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#### Pisoila

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If the car has reached a high speed, what electrical energy do you consume to maintain it at that speed in the 100Km straight line?

I mean a medium-sized car on the road.

If you set an example:
Audi A4 from the year 2000 whose engine has been replaced with an electric one,
It goes on a highway close to a straight line 100Km
With a speed of 100 km / h and travels the road in 1 hour

highway to (does not go up the hill does not go down).

You choose the power of the electric motor
It doesn't slow down, it doesn't accelerate. it goes at about constant speed.

Why this discussion? Because it is important to know how much an electric vehicle consumes when we choose an electric vehicle.

We are told to consume less. Not a good answer. I needed "numbers". How less do you consume?

#### brian_

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Government energy consumption tests include operation in various conditions. The highway tests are close to just driving at a constant speed. Canadian test results are published in various forms including kWh per 100 km, so at 100 km/h that's kWh per hour... or just kW.

A VW e-Golf uses 19.9 kWh/100km in the highway test, and other models use 16 to 30 kWh/100km. So close to 20 kW is a reasonable estimate of consumption at 100 km/h... but if you're interested in what EV uses less energy or just how much energy an EV uses, it's really the consumption averaged over the whole test cycle - not just at a fixed speed - that matters.

#### Pisoila

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~ 20KWh for 100Km

[TH]Petrol/gasoline[/TH]
[TD]44-46 MJ/kg[/TD]​
[TH]Diesel fuel[/TH]
[TD]42-46 MJ/kg[/TD]​

 Gasoline 715-780 Kg/m3
750Kg............ 1m3
750Kg............ 1000 liters

1 Liter =0.75Kg gasoline.... 0.85Kg diesel

20KWh with 30% otto efficiency ....need 20KWh*10/3=67KWh

Gasoline 45MJ/kg=12.5 KWh/kg

12.5KWh..............1Kg
12.5KWh...............(1/0.75)Liters
12.5KWh...............1.33Liters
9.4KWh.................1 Liter

10KWh/L

20KWh is the equivalent of 2 liters of gasoline consumed by the Otto engine.
It would be great if that were the case.

Can a car with gasoline consume 2 liters to go 100 km from the highway? No, consume at least twice as much!

We could "play" with this data. I guess that's why they're public.

#### MattsAwesomeStuff

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A VW e-Golf uses 19.9 kWh/100km in the highway test
19.9kwh/60miles = 331 watthours/mile. Not... great. I'd expect a bit better.

Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/

I was hoping my current stash (~32kwh) would be enough for 200km.

#### brian_

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19.9kwh/60miles = 331 watthours/mile. Not... great. I'd expect a bit better.
Perhaps because your expectations are based on the claims of people who test only in favourable conditions? Actually, the e-Golf is not great for its size, and it's an adaptation of a gas/diesel car, so it's probably not optimal. The best results in these tests are from Tesla, I believe not because they are more efficient but primarily because they have optimized the car's programming to the test cycle - it's a long standing practice in the automotive industry.

Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/

I was hoping my current stash (~32kwh) would be enough for 200km.
On the bright side, you have lower frontal area and probably narrower tires than the e-Golf.

#### Pisoila

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330Wh/mile
That is a 12V car battery at 60Ah (12V*60Ah=720Wh) It's good for 2.2 miles, a little bit.

When I proposed this subject, I thought of Newton's first law.
Ideally we should not consume energy to keep a vehicle moving in a straight line with constant speed.

This can be seen in space. There is friction on the ground, including air friction.

Let's say then what are the factors that determine this consumption in a straight line.
What can we do to reduce them?

#### Pisoila

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Actually, a better phrase would be that it worries me that I'm not planning to carry more battery :/
That is the problem, not the production of energy but its transport! energy storage.

It has to do with this topic, because the less you consume, the less batteries you need.
I'm not saying that what is written in that article is true or not, I'm just quoting it.

#### brian_

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...
20KWh is the equivalent of 2 liters of gasoline consumed by the Otto engine.
It would be great if that were the case.

Can a car with gasoline consume 2 liters to go 100 km from the highway? No, consume at least twice as much!

We could "play" with this data. I guess that's why they're public.
Wow, you shuffle a lot of numbers to get to the result.

I didn't try to follow all of your process, but the energy content of gasoline is assumed in this industry and its regulators to be 8.9 kWh/L, which would be 11.4 kWh/kg at 0.78 kg/L.

If gasoline contains 8.9 kWh/L and 20 kWh is required per 100 km, you can conclude that is somehow equivalent to about 2.3 L or 1.8 kg of gasoline, and that value (in L/100km) is published in the fuel consumption guide, so you don't need to work it out.

That comparison is common, but it is ridiculous because gasoline is not electricity: the gasoline-fueled car must convert the chemical energy in the gasoline to usable mechanical energy, while the electric car is working with very usable electricity. That energy conversion from fuel can be done at 40% efficiency at best with current engines, so the corresponding fuel consumption is at least 2.5 (and more realistically 3 to 4) times as high as that. Indeed, gas-engine cars much cheaper than their EV equivalents are more than that efficient.

If you want a fair comparison of gasoline or diesel to an EV, measure the amount of gasoline or diesel which you would need to feed into an engine running a generator to produce the electricity to distribute via the power utility grid to charge the car. Yes, that's about what gasoline and diesel cars use.

Data note: the heating value of a hydrogen-containing fuel (such as gasoline or diesel) depends on whether the water in the combustion products (the engine exhaust in this case) is a liquid (that's the Higher Heating Value or HHV) or still a vapour (that's the Lower Heating Value or LHV). High-efficiency furnaces can condense the water in their exhaust to collect the heat of vapourization, but a practical engine cannot, so you should use the LHV for engine calculations... for gasoline, that's the 44 MK/kg value.

By the way, I linked to the current (2021) Natural Resources Canada fuel consumption guide, which no longer includes the e-Golf; my e-Golf data was from the 2020 version.

#### brian_

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330Wh/mile
That is a 12V car battery at 60Ah (12V*60Ah=720Wh) It's good for 2.2 miles, a little bit.
In practical terms that doesn't work, because the "60 Ah" car starting/accessory battery can only produce that much output if discharged very slowly (the common test condition is at steady rate for 20 hours). When you try to get it all out in less than an hour, much of the energy is consumed by internal resistance of the battery. But sure... the battery is an energy container.

When I proposed this subject, I thought of Newton's first law.
Ideally we should not consume energy to keep a vehicle moving in a straight line with constant speed.

This can be seen in space. There is friction on the ground, including air friction.

Let's say then what are the factors that determine this consumption in a straight line.
What can we do to reduce them?
By "ideally", in terms of Newtonian laws, you mean in the absence of an external force. Obviously drag is the external force and reducing it reduces the energy required to move the vehicle. The energy required to overcome drag through a distance is simply the total drag force multiplied by the distance, or for a non-steady-state situation the total instantaneous drag force integrated over the distance.

Millions of people have spent more than a century working on the reduction of all forms of drag; of course we should use the resulting knowledge... as every production car does.

#### brian_

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19.9kwh/60miles = 331 watthours/mile.
100 km is 62 miles. If you must use primitive units, it's easier to just convert 199 Wh/km to Wh/mile by converting 199 miles to kilometres (because you need the inverse of km to mile conversion): 320 Wh/mile

#### brian_

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Note: Ayrton.980 is a an advertising spambot, not a real person. There is no point clarifying the nonsensical post by that account.

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If the car has reached a high speed, what electrical energy do you consume to maintain it at that speed in the 100Km straight line?

I mean a medium-sized car on the road.

If you set an example:
Audi A4 from the year 2000 whose engine has been replaced with an electric one,
It goes on a highway close to a straight line 100Km
With a speed of 100 km / h and travels the road in 1 hour

highway to (does not go up the hill does not go down).

You choose the power of the electric motor
It doesn't slow down, it doesn't accelerate. it goes at about constant speed.

Why this discussion? Because it is important to know how much an electric vehicle consumes when we choose an electric vehicle.

We are told to consume less. Not a good answer. I needed "numbers". How less do you consume?
i have a self built mitsubishi express van, a 24kw/144v dc motor is used but i have higher voltage batteries for more km per charge, to answer your question, flat forward 100kmh needs 120amp @ 155v, following behind a truck can go down to 80 amps @ 100kmh, strong wind against my van will be 160 amps, your driving behaviour will determine the distance you can drive per charge

#### brian_

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i have a self built mitsubishi express van, a 24kw/144v dc motor is used but i have higher voltage batteries for more km per charge, to answer your question, flat forward 100kmh needs 120amp @ 155v...
That's 18.6 kW - good for a DIY van.

#### TT-Man

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Hi All,
Did some calculations as follows
Hi Girls, This imaginary vehicle going along a theoretical horizontal 100 kM road, what will the power burn be? So let us look at the theory!
The power burn will be mostly to overcome wind resistance (drag) and rolling resistance. If we take an approximately standard car, work out drag, make an allowance for rolling resistance (not a lot) we can estimate a value to maintain the desired speed.
The standard car that I have chosen is 2M wide by 1.5M high making 3 sq M, and fudge a factor of 0.85. A modern reasonably aerodynamic 4 door saloon has about a 0.3 Cd value. Using these in the following calculations will give a drag force, from which we can determine the power ‘burn', and energy for the 100 kM drive.
Power = Force X Velocity. You want to know the power, you have selected the velocity at 100 km/hr so we need to know the force required.
At a constant speed that force is mostly drag. Drag Force F = Cd . p . ½ V . V .
A Cd factor ranges from 0.25 low, 0.3 mid modern 4 door, 0.4 SUV.
Density of air p = 1.225 Kg/ M3
Area of vehicle normal to direction A = 0.85 x Width x Height (an approximation).
100 Km/hr is 27.78 m/sec.
So we have might have: Drag = 0.3 x 1.225 x 0.5 x 27.78 x 27.78 x 0.85 x 2 x 1.5 Drag force = 361.6 Newtons
Power = F x V = 361.6 x 27.78 = 10045 Watts .... ????
10.045 kW Or 13.5 Hp or consumption per hour, nominally 10 kWhr
This does not take into consideration rolling resistance which is small, but exists. This is perhaps a guide for an ideal constant speed running. You need to consider a factor for head winds, and the power to accelerate the mass of your vehicle up to the 100 kM. I was surprised how much power was needed, unless you accept a very long acceleration time. In my previous calculations a decent hill climbing rate also was a surprise.
If we do an out and back against a 15 kM head wind in one direction and a 15 kM tail wind in the other we can see what happens to consumption. I will be interested! If I put all the common factors into a constant K, we get:- 0.4686
So burn against the head wind = k x (27.78 +4.17)squared = 0.4686 x 31.95 x 31.95 = 478.35 N The velocity is still the same so power is F x V = 478.35 x 27.78 = 13,288.6 W (13.3 kWhr)
The return journey will be : K x (27.78 – 4.17) squared = 261.2 N.
As above Power = 261.2 x 27.78 = 7,256.5 W (7.26 kWhr)
The average = 20545.1 /2= 10272.5 W. (10.27 kWhr) Interesting a surprisingly small increase. Note that the weight of the vehicle is not included in the calculation. This would potentially effect rolling resistance. I suppose adding a nominal 10% or less would cover that.
Adjusting for other variables would be a linear calculation. Cd, and cross-sectional area would be to divide the power by the existing variable and multiply by the new.
For example if the Cd was say 0.4 then new power burn = 10 x 0.4/0.3 = 13,333 W
If CS Area also increased from 3 sq mtr to 4 sq mtr then new power = 13,333 x 4/3 = 17,777 W Now I am rusty on my maths and all the above may be cobblers, but hey Guys, trying to help. Cheers, Joe
P.S. Hill climbing and acceleration are different sums, and weight (mass) do come into the picture! I take it that someone lives on the flat lands!

#### MattsAwesomeStuff

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Hi All,
Did some calculations as follows
Hi Girls,
hey Guys, trying to help. Cheers, Joe

...

Rolling resistance accounts for about 40% of total power required on a typical EV. You can almost ignore it for bicycle weight things, but not cars.

Here's a calculator:

Just remember to zero out Acceleration unless you're specifically solving for that.

I don't know what "fudge factor" is or why you would include that number versus any other number, but this calculator predicts ~11kW for air resistance (at zero weight) and ~22kW for total power (so, another 11kw for rolling resistance) presuming a 2000kg 4-door car.

If you want to climb a modest hill you'll need to roughly double your power requirements. If you need to accelerate up a hill, triple them.

#### brian_

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You can almost ignore it for bicycle weight things, but not cars.
Perhaps, because those things have terrible aerodynamics and rock-hard skinny tires.

#### TT-Man

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Hi Matt & Brian,
The calcs were all my own effort. The formulas were refreshers from the net. I copied from my file on MS Word so limited in math functions ... hand held calculator ... steam tech! ...No a slide rule and logs would be steam tech.
The suggested estimate for rolling resistance came from an observation on the net. I do not believe it to be as large as you suggest, unless on a large ut with fat chunky tyres (tires).
The area calc fudge factor allows for rounding of the nominal rectangle. I chose a 2 x 1.5 giving a nominal 3 sq mtrs. The 0.85 allows for the fact that it is't a rectangle, but rounded.
Remember this is all a bit of a gestimation.
Just a comment about me. Health wise I am now limited in physically activity, plus we have gone back to lock up. So this is an interesting diversion rather than TV..
Cheers all,
Joe

#### brian_

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The suggested estimate for rolling resistance came from an observation on the net. I do not believe it to be as large as you suggest, unless on a large ut with fat chunky tyres (tires).
Either you misunderstood what you read on some web site, or it was not applicable to typical cars.

Using the most optimistic value of CRR = 0.01 from the Engineering Toolbox, the rolling drag force is 98.1 N per 1,000 kg of vehicle mass, or 2.73 kW per 1000 kg at 100 km/h. Realistic values are up to twice as high. That's not negligible.

#### TT-Man

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Hi Brian,
OK, I shall accept your better sources. So what R.R. at 100 k would you suggest for the example?
Cheers, Joe

#### brian_

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OK, I shall accept your better sources. So what R.R. at 100 k would you suggest for the example?