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#### TheJammiestGeezer

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Hi there,

I am attempting to use a formula which I found in the forums to calculate the power required to travel at 60mph by a Renault Zoe. I have attached the thread link and formula below with what values I plugged into the formula. When using this formula, I ended up with a figure of 10423.82 however, I am unsure whether this is correct. Can anyone help to clarify?

Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + (0.5)(1.293 kg/m³) (Coefficient of Drag) (Area in m²) (Velocity^3))

1468 (mass of Zoe in kg) x 9.8 x 26.8 (60mph in m/s) x (0.02) + (0.5) x (1.293) x 0.29 (drag coefficient of Zoe) x 0.75 (frontal area of Zoe in m^2) x 26.8^3 (not too sure if I used the correct value here)

Thank you.

#### kennybobby

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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.

#### TheJammiestGeezer

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

#### TheJammiestGeezer

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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?

#### MattsAwesomeStuff

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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 ?

#### brian_

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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?
In the formula "P = maV"
P is power (in watts if using consistent rational units)
m is mass (in kilograms)
a is acceleration (in metres per second squared)
V is velocity (in metres per second... and usually "v", but it doesn't matter what symbol you use)

26.8 m/s / 13 s = 2.06 m/s2
is an average acceleration rate to reach 60 mph in 13 seconds

The problem is that the formula, which is really just power = force * velocity, is the power needed to accelerate at that rate while travelling that speed. You won't maintain the same rate of acceleration for the whole 13 seconds, but if you did the power required would rise with the speed to that 81 kW (not 181 kW) at the end (1468 kg * 2 m/s2 * 26.8 m/s = 81 kW). Since the acceleration is constant, the power at the end is twice the average, which Matt noted. This illustrates the difficulty with using a formula - no matter how carefully - which you don't understand; hopefully this helps you understand it.

Matt's energy-over-time approach works better in general, because it corresponds more closely to what the car would do. It's still not right, because the motor will be limited (by current capability of the controller and of the motor) to a constant torque at low speed, so higher-than-average power will be needed at higher speed to reach the target speed in the target time.

#### MattsAwesomeStuff

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It's still not right, because the motor will be limited (by current capability of the controller and of the motor) to a constant torque at low speed, so higher-than-average power will be needed at higher speed to reach the target speed in the target time.
After that, still worse, because this is presuming we're accelerating in a gravity-less vacuum.

To travel highway speeds, you are pushing air and squishing tires at highway speeds. So, some non-linear portion of that 10.4kw to travel highway speed needs to be added as well.

#### TheJammiestGeezer

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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!

#### brian_

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To travel highway speeds, you are pushing air and squishing tires at highway speeds. So, some non-linear portion of that 10.4kw to travel highway speed needs to be added as well.
Yes, I assume that with the "P=maV" formula the intent was to calculate the power required to accelerate in addition to the power required to keep the vehicle moving.

#### TheJammiestGeezer

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Yes, I assume that with the "P=maV" formula the intent was to calculate the power required to accelerate in addition to the power required to keep the vehicle moving.
Yes, it was in addition. Just to clarify based on Matt's explanation, would the 40kW value need to be converted into energy units to allow me to add this to the value required to keep the vehicle moving? slightly confused on this part.

#### MattsAwesomeStuff

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Yes, it was in addition. Just to clarify based on Matt's explanation, would the 40kW value need to be converted into energy units to allow me to add this to the value required to keep the vehicle moving? slightly confused on this part.

Energy is a quantity, it's an amount. You have an amount of energy you can store in a battery. You have an amount of energy you have to give an object to make it move faster.

Power is the rate at which energy flows, it's how fast you are consuming your energy. Your motor will consume energy at a certain rate, depending on what you are demanding it to do.

Joules and watt-hours are a unit of energy.
Watts are a unit of power.

So 40kW is power, it's the average rate at which you need to be using energy to accelerate, if you want to reach your top speed in 13 seconds (you have 13 seconds to add that much kinetic energy to that mass, so you need to be adding kinetic energy at a certain rate).

The 10.4kW is also power, it's how much power it takes to continue travelling at 60mph.

In terms of peak power, you'd need at least 40+10.4kW = 50.4kW to have a 13 second 0-60 car.

It's actually worse, as Brian pointed out, because you won't be able to actually put 40kW into your motor at first, and then to catch up you'll need to put more than 40kW towards the end of that 13s, but it's at least roughly appreciable.

Then there's inefficiencies and driveline losses. These are the end result of the physics, you waste some energy as heat in the motor and controller and every gear and linkage.

So I presume you were talking about power the whole time. If you were talking about energy then I'm not sure what your question is.

#### TheJammiestGeezer

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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)

#### kennybobby

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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.

#### brian_

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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.

#### brian_

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Just to clarify based on Matt's explanation, would the 40kW value need to be converted into energy units to allow me to add this to the value required to keep the vehicle moving?
As Matt explained, power and energy are not the same thing - you can't just convert one to the other.

If what you are asking is whether or not you need to consider the energy needed to accelerate when determining the range you can drive on a battery charge... yes, probably.

If the vehicle didn't need to ever change speed, then the energy used to go any distance would be just the power used to keep it moving multiplied by the time it was moving. In actual use you will need to use energy to accelerate the vehicle, and you'll get some of that back either by coasting down in speed (continuing to move without applying any motor power), or by regenerative braking. Every time you use the normal (friction) brakes in a vehicle, you are throwing away (as heat) the energy that you put into increasing the vehicle's speed.

#### MattsAwesomeStuff

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the power required to have a 13-second car
Just a nitpick, but, that is not what people mean when they say "a 13-second car". A "13 second car" is one that can travel 1/4 mile in 13 seconds. You want a car that can do 0-60 in 13 seconds, something an order of magnitude easier to accomplish.

But don't be thinking you'll end up with a 13 second car would be 10.4kW + 40kW would this exceed the energy amount?
As others have pointed out, that's a nonsensical question, and I'm not sure what the actual question you're trying to ask might be.

You would need to be capable of providing 50.4kW of power, to accelerate from 0-60 in 13 seconds.

If you continued using power at that rate for an hour, you'd use 50.4 kiloWatt-hours of energy. But since you're only doing it for 13 seconds, you're only using a small amount of that.

A battery pack stores energy, it could be whatever size you want.

If your battery was charged, and you wanted to drive 100km/h for 2 hours, you'd use 20.8kWh to do that.

To accelerate from a stop, just once, you'd use energy at ~6x that rate, but you're only doing so for 13 seconds, so it's not a big impact. If you're doing it every mile, you'll notice a bigger impact.

#### TheJammiestGeezer

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If you continued using power at that rate for an hour, you'd use 50.4 kiloWatt-hours of energy. But since you're only doing it for 13 seconds, you're only using a small amount of that.

A battery pack stores energy, it could be whatever size you want.

If your battery was charged, and you wanted to drive 100km/h for 2 hours, you'd use 20.8kWh to do that.

To accelerate from a stop, just once, you'd use energy at ~6x that rate, but you're only doing so for 13 seconds, so it's not a big impact. If you're doing it every mile, you'll notice a bigger impact.
My bad, I was assuming that power would be used at that rate for an hour so 50.4kWh of energy would be used, but I can see now 13 seconds a 50kW is nothing close to that. I can see now my question didn't make much sense!

Thanks all for pointing out my errors, I have gained a much better understanding regarding the topic now #### MattsAwesomeStuff

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Thanks all for pointing out my errors, I have gained a much better understanding regarding the topic now I'm glad you stuck with it until things clicked.

It's not complicated stuff, but you do have to have it sorted properly before you can start to use it to solve problems.

#### jonescg

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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.

#### kennybobby

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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.

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