DIY Electric Car Forums banner

1 - 20 of 25 Posts

·
Average Joe
Joined
·
705 Posts
Discussion Starter #1
Range Estimation:
Code:
range[km]=250 x capacity[kWh] / (mass[kg]^0.6) 
-or-
range[miles]=250 x capacity[kWh] / (mass[lbs]^0.6)
Wire Temperature? :
Code:
Thot = Tcold + (K x ((Rhot - Rcold) / Rcold)) 
 
where K = 256.4 for copper 
       Thot = hot temperature in deg.C 
       Tcold = cold temperature in deg.C 
       Rhot = resistance at hot temperature in ohms 
       Rcold = resistance at cold temperature in ohms
Horsepower:
Code:
Hp=(torque*RPM)/5252
[wiki]AWG to Metric Conversion[/wiki]

[wiki]Current Limiting Formula for LEDs[/wiki]

[wiki]Amps, Volts, Watts and Watt-Hours 101[/wiki]
 

·
Registered
Joined
·
117 Posts
In the range equasion (range[km]=250 x capacity[kWh] / (mass[kg]^0.6) ), what is the "0.6"? It looks like me to be CoD, but there's no real reason for that, just a hunch.
 

·
Registered
Joined
·
37 Posts
I am not familiar with the formulas, but I can tell you that it is not from Coefficient of drag. The fact that the exponent is under 1 means that the mass has a relation that factors it in less than what it would factor in as if you were looking purely at acceleration--where weight matters.

This equation is very simplified. If you were to develop an equation for range that assumed that the car were traveling on a flat surface at a constant speed, weight would not matter at all, but drag area would drive the results.

Conversely, if you were to base range on a stop and go city type environment where it is mostly acceleration and aerodynamic losses are relatively minimal compared to acceleration and rolling resistance, then the weight of the car would matter more, and that exponent would be 1. so it looks like some assumptions about an average driving cycle have been made and the 0.6 exponent has been used to reflect that compromise. It looks as though it uses some stop and go or some hills.

The fact that Cd is not in the equation suggests to me that this assumed driving cycle doesn't have much traveling over 45 mph.

So I am really curious. What is the assumed cycle? Is there an assume value for drag area? (Drag area is just Cd times frontal area, the two variables specific to a car's design.)
 

·
Registered
Joined
·
51 Posts
^ is powers. You need a scientific calculator, it will have a ^ key.

BTW the formula is very optimistic. for example a 2500 lb car with a 10KW power pack would go 91 miles according to the formula.
 

·
Registered
Joined
·
122 Posts
I don't think that range formula is all that accurate...

I was trying to figure out the range a 3000Lbs Corvette with a 40.2kWh pack would go and it's saying something around 82mi... That doesn't sound right at all unless I'm doing my math wrong:

range = (250 * 40.2) / (3000 ^ 0.6) = 82.394mi

That really doesn't sound right.

Tesla Roadster is 2723Lbs with a 53kWh battery pack (I know this has changed lately, but lets go along with the old numbers for now.)

Using the same formula I get 115.130 which is 135miles less than what Tesla claims.

I think that 0.6 is the wrong number. It should probably be closer to 0.5. 0.5 would actually put the range for the Tesla closer to 250mi.

Running the numbers with the 3000Lbs Corvette and a 40.2kWh pack comes out to 183mi. That makes more sense to me. Maybe still a little high of an estimate, but sounds more accurate than 82 miles out of a 40kWh battery pack.

EDIT: I forgot to mention, the original formula is suggesting that my car would use 490wh/mi.
 

·
Registered
Joined
·
3,141 Posts
I have made some corrections and improvements to my on-line calculator. It is still located at:
http://enginuitysystems.com/EVCalculator.htm

It is now visually more appealing, with read-only calculated text boxes in a different color. I recently added slope and angle calculations, and just now I added inertia (kinetic energy) as well as gear ratio based on the motor and wheel RPM. :)
 

·
Registered
Joined
·
8 Posts
I am wondering about few things.
First, C rate. Does it counts in calculation of the range?
If I assume the range = Usable battery size/draw ratio.

I mean if i have battery pack of 42 cells with 3,2 Volts and 100Ah @ 0.3 C discharging rate. Without C rate the capacity would be 13,44 kW, (times 0,8 = 10,7 kW usable power) , but then what the role of C rate is? Does it mean the real battery capacity is 10,7 kW, but if I look at available power draw from the battery it getting less 30Ah* 42*3.2 = 4,3kWh?

But lets assume the capacity of 10,7kW and continue, I assume the draw ratio will be 100Ah at 144 Volts to maintain constant 80km/h speed, so it is 180 wh/km.
10,7kWh/(180wh/km) = the range of ~60km. It all would be okay, but doesn't the controller also has some energy draw?
 

·
Registered
Joined
·
7,838 Posts
I am wondering about few things.
First, C rate. Does it counts in calculation of the range?
If I assume the range = Usable battery size/draw ratio.

I mean if i have battery pack of 42 cells with 3,2 Volts and 100Ah @ 0.3 C discharging rate. Without C rate the capacity would be 13,44 kW, (times 0,8 = 10,7 kW usable power) , but then what the role of C rate is? Does it mean the real battery capacity is 10,7 kW, but if I look at available power draw from the battery it getting less 30Ah* 42*3.2 = 4,3kWh?

But lets assume the capacity of 10,7kW and continue, I assume the draw ratio will be 100Ah at 144 Volts to maintain constant 80km/h speed, so it is 180 wh/km.
10,7kWh/(180wh/km) = the range of ~60km. It all would be okay, but doesn't the controller also has some energy draw?
Hi fraitas,

C is the amount of charge in the battery in Ah, Ampere hours.

C-rate is how fast you take out (or put in) that charge. So Ampere hours divided by hours is Amperes. Ah / h = A. C-rate is current normalized for the purpose of describing the battery capability.

If you have a 80Ah battery and discharge at 2C then the current would be 160A and the time would be one half hour (30 minutes). So you can use C-rate to figure your range if you know that 160A will be exactly the current for a certain speed. Say 60 mph. So you use your stored charge at a 2C rate and use it all in 30 minutes at 60mph so travel 30 miles.

The above method is not the recommended procedure. It is more accurate to use energy in stead of charge to do the range calculations. There, use the energy stored in the battery in terms of kiloWatt hours, kWh. And used the power (kW) required to travel at speed or the energy required to travel a specific distance (Wh/mile).

Power (kW) is the rate at which work is done. Work is energy (kWh). Rate is inverse time (1/h). Power is energy divided by time. kW = kWh / h. Or energy is power times time. kWh = kW * h.

C-rate is generally used in the battery specification to relate the capability of the battery to its charge and therefore describes how powerful it is.

I mean if i have battery pack of 42 cells with 3,2 Volts and 100Ah @ 0.3 C discharging rate.
If your 100Ah battery was rated for 0.3C continuous, then it is recommended that you not exceed 30A current for long durations. It would still deliver the 100Ah (or close to it).

I recommend to not use C-rate in range calculations. Also please get a handle on the units of power and energy. Your post is all over the place with many mistakes. Read over the earlier part of the sub forum like: http://www.diyelectriccar.com/forums/showthread.php?t=11708

Hope that helps,

major
 

·
Registered
Joined
·
8 Posts
Thanks for the answer and I am really sorry for my mistakes. I have read that part few times, but some things still cannot find right places in my mind.

I would like to avoid C-rate, but it stays in my mind, so I want to realize so I can go on without it.
'If your 100Ah battery was rated for 0.3C continuous, then it is recommended that you not exceed 30A current for long durations.'
I am talking about this part. If it is not recommended to exceed 30A for long duration, how it can be right to calculate battery energy with 100Ah? :confused:



Paulius
 

·
Registered
Joined
·
7,838 Posts
'If your 100Ah battery was rated for 0.3C continuous, then it is recommended that you not exceed 30A current for long durations.'
I am talking about this part. If it is not recommended to exceed 30A for long duration, how it can be right to calculate battery energy with 100Ah? :confused:
30A for 3.33hours is 100Ah. That would be the continuous rating for a 0.3C, 100Ah battery. What we would call a low rate battery. One that is not recommended for use which requires discharging it in less than 3.33 hours. It is not a good choice for an EV unless it is a very large battery capable of driving the EV for 3.3 hours or longer.

Where did the 0.3C number come from in your first post? It is a battery which you are considering?
 

·
Administrator
Joined
·
6,157 Posts
Hi fraitas,

The number of Ah (Ampere hours) is like the size of the fuel tank

The "C" rate is like the size of the pipe going to the fuel tank
and like the size of a pipe tells you the maximum flow - you can flow less that that if you want

3C - means that you can empty the "tank" in 1/3 of an hour

0.5C - means that it will take 1/0.5 = 2 hours to empty the tank

Batteries for EV's normally have a capability of operating at 3C continuous - or even more for a short time (10 second)

Which means that a 100Ah cell (100 ampere hours)
can give 300 amps for 1/3 of an hour (20 minutes)
Which will flatten the battery
Or if it is rated at 10C (burst) - it can give 1000amps for 10 seconds
The 10 seconds is not to "empty" - but it will be heating up under that high current and doing that for more than 10 seconds may damage the cell

Low values like 0.3C are more usually seen on the charging cycle
 

·
Registered
Joined
·
8 Posts
I found it in technical parameters for the battery what is installed in the ev I am talking about. But I got it know. My fault. It is written "Nominal capacity 100 Ah @ 0,3 Discharging" and i have stock at this number, but know I also see "Maximum discharging current 2C", so it makes sense now.
Thanks a lot for explanations!
 

·
Administrator
Joined
·
6,157 Posts
Hi fraitas,

"Nominal capacity 100 Ah @ 0,3 Discharging"

OK I may have been guilty of an incorrect assumption
That sounds like a Lead Acid Pack
Lead is a pain, its heavy, it only lasts five minutes and if you take the power out at a reasonable rate you don't get as much

So you should be able to get 100Ah out of that cell (when new) if you draw at less than 30 amps

If you draw 100amps then you will only get about 50Ah out of the cell
Look up "Peukert coefficient"

This is NOT the case for Lithium
 
1 - 20 of 25 Posts
Top