Electric Vehicle average power calculations 1

[bill318]

I have used numerous EV battery spreadsheets and calculators and always came up with very different conclusions. What I really wanted to find out is how the average EV conversion performs in real life under average driving conditions/terrain. I have entered over 40 real EV conversion stats into a spreadsheet and they do seem to have a trend when it comes to Watt-Hours/weight/distance.

I’m a bit unsure of the numbers I’m getting; do the following calculations look reasonable?

Converted Electric Vehicle Gross Weight - 4200 lbs. (1905 kg)
Batteries – Qty 24, Trojan T-125, 6V FLA, 240 Ah (20hr rate)
Known range of 50 miles (80.5 km) over average terrain with combination city and highway driving.
Average speed over 1.25 hour trip = 40 MPH (64.37 km/hr)
Battery bank depth of discharge (DOD) at end of trip = 65%

Due to shorter discharge period (1.25 hrs) and Peukert effect, the battery amp hours need adjustment. A flooded lead acid battery has about 52% the Ah capacity when discharged at a 1.25-hour rate vs. the normal 20-hour battery rating.

240Ah * 52% = 124.8 Ah per battery at 100% DOD

Since we only drained the battery to 65% DOD, then the actual Ah used from each battery would be.

124.8 * 65% = 81.12 Ah per battery

So the total Watt-hours (Wh) used from each battery would be…

81.12Ah * 6volts = 486.72 Wh per battery

Since we drained all 24 batteries during the trip, the total power consumed would be:

486.72 Wh * 24 batteries = 11681.28 Wh

To get the Wh/pound/mile (Wh/kg/km)

11681.28 Wh / 4200 lbs / 50 miles = .0556 Wh/lb/mi
11681.28 Wh / 1905 kg / 80.5 km = .0762 Wh/km/kg

When collecting the data, I had no way to tell how deep each owner discharged their battery banks for the mileage they reported. Since most discharge anywhere from 50% to 80% DOD, I chose 65% DOD for all vehicles as the average. Out of the 40 vehicles I entered into the spreadsheet, the values seem to lie between .045 Wh/lb/mi (.0616 Wh/km/kg) and .065 Wh/lb/mi (.089 Wh/km/kg) with the average being .055 Wh/lb/mi (.0753 Wh/km/kg).

There are so many variables to average out like terrain, weather, vehicle losses, depth of discharge, optimistic mileage & weight specs, battery life, average speed, etc… Anyways, perhaps it could tell if the vehicle is performing above or below average and be used in reverse to see what different battery chemistries may do to performance.

Any thoughts/corrections would be appreciated. –Bill-

 

[itsmoked]

Well let's make a comparison or two.

1000Whr = 1kWhr
So 11,681Whr = 11.6kWhr.

Next: 1kWhr = 1.34 hp-hr = 3,413Btu

Hence: 11.6kWhr * 1.34hp-hr = 15.5hp-hr.

Does the average 50mile, 1.25hr trip seem to require only an average of 15.5hp-hr which when spread over 1.25hrs actually means 12.44hp?

Seems like too little, to me, to drag over two tons with windage and mushy tires.


Or looking at it another way:
11.kWhr * 3,413Btu = 39,867Btu.

If we think about it I can imagine getting typically 18MPG in a two ton vehicle. (Bigger than say a Toyota Camry).

50Miles/18MPG = 2.78 Gallons for your 50 mile 1.25hr trip.

Well one gallon of gasoline is 124,000Btu.

So 2.78Gal * 124,000Btu = 344,444Btu.

BUT! We only get a conversion factor of about 15% in car.
So we only get about 0.15 * 344,444Btu = 51,666Btu.

51,666Btu >> 39,867Btu

Still seems like too little to me. Seems a bit too optimistic. But certainly your numbers aren't ridiculous and can be used as a comparison between vehicles if not an absolute predictor.

 

[bill318]

Yes, I thought it might be a little light too, your gasoline example comes out to about 16hp average which certainly seems more reasonable than 12.5hp.

One other spread sheet I downloaded came up with 54hp for this 4200 lb vehicle. This would make each of the 24 batteries have to put out 351Ah or 50,544 Wh for the 1.25 mile trip. To me, this was way on the other end of the scale.

I guess what is really needed is better (more accurate) data to work from. As they say, garbage-in, garbage-out!

Thank-you for the feed-back, I think this needs a bit of massaging yet!

Bill

 

[VEBill]

Although 'terrain' was mentioned, unless the course is a closed loop, it would seem to be absolutely critical to consider starting and ending height (altitude). Over some downhill courses, one could imagine that the battery might be more charged after than before.

There are so many rough estimates in your input parameters, that you need to consider the significant figures in your answer (looks like maybe one significant figure to me).

 

[itsmoked]

Very good point VE1BLL! One sigfig would probably do far more Justice to the whole thing.

 

[bill318]

By significant figure do you mean using the upper range of .065 Wh/lb/mi instead of .055 Wh/lb/mi?

This would be about 13.65 kWH /1.25 Hr = 10.92 kW * 1.34 = 14.63 Hp for the 40 MPH average. This does seem about right.

I think most EVs are used for commuting to and from the store/work. In which case most people make a round trip using the same path each way. I know this would be true in my case. Where you use more power going up a hill, you will use less on the trip back down the same hill.

I think in the data I collected that people were very optimistic in their mileage reports. So perhaps I should use the 80% DOD for each case.

Thanks for the input! Bill

 

[itsmoked]

Significant figures as in significant figures.

When working a bunch of equations with numbers you inject, your answer can only be as accurate as the least accurate number you supply.

You can take this further if you look at what you are doing and knock off accuracy you know is not valid.

For instance your 0.0556 Wh/lb/mi should probably be trimmed to 0.05 or 0.06Wh/lb/mi

 

[VEBill]

Specifically [One example]" "240Ah * 52% = 124.8 Ah per battery at 100% DOD .... 124.8 * 65% = 81.12 Ah per battery"

The '124.8' should be 120 (although we might allow 125 if we're generous). The '81.12' should probably be 81 (or even just 80 if we trace back all the other data to their sources). There's certainly no way to carry forward the '81.12' that is based on 52% and 65% and so on.


Another question: Is the consumption per unit distance really proportional to mass? In radio, sometimes people compare miles-per-watt not realizing that it naturally gives the win to the guy that can send a picowatt one cm. In your case, someone might add ballast to take the win (on level ground with very, very hard tires).

 

[itsmoked]

You last point VE1BLL, is an interesting one!

 

[VEBill]

Scaling Laws.

It's why ants are so strong (per unit body mass), and why elephants would probably die if dropped from a height of twelve inches.

 

[abcd3286]

I have studied the same questions and read book by Brant - good book. I have looked at only two predictions. Yes they vary but to my surprise not much.

You must remember they are only predictions which fall apart when prototyp or first production unit are tested.

ONe of the difficulties is what are the imput parameters and who is defining them. What is average terrain - where - it will differ from San Francisco to Seattle to Iowa.

How is vehicle driven (whoops I have yet to see an input paramteter for the nut behind the steering wheel). Jack rabbit starts?

What are they using for average wind velocity and relative wind velocity?

What are they using for aero drag? At 40 not a huge consideration but it is definitely there and cannot be discounted as it can at say 20 or less.

I think take the prediction you trust most use those values and add 10 or 20%.

I also do not believe you need to derate the AH capacity of a battery if you are using the factory 20 hr rate. I remember in teh Navy (submarine) we used the factory 4 hr rate to gage battery consumption and test discharge capacity.

 

[bill318]

Thanks guys, I'm with you now. Dropping decimal places would only affect the perceived accuracy of the estimate.

Mass and acceleration seem to be the largest determining factors. Since most everybody follows the same rules of the road concerning acceleration to regulated speeds, I chose mass for the estimate. Most of the vehicles are later models with similar aerodynamics and radial tires. As far as I know, none of the vehicles had magnetic bearings and steel wheels on a polished level track. ;-)

In charting out the vehicles, it was kind of fun to see 2,000 and 5,000 lb. vehicles using about the same estimated energy per pound. I fully expected a slope or curve, but the graph shows basically a flat band of points.

Thanks again for the feedback, Bill

 

[itsmoked]

abcd3286: Pulling more than the 20hr rate out of LA batteries definitely blasts the total energy available and must be considered.

Bill318; Seems to me, what is missing is the actual info on what your average car goes thru. Why don't you stuff a junker laptop into your car and an accerometer. Drive to the nearest gravel yard with a scale and weigh you and the car. Then log all daily driving. This will give you exactly the the horsepower required to move that load of metal and plastic around.

From that, weight scaling would be possible.

You would then have a great baseline for the EV work.


Also do you realize that you can buy dash-dynamometers which racers use? They directly calculate the absolute horsepower delivered to the ground as required to accelerate the vehicle, knowing its weight?

 

[VEBill]

"...absolute horsepower delivered to the ground"

For laughs, I once calculated (basic physics) the average horsepower that ends up as kinetic energy over a quarter mile for a top fuel dragster with 'thousands of horsepower'. It was around 400.

[Another factoid about top fuel dragsters: they use fuel at about the same rate as you can turn over a bucket.]

It would seem likely that mass would have a direct effect on your EV 'city' score (as you've proven), but mass should have less effect on an EV 'highway' score. YMMV.

 

[abcd3286]

itsmoked (Electrical) 2 Nov 05 2:31
abcd3286: Pulling more than the 20hr rate out of LA batteries definitely blasts the total energy available and must be considered.

Yes I know - as submarine electrician I learned early on that if on battery and you want the lights to stay on longer drop the discharge rate.
Total AH out at
1 hr rate < 4 hr < 10 < 20 hr rate.

THe point I was trying to make is that 20 or 4 or 1 or whatever FACTORY amp hour rating you use - that is the guage battery performance is measured to you do not derate this. When you test a battery capacity you discharge at the FACTORY rate (commonly 20 hr).

 

[bill318]

itsmoked I was debating doing just that only using a GPS to track elevation, distance and speed. Then dump the data out of the GPS. This method is used by the author of one of the spread sheets I downloaded as suggested by abcd3286 in an earlier thread. Guess this means I need to break down and get an interface cable for my GPS.;-)

VE1Bll 400Hp to the ground... With our whopping 14 HP at the flywheel it is amazing that any makes it to the ground! At least the electric motors can generate the needed torque to get things rolling. I see your call sign and it reminds me of how long it has been since I've been on the radio.

abcd3286 Maybe we are not talking about the same thing here. In any event, the Ah de-rating we are doing above estimates the battery performance under a load far in excess of the factory 20Hr load test. A fully charged Trojan T-125 can sustain a 12A load for 20 Hrs before it reaches 100% DOD (12*20 = 240 Ah). This same fully charged battery can only sustain a 100A load for about 1.25 Hrs (100*1.25 = 125Ah) before it reaches 100% DOD. This would be true if the battery was at 77 deg. F during the load tests. As a comparison, an AGM or Gel-cell lead battery can retain about 68% of its 20Hr AH rating under a similar 1.25 Hr discharge period.

Bill

 

[abcd3286]

Maybe we are not talking about the same thing here. In any event, the Ah de-rating we are doing above estimates the battery performance under a load far in excess of the factory 20Hr load test. A fully charged Trojan T-125 can sustain a 12A load for 20 Hrs before it reaches 100% DOD (12*20 = 240 Ah). This same fully charged battery can only sustain a 100A load for about 1.25 Hrs (100*1.25 = 125Ah) before it reaches 100% DOD. This would be true if the battery was at 77 deg. F during the load tests. As a comparison, an AGM or Gel-cell lead battery can retain about 68% of its 20Hr AH rating under a similar 1.25 Hr discharge period.

It is like I said you get more power at the longer discharge interval.
20 hr rate 12 A = 240 AH
1.25 hr rate of 100 = 125 AH

 

[bill318]

I went back to the spreadsheets and tried to determine why there was a small wh/lb/mi spread between vehicles regardless of vehicle weight or type. The only reason I came up with was the type of driving each vehicle is seeing. The shorter-range vehicles are probably doing mostly stop-and-go city driving (most energy used to accelerate) while the longer-range vehicles are doing more highway miles (most energy used overcoming drag and friction). This is what VE1BLL alluded to earlier (a star for that one!).

So let’s re-do the equations a bit using the same example vehicle…
Vehicle weight = 4200 lbs
Battery Type = Trojan T-125 , 240 Ah each (20 Hr rate), 6V
Battery Qty = 24
Battery Peukert exponent = 1.23
Pack Voltage = 144V
Pack Ah = 240 Ah
Pack Watts Avg. = (240Ah/20hrs) * 144V = 1728 Watts average over 20 hours

For city driving, I’m using .09 wh/lb/mi with an average speed of 27mph.
Lets use 15 miles worth of city driving.
Travel time = 15miles/27mph = .556hours
.09*4200lbs*15miles = 5,670Wh
Average watts = 5670Wh/.556 hours = 10,198 Watts
Apply Peukerts equation to estimate battery capacity used for the 15 miles
((10198W/1728W)^1.23) * (.556H/20H) = 24.7% total capacity used

For highway driving, I’m using .06 wh/lb/mi with an average speed of 50mph.
Lets use 35 miles worth of highway driving.
Travel time = 35miles/50mph = 0.7hours
.06*4200lbs*35miles = 8,820Wh
Average watts = 8,820Wh/0.7 hours = 12,600 Watts
Apply Peukerts equation to estimate battery capacity used for the 35 miles
((12,600W/1728W)^1.23) * (0.7H/20H) = 40.3% total capacity used

Total distance driven = 50 Miles
Battery pack Depth of Discharge = 24.7% + 40.3% = 65% DOD

It’s not perfect, but may make a decent rough estimate for an average vehicle.

-Bill-