# Energy density and range calculation Electric Vehicle

 P: 6 I want to calculate the range of an Electric Vehicle (EV) with a Li-ion battery of 200 Wh/kg. Parameters v= 50 km/h = 14 m/s A= 0.8 m2 d = 500 m efficiency = η= 85% mc = 900 kg incl. 100 kg battery ρ = 1.3 kg/m3 First I have calculated the power of the EV with a speed of 50 km/h and 100 kg battery. Power car= 1/2η [m v^3/d+ρAv^3 ]=1/(2×0.85) [1000×〖14〗^3/500+ 1.3 ×0.8×〖14〗^3 ] Power car = 4.91 kW Energy density battery = 200 Wh/kg. With a 100 kg = 20 kWh Driving time on a full charge (hours)=(power of battery (kWh))/(power of car (kW))=20/4.91= 4.07 hours Range (km)=Speed (km⁄h)×time (h)= 50 ×4.07 = 204 km Is this correct? I am not sure if you can divide the power of battery with power of car.
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 P: 6 Well, the problem is that if I do the same calculation with a 200 kg of battery (mass car is then 1100) then the range would be less. I don't think that this is correct, the range should be more of course.
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Energy density and range calculation Electric Vehicle

 Quote by Bianca12 I want to calculate the range of an Electric Vehicle (EV) with a Li-ion battery of 200 Wh/kg. Parameters v= 50 km/h = 14 m/s A= 0.8 m2 d = 500 m efficiency = η= 85% mc = 900 kg incl. 100 kg battery ρ = 1.3 kg/m3 First I have calculated the power of the EV with a speed of 50 km/h and 100 kg battery. Power car= 1/2η [m v^3/d+ρAv^3 ]=1/(2×0.85) [1000×〖14〗^3/500+ 1.3 ×0.8×〖14〗^3 ]
Where does 500m come from? You appear to have calculated the average power needed to go from rest to 50k/h over some negligible distance, then maintain speed for 500m. How does that equate to the power needed over the whole driving range? Is the assumption that you will have to stop on average every 500m?
You gave car mass as 900kg including 100kg battery, but you've added the battery weight on to get 1000kg.
 if I do the same calculation with a 200 kg of battery (mass car is then 1100) then the range would be less
Are you sure? Try getting the general algebraic expression before plugging any numbers in.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,470 If the original mass of the car was 900 kg including the 100 kg battery, how can putting in a 200 kg battery (in place of the 100 kg battery) make the car's mass 1100 kg?
P: 6
 Quote by haruspex Where does 500m come from? You appear to have calculated the average power needed to go from rest to 50k/h over some negligible distance, then maintain speed for 500m. How does that equate to the power needed over the whole driving range? Is the assumption that you will have to stop on average every 500m? You gave car mass as 900kg including 100kg battery, but you've added the battery weight on to get 1000kg. Are you sure? Try getting the general algebraic expression before plugging any numbers in.
The 500m is the distance between two stops. This is very common in the city. The car that I am talking about drives only in the city and you need to stop every 500m (traffic).
The mass of the car is 900 kg and 100 kg battery (thus 1000 kg). If the battery is 200 kg the total mass will be 1100 kg. In the calculation I have done it correct, but in the given parameters it is not correct.

There is something about this calculation that is not correct. But I can't quite figure out what the problem is. If the car stops every 500m there is a lot of energy lost due to braking, but how then can you drive about 200 km (calculated)?

Thank you, I hope it is clear what I have done.
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You seem to be taking the drag coefficient as 1, whereas at http://en.wikipedia.org/wiki/Drag_(p..._high_velocity it says .25 to .45. Using your numbers (including 14m/s instead of 50kph) in a spreadsheet I get the same result as you for 100kg of battery, but when I change it to 200kg I get a range of 385km. That seems reasonable in proportion.
Putting in a drag coefft of .35, and an exact 50kph, the range (for 100kg battery) increases to 268km. On top of that, you're taking the driving pattern to be full speed or nothing. Allowing for likely patterns of acceleration and deceleration might add 5%.
So I can't find a good reason to say you're overestimating the range.
P: 6
 Quote by haruspex You seem to be taking the drag coefficient as 1, whereas at http://en.wikipedia.org/wiki/Drag_(p..._high_velocity it says .25 to .45. Using your numbers (including 14m/s instead of 50kph) in a spreadsheet I get the same result as you for 100kg of battery, but when I change it to 200kg I get a range of 385km. That seems reasonable in proportion. Putting in a drag coefft of .35, and an exact 50kph, the range (for 100kg battery) increases to 268km. On top of that, you're taking the driving pattern to be full speed or nothing. Allowing for likely patterns of acceleration and deceleration might add 5%. So I can't find a good reason to say you're overestimating the range.
I understand the calculation errors you've mentioned above, I have corrected for that. But I don't get more range for more kg battery.
If you look at the formula's more mass of car means more power needed and therefore less range, because the driving time will be less (and thus less distance).
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 Quote by Bianca12 I understand the calculation errors you've mentioned above, I have corrected for that. But I don't get more range for more kg battery. If you look at the formula's more mass of car means more power needed and therefore less range, because the driving time will be less (and thus less distance).
That's not the result I get, using your formula. Pls post your working for that case.
 P: 6 Could not insert table.
 P: 6 Here is the table with the parameters (the parameters are OK, they are not the problem). v= 50 km/h = 14 m/s Effective area = C(d)A= 1 m2 d = 500 m efficiency = η= 85% mc = 1000 kg (100kg battery) or 1100 kg (200 kg battery) ρ = 1.3 kg/m3 100 kg battery (total mass 1000kg): power car (kW) 5,33 time (h) 3,75 range (km) 187,74 Formula's: Power (kW): ((1/(2*0,85))*(1000*(14^3/500)+1,3*1*14^3))/1000 Time (h) = 20/5,33 Range (km) = 3,75 * 50 For the 200 kg (total mass 1100kg) battery I get these answers: power car (kW) 5,65 time (h) 3,54 range (km) 177,01
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 Power car = 1/2η [m v^3/d + ρ Av^3 ]
Can you explain how you arrived at that equation?

0.5 ρ Av^3 looks like the instantaneous drag in Watts. However you can't just set v=50kph because the car is accelerating? Means V and hence drag isn't constant.

0.5 m v^3/d looks like KE / time ? eg 0.5mv^2 * v/d. Which again looks wrong because the velocity isn't constant.

I think you need to plot a graph of velocity v time for the 500m segment and integrate it to get the energy consumed. If we call that E500 (in Joules). Then

Range = 500 * Battery Capacity (in Joules) /E500

PS: You don't say how fast the car accelerates to 50kph and that will certainly alter the average speed and power consumption....just as it does for a gas car.
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 Quote by CWatters Can you explain how you arrived at that equation? 0.5 ρ Av^3 looks like the instantaneous drag in Watts. However you can't just set v=50kph because the car is accelerating? Means V and hence drag isn't constant. 0.5 m v^3/d looks like KE / time ? eg 0.5mv^2 * v/d. Which again looks wrong because the velocity isn't constant. I think you need to plot a graph of velocity v time for the 500m segment and integrate it to get the energy consumed. If we call that E500 (in Joules). Then Range = 500 * Battery Capacity (in Joules) /E500 PS: You don't say how fast the car accelerates to 50kph and that will certainly alter the average speed and power consumption....just as it does for a gas car.
Bianca's model considers that the car cruises at a steady 50kph for 500m intervals. The acceleration and deceleration stages in each interval are presumed unimportant. My estimate is they'll not make more than a 5% difference to the answer.
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 Quote by Bianca12 For the 200 kg (total mass 1100kg) battery I get these answers
Yes, but I don't, and I keep asking you to post your working that gets those answers.
Var	Item          	Units	Formula	sc 1	sc 2
eta	efficiency		                	0.85	0.85
mc	car mass     	kg	        	900	900
mb	battery mass	kg	        	100	200
k	energy density	J/g	        	720	720
A	x area        	m2	        	0.8	0.8
rho	air density    	kg/m3    	       	1.3	1.3
v	cruise         	m/s	        	14	14
c	drag coeff    		        	1	1
d	drag           	W	c*v^3*rho*A/2	1426.88	1426.88
s	cruise dist    	m	        	500	500
ec	energy/cruise	J	(mc+mb)*v^2/2	98000	107800
p	power demand	W	(ec*v/s+d)/eta	4907	5230
t	range time   	s	mb*k/p    	14673	27535
range         	m	t*v         	205424	385488
P: 3,135
 Quote by haruspex Bianca's model considers that the car cruises at a steady 50kph for 500m intervals. The acceleration and deceleration stages in each interval are presumed unimportant. My estimate is they'll not make more than a 5% difference to the answer.
In which case what does this term represent in his equation?

0.5 m v^3/d
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 Quote by CWatters In which case what does this term represent in his equation? 0.5 m v^3/d
Average power. d = interval distance = 500m. KE generated then lost over each 500m interval (.5 mv2) / time of each interval (d/v) = mv3/2d.
 P: 3,135 Edited: Humm ok I think I understand. He's including the energy needed to accelerate from 0-50 every 500m BUT he's assuming that it gets to 50 so quickly that for most of the distance it's doing close to 50?
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