July 11 2008

Basics of where the energy goes and how to make your 4th grader proud.

If you are interested in estimating drag force and power requirements for your own vehicle, it is simple to do without Matlab. If you can multiply and extract square root, you can ask your 4th grader's to do it for you - his or her math skill will be sufficient. Below is an example for my Audi A6 you can apply to any vehicle. But first I'll refresh your understanding of what exactly is going on when a vehicle is moving and where the energy is being spent, so we're in sync. This rudimentary physics applies to any vehicle, including EV.

When a vehicle moves at steady speed in a given direction it needs some force keep moving it to maintain that speed. That force is needed to overcome forces wanting to slow vehicle down: air drag and tire friction losses. These are two components worth discussing and they are by far the major contributors to the phenomena, especially air drag force. If you take these two into account, contribution of all other forces combined can practically be ignored as being negligible.

What is air drag? It is force slowing you down resulting from the pressure of air piling up in front of the moving vehicle thus pushing it back, and negative pressure of the air immediately behind the vehicle - air pulling in from all the directions filling the low pressure void, including pulling rear of the vehicle toward the void. Granted, abrupt end of the vehicle will create larger void than teardrop shaped end allowing air streams around to "close" behind without changes in pressure (thus without dragging vehicle back into low pressure zone). Similarly, in front, the more gently you can "cut" the oncoming air stream redirecting it around the vehicle, the less air "piles up" up front and so less pressure pushing the nose back. The "slipperiness" of a particular car shape is expressed in drag coefficient Cd. The lower Cd - the less air is disturbed - the less force needed to move the car at the same speed (all else being equal). Obviously, the less physical size of the vehicle - the less air it disturbs, so it affects the force.

The faster you go, naturally the more force it takes to overcome air drag. Turn out the force increases as square of speed. Double your speed and you need 4 times force to sustain it. Tripling speed will require 8 times force.

The tire drag is caused by flexing the rubber as you move. The harder the tire (more pressure) - the less it flexes resulting in less drag force. (Tire less "resists" to roll, so it is more common to hear about tire rolling resistance). Very hard pumped tires are the best as far as easy to roll, but the ride is harsh, so people make a reasonable compromise between desired riding comfort and energy wastefulness of softer and quieter tires. The tire drag force is not as dependent on vehicle speed and thus is by far the most significant portion of total drag at very low speeds, such as maneuvering on the parking lot. For modern vehicles and tires it is about 1.3% of the vehicle weight at low speeds (~50 km/h) increasing to about 1.8% at high speeds (~100 km/h).

So here are some formulas to determine the drag forces, power it takes to overcome them and energy an electric (or any for that matter) vehicle will spend fighting these forces on the way.

What I know about the vehicle and environment:

Coefficient of drag Cd | 0.31 |

Air mass density Rho | 1.25 kg/m^{3 }( @25C ) |

Gravimentric acceleration | 9.8 m/s^{2} |

Frontal area A | 2.4m^{2} |

Vehicle mass | 1782kg |

Aerodynamic drag force:

**F_drag = 0.5 * Rho * Cd * A * V ^{2}**
where V is the speed at which the vehicle is moving (m/s).

Let's figure this out for typical urban speed of 56 km/h (35 mph) and twice as fast freeway speeds of 112km/h (70 mph).

__1. __

(Have your 4th grader to do this math for your vehicle, only multiplications are involved. Explain to him/her that squaring is multiplying a number by itself):

56 km/h is 56 * 1000 / 3600 = 15.56m/s.

So, F_drag = 0.5 * 1.25 * 0.31 * 2.4 * 15.56^{2}
= ~113 N. This is continuous force dragging your vehicle backwards.

As discussed before, contribution of tire drag is ~1.3% of
total vehicle weight. The vehicle weight is 1782kg * 9.8m/s^{2} = ~17,464N
(you know the difference between mass and weight, don't you?) so F_tire = 17464 * 1.3% =
227N. This might be surprising conclusion: tire drag at 56 km/h is dominating factor and
is more than air drag! So the total drag force then is 113N + 227N = 340N.

How much motor power do I need to overcome this force and maintain 56 km/h? Well,

**Power = Force * speed, or P = F_drag *
V **(Because F_drag itself is function of V squared, then
multiplied by V once more, power becomes function of V^{3}; in other words

**P = F_drag * V = [0.5 * Rho * Cd
* A * V ^{2}] * V = 0.5 * Rho * Cd * A * V^{3}. **

So, P = F_drag * V = 340 * 15.56 = ~ 5290W (or 7.2 hp).

Note, if rolling resistance was near zero (for instance steel wheels on a railroad track)
it would only take 113 * 15.56 = 1758W (or = ~2.4 hp) to keep moving at 56 km/h;
less than a half!

How much energy do I spend moving at that speed, e.g. what is my Wh/km efficiency? Easy:

Obviously driving while applying 5290W of power for 1 hour I will consume 5290Wh out of the battery while I'd have gone 56km. Energy consumption rate then is 5290/56=~94 Wh/km (151 Wh/mile). Without tire drag that number would be only 1758/56 = ~31 Wh/km (50 Wh/mile)!

Conclusion 1: at low speeds I can increase my range by ~40% by halving rolling resistance alone. It really pays to maintain tires in top condition (and front wheel alignment as close to zero toe as practical). Never underinflate tires, use only low rolling resistance (LRR) models. I know, they are skinny and don't look as cool as wide ones, but the penalty for good look is miles and miles of lost range. The choice of priorities is yours

__Range estimation: __

If I allow to discharge my 28.4kWh battery to 80% DOD, I make 22.7kWh available for driving. My total range will be:

22,700 / 94 = ~241km (~151 miles)

__2. __

Let see what do we get moving twice as fast.

112 km/h (70 mph) = 112 * 1000/3600 = 31 m/s

Air drag force F_drag = 0.5 * 1.25 * 0.31 * 2.4 * 31^{2} = 447 N (4 times of that
for 56 km/h speed!) Contribution of tire drag increases to 1.8% of the vehicle weight of
17,464N, and becomes 17,464 * 0.018 = 314N. Note that at this speed air drag is dominating
factor. Total drag force then is equal to 447N + 314N = 761N.

So, how much motor power do I need to maintain 112km/h?

Remember, P = F * V, so P = 761 * 31 = 23591 W (32 hp, ~5
times of that for 56 km/h). This means if I need no less than ~25 kW rated power motor for
my EV if I want to move at 112 km/h (70 mph) continuously.

Energy spent in 1 hour is then 23591Wh and consumption rate is 23591/112 = ~211 Wh/km (337
Wh/mile) respectively.

__Range estimation: __

22,700 / 211 = ~108km (~67 miles)

Conclusion 2: for high speeds you need to start increasingly worry about air drag (shape of your vehicle) rather than tire drag. What can you do to reduce it? You can't reduce frontal area A. You can somewhat improve Cd by installing flaring, spoilers, belly pan, removing sticking out objects (large fog lights, mirrors), or covering a truck bed with liner, but the best is to choose a vehicle with low Cd (and frontal area A for that matter) to start with. In this respect the best production vehicle I'm aware of at the time of this writing is Honda Insight.

__3. __

Finally, fun exercise: how fast at given motor power will the vehicle possibly move? I know my battery is nowhere near 400kW capable (actually more because of system losses), but if in future I can get one... In other words, I want to find the limitation of my drive systems, assuming battery limitation is not impacting me first.

Because we don't know the speed we can't subtract the power
needed to overcome tire drag to see how much is left to move the car through the air - we
don't know how this drag value. But we can get reasonably close:

We know that P = 0.5 * Rho *** **Cd * A * V^{3} so V = [P/(0.5*Rho*Cd*A)]^{1/3} = [400000/(0.625*0.31*2.4)]^{1/3} = 95
m/s = 342 km/h = 214 mph if no tire drag is taken into account.

The tire drag will be at least about 2% of the vehicle
weight at these speeds (my guess), so 17,464 * 0.02 = 349N. P = F * V and we just
have determined V. So the power required to overcome tire drag is 349*95 = 33,155W. This
power is no longer available to overcome the air drag, so what's left to move the car and
overcome air drag is 400,000 - 33,155 = ~366,845 motor watts. Reiterating:
[366,845/(0.625*0.31*2.4)]^{1/3} = 92.4 m/s = 332 km/h (208 mph). The speed
dropped a bit and so is tire drag, so it will take few iterations to get to exact answer.
But even without iterations, I know the error margin is no more than 342 - 332 = 10 km/h
(6.3 mph); close enough. Enough for a steep speeding ticket anyway :-). So don't let your
4th grader get scared of your plans by asking to do your future EV's max speed estimation
math for you. Please promise you're not going to drive that fast...