[wolfekeeper]

Now let's make the trip with Hamdani at 150 mph, in the same car.

We can calculate the drag loss at 150  mph, using the standard formula F = ½ρAC_Dv²,  as 50 x (150/50)² = 450 kWh.

Add the original 50 kWh for friction losses and we now need 500 kWh for the same journey.

No. Really, that would be the same as you assuming that the friction losses have gone up quadratically as well, and then added them on again for no apparent reason. That's not in any way correct, the drag losses were lower to start with, and the rolling friction losses should have gone up linearly.

It's not that hard to do this properly, the Cd factor and frontal cross sections for Tesla cars are available.

But let's take that 450 kWh even though it is a gross overestimate, knowing full well that it's too high.

So we need to recharge the battery four times en route. As the battery will be hot, maximum charge rate will be about 40 kW, so the charge time will be 10 hours and the whole journey will take 12.3 hours.

Uhhhh no.

Your numbers are just garbage.

I thought you were an electrical engineer. You do know how batteries work???

They charge and discharge at the same maximum rate.

How is it that you think a car can output a particular high power for hours but then when it comes to charging, suddenly they have to charge four times slower than they were discharging minutes before?

These battery packs are ACTIVELY COOLED, even when charging. ESPECIALLY when charging.

The fastest Tesla chargers are 250 kW. Given that a 100 kWh Tesla will start fully charged it needs 350 kWh of charge. 350/250 = 1.4 hours. So the trip will take nearer 3.5 hours, an average of just under a hundred miles per hour.

Note that you would stop multiple times and do partial charges. Electric cars charge fast initially at low battery state of charge and then slow down, so it's fastest to partially charge the battery each time and do more stops. The exact time will depend on how far apart the super fast chargers are. But even so, your 12.3 hours is utter bollocks. You clearly know nothing about electric cars.

[alancalverd]

I was being generous by not scaling the friction losses. But since you insist, let's take the published 0.32 kWh/mile at 50 mph and scale the whole  lot quadratically to 150 mph, to get  2.9 kWh/mile. Now the coast-to-coast trip of 2800 miles or so at 150 mph requires 8120 kWh, or 82 charges of the 100 kWh battery. At 350 kW (max available from a superfast charger) that takes 23.2 hours.

So by going at 150 mph between charges we spend 42 hours on the road compared with 45 hours at 60 mph, assuming no cooling time or other charge rate limit.

I am quite familiar with the limitations of lithium batteries as I have an interest in electric aircraft. Pretty much the same as 150 mph cars, the recharge time with best present technology is at least as long as the flight time and can be more than doubled if the temperature is outside the optimum range of 5 - 45 °C (which it often is with airplanes). This makes a battery-powered selflaunching glider moderately attractive, or a trainer that returns to base for a battery swap after each sortie, but cross-country at 150 mph needs a lot of careful thought.

Sadly, active cooling requires power that would otherwise be available for motion, so there's another square-law effect to consider. Not a problem with IC engines that run at 100 °C or more, as heat transfer to a 20 °C ambient is very efficient (the problem with small aircooled planes is keeping the engine hot when the ambient  is -40°C !) but to maintain a battery at say 30 °C is quite hard work.

When the charging grid matures, and we have discovered some way to power it, I'll be happy to drive an electric car at  a sedate 50 - 70 mph, but the present technology doesn't lend itself to high speed transit.

[alancalverd]

PS I work with a number of technicians who get called from home to unscheduled site visits. They have been issued with electric cars which delight them as they can claim for recharging time on long trips!

Apropos partial fast-charge times, I note that the "catalogue" number is always the recharge time from around 10 to 80%, so the 100 kWh battery actually only delivers 70 kWh for optimum overall performance (on a warm day). 

[Petrochemicals]

We can calculate the drag loss at 150  mph, using the standard formula F = ½ρACDv2,  as 50 x (150/50)2 = 450 kWh

What is this formula, is it specific to something you know as velocity is rarely listed in mph or kph as they do not tally with joules etc. It's usually in metres a second.

[alancalverd]

The formula F = ½ρAC_Dv² is in every aerodynamics textbook. v² = v x v, whatever the units.

The numbers I used (0.32 kWh/mile at 50 mph) are culled from electric car manufacturers' websites - but what do they know about miles per hour, eh?

[Petrochemicals]

The formula F = ½ρAC_Dv² is in every aerodynamics textbook. v² = v x v, whatever the units.

The numbers I used (0.32 kWh/mile at 50 mph) are culled from electric car manufacturers' websites - but what do they know about miles per hour, eh?

Oh, a simple scaling operation giving 3 kwh for 3 miles (3 times the speed therefore 3 times the distance?)

[alancalverd]

Energy = force x distance

3 times the speed => 9 times the force (v x v) so 3 miles at 150 mph requires 3 x 0.32 x 9 = 8.64 kWh.

[wolfekeeper]

Except it's not because rolling friction is usually the biggest factor up to about 50 mph; and simple friction doesn't go up by a factor of 9 it is in fact linear, and the total energy needed to overcome it is independent of speed it's just distance. So at 50 mph, about half would aerodynamic drag. Your calculation assumes it's all aerodynamic drag and you've multiplied it by 9.

So 0.5 * 50 + 0.5*50*9 = 250 kWh. And you normally would have already would have fully charged to 100 kWh before you left. So 150 kWh left that you need to charge on route. That's 150/250 = 36 minutes (minimum) charging, and even the standard 150 kW chargers can do it in about an hour. You said: ten hours.

[alancalverd]

So at 50 mph, about half would aerodynamic drag..

Which is what I said originally. I estimated the frictional loss as half of the claimed power expended at 50 mph, just as you said.

 

According to the manufacturer, on a warm day you could drive a Tesla 350 miles at 50 mph, on a single 100 kWh charge......Let's generously assume that half the energy was expended on rolling friction, motor heat, etc. This means that we have expended 50 kWh on aerodynamic drag at 50 mph.

then I used the  standard formula to calculate the energy expended in aerodynamic drag to travel the same distance at 150 mph.
Energy = force x distance.
Drag force is proportional to (speed)².
So drag energy is proportional to (speed)² to cover a given distance at any speed.

It's a common enough calculation in aerodynamics - "Student Pilot 101" level, in fact.

And then I added the 50 kWh you allowed for frictional losses, to give 500 kWh required to travel 350 miles at 150 mph.

[wolfekeeper]

Only HALF the initial 50 kWh would be rolling friction. You didn't subtract that off first before multiplying it by the drag equation. The drag equation only applies to aerodynamic drag, NOT rolling friction.

[Petrochemicals]

Energy = force x distance

3 times the speed => 9 times the force (v x v) so 3 miles at 150 mph requires 3 x 0.32 x 9 = 8.64 kWh.

That is 3 times the drag for the same distance at 3 times the speed, that is very convenient.

[alancalverd]

Only HALF the initial 50 kWh would be rolling friction. You didn't subtract that off first before multiplying it by the drag equation. The drag equation only applies to aerodynamic drag, NOT rolling friction.

Let's start again.

The manufacturer says 100 kWh will take you 350 miles at 50 mph.

You say friction accounts for half the retarding force at 50 mph, and the friction force isn't speed-dependent.

So 50 kWh is lost to friction over 350 miles at any speed

And 50 kWh is lost to aerodynamic drag over 350 miles at 50 mph.

This is your arithmetic so far.

Now I calculate the energy dissipated by drag over 350 miles at 150 mph by multiplying 50 kWh by the square of the speed ratio. 50 x (150/50)² = 450 kWh.

And then I add your 50 kWh friction loss to get a total of 500 kWh.

How would you calculate it, still using your data, please?

[alancalverd]

That is 3 times the drag for the same distance at 3 times the speed, that is very convenient.

No, it is 9 times the drag force, over the same distance. Energy = force x distance.