DIY Electric Car Forums banner

1 - 3 of 3 Posts

·
Registered
Joined
·
19 Posts
Discussion Starter #1
I am working through my conversion project, and the motors and batteries are going in. Now comes part 2, bringing it all together functionally.

The battery-packs have water-cooling, the 2 motors have water-cooling and so do the 2 controllers. So, I need to figure out water cooling. Thing is, how much/how big is big enough?

The original ICE-version of the car was not super-efficient at maybe 10 mpg in 1959. Say I allow for 30 years of technological advancements in engine tech, and the same car with a 1990's engine would have been maybe 12-15 mpg.

At 75 mph for an hour, that means say 5.5 gallons of premium. What I found on Google says that 1 gallon roughly equals 8 kW, so that's how I get to 45 kW per hour of driving. If anyone has a better estimate, please leave a comment. Freeway driving is not city driving and all that, got that, but for the sake of figuring out the cooling it's probably not a bad number to go with??

Question #1 is whether all of the 45kW effectively translates into heat, or is there some portion that goes somewhere else?
Question #2 is how much a normal radiator "exhausts" per square foot (??) at 100 kmh/60 mph at say 30 degrees outside?

If a few people with working cars could reply with what they've seen for
- weight of their car
- running amps for freeway commute speeds
- cooling set-up
- driving conditions (uphill&down, stop&go) and operating temperatures
then I can try to find the common denominator and see where that puts me...

Thanks in advance for helping out, RK
 

·
Administrator
Joined
·
6,136 Posts
Hi Rob

That 45 kw is to drive your car through the air - sounds like a lot my car needs 26 kw at 100 kph

Anyway you will be losing about 10% of that as heat - so 4.5 kw - that is from the motor controller and all

An IC engine will be rejecting about the same as it is using to drive the vehicle so 45 kw

The radiators on a normal IC engine car will cope with the full power of the vehicle - even when they are 50% clagged up

So overall you need about 1/20th of the cooling capacity of a normal car

However you don't want to be quite as hot so aim at 1/10th
 

·
Registered
Joined
·
4,890 Posts
These discussions get really confusing and difficult if people are not careful to distinguish between energy and power.

At 75 mph for an hour, that means say 5.5 gallons of premium. What I found on Google says that 1 gallon roughly equals 8 kW, so that's how I get to 45 kW per hour of driving.
Okay, a volume of fuel contains a known amount of energy. But 8 kW (kilowatts) is an amount of power, so you probably mean 8 kWh (kilowatt-hours) of energy.

"kW per hour" means nothing at all. The power required to drive the car at 75 mph might be 45 kW (kilowatts).

If anyone has a better estimate, please leave a comment. Freeway driving is not city driving and all that, got that, but for the sake of figuring out the cooling it's probably not a bad number to go with??
That's 45 kWh / 75 mph = 600 Wh/mile, which is high for an EV, but perhaps a reasonable estimate due to the high speed and high-drag car.

In an electric vehicle, the energy consumed per distance travelled is higher at higher speeds, due to increase aerodynamic drag. There are lots of other factors which make urban fuel economy worse with an engine, but for an EV the highest rate of energy consumption while travelling at a steady speed (so, the highest power level) is at the highest road speed, so that's not a terrible scenario to consider for cooling.

What you miss by using this scenario is the higher heat production resulting from higher power while accelerating, or while climbing a grade. You might assume that those things only happen briefly, so that the cooling system only needs to keep up with continuous demands.

Question #1 is whether all of the 45kW effectively translates into heat, or is there some portion that goes somewhere else?
It all becomes heat, but most of that heat is in the air (due to aerodynamic drag) and tires (due to rolling drag). If you put 50 kW into a motor which is running at 90% efficiency, 45 kW comes out as mechanical power to drive the car, and 5 kW as heat in the motor's windings, stator iron, and rotor iron. It's that 5 kW that the cooling system needs to handle. Similarly, the heat production in the battery and controller are due to the inefficiency of those components.
 
1 - 3 of 3 Posts
Top