10.4 kW seems about in the right range for a small lightweight car at a steady speed of 60mph.

This is just the power to overcome friction and aero drag, doesn't include the power to accelerate up to speed.

 

Ah right I see, thank you. Any ideas what the equation would be to calculate the power required to accelerate the vehicle?

 

I have come across this equation.

P = maV

but substituting in my values gives me 181.5kw using these values.

1468 x (26.8(60mph in m/s)/(13) (the time it takes to accelerate from 0-60)) x 60 (final velocity)

Does this sound right as it is a large number?

 

I have come across this equation.

P = maV

I'll go at it a different way...

Kinetic Energy = mass * speed * speed / 2

1468 * 26.8 * 26.8 / 2 = 527 kJ

You need 527kJ to move that mass up to that speed (which is also how much energy the
car has at that speed, and how much energy you have to convert to heat to make it stop).

3600 joules per watt hour, so, that's 146 watt-hours.

You want to consume 146 watt-hours of energy in 13 seconds.

If you took 1 hour, you would need only 146 watts.

If you took 1 minute, you would need 60 x 146 watts = 8760 watts.

If you took 1 second, you would need 60 x 8760 watts = 525,600 watts.

But you can take 13 seconds, so 525,600 / 13 = 40 kW = 54 hp.


54 hp isn't that much, but a 0-60 in 13s isn't that quick either. Seems roughly correct.

Not sure what got you to

substituting in my values gives me 181.5kw using these values.

1468 x (26.8(60mph in m/s)/(13) (the time it takes to accelerate from 0-60)) x 60 (final velocity)

Does this sound right as it is a large number?

Does not sound right. Also, "60" as your final velocity should've been in m/s not mph. But that still doesn't account for it, you're at 81kW. You are oddly, exactly double my calculation then. Maybe forgot the /2 ?

 

Thank you, Matt and Brian, for your responses - extremely helpful. After reading your explanations I can see the errors made in my attempts to use the formula and the time explanation has helped me to understand why less time requires more energy.

I will adapt Matt's energy over time approach for my calculations. Currently in the process of calculating a rough estimate of how much energy would be used by a Zoe during a typical motorway journey, just to see how the addition of a fuel cell could have an impact.

Thanks again!

 

Right, I see where I have confused things. Apologies as it is my first time working with power and energy.

What I was attempting to get at is, if the car battery contains say 44kWh of energy and the power required to have a 13-second car would be 10.4kW + 40kW would this exceed the energy amount?

My understanding from reading the How Much Power Do I Need thread (https://www.diyelectriccar.com/forums/showthread.php/much-power-do-needi-15508.html) was that Energy in Watt-Hours = (Power in Watts)(Time in Hours)

 

What I was attempting to get at is, if the car battery contains say 44kWh of energy and the power required to have a 13-second car would be 10.4kW + 40kW would this exceed the energy amount?)

That's like asking "if I want to go to another city 100 miles away and I travel at 50 miles per hour is that fast enough?" The question makes no sense, unless you provide more information, because distance and speed are not the the same thing. Energy and power are not the same thing - the amount you have or need of one doesn't tell you how much you need or have of the other.

 

Energy of battery is 44kWh, which can be written as 158400 kW-seconds.

Using 13 seconds @ 50kW is only 650 kW-seconds. So you could accelerate like that about 240 times before you run out of "gas". Puzzle thru these numbers.

 

Only thing I will add is that the Renault Zoe takes substantially more power to maintain 100 km/h. We drove one from Perth to Esperance (long trip, even longer story) and it maintained just over 200 Wh/km at 100 km/h. Therefore, 20 kW just to maintain highway speed. I suspect it's slip-ring induction rotor is the source of most of this inefficiency.

 

The unusual (for an EV) motor type with slip rings is probably an efficiency issue, but it's not an induction (asynchronous) motor. As previously discussed, Renault uses a synchronous motor with power to the wound rotor via slip rings.

 

i think the OP underestimated the frontal area in the calculation at 0.75. Looks like it should be much higher, like 2.652 sq m using height x width.

 

Good catch - 0.75 m² is definitely wrong. With a more realistic area, the power to overcome drag at 60 mph is likely closer to 15 kW... and with mechanical losses and inefficiencies in the controller and motor the electrical power requirement the 20 kW that jonescg reported seems reasonable.

 

If I were trying to work this out I'd work out the force required for x acceleration at y velocity. Instead of x being a constant acceleration though, it needs a relationship to y. You can then convert to power and graph to find the peak power. You'll then need to consider the drive train losses, motor losses, controller losses etc at that RPM and power output.

So Force = Drag Force + Friction Force + Acceleration Force
Drag Force ∝ Velocity (and a bunch of stuff that isn't that easy to calculate, and I'd assume is derived experimentally)
Rolling Resistance Force = somewhat constant I think
Acceleration Force = Mass * Acceleration
Acceleration ∝ Velocity (not a calculation, but to represent the changing acceleration over time, ie starts high and ends at 0 at the top speed)

Then Power = Force * Velocity. Graph Power vs Velocity and find the peak power at y velocity. Add an additional factor for the drivetrain, and electrical losses.

However this assumes the motors peak power and the peak required power align, a better way would be to further shift the Power vs Velocity graph to be at the motor, eg Power vs RPM, and then overlay the motors power output graph onto the required power graph and ensure the output is always greater than the required input.

Instead of doing all this though, I took the motors output curve, worked out the force at the ground at a certain velocity, and compared it to a similar car with similar desired performance.

 

Rolling Resistance Force = somewhat constant I think

Linear, but not constant.


http://www.enginuitysystems.com/EVCalculator.htm

Zeroing out acceleration (it misleadingly defaults to 1g, instead of 0).

0.29 coeff of friction
2.65 frontal area.

I got 18kw.


Doesn't change the energy requirements to accelerate (except that it must be in excess of the power to maintain speed), those only depend on weight, speed and time.

 

Hi All, the calculations are very interesting, and educational. As noted the accelleration ability requires more power than constant speed running. Note also at the start 0 to 10 mph, nearly all power is available to accelerate, i.e. little drag or rolling resistance. Also to be born in mind is hill climbing ability. This is also effecively accelleration, but adding potential energy not kinetic. However, that is relatively accademic, if you can get to 60 quickly, you can climb most average hills OK. The only thing about hills over accelleration is the time energy burn. I did an estimate for a 3 mile hill locally at about 40 mph, so a high power requirement for 4.5 mins. A hot motor, hammered battery! Cheers, Joe