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How to calculate power required by vehicles 
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#1
Jan2710, 01:58 PM

P: 618

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. 


#2
Jan2710, 02:49 PM

P: 87




#3
Jan2710, 03:04 PM

P: 18

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?



#4
Jan2810, 12:40 AM

P: 777

How to calculate power required by vehicles
For example, depending on how fast it's going, a car might need 25 watts or it might need 10,000 horsepower. 


#5
Jan2810, 01:43 AM

P: 618

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


#6
Jan2810, 04:41 AM

P: 398

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 buttend friction and 50 horsepower irreducible sliding friction. 


#7
Jan2810, 07:14 AM

P: 777

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. 


#8
Jan2910, 10:19 PM

P: 618

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


#9
Jan2910, 10:47 PM

P: 4,663

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}·v^{3} 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/s^{2} Bob S 


#10
Jan3010, 05:07 AM

P: 618

Bob, thanks for the explanation.



#11
Jan3010, 07:10 AM

P: 777




#12
Jan3010, 11:21 PM

P: 4,663

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 


#13
Jan3010, 11:59 PM

P: 4,512

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. 


#14
Jan3110, 01:44 AM

P: 618

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. 


#15
Jan3110, 02:05 AM

P: 4,512

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 costour labours, not energy units. Isn't the energy density of Lithium polymere better than NiMH? 


#16
Jan3110, 03:20 AM

#17
Jan3110, 05:45 AM

Sci Advisor
Thanks
PF Gold
P: 12,269

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.



#18
Jan3110, 06:16 AM

P: 618

LiPo 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. 


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