
#1
Feb1213, 12:23 PM

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I want to calculate the range of an Electric Vehicle (EV) with a Liion 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. 



#2
Feb1213, 12:40 PM

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Looks good.




#3
Feb1213, 12:44 PM

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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.




#4
Feb1213, 06:59 PM

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Energy density and range calculation Electric VehicleYou gave car mass as 900kg including 100kg battery, but you've added the battery weight on to get 1000kg. 



#5
Feb1313, 12:05 AM

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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?




#6
Feb1313, 03:13 AM

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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. 



#7
Feb1313, 04:45 AM

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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. 



#8
Feb1313, 05:24 AM

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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). 



#9
Feb1313, 06:04 AM

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#10
Feb1313, 06:20 AM

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Could not insert table.




#11
Feb1313, 06:23 AM

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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 



#12
Feb1313, 08:28 AM

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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 E_{500} (in Joules). Then Range = 500 * Battery Capacity (in Joules) /E_{500} 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. 



#13
Feb1313, 03:55 PM

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#14
Feb1313, 04:27 PM

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Here's from my spreadsheet:




#15
Feb1413, 03:28 AM

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0.5 m v^3/d 



#16
Feb1413, 03:34 PM

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#17
Feb1413, 05:46 PM

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Edited:
Humm ok I think I understand. He's including the energy needed to accelerate from 050 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? 



#18
Feb1413, 06:18 PM

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The main issue in the post is that the calculated range seemed to go down when the battery size was increased. Using the same data and equations, I was not able to reproduce that, but I don't seem to be able to get Bianca to post the details of the second calculation. 


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