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How the weight of the batteries and vehicle affect Electric Vehicle range.
I"m
not going to try to explain how batteries work or the different kinds
of batteries you can use in Electric Vehicles in this article. There are
several other expert sources on this information. I will try to explain
the usage of the batteries and the affects of the amount of "Energy"
you can get out of them in relation to their weight and how that applies
to Electric Vehicles.
Battery terms commonly used are amp/hrs
(amps per hour), cold cranking amps, reserve capacity, etc. What we
really want to know is "How much power can I get out of them per their
weight to make my electric vehicle go far and fast""
I have seen
this battery terminology described as amps/kg or amps/lb. It would be
really better to understand the amount of "energy per pound" we can get
out a battery. Regular people mostly think in terms of power as
"horsepower" or "watts". We know how much light a 100 watt light bulb
will put out compared to a 60 watt bulb or how much heat a 1200 watt
hair dryer can put out and how fast a car with 300 horsepower can go or
that a 5 horsepower lawnmower can mow grass better than a 3 horsepower
mower.
If you want to compare 12 volt batteries, amperage works
OK, since the volts is taken out of the equation. and their weight is
not that important if you are using just one battery in a 3,000 pound
car. But in Electric Vehicles we care how much the "Battery Pack" will
weigh because we use more of them and that affects our speed and range.
The best way however, is to understand "Watt/hrs per Pound" (or
Kilowatt/hrs per Pound) of a battery or even "Horsepower per Pound"
since it takes 746 watts to make one Horsepower. Both of these can be
used to describe "Power Per Pound"
Electric Vehicles use
different "Battery Pack" voltages, depending on the battery voltage and
the number of batteries. Some can use 24 volts others could use 72, 96,
120, 144, and almost any battery combinations you can think of. There
are some advantages to higher voltages. Mostly in the efficiencies of
transferring the electricity from one place to another, but there are
disadvantages when it comes to the cost of components, weight and
charging capabilities. There are additional inefficiencies when you take
energy out of batteries at a higher rate verses a lower rate. (These
subjects will be talked about another time)
I"m going to leave
out the above efficiency factors in explaining the affects of what I
will call "Battery Power Per Pound". I"ll try to leave out some
variables and as much technical stuff that I can in explaining this.
Since
we are concerned about how fast we can go and what range we can have in
our Electric Vehicle, whether it be a boat, cycle, or car, we need a
"Battery Pack" that can accomplish this, just like a fuel tank that
holds the "available power" in a gas vehicle.
Each individual
battery contains "available power". The more batteries or bigger
batteries the more "available power" (we will leave out different kinds
of batteries here), then you take this power and turn it into speed or
range. If you want more speed you lose range, if you want more range you
may need to give up some speed. It all depends on how you use this
"available power", again just like the amount of gas in the gas tank of a
car and the size and power of the engine.
More batteries, more
power but you also have more weight, unlike the gas in a car, battery
weight is proportionally more of the total vehicle weight and has much
more significance in a electric vehicle as to your range and speed. The
power per weight ratio comes into play here.
To best describe the
affects of the power to weight, I"ll use this example that leaves the
weight of the vehicle out of the equation. Just using Battery weight it
takes a certain amount of energy to accelerate each time you make a
stop, go up a hill, turn corners and just driving along which is called
rolling resistance and Aerodynamic resistance.
Power Equations:
1)
One pound of weight requires .0166 horsepower to accelerate to 60 mph
in 1/4 mile and takes 25 seconds at this acceleration rate.
2) 25 seconds is .0069 of 1 hour
3) One horsepower = 746 watts
4) .0166 horsepower = 12.38 watts or .087 watt/hrs to accelerate each pound to 60 mph in a quarter mile.
The
example below will use just the power it takes to accelerate one time
to a speed of 60 mph in a 1/4 mile which will take about 25 seconds
using just the weight of the batteries as the total accelerated weight.
If you have 500 pounds of batteries.
8.3 horsepower is required.
If you have 1.000 pounds of batteries
16.6 horsepower is required
You
can see that the power required is doubled when the battery weight is
doubled. So 0.0166 horsepower is required to achieve a 25 second 1/4
mile acceleration time with each pound of weight.
If a 500 pound
battery pack consisted of ten fifty pound 12 volt batteries at 83.33
amp/hr each or 1000 watt/hrs then you have 10,000 watt/hrs of power
available in the "battery pack". Using 8.3 horsepower for the quarter
mile requires 6,190 watts for 25 seconds or 43.5 watt/hrs. That would be
about .435% of the total battery pack energy . If you were to apply
this to accelerating from 0 to 60 mph ten times you would use 4.35% of
your battery pack just to accelerate your batteries.
Of course we will need to factor in the weight of the vehicle as well.
A
2,000 pound vehicle with 500 lbs of batteries would require 41.5
horsepower to achieve a 25 second 1/4 mile acceleration time, in a 2500
pound car. That would be 30,950 watts or 217.5 watt/hrs which is 2.175%
of your total battery pack "available energy". So that means you could
start and stop your vehicle about 37 times and then your batteries would
be 80% depleted. (Which is as much as you can go using lead acid
batteries without harming them)
A 3,000 pound vehicle with this
same battery pack would use 304.5 watt/hrs which is almost 3.045% of the
available battery pack energy and you would be able to accelerate at
this rate about 26 times which is 30% less times or you could apply it
to range and say 30% less range.
So this vehicle weighs about 30% more and will get about 30% less range with the same battery pack.
Now
what if we take the 2,000 LB vehicle and doubled the battery pack. Do
you get double the range" No, you actually get less than double. With
the 3,000 pound car (with the batteries included), it requires 261
watt/hrs or 1.3% of the total available energy so you can make 62 stops
which is just 60% more range.
If you triple the battery weight
you end up with a 3,500 pound vehicle which will require 304.5 watt/hrs
or about 1% of the available energy, we have tripled the available
energy (and the cost of batteries) but just about doubled the range.
Now
lets cut in half the vehicle weight from 2,000 pounds to 1,000 pounds
or half the starting weight, with the same 500 LB battery pack it then
requires 130.5 watt/hrs, which is 1.3% of the total battery pack which
would be about 60% more range. This is without adding any more
batteries.
Now if you take a 1,500 pound car and double the
battery pack you end up with 2500 pounds, just like our example above,
but you have doubled the available energy, you then still need 41.5 hp
or 217.5 watt/hrs which is now just 1.09 % of the available energy, then
you have doubled your range. Just by starting with a 500 pound lighter
vehicle and doubling the battery pack. If you start with a 1,000 pound
vehicle and triple your battery pack then you would triple your range.
You
can see that by starting with a lighter vehicle you can more
efficiently increase your range. So any lighter vehicle will have a
better range when you add more batteries.
We have this saying, "Less iron more Lead", Translation: "Use a lighter vehicle and add more batteries"
There
are a lot of other factors that affect Electric Vehicle "range". In all
these cases the lighter, smaller, more aerodynamic vehicles will give
you more range per your battery dollar value.
Starting with the
same vehicle weight and using batteries that have a better "Power to
Pound" ratio will dramatically improve your range as well. Then the
starting vehicle weight becomes even more important. More information on
types of batteries and the "Power to Pound" ration will be in another
article.
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Range Vs Weight of an Electric Vehicle
Relation of the energy required to obtain distance for a given vehicle weight.
I
will generalize this explanation to allow a better understanding of
just the affects of vehicle weight on the potential EV range using a
given Battery Pack.
Some of the variables that will not be in this comparison include.
* Air temperature
* Barometric pressure
* Wind speed
* Aerodynamic drag (although this may vary with a heavier Vehicle)
* Rolling resistance (and this probably would vary with heavier Vehicles)
* Road surfaces and Terrain
Assumptions that we will include are:
* Equal battery pack voltages
* We are not using a "Regen Braking" system
* System efficiency of the compared vehicles equals 80% (this can vary with system voltage, motors used, controllers used etc.)
First
to establish the term "Power". We have been accustomed to think of
power as horsepower in the general public, because of the use of the
term in Gas Automobiles. Electric Vehicles could use that term also, but
it would be better to think of "Power" as "Watts" or "Kilowatts"
because we can determine the available "Kilowatt/hours" that a "Battery
Pack" can provide.
To understand the relationship, it takes 746
watts to equal 1 horsepower or .746 Kilowatts. If the battery pack
system has a potential energy of 105 kW/hr then you can expect to have a
total of 105/.746 = 141.55 hp/hr available for use. This means that if
we run the vehicle using 141.55 hp * 80% (system efficiency) = 113.24
horsepower for 1 hour the battery pack will be discharged.
If all
the assumptions above are equal and we also assume that the two
vehicles of different weight were already traveling at 60 mph with a
full battery pack then they would both travel the same distance before
the batteries were discharged. Even if one vehicle weighed twice as much
as the other. (remember we are ignoring other resistance factors)
In reality we need to accelerate and stop, go up hills and then back down, make turns and sometimes go in reverse.
Each
time you accelerate from a stopped position to a given speed you
transfer electrical energy into kinetic energy, and when you stop you
waste that available kinetic energy into heat in the brakes (you may
think you can use a "Regen Braking"
system here, but that is another subject). So the affect of each stop
is the ?energy used factor?. The affects of the mass (weight) of the
vehicle comes into the equation at this point.
Equations comparing a 2000 LB vehicle and a 3000 LB vehicle using an constant distance to stop, and a speed of 60 mph:
F * X = 1/2 mv2 (From "Modern College Physics Sixth Edition" ) or
Force times the Distance = 1/2 the Mass times the speed squared
Example: (Since we are using the same distance for both vehicles "X" can be ignored)
Car "A" = 2000 LB car = F * X = 1000 * 3,600
The used energy factor is 3,600,000
Car "B" = 3000 LB car = F * X = 1500 * 3600
the used energy factor is 5,400,000
Comparing
the two 3,600,000/5,400,000 = .67, or in other words car "B" used 33%
more energy to accelerate to 60 mph and then stopped. For every stop car
B would waste 33% more energy.
Example:
Using our above
assumptions only, and both cars travel the same distance at the same
speed with 10 stops, and car "A" accelerated to speed using 14.92 kW
then Car "B" would need to use 22.27 kW just to accelerate at the same
rate as car "A".
For the 10 stops car "A" would use 149.2 kW and car "B" would use 222.7 kW)
If
they both had a battery pack of 105 kW and car "A" went 50 miles then
car "B" could have a range of only 33.5 miles before the batteries would
go dead.
This all goes back to the original weight, Car "B"
weighs 50% more than car "A" and gets 33% less miles, using equal speed,
and acceleration rate.
This is just a generalization, but shows the significance of weight on range.
Of course this is not taking into all the other considerations that affect range with different size and weight vehicles.
Some of these are:
Heavier vehicles have a higher rolling resistance.
The terrain affects heavier vehicles more - Every hill you go up and every turn you make.
Heavier vehicles usually have higher aerodynamic resistance.
Usually the power systems in bigger heavier vehicles have more electrical resistance.
You
may think you can just add batteries to increase the range, but the
more batteries the more weight, the more weight the less range per
battery weight plus the additional cost.
Conclusion:
Use the lightest vehicle possible for an Electric Conversion!
I
will publish more information on how each vehicle attribute affects the
range of an EV. You can see how Aerodynamics, Rolling Resistance and
other variables affect range. But in all these cases, small lighter
vehicles make the best conversions. |
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According
to the "Quarter Mile Acceleration Elapsed Time and Terminal Speed"
calculator found on the Web Site link below, the horsepower required to
accelerate a given vehicle weight is directly proportional to the
weight.
25 second 1/4 mile = 60 mph end speed
Car Weight....horspower used....kilowatts used
1000 lbs.............16.632...............12.41
1500 lbs.............24.948...............18.61
2000 lbs.............33.264...............24.81
2500 lbs.............41.580...............31.02
3000 lbs.............49.896...............37.22
3500 lbs.............58.212...............43.43
4000 lbs.............66.528...............49.63
Quarter Mile Acceleration Elapsed Time and Terminal Speed | |