# Battery estimation for an Electric Race-Car



## enzo (Nov 15, 2012)

Hello people,

I am currently doing a project on electric cars. I am designing the powertrain of an electric race-car on MATLAB/SIMULINK for the Formula Student UK competition. 

I am struggling with the claculation of the power required from the power source (battery).
I simulated my design and i found that, the power required for the race-car to complete one lap (1:13 secs) is around 220KW. Does this make sense?
The equation that i used is the following: Pbatt=Torque*ω/efficiency for T>0 and Pbatt=Torque*ω*efficiency for T<0. Then i did the same simulation for the endurance race (it has to complete 22 laps) and it dissipates 3586KW. 
Thus, considering the endurance race, i have to provide 3586KW to the motor to complete the race???
I need these values to estimate how much power is needed for the race-car to complete the endurance race wich lasts for 30 mins.

For my first concept i have chosen a YASA 750 DC motor. 
Peak Torque @ 360A= 750Nm
Continues torque= 450 Nm
Peak power @ 380V=100KW
Continues power= >50KW
Peak efficiency=95%
Total weight=25Kg

And these are the batteries that i will use: http://www.boston-power.com/sites/default/files/documents/2012-03 Swing5300 DS Rev1 0.pdf

Can you please help me with the calculations? I need to estimate the maximum power needed to run the car for 30 mins (for the endurance race) and how many batteries have to be connected in series and in parallel.

Thanks in advance


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## major (Apr 4, 2008)

enzo said:


> Hello people,
> 
> I am currently doing a project on electric cars. I am designing the powertrain of an electric race-car on MATLAB/SIMULINK for the Formula Student UK competition.
> 
> ...


Hi enzo,

Does the figure of 3586 kW sound right to you? That is 4813 hp  Off hand I'd say you need to study the physics lesson on power and energy.

I don't mind helping a guy out and hope I am doing so with this post. But I do object to doing homework assignments, if you know what I mean. There have been numerous threads and posts, some currently on-going, on your topic of interest. Please take the time and effort to research and then come back with some specific inquires.

Regards,

major


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## Karter2 (Nov 17, 2011)

Are you compelled to use those battery cells ?
or have you chosen to use them ...if so , why ?


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## Joey (Oct 12, 2007)

enzo said:


> Hello people,
> I am struggling with the claculation of the power required from the power source (battery).
> 
> I simulated my design and i found that, the power required for the race-car to complete one lap (1:13 secs) is around 220KW. Does this make sense?
> ...


I think you are not clear on the difference between power and energy. Batteries store energy, Joules. Power is the rate at which energy is applied, Joules per second = Watts. 

As an example, if your motor is rated at 50,000 Watts continuous, your battery should supply (after losses at the controller) 50,000 joules of energy per second. If you want to deliver this much power for 30 minutes, your battery needs to store 50,000 J/sec*1800 s = 90,000,000 Joules of energy. There is another unit for energy that is a little easier work with for electric vehicles - the Watt hour. If you multiply power and time in hours you get: 50 kW * 0.5 h = 25 kWh. (1 Wh = 3600 J, both are units of energy). Continuous power is but one constraint of your motor. You need to look at your whole system to design something that is within the constraints of all your components (and there are a fair number of contraints, but that's what data sheets will tell you). 

You have quite a project ahead of you. Try to ask specific questions, as the general ones are usually already answered with a little bit of looking (thank you google), and a lot of people will be glad to help you.


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## enzo (Nov 15, 2012)

@major, Yes you are right. I must do more research and come back with more specific questions.

@Karter2, No i am not forced to use them. Actually i want to use Lithium Polymer batteries. There are too many manufacturers, so i have chosen one of them. This type of batteries are widely used in Formula Student.

@Joey , According to my motor's specs, the continues peak power is 50KW. It will be run for 30 minutes therefore i end up with 25KWh. Therefore i need to supply 25000/380 65.79Ah

Thank you very much for your rapid response and help


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## enzo (Nov 15, 2012)

Joey, thank you for your useful coments.
I run the model for 1 lap, and i got the motors power. I use an integrator and i got the energy in Joules.
Energy=35000 Joules
Then i multiplied that energy by 22 which is the number of laps that the car have to complete. 
Total Energy=22X350000=7700000 Joules
Then i divided by 3600 to get the Energy in wh.
Energy=7700000/3600=2138 Wh
My motor's voltage is 380V. 
Thus, Capacity (Ah) = 2138/380 = 5.6Ah

I end up with 104 cells in series and only one in parallel. Does this make sense?
I used lithium batteries, 3.65V and 5.3Ah


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## Karter2 (Nov 17, 2011)

enzo said:


> I end up with 104 cells in series and only one in parallel. Does this make sense?
> I used lithium batteries, 3.65V and 5.3Ah


  sorry !..NO .. back to class for you .


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## Jordysport (Mar 22, 2009)

FS *shakes head* been there done that got the t shirt.

right look at it another way, 360A is your peak motor amps, assume worst case it pulls that continuous (although other components won't take that continuous). you want it to run for 30 mins... 180Ah capacity required. (in very basic form).

Also assume the cont amps is 200A, that's 100Ah capacity. you will obviously be running higher than cont amps rating some of the time, and some time below it. 

also Bin Matlab its more hassle than its worth and most stuff can be done in excel, when i was at your stage i used an excel model to which was inputted into matlab (for matlab assignments) 

If i was in your seat right now, i'd say to forget being clever with integrating etc, and just Read up on it and not to Assume formula's which don't apply (again one of my mistakes at uni). 

I could go into it further on how but why do your job for you!


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## Joey (Oct 12, 2007)

Enzo, I would have to say that your estimate sounds quite low to me. Read this web page (http://physics.ucsd.edu/do-the-math/2011/07/100-mpg-on-gasoline/), and look at this equation: _E_drag=½_c_D_ρADv_².

Another reference to consider - most street cars use 250 to 500 whr/mile. I could see a race car easily using twice this amount, or more.

Then you will also need to consider the limitations of your batteries. If your battery discharge is rated 12C, a 5 Ah battery will only be able to give you 60 amps.


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## PStechPaul (May 1, 2012)

If you run the motor at continuous rating of 50 kW for the duration of the race, 1/2 hour, that's 25 kWh. If the battery pack is 380V it's about 66 Ah. Your post #5 seems to be correct, but you shifted a decimal place in #6. No need for complex calculations. But to win a race you need to consider many other factors. I'm just assuming the chosen motor is appropriate. "YMMV"! 

Interesting discussion in the UCSD link. My 1999 Saturn SL1 gets 40-45 MPG on the freeway. If gas has 37 kWh/gallon, then I'm getting about 820 wH/mile. If an ICE is limited to about 25% efficiency, then my actual usage is 205 wH/mile. Seems like that shows a good potential for this model as an EV conversion. I might expect about 50 miles range on a 10 kWh battery pack, which would be about $5000 for LiFePO4.


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## xmbaleaf (Dec 13, 2012)

enzo said:


> Hello people,
> 
> I am currently doing a project on electric cars. I am designing the powertrain of an electric race-car on MATLAB/SIMULINK for the Formula Student UK competition.
> 
> ...


in fact,as pe lifepo4 battery packs supplier,you can tell to they the motor max power/rated power,max power continous time,motor working time,packs dimension and so on,they will can supply one solution to resolve your DC sources.

it seems to that your 5# post was right.


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