Cabin Heat / Speed Interplay on kWh/mile

[hugh_jassol]

This is sim

So the best efficiency is going to be at the same speed, no matter if full A/C, no A/C, or full heat.

This is true. The power required to heat the cabin is a function of outside temp, not the speed of the car, state of charge of the battery or anything else. Conversely, the power required to go 60 mph doesn't depend on how warm the cabin is. Those are two independent power needs, so they can be optimized independently.

In other words, the least energy you need for any given drive will always be the optimal speed for a Tesla (~40ish mph). Because for a drive of more the just a few min, the cabin heat requirement is qasi-constant.

And yes, I am aware that the air density changes with ambient temperature and pressure, so the optimal speed varies a bit.

 

[ralph142]

The heater may be capable of 4 kW or so but it will seldom run at that rate. Think about it for a minute. A kW is over a quarter of a ton of heating. Do you need a quarter of a ton to keep the tiny cabin of your car warm? For perspective, it's just freezing here at the moment and my 8 ton heat pump is running at about 35% capacity as it brings my house up to its its daytime settings. There's another 9 tons of gas backup behind that so even though the house is in "recovery" only 16% of what I've got is being used.

So also is it with the car. It may draw a couple of kW (put a watt meter or clampon ammeter on the UMC to see) when you initially warm the car but within minutes it will be down to under a kW and stay there as the PID or fuzzy controller maintains the temperature. So preheat the car on shore power, if you can, and you will find that the heater isn't the great sink for battery power that people seem to think it is. Yes, it will add a few percent (10% ?) to your Wh/mi and reduce your range and, of course, if you are going across the prairies in -30 °F weather at 70 mpH heater load will go up. But for more nominal cold weather driving it won't make that much difference - certainly not in terms of the question posed by the OP. As I keep saying whenever this subject comes up the big attention getter with cold weather driving is stuff (water, snow, sand) on the road.

A small resistance heater that heats up a room or a large boat cabin perfectly well usually draws 1500 to maybe 3000 watts, no way is the car heater pulling 5 k watts on anything but a burst mode.

 

[ajdelange]

So the best efficiency is going to be at the same speed, no matter if full A/C, no A/C, or full heat.

This is true. The power required to heat the cabin is a function of outside temp, not the speed of the car, state of charge of the battery or anything else. Conversely, the power required to go 60 mph doesn't depend on how warm the cabin is. Those are two independent power needs, so they can be optimized independently.

No, it is not true. One can drive any speed he wants and have at least some control over what the heater does independently of that but the Wh/mi, inversely proportional to the efficiency couples those two i.e.

Wh/mi = P/v + v^2/30 + 5*v/6 + 120

in which P is the demand of the heater in watts and v is the speed in mph is a function of both. A plot may be of interest

as it clearly illustrates this point and affords an alternative way to find the optimum speed for a given heater load (assuming the model for the car's traction load is accurate which, as we know, is not the case for the model in the equation taken from another post).

Also for a given heat set point the demand from the heater clearly is a function of the car's speed due to convective cooling but attempting to add that to this simple model would not, IMO, improve OP's understanding of what I believe he wants to know.

In other words, the least energy you need for any given drive will always be the optimal speed for a Tesla (~40ish mph).

Clearly not true from the model used here and not true for any other model either though with more accurate models and low heater draw it may be approximately so.

 

[StealthP3D]

The power required to heat the cabin is a function of outside temp, not the speed of the car, state of charge of the battery or anything else.

This statement is mostly true (ignoring wind chill) but it misses the purpose of the exercise which is to determine *trip* efficiency at different speeds. Because a slower trip takes longer the heater is on for a longer period of time. So what is of interest is not the *power* needed to maintain a certain temperature but the *energy* used to heat the cabin for the duration of the trip. *Power* is measured in kW while *energy* is measured in kWh. At a given heater power level, the *energy* used by the heater for the duration of the trip will be higher if the trip takes longer with a slower cruising speed.

In other words, the least energy you need for any given drive will always be the optimal speed for a Tesla (~40ish mph). Because for a drive of more the just a few min, the cabin heat requirement is qasi-constant.

This statement is false. In cold weather, when cabin heating is required to maintain comfort at a specific cabin temperature, a slightly faster cruising speed will return better overall trip efficiency (as measured in kWh/mile) because it will reduce the total time the heater needs to run to arrive at your destination. Partially offsetting this factor will be the higher windchill the cabin will experience at slightly higher speeds. Yes, a heated cabin does experience the effects of windchill and those effects increase with increasing speeds.

 

[thedanl]

I made an account to post this b/c this seemed like a fun Saturday morning exercise.

I extracted Wh/mi consumption data of the model 3 LR RWD from A Better Route Planner (ABRP) post here:
Tesla Model 3 Performance vs RWD consumption - Real Driving Data from 233 Cars

I found a tool to extract data from figures so the data may not be perfect.

I took this as the baseline and used the formula: new_consumption = baseline[Wh/mi] + power_in_watts/speed_in_mph

The baseline optimal consumption is 155.2 Wh/mi at 26.9 mph.
If there is a 1500W continuous overhead due to heating, the optimal consumption is 204.5 Wh/mi at 33.5 mph.

I'm not sure if there is any useful information to be gleaned, but it's an interesting exercise and it does result in a noticeable difference. Note that while noticeable, the difference might not be substantive in real would applications.

Here is a plot showing the comparison:

 

[hugh_jassol]

This statement is mostly true (ignoring wind chill) but it misses the purpose of the exercise which is to determine *trip* efficiency at different speeds. Because a slower trip takes longer the heater is on for a longer period of time. So what is of interest is not the *power* needed to maintain a certain temperature but the *energy* used to heat the cabin for the duration of the trip. *Power* is measured in kW while *energy* is measured in kWh. At a given heater power level, the *energy* used by the heater for the duration of the trip will be higher if the trip takes longer with a slower cruising speed.

Yes, but given the disparity in power draw by the motive force vs the heater, I'm estimating that the heater is going to be a small fraction. For example, if you get 250 Wh/mi and drive 60 mi at 60 mph, that's 15 kWh of energy. If you have 1 kW of heat on for that whole time (I doubt it's even that high), that's only about 6% of the total energy used. So I find it highly unlikely that that whether or not you take the heat into account changes the answer appreciably from "optimal speed"

 

[Big Earl]

A small resistance heater that heats up a room or a large boat cabin perfectly well usually draws 1500 to maybe 3000 watts, no way is the car heater pulling 5 k watts on anything but a burst mode.

Keep in mind the car's HVAC system is almost always using outside air when in heating mode unless you manually override it. Your example of a space heater in a room is using recirculated, conditioned air. It takes a lot more energy to heat 0F air up to 100F than it takes to heat 65F air up to 100F.

Steady-state heating in auto mode on a Model 3 at 30F is around 2,000 Watts. With an outside temperature of -10F, it's closer to 4,000 Watts.

 

[ajdelange]

Yes, but given the disparity in power draw by the motive force vs the heater, I'm estimating that the heater is going to be a small fraction.

That is true at higher speeds as the plot in No. 23 shows. Note that the curves become farther and farther apart as speed increases. For example, at 65 mpH they are about 700 W apart. At 60, only 600. As these curves are 10 Wh/mi apart that means we turn on the heater to 1000W at 65 mph we would expect an additional 10*1000/700 = 14.3 Wh/mi consumption whereas at 60 it would be 10*1000/600 = 16.7 Wh/mi. Now at 40 mph they are only 400 W apart and turning on a 1000 W heater is going to cost 25 Wh/mi which is not only larger absolutely but a larger percentage of the smaller traction demand.

 

[StealthP3D]

Yes, but given the disparity in power draw by the motive force vs the heater, I'm estimating that the heater is going to be a small fraction. For example, if you get 250 Wh/mi and drive 60 mi at 60 mph, that's 15 kWh of energy. If you have 1 kW of heat on for that whole time (I doubt it's even that high), that's only about 6% of the total energy used. So I find it highly unlikely that that whether or not you take the heat into account changes the answer appreciably from "optimal speed"

You can argue the *degree* to which the heater impacts range but I was responding to the claim that the heater has a specific average power consumption and therefore the speed of the trip is not relevant to trip efficiency.

That has been shown to be false. IMO, the speed of optimum efficiency will vary a small but not a negligible amount depending upon heater usage.

It's mostly an academic exercise because in practice we all drive whatever speed we feel like. But I can see how it could be useful if you were in an arctic blast in N. Dakota and it wasn't certain the next Supercharger was in range. In this case, it would be prudent to turn the heater as low as possible without letting the windshield ice and without becoming hypothermic. But you probably wouldn't want to drive 35 mph because your optimum speed would likely be more like 45-55 (depending upon how much electricity the heater was drawing).

 

[thedanl]

You can argue the *degree* to which the heater impacts range but I was responding to the claim that the heater has a specific average power consumption and therefore the speed of the trip is not relevant to trip efficiency.

That has shown to be false. IMO, the speed of optimum efficiency will vary a small but not a negligible amount depending upon heater usage.

It's mostly an academic exercise because in practice we all drive whatever speed we feel like. But I can see how it could be useful if you were in an arctic blast in N. Dakota and it wasn't certain the next Supercharger was in range. In this case, it would be prudent to turn the heater as low as possible without letting the windshield ice and without becoming hypothermic. But you probably wouldn't want to drive 35 mph because your optimum speed would likely be more like 45-55 (depending upon how much electricity the heater was drawing).

In my earlier post (though it was my first post and its not clear to me that anyone else can see it) I had calculated the baseline optimal speed at 26mph but that value rises to 33mph assuming an average 1500W power draw from the heater.

 

[StealthP3D]

Keep in mind the car's HVAC system is almost always using outside air when in heating mode unless you manually override it.

Do you have a source for that info? Just based on my own experience with our two Model 3's, I was under the impression that "Auto" mode made fairly regular use of "Recirculate" mode. And this would make sense from the standpoint of winter driving range.

 

[Big Earl]

Do you have a source for that info? Just based on my own experience with our two Model 3's, I was under the impression that "Auto" mode made fairly regular use of "Recirculate" mode. And this would make sense from the standpoint of winter driving range.

Sure - look at the screen. If it's recirculating, it'll be lit up in solid blue (if manual recirculate) or faint blue (if automatic and recirculating). Fresh air is solid gray, while faint gray is automatic and fresh. I don't think I've ever seen it recirculate while heating - only while cooling.

 

[StealthP3D]

In my earlier post (though it was my first post and its not clear to me that anyone else can see it) I had calculated the baseline optimal speed at 26mph but that value rises to 33mph assuming an average 1500W power draw from the heater.

It may be that low in some EV's. I was surprised to find my Performance Model 3 can achieve very low numbers around 40 mph (with all electrical accessories off). I cannot replicate these numbers at speeds under 30 or even 35 mph. In my wife's LR RWD Model 3 I don't have as much experience messing with efficiency (and I do think the speed of highest efficiency might be a little slower) but I've also noticed it's difficult to get as good of numbers when going very slow. I think it's mostly due to the rpm/torque efficiency curves of the electric motors. Every model of car (and even the specific configuration (AWD vs. RWD) will have different efficiency curves.

At speeds above 50 (for sure) the air resistance definitely overwhelms the variability in the motor efficiency curves but keep in mind, these are subject to change as Tesla is continually optimizing the programming of the drive inverter based on customer data collected by their over-the-air-network. That's how we get updates that improve our power and range. It's pretty amazing and exactly the opposite of what happens to a gasoline car once they are well broken in.

 

[StealthP3D]

Sure - look at the screen. If it's recirculating, it'll be lit up in solid blue (if manual recirculate) or faint blue (if automatic and recirculating). Fresh air is solid gray, while faint gray is automatic and fresh. I don't think I've ever seen it recirculate while heating - only while cooling.

I'll take a look at that, thanks for the heads up!

Do you take many long trips? I ask because it would make sense for recirculate to be "Off" for the first 15 minutes or so but to be utilized more heavily on a longer trip after the car is defrosted and fully warmed up.

 

[Big Earl]

I'll take a look at that, thanks for the heads up!

Do you take many long trips? I ask because it would make sense for recirculate to be "Off" for the first 15 minutes or so but to be utilized more heavily on a longer trip after the car is defrosted and fully warmed up.

You can cycle through the three recirculation modes (auto / recirc / fresh) by tapping the button repeatedly. You'll see the various colors. When you're in auto mode (faint gray), the system will either recirculate or provide fresh air depending on conditions - it will be either faint gray or faint blue depending on what it's doing.

Do I take many long trips? Yes - 45,000 miles worth of driving in 15 months of ownership, with 75% of those miles being trips >300 miles. Any amount of recirculation in the winter will result in unacceptable humidity buildup in the cabin. I leave it in full auto and let it do its thing.

 

[ajdelange]

It takes a lot more energy to heat 0F air up to 100F than it takes to heat 65F air up to 100F.

100/35 = 2.85 times as much but who would heat his car (or room) to 100 °F?

Steady-state heating in auto mode on a Model 3 at 30F is around 2,000 Watts. With an outside temperature of -10F, it's closer to 4,000 Watts.

Let's look into that a little further. Lets suppose the cabin contains 200 ft^3 of air (I expect the cabin of a 3 is smaller than that) and that we change it 10 times per hour. That's 2000 ft^3/h. The formula for the amount of heat required is

BTU/h = 1.08*CFM*∆T

If we look at heating air from 0 °F to 70 °F then we'd need 1.08*70*2000/60 = 2520 BTU/h. As 1 kW = 3412 BTU/h we'd need about .740 kW. If we are dealing with -10 °F then the drop is 80 °F and we'd need 8*740/7 = .845 kW.
Wonder where the 2 and 4 kW numbers came from. ?

But the reason I leave scenic Quebec in the winter is so that I don't have to drive in 0 or sub 0 weather. Here in No. Va its pretty cold - around freezing in the mornings so lets do the number for a 40 ° drop (32 °F out but the cabin set for 72 °F. That would require 4*740/7 = 0.423 kW. Supposing one to be doing 60 in 32° weather that would be adding 7 Wh/mi to the consumption which is about 3% of a traction demand of 230 Wh/mi (no idea if that's reasonable for a 3).

Keep in mind the car's HVAC system is almost always using outside air when in heating mode unless you manually override it.

Clearly, then, the strategy for a driver who wants to maximize range is to at least make sure external air isn't being introduced or, better yet, use the seat heaters.

 

[thedanl]

It may be that low in some EV's. I was surprised to find my Performance Model 3 can achieve very low numbers around 40 mph (with all electrical accessories off). I cannot replicate these numbers at speeds under 30 or even 35 mph. In my wife's LR RWD Model 3 I don't have as much experience messing with efficiency (and I do think the speed of highest efficiency might be a little slower) but I've also noticed it's difficult to get as good of numbers when going very slow. I think it's mostly due to the rpm/torque efficiency curves of the electric motors. Every model of car (and even the specific configuration (AWD vs. RWD) will have different efficiency curves.

At speeds above 50 (for sure) the air resistance definitely overwhelms the variability in the motor efficiency curves but keep in mind, these are subject to change as Tesla is continually optimizing the programming of the drive inverter based on customer data collected by their over-the-air-network. That's how we get updates that improve our power and range. It's pretty amazing and exactly the opposite of what happens to a gasoline car once they are well broken in.

Since my second post is showing up (but not the first), I'm reposting here below. Since the referenced ABRP figures include the performance version, I could probably repeat the procedure for the performance model 3.
---
I made an account to post this b/c this seemed like a fun Saturday morning exercise.

I extracted Wh/mi consumption data of the model 3 LR RWD from A Better Route Planner (ABRP) post here:
Tesla Model 3 Performance vs RWD consumption - Real Driving Data from 233 Cars

I found a tool to extract data from figures so the data may not be perfect.

I took this as the baseline and used the formula: new_consumption = baseline[Wh/mi] + power_in_watts/speed_in_mph

The baseline optimal consumption is 155.2 Wh/mi at 26.9 mph.
If there is a 1500W continuous overhead due to heating, the optimal consumption is 204.5 Wh/mi at 33.5 mph.

I'm not sure if there is any useful information to be gleaned, but it's an interesting exercise and it does result in a noticeable difference. Note that while noticeable, the difference might not be substantive in real would applications.

Here is a plot showing the comparison:

---

 

[Big Earl]

100/35 = 2.85 times as much but who would heat his car (or room) to 100 °F?



Let's look into that a little further. Lets suppose the cabin contains 200 ft^3 of air (I expect the cabin of a 3 is smaller than that) and that we change it 10 times per hour. That's 2000 ft^3/h. The formula for the amount of heat required is

BTU/h = 1.08*CFM*∆T

If we look at heating air from 0 °F to 70 °F then we'd need 1.08*70*2000/60 = 2520 BTU/h. As 1 kW = 3412 BTU/h we'd need about .740 kW. If we are dealing with -10 °F then the drop is 80 °F and we'd need 8*740/7 = .845 kW.
Wonder where the 2 and 4 kW numbers came from. ?

But the reason I leave scenic Quebec in the winter is so that I don't have to drive in 0 or sub 0 weather. Here in No. Va its pretty cold - around freezing in the mornings so lets do the number for a 40 ° drop (32 °F out but the cabin set for 72 °F. That would require 4*740/7 = 0.423 kW. Supposing one to be doing 60 in 32° weather that would be adding 7 Wh/mi to the consumption which is about 3% of a traction demand of 230 Wh/mi (no idea if that's reasonable for a 3).



Clearly, then, the strategy for a driver who wants to maximize range is to at least make sure external air isn't being introduced or, better yet, use the seat heaters.

You aren't heating the car to 100F, but you are heating the supply air from the vents to over 100F. I was just using a round number as an example. Depending on how wide the delta T is between the setpoint and the actual interior temperature, supply air temperature could be as high as 140F.

So going back to the previous example where someone compared their car to a home space heater, my point is that it takes more energy to heat outside air in a vehicular HVAC application than it takes to heat recirculated air in a home HVAC application.

 

[ewoodrick]

No, it is not true. One can drive any speed he wants and have at least some control over what the heater does independently of that but the Wh/mi, inversely proportional to the efficiency couples those two i.e.

Wh/mi = P/v + v^2/30 + 5*v/6 + 120

in which P is the demand of the heater in watts and v is the speed in mph is a function of both. A plot may be of interest



as it clearly illustrates this point and affords an alternative way to find the optimum speed for a given heater load (assuming the model for the car's traction load is accurate which, as we know, is not the case for the model in the equation taken from another post).

Also for a given heat set point the demand from the heater clearly is a function of the car's speed due to convective cooling but attempting to add that to this simple model would not, IMO, improve OP's understanding of what I believe he wants to know.

Clearly not true from the model used here and not true for any other model either though with more accurate models and low heater draw it may be approximately so.

Honestly, in my view, it seems as if you've posted random information that looks really official. Your equation, without any units or reference to what it is doing just seems random to me.

Let's do a quick verification with no heat or A/C required.

It's an ever increasing formula that says the optimal speed is at 1 mph.

There seems to be nothing in the equation to take care of the car's friction and other function that result in the car's optimal speed to be in the 30's. And in this case, that's the whole intent.

Now the graph that you displayed seems to have more truth in it. About all I can say is that it seems to just confuse the original poster's intent.

As far as I can see, there are a few factors here.
​

• The usage of the car at speed.​
• The amount of time of the trip, which determines total power required for heat​
• Impact of speed on the amount of heat needed (i.e. watts @ 10C vs watts @ 0C)​

So, after spending a few hours and working through some numbers, I stand corrected, there is an impact, from some really rough calculations, the most efficient speed is about 10 mph faster, not considering:​

• Pre-conditioning the battery​
• Additional heating requirements at different speeds​
• This is a not a short trip. (i.e. car can come to temperature and stay at temperature for more than 70% of the trip)​
• Needing to stop to charge (which interesting suggest that you go a LOT faster)​

SO, in short, unless you are willing to drive 40 mph, this whole discussion is still moot.

If you drive over 30-40 mph, you pay the price in energy cost.
Which is more, your time or your money.

​

 

[ajdelange]

You aren't heating the car to 100F, but you are heating the supply air from the vents to over 100F. I was just using a round number as an example. Depending on how wide the delta T is between the setpoint and the actual interior temperature, supply air temperature could be as high as 140F.

The temperature of the air that comes out of the vent doesn't enter into the calculation. Energy is conserved. The relatively small volume of quite hot air transfers its heat to the larger volume of air in the chamber being heated. What counts is how much room air is heated through the rise, IOW the difference between the load temperature and the set point. The stream of hot air is merely the vehicle to transfer the heat to the larger volume of cabin air. One could measure the amount of heat transferred to the cabin by measuring the flow and rise of that air but then one has to understand that that heat is diluted as this mixes with the cabin air. It is the same problem as determining what happens when you add a quart of hot water to a pot with 5 gallons of cold water in it.

So going back to the previous example where someone compared their car to a home space heater, my point is that it takes more energy to heat outside air in a vehicular HVAC application than it takes to heat recirculated air in a home HVAC application.

Actually it doesn't. The same physics apply. In the home one is required to replace a certain volume of the air per hour depending on room type. I got the 10 changes per hour number I used for the car calculation from the requirement for a kitchen in a home. You can argue that the Tesla changes cabin air more frequently (or less frequently) than 10 times per hour if you have any evidence to support that but you can't argue that mass and energy are not conserved.