Anyone know why my Model 3 charges at 5 miles per hour in a standard 110 Volt outlet whereas my Model S only charges at 3 miles per hour?
I have a 70D in the S and the LR battery in the 3
More efficient at charging and different batteries.
5 miles per hour is pretty significant. If one drives less than 50 miles per day on average, there is no point in going through the expense of installing a level 2 charger in ones house/garage.
The Model 3 will get more miles per hour of charge because it takes less charge for it to travel a mile.
Because the W/mile is much less for model 3 than model S.
So if model 3 charge 5 miles an hour the W/miles is 200. For S it would be 333 W/miles in your case.
Is charging with 120 volts just as inefficient on the Model 3 as it is on other Teslas?
I think that there is around a 30% wastage in heat loss charging with 120 volts (110) compared with a 240 volt 50 amp plug.
I was expecting 3 miles/hr, the computer states 5 and sometimes 4. When I sit down and worked out the numbers a couple of times the next morning, it has actually been around 4.33
johnmann nailed it. Using miles gained as the measure, M3 charges "faster" than the MS which charges "faster" than the MX. Using kWh gained, they are probably all about the same.
Would you ask why my Accord goes 22 miles per gallon but the Civic goes 30 miles for the very same gallon? Imagine your 110 v outlet is feeding one pint worth of gas per hour, which will get more drivable range Accord or Civic?
BTW the spec sheet says Model3 will charge at 3 miles per hour of charge. Reality it seems to be doing better than that.
So does the Model X charge at only 2 miles per hour on a 110 outlet given that it is less efficient than the S?
"5 miles per hour is pretty significant. If one drives less than 50 miles per day on average, there is no point in going through the expense of installing a level 2 charger in ones house/garage."
Charging at 110V is inefficient compared to charging at the higher rates available with a 240V connection. If you care about how much total energy you are using I would try to avoid charging exclusively form 110V unless you don't have another choice. There's also data indicating that faster charge rates are better for the battery's longevity.
Not sure why, but we have been charging our M3 using UMC and 12A 120V outlet for past 2.5 months and it has worked out very well for us. I do have HPWC 100A 240V charger, which I use to charge our MS, but have not felt the need to charge >5 mph.
@ jb1120 | April 17, 2018: I don't own any Tesla, so I am talking purely theoretical point of view.
I would guess since X is going to use more energy per mile, you are going to get less miles per charging hours than S. Be it 110 v 15 amp circuit, be it 240 v 60 amp circuit, be it the supercharger.
Just a fun fact: charging at this rate is the energy equivalence of getting 5oz of gasoline per hour Also, the entire LR battery holds about the same energy as only 2.3 gallons of gasoline. It’s pretty amazing what our cars can do with it.
At officially 260 wh/mile, it takes 1300 wh to go 5 miles. That’s about out a 110 volts/12 amps outlet will produce. The model s is rated higher than 260 wh/mile.
Handy charging table comparing the three vehicles. The difference between them is more easily seen at the higher charge rates. Based on forum comments, it seems Tesla's 120V estimates for M3 are on the low side.
@djharrington It's really a function of efficiency in converting the stored energy to useful work. If I remember correctly, most gas engines only achieve 20-25% efficiency. Most of the energy stored in the gasoline is expelled as waste heat. I believe battery electric drive trains are over 90% efficient. Quite a difference, plus electric drive trains can recapture some of that energy using regen (something an ICE cannot do).
Exactly right. The super high efficiency does have some downside though.
One easy example that catches people new to EVs off-guard is the effect of wind. Let’s use efficiencies close to what you gave above for the example (90% EV and 30% ICE from stored energy to motion). Let’s assume you’re making a road trip at a speed of 70mph. On one day there is no wind. On another day, you have a 20mph headwind. The difference in power required to overcome the drag in these two scenarios is around a factor of 2 (it’s a cubed function for velocity). Let’s also assume that a third of the power needed to maintain the car’s speed is for aerodynamic drag, and then other two thirds are rolling resistance, etc (this relationship will change for diffeeent speeds, but we’re simplifying here).
What all this means is that for the EV, one third of the energy required once you’ve accelerated to speed is due to drag. In our 90% efficient EV, this means 30% is for drag. The difference with the 20mph wind means that we double that drag, so we’ve added another 30% of energy required to travel a section of the freeway because of the increase in wind.
For ICE, at only 30% efficient (30% of the total energy consumed is used for motion, rest is waste), the drag component we’ll assume the same one third of overall kinetic energy requirements(the 30%), so 10%. With the added 20mph wind, that doubles, so we’ve added another 10% of energy required to to travel a section of freeway.
While these number are used as an example, the important take home is that the same wind that would only have a 10% reduction in fuel economy in an ICE car, has a 30% reduction in “fuel economy” in the EV. When moving to EVs, the effect of things like wind seem to catch people by surprise. There should be a class on it (and really, there is ... physics), but I think most people don’t think of the application (or remember what they learned).
Ok class, time to wake up and go to your next class.
dj, mostly right, but you fell off a cliff at the end :-) The ICE car has the same increase in energy needed as the EV, regardless of how efficient / inefficient the different sources are. A 30% increase in drag means a 30% increase in electric or gas usage. It's just that for ICE cars you don't have sophisticated instrumentation showing you accurate consumption rate nor do you worry so much about range to the next gas station. You just fill up when you are near empty, not even realizing it was 50 miles shorter than usual.
Incorrect, Johnyi. In the ICE, 70% of the gas energy is being wasted as heat. The increase in energy needed to overcome the extra wind does not affect this 70%, but only the 30% that’s actually being converted to motion.
The ICE has the advantage if cabin HEATING is required, because it has lots of waste heat.
But aerodynamic inefficiencies will hit both types of cars the same.
And barometric inefficiencies will hit an ICE harder.
djharrington | April 17, 2018 said:
Incorrect, Johnyi. In the ICE, 70% of the gas energy is being wasted as heat.
The increase in energy needed to overcome the extra wind does not affect this
70%, but only the 30% that’s actually being converted to motion.
That is true. But to get one more unit of energy converted into motion, you need to waste three more units to heat.
Let us say you spent 25 joules in motion and 75 joules in waste heat in an IC engine.
Wind resistance makes you spend 50 joules. To get 50 joules converted to motion, you need to give 150 joules to Agni the God of Fire. So net net the effect on MPG or MPWh would be the same, no?
Haha, math fail. I started writing a “let me put it another way” example, only to realize my error. I sincerely apologize @Johnyi. It was apparently me that needed to stay in school!! :)
@DJ It’s ok, you were making a good point. :-) And it sort of feels true even if the physics weren’t quite right.
One reason could be the efficiency at different RPMs of an ICE. Most of my old cars have had a sweet spot where the engine was more efficient. If I could keep it in that range, regardless of wind resistance or elevation gain, my mileage was pretty good. So it may very well be that, in your example, a 30% drag increase is forcing the engine to work in a range where it’s more efficient than at lower rpms so the net increase is only ~10%, like in your example. Which, as Johnyi said, you might not even notice at your next fill up if you don’t actively track your mileage.
Or maybe I need to go back to school, too?
All I know is I’m putting myself in detention for a while. The kicker is: physics is my day job. Luckily, I only work with radiation ;-)
A bit of a tangent, but a lot of the Model 3s efficiency comes from its sleek aerodynamics.This advantage (over a boxy vehicle) is lost when going up hill as the energy used to overcome gravity far outweighs that gained from being slippery. This applies to all vehicles not just EVs. More aerodynamic vehicles lose more efficiency when going uphill. Of course an EV can gain some back when descending.
Steven.haver | April 17, 2018: " One reason could be the efficiency at different RPMs of an ICE"
That is true, and there is more to it. ICE designed for wide range of RPMs are less efficient even in their peak efficiency rpm zone. Car engines that need to pull the car from rest are less efficient than the ones that don't have to. Most of the efficiency of Prius comes from an engine that will de-clutch and idle below 9 mph. Without an electric motor taking the car from 0 to 9 mph, Prius would stall. All the regenerative braking, engine going off at traffic lights are all just the icing. The cake is the engine not tuned for low end torque.
Diesel-electric locomotives are like the Prius in that respect. But the load on their generator varies and so those engines are tuned for a range of throttle settings. ICEngines that are fixed in rpm, throttle setting and load are even more efficient.
But, even the most efficient heat engines, built flawlessly, with zero friction are not as efficient as electric motors. Take a look at the cooling fins of the air cooled engine of your lawn mower. Fixed rpm, fixed throttle, fixed load, most efficient conditions. It is pumping out 6 BHP. Find a 6 HP air cooled electric motor and compare the size of its fins. What? you have never seen cooling fins on electric motors? Yup! that tells you the story! How much of heat is being dissipated away.
A 25 HP electric motor: (the stubby ribs are the cooling fins, cast iron)http://www.galco.com/buy/Baldor/EM4118T?source=googleshopping&gclid=Cj0K...
A random motorcycle from my memory, 20 HP engine. Aluminium cooling fins.https://www.google.com/search?q=royal+enfield+bullet+350&rlz=1CATAAB_enU...
Is it fact?
“ There's also data indicating that faster charge rates are better for the battery's longevity.”???
Somem studies have shown that some of the time. :-)
I thought it was the opposite - trickle charging is better for the battery. I read somewhere that using Superchargers too much is bad for the battery, and that you're supposed to slow charge it once in a while.
@johnyi | April 17, 2018: Battery chemistries are so vastly different between lead-acid, ni-cd, lithium ... The trickle charge idea goes back to lead acid days. Not sure how much of that lore is still valid.
I'm sure there's both a "too fast" as well as a "too slow" in terms of what's healthy for the battery. But in some pretty extensive high-precision testing of various lithium-based battery technologies the most significant factor in overall longevity was charge rate. Cells that were cycled at a higher current had significantly less capacity degradation than cells cycled at a lower current.
Supercharging is an outlier relative to home charging rates and is likely faster than ideal for battery longevity.
Here's the video that shows the data and explains why charge rate affects cell longevity:https://www.youtube.com/watch?v=9qi03QawZEk
The guy in the video is an expert on battery chemistry and a Tesla research partner.
tl;dr: The act of charging maintains a reactive state causing the electrolyte to degrade. When charging stops the reaction forms a protectively film that stops the degradation process. Hence a shorter charge period decreases the amount of time during which the electrolyte is subjected to degradation.
IIRC, one of the studies in question found that in Li batteries, “dendrites” would form between cathode and anode, and if completely formed, would short the cell. Fast charging (~1C) would help to dissolve the dendrites. Of course, after my earlier mishap, look that one up for yourself before trusting me :)
I believe Idaho Natl Labs had some stuff on the topic.