The Case for an EV Semi Truck

[Curt Renz]

ARK Invest - Nov 8: EV Semi Trucks to Take Share From Traditional Long-Hail Diesel Trucks

 

[mongo]

ARK Invest - Nov 8: EV Semi Trucks to Take Share From Traditional Long-Hail Diesel Trucks

Interesting, but I think they may be giving regen too much credit. It is highly dependent of the drive cycle (highway vs city).

Based on my evening calculations, an 80,000lb semi traveling at 60 MPH has a kinetic energy of 3.63 kWh. A semi that gets 12 MPG per gallon of diesel @ 50% efficiency uses 1.7 kWh per mile. So the energy regained in one stop from 0-60 is equivalent to ~2 miles of range.

To go from the 85 mile range to 200 as in their graph, the drive cycle would need to be:
200-85 = 115 miles gained due to regen
115 * 1.7 kWh = 195.5 kWh post regen / 90% = 217 kWh kinetic
That is equivalent to 217/3.63= 60 stops from 60 MPH
So basically, in the 85 mile cycle, they are stopping every 1.5 miles on average.
However, the additional 115 mile stretch should have the same stop interval, which I'm not bothering to calculate.

Data check: 60*3.63/90% acceleration + 85*1.7kWh/mile = 385kWh which lines up to their 335 kWh pack size.
200*1.7 + 60*3.63*(1-.9*.9) = 381, also close.

So for routes in areas with high speeds and many stops, regen helps a lot. For highway/ cruise control routes, there are not enough regen events to make a large difference.
The graph of regen vs motors must have some other assumptions baked in (gearing?) since there is no reason the second motor could not produce the same level of regen as the first (two rear axles). The front/ third motor is limited by tire friction * cab weight.

In regard to stopping distance, a fully loaded semi at 65 takes ~525 feet to stop. Assuming constant deceleration, the time to stop is d=.5t*a^2 and t*a=v, t=v/a so d=0.5*v/a*a^2, d=0.5*v*a, d=525 ft, v = 95 ft/s , 525=.5*95*a, a= 525/(.5*95)=11 ft/s^2 or roughly 1/3 g. Time is then 95/11 = 8.6 seconds. Truck had 3.63kWh of energy (@60) removed in 8.6 seconds, that is a rate of 3.63/(8.6/60/60) = 1.52MW. I think the current Ludicrous limit is around 1500 Amps @ 400V (assume no sag) = 600kW. However, in this case, the rear axles on the trailer would be contributing to braking (about 40%) and the front axle would be the remainder. So the capturable energy would only be 50% or so, but feasible. 1.52MW *50% /2 axles = 350kW per motor, 700kW to the battery. Sounds inline with Elon's supercharger comment

Going the other way, the semi could have a 0-60 time of 17 seconds or so.

If someone would like a specific scenario run, or if my assumptions/ formulas are off, let me know.

 

[roblab]

Data check: 60*3.63/90% acceleration + 85*1.7kWh/mile = 385kWh which lines up to their 335 kWh pack size.
200*1.7 + 60*3.63*(1-.9*.9) = 381, also close.

So for routes in areas with high speeds and many stops, regen helps a lot. For highway/ cruise control routes, there are not enough regen events to make a large difference.

Obviously not from the western USA. We have "hills" here, often with a 7000 foot rise or drop. Now, THERE'S regen!

 

[mongo]

Obviously not from the western USA. We have "hills" here, often with a 7000 foot rise or drop. Now, THERE'S regen!

Nope, but have driven through its beauty. 80,000 pounds at 7,000 ft is 211kWh of potential energy. Definitely something to claw back on the way down, and save the brakes!

 

[brucet999]

Obviously not from the western USA. We have "hills" here, often with a 7000 foot rise or drop. Now, THERE'S regen!

The author also did not specifically calculate the savings on brake jobs due to regen. Maybe that was included in the first chart maintenance.