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Discussion Starter #1 (Edited by Moderator)
One of the difficult challenges in planning an EV conversion is choosing the voltage and size of the battery pack you plan to use. This following page aims to simplify that process explaining how each aspect of the pack will affect the performance of the EV. If you find these concepts difficult you may want to read about Power (kW) and Energy (kWh), as a background to this article. Here are 4 steps that should make the whole process of sizing your pack nice and simple:

Step 1: Top speed.
The top speed of your conversion is primarily governed by the total voltage of the battery pack (we will ignore gearing for this discussion). The more batteries you string together in series, the higher speed you will be capable of achieving. Voltage determines the maximum rpm of the motor, provided there are enough amps to overcome the load and the physical limitations of the motor are not exceeded. The voltage should be chosen to match the power requirement of the vehicle at the required top speed. For example most DC motorbike conversions will require a minimum voltage of approximately 72V to be able to travel at 60+mph (100km/h) while a pick-up conversion would most likely require 144V to maintain the same speed.

Step 1 summary: The first step in sizing your pack is to determine your top speed requirements and look at the voltage of other similar sized conversion needed to achieve that speed.

Step 2: Range.
The range of your conversion is determined by the efficiency (measured in Wh/mile or Wh/km) and the total energy stored in your battery pack (measured in kWh). As a general guide motor bikes will generally use 75-150Wh/mile, cars may use around 200-400 and pick-ups and heavy vehicles around 400-600Wh/mile depending on the weight, rolling resistance and aerodynamics of the vehicle (also higher voltage systems are generally more efficient than lower voltage and AC systems more than DC). Since you have already determined the voltage of the pack, the range will determine the Ah rating of your battery pack so that the pack has enough energy, which is the product of voltage and amp-hours (measured in kWh), to travel the required distance. Lets say we have a round trip commute of 40 miles (65km) which we wish to travel in a standard car conversion at 120V, by looking at other conversions we estimate our own will have an efficiency of 250Wh/mile. The product of our efficiency (energy/mile) and our required range (miles) will give us total energy required for that distance. In our example, the efficiency of 250Wh/mile multiplied by the range of 40 miles gives us a total energy of 250x40=10,000W or 10kWh. Since we already determined our voltage was 120V (see step 1) this means our Ah rating will be the total energy divided by the voltage. In this case 10,000W / 120V = 83.3Ah.

Step 2 summary: The second step is to work out your maximum range and estimate the efficiency of your conversion based on others the same size, multiplying the two to get the total pack energy. This will determine your amp-hour rating.

Step 3: Make allowances for your battery type.
Electric conversions use deep-cycle batteries which have some important characteristics that affect how they can be used in EVs. The two major factors are the Depth of Discharge (DoD) limitations and the Peukert effect:

EV batteries do not like being emptied down all the way and so emptying them completely will drastically shorten their life (the number of times you can use them). In order to counter this most EV conversions arrange things so that their battery pack never goes below 20% full. This is usually known as 80% DoD, or depth of discharge. So for our battery pack we need to make sure that when we have traveled our full range we still have 20% of our energy still in the batteries. To make sure this happens we take the the Ah rating we worked out in part 2 and multiply it by 1.25. This will mean when we have traveled our required distance we still have 20% left in the batteries. So for our previous example we worked out the Amp-hours to be 83.3Ah so we multiply by 1.25 to give us about 104Ah.

The Peukert effect sounds fancy but it simply means that the faster you use up the energy in the battery the less you will get out in the end. Battery Amp-hour ratings are usually given for a pretty slow discharge over 20 hours. However most EV conversion will use up their power much faster than that, usually in about 1 hour. Because the faster you use the energy the less you get altogether most EVs using Lead Acid batteries will only be able to use about 55% of the energy of the 20hr rate and we need to again compensate for this in our total pack size, by multiplying by 1.8. So our the amp-hour value in our example of 104Ah becomes 187Ah. When sizing the battery pack we need to make sure that the batteries we choose have an Amp-hour rating of 187 or better to achieve our range of 40 miles.

Lithium based batteries perform much better under high strain loads so you should be able to use 95% of the 20C energy rating. This means for a lithium pack you only need to multiply by about 1.05 to compensate, so a smaller pack is needed compared to a Lead Acid pack.

Step 3 summary: The third step in sizing a battery pack is compensating for the characteristics of the batteries we choose, for Lead acid batteries this can be achieved by multiplying our Amp-hour rate from step 2 by 2.25, For Lithium batteries this can be achieved by multiplying our Amp-hour rate from step 2 by 1.32.

Step 4: Acceleration
Acceleration is a more complex requirement to work out and quantify. One simply way to determine this will be the power to weight ratio of your conversion. We have seen above that energy = volts x amp-hours. Power is given by power = volts x amps. It is important to note the distinction between these two formulas. Amp-hours determine the energy and range of the EV, Amps determine the power and acceleration of the EV. As well as having an amp-hour rating, most batteries will also have a maximum discharge rating (sometimes called 'cold cranking amps'), which is the most amps that batteries can push out. This value is often harder to find than the voltage and 20hr Ah rating since deep cycle batteries aren't necessarily made to have high discharge rates. Generally speaking the better a battery is at high discharges the shorter its lifespan will be in EV use but the higher the acceleration. The following factors will improve acceleration: higher pack voltage, lower weight of the car and high max discharge rates of the batteries. Also, generally 12V batteries are better than 6V, and AGM batteries are better than flooded but both will have shorter life (number of cycles) than 6V or flooded batteries).

Step 4 summary: The final step in choosing a battery pack is to balance the acceleration and life of the batteries. Higher voltages give better acceleration but costs more (for controller, motor etc.), high discharge rates, 12V batteries and AGMs are better but mean more battery changes, higher Ah generally means higher discharge rates but also increases weight so may hinder acceleration. You will need to balance these factors depending on your priorities for your conversion.

Conclusion:
Once you have completed these four steps the rest of your conversion will be fairly easy. The voltage you picked will determine the voltage of your motor, controller and charger and the max discharge rate will give you the max amps for the controller.

In our example above, a 120V system with 190Ah would give a range of 40 miles. This could be made up of 10 12V batteries which might put out 800A peak making 120x800=96000W or 96kW of peak power. The controller would be 120V and anywhere up to 800A though this might be limited to increase range or protect the motor (perhaps 500A would do). The battery pack would probably weigh around 1100 lbs (500kgs).

*A Cautionary Note: The Wh/mile figures are the biggest unknown in these calculations and generally people will determine their Wh/mile with their existing batteries already factoring in Peukert's effect (often without knowing they are doing so). In order to work out the Wh/mile of your vehicle take a look at similar makes/sizes on the EValbum. So if you are using the Wh/mile of a previous Lead Acid conversion you can probably ignore compensating for Peukert's effect (part of step 3) if you too will be using Lead Acid. If you plan on using lithium you could possible retroactively work out the true Wh/mile without Peukert's and then readjusting it for lithium. Nevertheless these calculations will give you a good starting point that can be tweaked for your vehicle.
 

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Re: Sizing a battery pack

mattW

added a couple of commas, hope you don't mind, just thought it helped for clarity. nice article, wish i had something similar awhile back, should be helpful

b.k
 

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Re: The Four Steps To Sizing a battery pack

That was quite helpful but it was odd how steps 1-3 were a straight forward calculation and then step 4 is not. Is that because step 4 depends so much on the chosen battery?

My calculations looked like this for a theoretical 80 mile range with a motorcycle (using some numbers from the example calculation):

Step 1 (Top speed):

72v / 12 = 6 batteries
70-100 Wh/mile
80 miles range


Step 2 (Range):

Watts: 100 Wh * 80 = 8000 Watts = 8 kWh
Amps: 8000w/72v = 111.111 Ah


Step 3 (Allowances for battery type):

80% DoD so:

111.111 Ah * 1.25 = 138.9 Ah (so after 80 miles, 20% left in batteries)

Peukert effect (fast drain means drain rating is really 55% of actual for fast drain):

138.9 Ah * 1.8 = 250.02 Ah

or

111.111 Ah * 2.25 = 250 Ah


Step 4 (Aceleration):

power = volts * amps

Ah = determine energy and range of the EV
Amps = determine power and acceleration of the EV

Say 800 A peak so: 72v * 800A = 50,400 watts = 50.4 kW of peak power

Weight of pack: 110 pounds/battery * 6 batteries = 660 pounds
As mentioned, this is theoretical. 660 pounds of batteries on a motorcycle frame seems a bit much but I'll keep reading. I can see how battery technology is the #1 concern. My commute is 37 miles one way :(.
 

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Discussion Starter #4
Re: The Four Steps To Sizing a battery pack

motorcycle range is pretty limited by the size and weight restrictions. I am going with a 72V 60Ah lithium pack and am hoping to get 50 miles (80km)
 

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I'm new here so forgive me.

Step 3. says

" When sizing the battery pack we need to make sure that the batteries we choose have an Amp-hour rating of 187 or better to achieve our range of 40 miles."

Does this mean that each battery in your pack should have an Amp-hour rating of 187 or does this mean that your total pack should have an Amp-hour Rating of 187?

Also, can you explain batteries in strings, I noticed if I use the calculator on EVconvert.com and I have a 144v system if I have 2 strings my range goes up.

Does this mean that I need 2 strings of 72v or 2 strings of 144v?

Also, I plan on using the Odyssey PC1500 (specs found -->) for my conversion. Am I supposed to be looking at the 20hr rate ah http://odysseybatteries.com/batteries.htm

Obviously I'm new and I'm trying to figure out my batteries before I go off and drop a load of money on them

Thanks for the help,

James
 

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Alternative Rule Of Thumb for LiFeP04 batteries:

It seems, based on the rather scant available data, that the new Lithium Iron Phosphate prismatic cells have a rough energy equivalent in gasoline:

8kWh of LiFeP04 cells = 1 gallon of gasoline

This is not an actual energy comparison, but a practical guideline number to compare ICEV and BEV efficiencies. Its really useful, because in most cases, the fuel efficiency of an ICE powered glider is known- if your Honda got 30mpg before converting, and you didn't add 800lbs to it in the process, then it probably will take 8kWh of Lithium pack capacity to go 30 miles. An interesting, simple and useful conversion tool. Remember to calculate usable pack capacity, which in most cases is really 80% of the actual pack capacity as usable capacity (above whatever your DOD cutoff is.)

Not a precise calculation here, but its pretty accurate, and very useful in sizing these packs, particularly where you know how much fuel the car used as a fossil fuel burner. It also kind of elegantly self-adjusts for driving conditions and style- if you got only 20mpg driving your Honda before the conversion, don't expect to get more than 20 miles from it per 8kWh of usable pack as a LiFePO4 EV.

If nothing else, its a nice way to triangulate your other range and speed calculations with battery capacity.

Does not apply to other battery chemistries...

TomA
 

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I have read and re-read your explaination. Just want to know if your calculations are all based on the top speed. Range would be a function of how fast you are discharging the batteries. If you only do one moderate acceleration .. and only do 45mph .. the range should be a lot higher than if you are doing frequent quick starts and higher speeds, right? If you have a car capable of 0-100 in 12seconds .. but you dont really drive that way.. how do you adjust the calculations for the "normal" way you drive?
 

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Have a look at http://www.diyelectriccar.com/forums/showthread.php/yes-another-ev-calculator-45278.html dmac257.

What you've said is correct, and the it's easy to understand with a graph.

In this calculator, the left part gives you what's required to move the car at the selected speed, and how far you can go if you stay at that speed.

The right part is how fast you accelerate if you use full power.

I also manage the gearbox, and take into account shift time. If you want to stay simple, you can use, a reduction gear instead of a gearbox.

You can modify any blue text parameter in the spreadsheet.
You can also modify the gearbox, motor, batteries tables to fit you needs.
 

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Assumed Gross weight of vehicle = 1500 kg
Desired top speed = 30 Km/h
Desired Range = 100 Km
Coefficient of rolling friction (μr) = 0.015
Density of air at 20° C = 1.2 kg/m3
Acceleration due to gravity (g) = 9.8 m/s
Voltage of the electric system = 48V



Force of rolling friction Froll = μr n
= 0.015*1500*9.8
= 220.5 N

Air resistance force (at 30 Km/h) Fair = ½ CdAρv^2
= ½ *0.5*1.8*1.2*8.34^2
= 37.56 N

Power required to maintain the constant speed of 30 Km/h
= (Froll + Fair) v
= (220.5 + 37.56)*8.34
= 2152.22 W

Time taken by the vehicle to cover the desired range at 30 Km/h
= 100 / 30
= 3.34 hrs

Current consumed to generate a constant power of 2152.22 W (IDC)
= 2152.22 / 48
= 44.84 Amps

K-factor for 3.34 hours of back up = 4
Depth of discharge of batteries = 80%
Depth of discharge Factor = 1.25
Temperature deviation of battery performance = 10%
Temperature factor = 1.1
Design Margin = 10%
Design Factor = 1.1

Final Ah rating = IDC*K-factor*Discharge factor*T-factor*Design Margin
= 44.84*4*1.25*1.1*1.1
= 271.282 Ah


The requirement is either 8 X 6V or 4 X 12V batteries with a capacity of 280 Ah or more.

Is my calculation correct? or did i put too much power for the requirement? Please help me, as i think this capacity batteries are not available (costly) and if i want to make one by connecting batteries in series, it will be too heavy. I'm very new to this. I started working on this only a week back.

The subscript and superscript thingy is showing up there.
for your clarification,

Cd = coefficient of drag = 0.5
A= cross section area = 1.8 m2
v = velocity in m/s = 8.34
rho = density of air = 1.2


Thank you,
Uday
 

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Does the calculated amp/hour rating of the pack mean each battery must have that rating, or the the total of the batteries in the pack must equal that?

I have a hard time imagining a lead-acid battery pack with 187ah rating per battery.
 

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The amp-hour (not amp/hour) rating of a battery is independent of its voltage. Current (amps) is a rate at which electrons pass a given point over a given time period. (See http://en.wikipedia.org/wiki/Ampere for more detail.) The voltage at which this happens doesn't matter. If a battery is 100% efficient then the Ah it has will be the same regardless of how much current is pulled out of it. For example, if our fictitious 100% efficient battery is a 100Ah battery then it could be drained in 1 hour if we pull 100A of current out of it or it could be drained in 2 hours if we only pulled 50A out of it. An amp-hour is derived from the current (A) multiplied by the time interval. With the 100Ah battery it doesn't matter what current is pulled or how long it is pulled as long as the product of the two add up to 100Ah.

Now suppose there are two of these perfect batteries, each a 100Ah battery but one is a 5V battery and the other is a 10V battery. The amount of energy stored in each battery is quite different but the Ah capacity is the same. In both cases if we pull 20A out of each it would take 5 hours to drain each one. But that 20A current will be at 5V for one battery and 10V for the other. (Remember we are talking about a perfect 100% efficient battery so the voltage doesn't sag.) Power = Volts * Amps so the power out of the 5V battery is 5V * 20A = 100VA or 100 Watts. The power out of the 10V battery is 10V * 20A = 200W. Similarly the energy stored in each battery is 5V * 100Ah = 500Watt-hours and 10V * 100Ah = 1000Wh = 1kWh.

Does the calculated amp/hour rating of the pack mean each battery must have that rating, or the the total of the batteries in the pack must equal that?
The answer to your question is YES. Both the pack and each battery has the same Ah rating but the energy stored in each battery is naturally less than the amount of energy of the whole pack.

For lead acid batteries they have a significant internal resistance so the higher the current drawn from them the fewer Ah you can get out of them. For example, I helped the local community college convert a Mazda Pickup and the battery pack is made up of 24 Trojan T-105 batteries. These have a 225 Ah rating at the 20 hour rate meaning if you draw a current of 11.2A it will take 20 hours to drain the battery. If the current is 37A, however, it will only take 5 hours to drain the battery it will only deliver 185Ah. Increasing the current to 75A will drain the battery in 115 minutes (1.917hr) so it will only deliver 143.75Ah. Since lead acid shouldn't be drained beyond 80% there are really only 115Ah available to drive with. As you can see if you take half of the 20hr amp-hour rating of a lead acid battery and use it as the capacity available to drive an EV you will be reasonably close to what you get.

From my experience and many others with LiFePO4 cells the Ah rating of the battery is really close to the same as long as the currents are not extreme so for all practical purposes if a LiFePO4 cell is a 100Ah cell you have all of it available but should only use 80% or less to prolong the cell and not kill it by a full discharge.

Hope that helps.
 

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Hmmm, thanks, that answers my question. :) Now I'm only confused by the fact that I've checked quite a few lead-acids and about the strongest I've found is an Optima Yellow Top with only 55ah. According to this that would be nearly useless. Back to the drawing board.
 

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Assumed Gross weight of vehicle = 1500 kg
Desired top speed = 30 Km/h
Desired Range = 100 Km
Coefficient of rolling friction (μr) = 0.015
Density of air at 20° C = 1.2 kg/m3
Acceleration due to gravity (g) = 9.8 m/s
Voltage of the electric system = 48V



Force of rolling friction Froll = μr n
= 0.015*1500*9.8
= 220.5 N

Air resistance force (at 30 Km/h) Fair = ½ CdAρv^2
= ½ *0.5*1.8*1.2*8.34^2
= 37.56 N

Power required to maintain the constant speed of 30 Km/h
= (Froll + Fair) v
= (220.5 + 37.56)*8.34
= 2152.22 W

Time taken by the vehicle to cover the desired range at 30 Km/h
= 100 / 30
= 3.34 hrs

Current consumed to generate a constant power of 2152.22 W (IDC)
= 2152.22 / 48
= 44.84 Amps

K-factor for 3.34 hours of back up = 4
Depth of discharge of batteries = 80%
Depth of discharge Factor = 1.25
Temperature deviation of battery performance = 10%
Temperature factor = 1.1
Design Margin = 10%
Design Factor = 1.1

Final Ah rating = IDC*K-factor*Discharge factor*T-factor*Design Margin
= 44.84*4*1.25*1.1*1.1
= 271.282 Ah


The requirement is either 8 X 6V or 4 X 12V batteries with a capacity of 280 Ah or more.

Is my calculation correct? or did i put too much power for the requirement? Please help me, as i think this capacity batteries are not available (costly) and if i want to make one by connecting batteries in series, it will be too heavy. I'm very new to this. I started working on this only a week back.

The subscript and superscript thingy is showing up there.
for your clarification,

Cd = coefficient of drag = 0.5
A= cross section area = 1.8 m2
v = velocity in m/s = 8.34
rho = density of air = 1.2


Thank you,
Uday
You might convert thus into a MATLAB script
 

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Can what be calculated?

How do we know that a battery pack with 120V and 187Ah pack can produce 800A. How do we get to the 800 Amp value? Do we know from experience that a pack this size makes 800A or is there an math equation.

Does a 144V and 187Ah pack to behave the same as a 120V and 187Ah pack? I think the 144V would be higher peak amperage but I don't know where to start.
 

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How do we know that a battery pack with 120V and 187Ah pack can produce 800A. How do we get to the 800 Amp value?
Experience of someone mostly. For example my batteries are rated to 3C continuous. This means I can pull three times the Ah capacity value in Amps and remain in spec. Testing has shown that they can put out much more than this. I set my pack up to be a 200Ah pack so 3C would be 600A. Someone once tested one of these 100Ah cells with a dead short and got something like 2500A. I know Jack Rickard of EVTV has seen 1000A from a 180Ah pack so it stands to reason that 800A out of a 187Ah pack would be quite feasible. Note that the voltage of the pack is largely irrelevant.

800A out of a 187Ah pack is only 4.3C which is a relatively low current for the pack. CALB cells are rated to 4C continuous and something like 10C pulse even though they can do more.
 

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Experience of someone mostly. For example my batteries are rated to 3C continuous. This means I can pull three times the Ah capacity value in Amps and remain in spec. Testing has shown that they can put out much more than this. I set my pack up to be a 200Ah pack so 3C would be 600A. Someone once tested one of these 100Ah cells with a dead short and got something like 2500A. I know Jack Rickard of EVTV has seen 1000A from a 180Ah pack so it stands to reason that 800A out of a 187Ah pack would be quite feasible. Note that the voltage of the pack is largely irrelevant.

800A out of a 187Ah pack is only 4.3C which is a relatively low current for the pack. CALB cells are rated to 4C continuous and something like 10C pulse even though they can do more.

Thanks, that helps. Summing up so I understand; Capacity rate dictates controller size to an extent.

The Gizmo is impressive, excellent work.

Vince
 

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Thanks, that helps. Summing up so I understand; Capacity rate dictates controller size to an extent.
Many controllers have settings to limit, among other things, battery current. For example, you could have a 1000A controller but limit battery current to 500A. The controller could still send 1000A to the motor but at a voltage lower than your battery voltage. To make it easy assume you have a 200V battery that doesn't sag in voltage. 200V * 500A = 100,000W or 100kW. At the motor it could see 100V * 1000A = 100,000W or 100kW also. This is ignoring losses in the system.

You don't want to size a controller smaller merely because your battery pack isn't capable of (or you don't want to push it) putting out the max current the controller can. Most of the time the battery current will be below the motor current, especially when you have a pack voltage higher than what the motor is set to take and you have the controller limit motor voltage. The Warp9 is rated to 170V, IIRC. If you have a 200V pack and have the controller limit the motor voltage, the motor current will always be above the battery current.

The Gizmo is impressive, excellent work.
Thank you. I've put quite a few upgrades in it since I bought it in 2006. It has been a fun, and useful, learning experience.
 
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