Will Tesla’s semi truck need the Mother Of All Batteries?

We all know Elon Musk and company are used to thinking big. Even so, the recent announcement that a Tesla Semi truck would be unveiled in September caused jaws to drop and heads to shake. In the trucking industry, there was plenty of skepticism, but also a healthy concern for the future. “Commercial vehicle makers – and their suppliers – would be wise to stymie their laughter and take [this news] seriously,” wrote analyst Alex Potter.

Is an electric semi truck even practical? The hurdles are huge. On-road trucks must haul tons of cargo over long distances in all weathers, so a viable EV would need massive power, long range and durability, not to mention that there is no highway charging network suitable for heavy-duty vehicles, and little incentive for operators to switch from cheap diesel fuel.

However, Tesla’s truck is no idle threat – it’s a working prototype. During a recent conference call with stock analysts, Musk said he had taken a test drive around the parking lot, and that the truck “feels like a sports car.” He assured us that the e-truck will be capable of long hauls, and that the cost of operation will be low enough to make it an attractive replacement for legacy diesel trucks.

CTO JB Straubel said the Tesla Semi wasn’t especially complicated, technically speaking, and Musk explained that it shares “a lot” of parts with Model 3, including “a bunch” of the new Model 3 electric motors. Neither said much about the battery pack, but there’s no doubt that it’s going to be big – perhaps the largest pack ever installed in a production vehicle. Did we mention that it’s big? [“How big is it?” shouts the audience in unison].

Well, there’s been a lively discussion about that over at InsideEVs, with writers and commenters doing some “napkin math.” A chart from engine builder Cummins illustrates the various factors that affect the fuel economy of on-road trucks. Together with truckmaker Peterbilt, Cummins has been developing a fuel-efficient SuperTruck that has demonstrated fuel economy of an impressive 10 mpg.

Tesla-semi-battery-sizing-GB1 copy

Assuming optimal aerodynamics, rolling resistance, etc. and accounting for the greater efficiency of electric motors, as well as regenerative braking, the envelope engineers estimated that an electric semi would require a 1,200 kWh battery pack to achieve a range of 600 miles. At a cost of $100/kWh, which Tesla hopes to reach soon, that would be a $120,000 battery. According to a 2016 article from AskTheTrucker, the cost of a new diesel-powered tractor starts around $113,000, and annual fuel costs are “easily more than $70,000.”

Perhaps a simpler way to go about estimating the battery size for a Tesla truck would be to look at heavy-duty vehicles already on the market. Electric buses from several manufacturers are running in revenue service in cities around the world (in China, cities are ordering them by the thousand). The biggest have battery packs of around 400 kWh. However, e-buses aren’t directly comparable to trucks – their human cargo is much lighter than the 34,000 pounds that a tandem-axle trailer may carry, and they typically have a range of less than 100 miles.

Tesla won’t be the first company to produce an electric semi truck. In fact, electric Class 8 electric trucks from BYD have been in pilot operation for some time, and Orange EV Class 8 trucks have been commercially available for more than 3 years. However, these are terminal trucks (aka hostlers or yard goats), which are used for cargo handling in and around ports and other freight facilities. They haul the same loads as highway trucks, but do not need to have a very long range.

Chinese EV-maker BYD also offers a Class 8 semi truck for on-road use. The BYD T9 has a 188 kWh battery pack, and a stated range of 92 miles. Andy Swanton, Vice President of BYD’s Truck division, told Trucks.com that medium- and heavy-duty electric trucks make up an increasing part of BYD’s commercial vehicle business, and that the company has sold about 150 electric trucks to US customers (the company also sells electric delivery vans to UPS and others). We don’t know if the company has any actual orders for its on-road semi yet – Swanton said only that it is “already in service in China.”


Source: InsideEVs

  • Lance Pickup

    I will be incredibly surprised if the Tesla semi doesn’t incorporate some kind of battery swap technology. Even though it didn’t really take off (deemed necessary) by the sedan crowd, fast charging a 1.2MWh pack is pretty much impractical.

    • Michael Walsh

      Unless they were to use ultra-fast charging next-gen Solid State batteries?

      • Lance Pickup

        That hardly even matters. To charge that battery in an hour you’d need around a 1.5 MEGAWATT charger! That is insane! Even with stationary storage so you weren’t actually pulling that from the grid the heat generated from even a measly 5% efficiency loss (and it would likely be greater than that) would be 75kW. Just for comparison an electric oven can heat up the cooking area with only 5kW!

        • Vincent Wolf

          You only get heat if there is extreme inefficiency in the charging event. Heat represents lost energy via resistance, etc. You don’t see our telephone pole lines carrying 800 Volts melting because they overheat and carry gazillions of AMPS to all the residences/businesses. It’s all dependent on resistance following the law V=IR.

          • Lance Pickup

            Several things wrong with what you’ve said…
            1) No, you don’t need extreme inefficiency to get heat if you are talking about 1MW+ of power transfer! 5% loss of 1MW is 50KW, and 50% loss is 500KW. Both of those are huge amounts of power!
            2) Phone lines don’t carry 800V…it’s more like 48VDC when static and 90VAC when ringing. And the current would not be a “gazillion” amps but rather just enough to power the old fashioned phone’s bell ringer electromagnet. Remember, in POTS each customer’s line is dedicated (from the CO anyway).
            3) Yes, V=IR, but P (Power)=IV. How much current flows through the telephone wire? It’s on the order of tens of milliamps which makes sense given that the voltage drop (I*R) on a wire that could be several miles long would be significant if you ran any serious amount of current through it. So if V is 50V, and I is let’s say 20mA, then P=I*V = 1W. How much loss assuming all losses are due to resistance in the telephone wire itself? For a wire length of 3 miles of AWG 22 gauge wire, this is about 260 ohms. At 20mA of current, this gives I*R = 5.2V of IR (voltage) drop over the 3 mile line. This is the loss. What is the power lost? Substitute in Ohm’s law for the power equation (P=IV) to get P=V^2/R or I^2*R, take your pick. Either way you get 5.2^2 / 260 or 0.02^2 * 260 = 0.104W of loss, or about 10%. But no, 104mW is not going to melt the wire by any means!
            4) What does this have to do with the 1.5MW charging station? Okay well yes, V=IR as you said. I don’t know how the cells in the semi would be arranged, but let’s assume they would be arranged in a 1200V pack (i.e. 3 of today’s 400V packs in series). So we know V. Now let’s say we have a charger capable of delivering 1.5MW of power. Now we know P. Solving for I in the power equation, we have I=P/V = 1.5MW / 1200V = 1250A! That’s a lot of amps! Now let’s assume that most of the losses are in the form of resistance which generates heat–our other outlets for losses would be noise, light, actual physical movement, etc. I’m sure this thing would make quite a racket, but I think assuming most of the loss would come from wire resistance or internal resistance of the driver components and be transformed into heat. Let’s assume that we use REALLY big and as short as possible wires and we can get the resistance all the way down to 50 mohm (I have no way of know whether this is practical or not, but I would think this would be quite the challenge to get through all the driver circuitry and whatever charging apparatus with only 50mohm of resistance. P=I^2 * R, even assuming R=50mohm still gives you 78kW of power loss, which would be in the form of heat, which would have to be dissipated somehow. Wow! And this only represents a measly 5.3% loss, which is likely not achievable.

            So I stand behind my assertion that you’re going to have only a few choices here:
            1) The range of the truck is significantly less than 600 miles and the battery pack size is reduced to scale.
            2) The range of the truck is 600 miles (with a 1.2MWh pack), and after that 600 miles you’re going to park it overnight where it will charge at the “snail’s pace” of 250kW and still take 5-6 hours to charge. And oh by the way, you’d better locate your depot next to a dedicated power station to generate all that electricity–only 20 trucks each charging at 250kW = 5GW of power, which is basically an entire power station (although I presume you would have some stationary storage to offset this).
            3) There is some kind of battery swap capability (which should be REALLY easy on a truck by the way).
            I’m still putting my money on #3.

          • Vincent Wolf

            Telephone pole lines carry the electrical grid power to homeowners and businesses. I’m not talking about phone lines jeeze.

          • Lance Pickup

            Hopefully I don’t have to repeat my explanation of the fact that losses are losses. If you are transmitting 500kW of power over a line that has a 10% loss, that is 50kW of power that is lost and is dissipated as heat. There is no magic number where the loss is “significant” or not. It’s just straight math.

            Now 10% loss is a widely accepted value for the electrical grid, so I’ll just stipulate that.

            Now the grid is not transmitting 500kW across the whole grid. If you look close to the power plant, we are talking somewhere near 1-2GW or more. If we are looking close to your street, you might have a neighborhood transformer that feeds a few houses that is only feeding 50kW. But 10% is still 10%, so either the losses are 200MW near the utility or 5kW near the home.

            But why is that not a problem? Especially in that first run from the power plant that is losing 200MW? Well if you look at a power plant, there is not one wire leaving it. There are MANY! So you immediately divide that 200MW by however many wires you have leaving the power plant (maybe on the order of 50-100?)

            Now each of those wires is an extremely high voltage wire (100kV or more–running at that voltage to be able to minimize current (remember P (fixed) = V * I, so higher V means lower I). And since loss is dependent on I (I * R (fixed for a given type of wire)) you can have very long runs (typically on the order of 100 miles or more–and the longer it is, the higher the R) so you size your transmission line and up your voltage so you can achieve the 10% loss you are trying to achieve.

            So let’s say you have 100 miles of line, carrying 1MW, and have losses of 10%, or 100kW that needs to be dissipated as heat. Well that heat is not dissipated at a single point–it is dissipated over the entire 100 mile run! So that’s about 1kW per mile, or 0.2W per foot of wire.

            Now these numbers are not necessarily real, but they are in that range. And 0.2W over a foot of wire is not going to melt it. But you better believe that it’s going to heat it up, especially on a hot day when everyone’s air conditioners are running. This is a big problem with high tension wires heating up and expanding because when they do, causing them to droop lower to the ground. This has been the cause of several blackouts where the wires droop enough to touch undergrowth below them and short out.

            Now you have the same issue closer to the home as well, but again, the power levels are far lower, and again, they are spread across distance and you don’t have a situation where you are trying to dissipate an enormous amount of power in a small space.

            But now getting back to the original reason for my comment about heat dissipation. If you are transferring 1.5MW over a relatively short distance (let’s even say 50 feet in the case of a really thick cord, but my guess is that it would be some kind of direct hard-wired overhead or underneath connection of some kind like electric busses use) and your 10% loss (150kW) is only spread out over 50 feet. That’s 3kW per foot! Yes, that’s going to significantly heat up that wire! That’s going to significantly heat up all the components that that current is flowing through. You would need crazy amounts of cooling to dissipate that 150kW of heat loss in a fairly confined space.

          • Lance Pickup

            And to your comments about up and coming technologies…

            First, let me say that I don’t deny that future technologies will enable much faster charging rates than we see today (although I caution you that some of this technology is still very much laboratory work and almost certainly not likely for a semi-truck that may debut in 3 years or so). My objection is more on the topic of whether or not there will be a charger available that can actually DELIVER this amount of charge, at this scale, in a reasonable time.

            So the question is: will these technologies be able to scale up to semi-truck size.

            The answer is, at this time, no.

            Let’s take supercapacitors for example. Sure, they absolutely exist today (in far smaller capacities than would be necessary for even an EV, much less an electric semi). And yes, they absolutely can be charged/discharged incredibly fast.

            But in quantities needed for an EV, they are still enormously expensive. No worries, you say, because once Tesla starts making them in a Gigafactory somewhere, the cost will come down dramatically, just as it did for Li-ion batteries. Well, that’s a fair enough statement. BUT, there is a long way to go. But remember that Li-ion batteries did not become practical (cost-wise) for automotive use because Tesla set out to build a Gigafactory…rather it was the consumer electronics industry that ramped up the volumes that made it practical to start thinking about putting these in cars. Li-ion batteries in laptops and smartphones are what drove down the cost to the point where it started to make sense to put these in cars, and yes, now the Gigafactory will drive down costs that much more.

            So my comment to you is this: where are supercapacitors in today’s consumer electronics? Certainly laptops, tablets and smartphones could benefit from being able to fully charge in 30 seconds, right? If the Samsung Galaxy S9 were able to be full charged from 0 to full in 30 seconds, and be able to recharge tens of thousands of times (unlike Li-ion batteries that pretty much make your phone worthless after 2-3 years), that would fly off the shelves. And we wouldn’t have to worry quite as much about heat dissipation since we’re not talking about charging a MWh class battery, but maybe a 3Wh supercapacitor. Although to charge that in 30 seconds is still going to require 360W which is an impressive amount of heat (consider touching a 100W incandescent bulb and multiplying that by 3 or 4!) But nonetheless, it could be managed.

            I will say that if supercapacitors are ever going to end up in vehicles of any kind (much less a truck), we will see them in laptops and phones probably at least 5 years prior.

            So: do we see them today? No. Why not?

            Well, in addition to being horribly expensive (although at the scale needed for a smartphone you might just be able to fold that into the cost of the phone), they have an extremely low energy density compared to Li-ion batteries. This means that for the equivalent amount of energy stored, supercapacitors would be incredibly bulky and heavy, and who wants that in a phone. Or a tablet. Or a laptop.

            Even putting them into a car would be problematic as EVs already use about as much space as they can for their batteries. You’d have to make the car significantly bigger to hold all the supercaps.

            But aha, I hear you say! A semi has plenty of room. Okay, fair enough. But how much room would you need? Well the energy density of Li-ion is about 20-40X that of supercaps. So imagine the Tesla semi that probably has a significant portion of the cab dedicated to the battery pack. Now multiple that by 20 or 40! You’d basically have to be pulling a small trailer with your supercap bank, and then have the main trailer behind that.

            Just doesn’t seem that practical to me. Maybe someday, but like I said, at this point improving this technology to be practical for this scale is pretty much a laboratory exercise. And that will make no progress unless there is funding. And in order for there to be funding, there has to be an eventual application. And like I’ve been saying, there are serious issues with the charging infrastructure alone that would make a supercap based solution for semis impractical. So why would anyone fund supercap research, given that the main benefit of supercaps is high charging rates, when I don’t believe we will ever have a solution for fast charging even the relatively glacial Li-ion batteries?

      • Michael Walsh

        Furthermore, the lesson should be learned by now that “Battery Swap Stations” as a concept are expensive, impractical==> and totally unnecessary if the batteries can be fast charged. Infrastructure that costs much, much less.

    • Vincent Wolf

      I agree. But long distance truck drivers must obey the 11/24 rule for truckers–which means in a nutshell they can only drive 11 hours in a 24 hour period (http://askthetrucker.com/the-1114-hour-truck-driving-rule/) and that means that if battery swapping is used then truckers only need about a 6 hour charge capability and then when they stop for lunch they can swap the batteries out and continue on for 5 more hours (maximum by law). Thus you only need about 650 kW and not 1200 kWh of battery charge. Of course if the trucker is using a buddy and making use of the sleeper berth it changes things a bit but still with quick battery swaps they really don’t need more than 600-700 kWh–not 1200.

      • Lance Pickup

        I’m only going on the statement that the article concluded that 1.2MWh packs would be needed…For now anyway I’m taking that at face value.

  • brian_gilbert

    Why the surprise? It looks ceratin to me that flashcharging will become viable within very few years so a smaller battery charged frequently will be feasible. Vehicles could probably recharge while moving but if necessary do it at existing truck stops.

  • http://www.efest.ca Robert (Electricman) Weekley

    So – “The BYD T9 has a 188 kWh battery pack, and a stated range of 92 miles. ” – or about 0.4894 Miles Per kWh; also 2.0435 kWh per Mile. So assuming a 20% improvement minimum for 18650 Cells, and 30% improvement over that for Tesla’s new 21-70 Cells, we get at least – 0.7634 Miles Per kWh; also 1.3099 kWh per Mile.

    At this rate, or there about, Tesla Could get a 400 mile range with a 524 kWh Pack (523.968 kWh); also – for 600 miles range, it could be done with about 785 kWh ( 785.953 kWh)!

    Sure – a 1200 kWh pack would make for interesting ranges, for an autonomous truck, but with Drivers driving 11 hours a day – with at least 1-2 breaks, 600 miles just about does it!

    Making a Cab better Shaped Aerodynamically, distributing traction, and therefor – friction and maybe rolling resistance, over more wheels, might reduce some more losses, Adding in making better Aerodynamic Trailers, and they could maybe get that down to or below 700 kWh for a 600 mile range!

    • eveee

      Congratulations Robert. You win the prize for closest goes to Teslas actual numbers.

      Making better Aero? Better rolling resistance? You betcha.

  • freedomev

    Not going to happen like that unless they use metal/air batteries which they are not going to.
    The only way is battery swapping using 4-600kwh packs or so.
    Swapping can be easy, less than 1 minute with it sliding in one side and out the other or saddle packs where the fuel tanks are now.
    They can be designed to use forklifts for places without much business and robot units where there is.
    Why is charging demand rates and time needed make charging not economic unless driverless and even still.
    What needs to be done is solar stations, maybe wind with maybe RE fueled generation using the batteries for not just semi’s but grid services too being a power dump or supply as needed it can get cheap power and make money from grid services instead of paying $.50-1/kwh for demand charges, prices from the grid.
    Since an EV tractor without the battery costs less than one with a diesel to build, sell and tesla really makes their money selling electricity/mile at about 50% of diesel/mile rate makes Tesla about 50% of that after costs.
    As this rates would cut running costs by 50% trucking companies are going to be all over this from the savings, thus profits.
    Any company not using EV trucks in 10 yrs won’t be competitive with them.
    Way to go Elon!!

  • JP

    The range of an OTR truck DOES NOT need to be 600 miles. It doesn’t even need to be 300 miles. Due to the 11/24 rule; https://www.fmcsa.dot.gov/regulations/hours-service/summary-hours-service-regulations If a driver starts his/her 24 hour period at ANY hour, they can get 11 hours of “drive” time in 16 hours if using a 2 to 1 ratio of driving to charging. For example, 6 am to 8 am drive. 8 am to 9 am charge. At 9 pm, driver drives one additional hour to max out 11 hour drive max. Next 24 hour period could start 10 hours later. Using my numbers, the range of the EV semi would only need to be about 140 miles plus a nice buffer. Let’s go with 60 miles. My battery would only need 200 miles of range AND could take up to an hour to charge. My idea is to have Tesla parking spots at truck stops that have 4 plugs with liquid cooling for the hoses. This would reduce necessary current by four times AND the temperature drop would reduce resistance. Thereby, the 1.2 M battery is reduced to 400k battery. Using 4 plugs to charge 4 100k batteries fulfills my numbers. Any questions?