack would consist of 72, 2V 200Ah cells.


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