How to calculate power required by vehicles

[likephysics]

I am trying to compare electric power vs gas power.
How do you calculate the power required by gas vehicles, say to go 50miles at constant speed.

 

[physixlover]

I am trying to compare electric power vs gas power.
How do you calculate the power required by gas vehicles, say to go 50miles at constant speed.

insufficient data.

 

[quantumdoodle]

yes. we need the mass of the car. the friction index, fuel efficiency and all that other stuff and may be the cost of the two type of fuels? r u trying to do some sort of reserch project?

 

[Lsos]

I am trying to compare electric power vs gas power.
How do you calculate the power required by gas vehicles, say to go 50miles at constant speed.

Speed is the most important part you missed here. Any discussion of power needs a mention of speed.

For example, depending on how fast it's going, a car might need 25 watts or it might need 10,000 horsepower.

 

[likephysics]

Quantum, not for a project. Just curiosity.
Lets say the car is Honda civic. Let's assume its a 2000lbs, fuel efficiency is 30MPG.
Can we ignore the friction index?
How much power would be required to go 100miles?

 

[conway]

At some point air resistance dominates rolling friction. That's because it is quadratic with velocity. A quick measure of air resistance is given by the Bernoulli pressure 1/2*(rho)*(v^2). For 40 m/sec this gives approx 1000 N/m^2. Take this over the cross section of the vehicle (approx 1 sq meter) at the velocity in question and you get a power of 40 kW or about 50 horsepower.

Streamlining helps only up to a point. That's because the friction of sliding sideways through the air eventually equals the amount of friction of butting headlong into the air. This point occurs for a cylinder approx 40 times its diameter in length. In the above example, then for a cylindrical car 40 meters long and 1 meter in diameter, you would have 100 horsepower consisting of 50 horsepower butt-end friction and 50 horsepower irreducible sliding friction.

 

[Lsos]

Quantum, not for a project. Just curiosity.
Lets say the car is Honda civic. Let's assume its a 2000lbs, fuel efficiency is 30MPG.
Can we ignore the friction index?
How much power would be required to go 100miles?

You can't ignore friction, becase that's the main reason a car burns gas in the first place. If there was no friction, you can just get it up to speed, turn off the engine, and coast forever or until you get to your destination. Power required = 0.

And again, for a proper discussion of power, you need to know the speed. conway even had to make one up to attempt to solve the problem. But it can vary anywhere between 0 to millions of horsepower.

Perhaps what you're asking about is energy required. If that's the case, you almost answered your own question when you said "fuel efficiency is 30mpg."
Simple math tells us that at 30mpg, we would consume 1.66 gallons of fuel in 50 miles.
That's 219MJ of energy.
Assuming a car is 25% efficient, it actually uses 55MJ of that energy.
I'll assume an electric motor would be 80% efficient, which means a battery would probably need about 69MJ to go 50 miles under the same conditions.

 

[likephysics]

Thanks everyone.
Lsos, yes, I was trying to ask about energy.
Can you tell me how you arrived at 219MJ?

 

[Bob S]

Gasoline contains about 44,000 joules per gram, or about 120 MJ per gallon, so 1.66 gallons represents about 200 MJ of energy.

The two biggest kinetic energy power losses are due to air drag and tire rolling friction.

Air drag power loss is See http://en.wikipedia.org/wiki/Drag_(physics)

P_drag= ½ρ·A·C_p·v³

Where A = frontal area of vehicle, ρ = air density, C_p = drag coefficient, v = velocity.

The other kinetic power loss is due to tire rolling friction see http://en.wikipedia.org/wiki/Rolling_resistance

P_tire = RRC·m·g·v ≈0.01 m·g·v

where RRC= tire rolling resistance coefficient, m= vehicle mass and g = 9.81 m/s²

Bob S

 

[likephysics]

Bob, thanks for the explanation.

 

[Lsos]

Thanks everyone.
Lsos, yes, I was trying to ask about energy.
Can you tell me how you arrived at 219MJ?

I just looked up "gasoline" on wikipedia. It said "132MJ/gallon". I multiplied it by 1.66

 

[Bob S]

This Wiki table shows automotive regular gasoline to be 114,000 BTU per gallon, or 120.3 MJ per gallon, using 1055 MJ per BTU.
http://wapedia.mobi/en/Gasoline_gallon_equivalent

This table from Oak Ridge national Laboratory in USA

http://bioenergy.ornl.gov/papers/misc/energy_conv.html

under Fossil Fuels states Gasoline: US gallon = 115,000 Btu = 121 MJ = 32 MJ/liter (LHV). HHV = 125,000 Btu/gallon = 132 MJ/gallon = 35 MJ/liter

HHV (higher heat value) is with all energy used, including condensing water vapor after combustion.

LHV (lower heat value) is with steam (water vapor) exhausted after combustion.

Bob S

 

[Phrak]

I am trying to compare electric power vs gas power.
How do you calculate the power required by gas vehicles, say to go 50miles at constant speed.

As you have asked, the numbers favor electic over gasoline. But this is misleading. Electric power, for the most part comes from coal, fuel oil, hydroelectric and other sources.

Conversion of coal, fuel oil, hydroelectric and other sources to charge electric car batteries involves significant energy loss.

There are losses in conversion to electricity. There are power transmission losses. There are losses in converting the voltage of the power mains to the voltage required to charge the batteries. There are incured charging losses in the batteries; it takes more energy to charge the batteries then they acquire.

And in the end, the bottom line is not BTU comparison but cost.

 

[likephysics]

As you have asked, the numbers favor electic over gasoline. But this is misleading. Electric power, for the most part comes from coal, fuel oil, hydroelectric and other sources.

Conversion of coal, fuel oil, hydroelectric and other sources to charge electric car batteries involves significant energy loss.

There are losses in conversion to electricity. There are power transmission losses. There are losses in converting the voltage of the power mains to the voltage required to charge the batteries. There are incured charging losses in the batteries; the batteries takes more energy to charge the batteries then they acquire.

And in the end, the bottom line is not BTU comparison but cost.

Phrak, I am keeping all this in mind.
Also, extracting gasoline involves loss. I am not sure if its comparable with the electric power conversion loss. Probably would be.
This whole thing started bcoz I got excited about supercapacitors. I saw 120F and got excited. But the energy stored is much less than NiMh battery. Only advantage is instant charging.

 

[Phrak]

Like you I'm very keen on electric vehicles. I'm not sure how this will develop.

The extraction of gasoline from crude costs about 15% energy, if memory serves me. I don't know what pumping and other energy costs are. But, really, what will drive us as the human beings that we are, is the incured cost--our labours, not energy units.

Isn't the energy density of Lithium polymere better than NiMH?

 

[OmCheeto]

Phrak, I am keeping all this in mind.
Also, extracting gasoline involves loss. I am not sure if its comparable with the electric power conversion loss. Probably would be.
This whole thing started bcoz I got excited about supercapacitors. I saw 120F and got excited. But the energy stored is much less than NiMh battery. Only advantage is instant charging.

If you are comparing gas to electric, it is good to know that gas cars get really bad gas mileage at low speeds.
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Graph of a typical ICE

The graph of a typical electric vehicle would start out at extremely high equivalent mpg's at zero mph and drop according to the graph described by the two equations in Bob_S's post #9.

 

[sophiecentaur]

To make a fair comparison between the two systems, you would need to introduce regenerative braking. It would be applicable to electric cars or to hybrids. The hybrid could be an optimal solution, so that would make the answer to the original question more complex, in as far as there wouldn't, ideally, be an either / or but a bits of both answer.

 

[likephysics]

Like you I'm very keen on electric vehicles. I'm not sure how this will develop.

The extraction of gasoline from crude costs about 15% energy, if memory serves me. I don't know what pumping and other energy costs are. But, really, what will drive us as the human beings that we are, is the incured cost--our labours, not energy units.

Isn't the energy density of Lithium polymere better than NiMH?

The Li-poly battery in your cell phone(3.7V, 1000mAH) has comparable energy as a NiMh used in digital camera(1.2v, 2500mAH).
Li-Po has 13.3KJ
NiMh has 10.8KJ

I remember reading that LiPo cannot be charged at a fast rate(1C). That may not be true.

 

[uart]

The Li-poly battery in your cell phone(3.7V, 1000mAH) has comparable energy as a NiMh used in digital camera(1.2v, 2500mAH).
Li-Po has 13.3KJ
NiMh has 10.8KJ

I remember reading that LiPo cannot be charged at a fast rate(1C). That may not be true.

Hi likephysics. You don't compare energy density by comparing the total energy of two random batteries. Energy density can refer to either volume energy density (kJ/L) or mass energy density (kJ/kg). For electric vehicles the latter (mass) is usually the most important consideration.

The mass energy density of NiMH is about 250 kJ/kg compared with up to 700 kJ/kg for Li Ion. So Li Ion is quite a long way ahead in this regard. NiMH's main advantage in the past has been that it's more tried and tested, with proven long term reliability in both early EV prototypes and production vehicles alike. LiIon (and it's derivates like Li polymer and Li Iron Phosphate etc) were later comers and so are a bit less tried and tested in this area, though their adoption and reliability in EV application is rapidly progressing.

 

[OmCheeto]

The Li-poly battery in your cell phone(3.7V, 1000mAH) has comparable energy as a NiMh used in digital camera(1.2v, 2500mAH).
Li-Po has 13.3KJ
NiMh has 10.8KJ

I remember reading that LiPo cannot be charged at a fast rate(1C). That may not be true.

Someone here at the forum pointed out to me that there is development on fast charging LiPo batteries:

http://web.mit.edu/newsoffice/2009/battery-material-0311.html", though the point of this thread has been that fast charge batteries won't do you much good as the charging infrastructure required is impractical.

Using their new processing technique, the two went on to make a small battery that could be fully charged or discharged in 10 to 20 seconds (it takes six minutes to fully charge or discharge a cell made from the unprocessed material).

Though I'm not sure how they come up with the "six minutes". The battery in my laptop takes hours to charge:

http://www.apple.com/batteries/"

I'd pursue your super capacitor idea for now, lest fast charge LiPo's are just vaporware.

 

[likephysics]

omcheeto, I doubt the 6min charging claim.
Supercaps are still in infancy. Right now they don't seem to be the solution.They are great for those rechargeable (wind it yourself) flash lights.
The energy stored in them compared to a NIMH cell is really less(1/10).
I did read about an invention at MIT that reduced charging time by using nanomaterials etc.
They are supposed to be working on making it commercial.

 

[OmCheeto]

Supercaps are still in infancy. Right now they don't seem to be the solution.They are great for those rechargeable (wind it yourself) flash lights.

How quickly the excitement wears off...

This whole thing started bcoz I got excited about supercapacitors. I saw 120F and got excited.

Supercaps are a solution. You just have to understand which problem they solve.

So I'll go back and answer your original question with the piece of the puzzle they solve:

A $1400, 50 watt hour supercapacitor will accelerate a 2000 lb vehicle to 45 mph, and then it will be dead.
But more importantly, they can be recharged to absorb most of that energy when coming to a stop.
And that is something no current battery can do, in a practical manner.

 

[Phrak]

But more importantly, they can be recharged to absorb most of that energy when coming to a stop.
And that is something no current battery can do, in a practical manner.

Why is that?

 

[Bob S]

Unlike batteries, which are based on chemical energy storage, supercapacitors store energy using the classic electric capacitance equation E = ½C·V².

So at constant power, P = -dE/dt = -C·V·dV/dt = constant.

So the equation is quadratic in voltage drop vs. percent of capacity discharged until the battery is completely discharged. The supercapacitor discharge curve is compared to several battery types in the attachment.

Bob S

 

[Phrak]

Yeah, that's just my question, Bob. I don't see how capacitors are any more useful for regenerative braking than batteries unless I'm missing something. They seem even worse for the task with the larger voltage span that needs to be matched with variable breaking power.

 

[Bob S]

There is a special type of very versatile dc-to-dc switch-mode power supplies called a SEPIC converter. These power supplies can convert in both buck mode (reduce output voltage) and boost mode (increase output voltage) at high conversion efficiencies. So they can accommodate the large voltage swings in the supercapacitor voltage, as well as the voltage output from regenerative breaking. See
http://en.wikipedia.org/wiki/Switched-mode_power_supply
http://en.wikipedia.org/wiki/SEPIC_converter

Bob S

 

[likephysics]

There is a special type of very versatile dc-to-dc switch-mode power supplies called a SEPIC converter. These power supplies can convert in both buck mode (reduce output voltage) and boost mode (increase output voltage) at high conversion efficiencies. So they can accommodate the large voltage swings in the supercapacitor voltage, as well as the voltage output from regenerative breaking. See
http://en.wikipedia.org/wiki/Switched-mode_power_supply
http://en.wikipedia.org/wiki/SEPIC_converter

Bob S

Great. I was just reading about sepic converter 2 days ago. Good to know its application.

Phrak,
I think they use caps for regenerative braking, as they can be charged quickly compared to batteries.

 

[Phrak]

Phrak,
I think they use caps for regenerative braking, as they can be charged quickly compared to batteries.

OK. I don't know who 'they are, but if you dump braking energy into capacitors in a vehicle that is battery powered, what will you do with the energy once the capacitors are charged? Extra systems have to be involved in removing useful energy from the capacitors in parallel with drawing power from the batteries where each are at different voltages. This involves current sharing techniques. It's not out of the question, but it does make power control more complicated. But this is just the beginning.

Secondly, it will do very little good for braking purposes to dump charge into a discharged capacitor. The capacitor represents no load at all when discharged. To be of use, they have to be maintained at some minimum voltage level. An additional constraint on a minimum required voltage is a result of the circuit elements. Power is equal to voltage times current. For a low voltage load, such as a capacitor nearly discharged, the current will have to be high to absorb the same power. There will be some practical upper bound on the current carrying capability of the components charging the capacitors.

There will need to be a back-up system if the capacitors are below some threshold voltage.
Either the usual friction brakes come into play, or an alternative dump to the batteries or a resistor bank needs to be employed. Alternatively, the car doesn't go until its capacitors are charged enough to be capable of stopping the car. For extended downhill breaking, a backup system would be necessary after the capacitors have acquired their maximum charge voltage.

And when all is said and done, very little energy recovery can be obtained in most driving situations. The best case recovery is during stop-and-go traffic where driving speed losses from air resistance, rolling resistance and drive train losses are comparatively low compared to the energy required to accelerate.

 

[sophiecentaur]

Capacitors have some advantages (see https://www.physicsforums.com/showthread.php?t=371697&highlight=super+capacitor) in that they can undergo many cycles and can be made with very low internal resistance. They don't. of course, have a nice steady 12V, like a Lead Acid battery and they need a lot of 'help' from technology - switching circuits etc.. From what I've been reading here and elsewhere, they may well be the way to go. The low operating voltage of present super capacitors limits their application but they are being used successfully even now, in some fields.
Just because of their exponential voltage characteristic, they can't just be written off. I'm almost becoming a convert. . . .

 

[Phrak]

Umm. The voltage characteristics are not exponential but polynomial, isn't it?. But in any case the one and only advantage of these capacitors is the ability to absorb large amounts of braking energy on demand (when not under or over charged). Everything else complicates and adds cost to their advantage.

In the thread you quoted I see the statement "Braking power can easily exceed 150KW;..."

Assuming this number is good, that's a lot of Watts. That's a lot of current, even dumped into a stack of super capacitors precharged to 150V.

Energy dumped into a capacitor of variable voltage by a variable pressure on a brake peddle will need to be modulated to meet the braking needs. This requires regulating circuitry and BIG inductive elements. Inductors are resistive (energy wasters) and big at this power level. This represents a lot of copper, a lot of loss, a lot of weight, and a large up-front cost.

Search the internet. Find me a a toroidal inductor rated at 1000 Amps DC. Is it smaller than a bread basket?

 

[sophiecentaur]

I agree with most of that; at least, I did, initially.
But you can't just write them off:
At present, all motors and generators are designed with constant supply volts in mind and that is clearly the most 'convenient' but the fact that capacitors can dump and supply considerable dollops of energy, rapidly seems to be what appeals about them. The energy per kg figures are around that of many rechargeable cells.
The pedal pressure thing wouldn't be a factor if the back pressure were supplied by a servo; whatever you were dumping the energy into, you'd still have a similar problem and you'd still need some Forodo somewhere in the system - just to be safe.
If all the control of charge and discharge were carried out using (a suitable) high frequency AC, the values of inductors could be a lot lower. Aero AC systems use 400Hz, I believe; why not 40kHz, for instance? High power semiconductors are pretty impressive these days. The fact that the available Super Capacitors have only a few volts maximum working conditions needs some ingenuity but they could well get better. Conventional Electrolytics work at much higher voltages than when they were first available.

I was referring to exponential discharging / charging through a resistor. But, of course,
E = C Vsquared, if you're talking about stored energy. It's just not a friendly device to use, like a car battery.

 

[Naty1]

it takes more energy to charge the batteries then they acquire

I haven't read much about the latest in battery development but older style lead acid batteries lose about 15% to 20% of charging power to heat...the battery gets noticeably warmer when charged. AGM batteries with extremekly low internal resistance lose only a few percent, maybe 2% to 4% to heating when charged.

In addition AGM batteries can accept a charge up to their full rating in amp hours...so,for example, a 200 amp hour AGM battery can be charged at up to 200 amps...a tradional wet cell lead acid battery, like the ones in most cars, will only accept about 25% of their rating in amps...about 50 amps for a 200 AH rated battery.

In addition an alternator, likely loses another 15% or more of engine power in belt friction, heating, and internal frictional and air movement losses. Likewise battery chargers have their own losses...as do inverters if boosting battery voltages to higher ac levels...these also are only about 80 or 85% efficient.

I opted for a lot of battery sourced power aboard my boat when cruising both for quiet and to avoid even greater losses and noise in running a gas or diesel generator. I got my quiet for reding physics, but the losses were substantial no matter what when away from commercial dockside power.

 

[Phrak]

I agree with most of that; at least, I did, initially.
But you can't just write them off:

I find capacitors as applied to braking systems inferior to batteries for the numerous reasons I've already give. In nearly parallel argument they are inferior to batteries in supplying power. Without evidence to the contrary I don't find them worthy of consideration.

At present, all motors and generators are designed with constant supply volts in mind and that is clearly the most 'convenient' but the fact that capacitors can dump and supply considerable dollops of energy, rapidly seems to be what appeals about them. The energy per kg figures are around that of many rechargeable cells.

I understand that. This is the initial attraction to capacitors. But you have to carry the analysis beyond the favorably qualities to the disadvantageous qualities if you are interested in a fair comparison

The pedal pressure thing wouldn't be a factor if the back pressure were supplied by a servo;

Pedal pressure is roughly equivalent to braking power. This manifests as a potential across the drive motor now used as a generator. This power, in a dynamic braking system, in whole or in part, is to be sent to a storage device. Whatever voltage is developed in this process needs to be matched to the voltage of the device that will be doing the storing. The unmatched voltage times the current generated is simply more lost heat.

With capacitors, the variation in voltage is compounded as the product by both the generator's swing and the swing in capacitor voltage. I've already explained this above in shorter words.

So is the lossive inductive element that will withstand these variations bigger or smaller than a bread basket?

I have a way-to-long background in power conversion electronics.
Though I am not sufficiently motivated actually solve this particular problem numerically, given some theoretical values, I'm quite confident that all current power conversion electronics are dismally insufficient to make this a practically competitive alternative to batteries.

 

[sophiecentaur]

You're probably right in the end. Dissipating hundreds of kW is hard enough (with brakes) - storing energy at that rate sounds even harder. Bigger than a breadbasket by quite a bit, I should imagine.
I still don't get your point about brake pedal force. Of course it applies in basic brakes but, in all other systems - including your everyday servo - it is made artificially so because we like it that way. I have driven a truck with some air brakes which were more or less on/off and there was virtually no feel back. A nightmare when you are not carrying a load.

 

[likephysics]

Just a question about regenerative braking - are the caps charged directly from the braking mechanism, Brakes--->dynamo-->charging ckt-->caps
or is it like
Brakes-->some mechanical winding system(like a spring or a flywheel)--> the spring unwinds slowly -->dynamo-->charging ckt-->caps.

 

[Phrak]

sophiecentaur. Sorry, by variable brake peddle pressure I was referring to, in rough terms to the variable power presented to the conversion circuit. The actual pressure has nothing to do with it. In effect, your foot pressure is just a mean to tell the braking circuitry how much power needs to be absorbed to reduce the vehicle's kinetic energy at any given time.

likephysics. Good thinking! I don't know if it's a viable idea, but I've only been thinking in the electronics box. Maybe you found a way around it.

Say the flywheel can quickly absorb a vehicle's kinetic energy during a braking episode. Now you may not have to dump it so quickly into electrical storage. The longer the time you have to do this, the smaller the currents that are involved. The smaller the currents, the more efficient, and the smaller, simpler and lighter the electronics that is required to do this.

 

[sophiecentaur]

It's amazing to think that, in a very few years, we are all likely to be driving vehicles which use something like we're discussing. Something else to go wrong, I suppose!

I am reassured by my (sailing) boat engine (diesel) which is 35ys old, has a flywheel that you can hardly lift and one bit of electronics - the rev counter. It seems to be refusing to die. (Refuses to start, too, on occasions)

 

[mheslep]

And in the end, the bottom line is not BTU comparison but cost.

In which case the energy cost for an electric mile is well established to be 3-4X less expensive than a gasoline powered mile, given the average efficiency of the US gasoline vehicle and the current price of gasoline.

 

[mheslep]

Someone here at the forum pointed out to me that there is development on fast charging LiPo batteries:..

Almost, that was Lithium Ion Iron Phosphate, LiFePO, that has demonstrated high power, not the polymer version of Li ion which is inexpensive but lower power as mentioned above.

 

[mheslep]

Adding a bit more to your fine response...

At some point air resistance dominates rolling friction.

Yes, break even (drag vs rolling resistance ) is about 40 km/h (25mi/h) in a sedan, and interestingly only 10 km/h for a bicyclist.

That's because it is quadratic with velocity. A quick measure of air resistance is given by the Bernoulli pressure 1/2*(rho)*(v^2). For 40 m/sec this gives approx 1000 N/m^2. Take this over the cross section of the vehicle (approx 1 sq meter) at the velocity in question and you get a power of 40 kW or about 50 horsepower.

Yes, just so for 1 M^2 at 40m/s = 144 km/h = ~90 mi/h, moving right along. The average car effective drag area is actually a bit lower, Cd*A = ~0.8 M^2, and at the highway speed of 60 mi/h the tractive load for air drag alone is about 18HP. The rolling resistance coeffient for a sedan is about Crr=0.01. So for a one ton vehicle the rolling resistance force is MgCrr = 1000kg x 10 x 0.01 = 100N, a constant over any speed. The power load for rolling resistance is also F x V, so at 60 mi/h we have 100N * 26.6M/s = 2.66kW = 3.6HP. Total tractive load, drag and RR, at 60 mi/h level road, constant speed, 1 ton sedan is ~22HP. Another metric is energy cost per distance, in this case that is 0.27 kWh per mile ( about 3 cents per mile at my electric rates).

Streamlining helps only up to a point. That's because the friction of sliding sideways through the air eventually equals the amount of friction of butting headlong into the air. This point occurs for a cylinder approx 40 times its diameter in length. In the above example, then for a cylindrical car 40 meters long and 1 meter in diameter, you would have 100 horsepower consisting of 50 horsepower butt-end friction and 50 horsepower irreducible sliding friction.

True, though there's a 10x gap between the Cd of the average sedan (0.3) and the average airplane (0.03)

From D. McCay's excellent reference on this subject.
The energy cost per 100 km here includes a engine efficiency of 0.25. To get actual tractive load divide by 4.

http://www.inference.phy.cam.ac.uk/withouthotair/cA/page_254.shtml