Cabin Heat / Speed Interplay on kWh/mile

[electrongeek]

Here is the thought problem. Its a very cold day. I have to drive 60 miles to my destination. Assume, just as an example, cabin heater needs to draw 5 kW to maintain a given cabin temperature. What would be the ideal speed to drive at for the least kWh/mile for this trip? Obviously if you are traveling at, say 1 mph, the cabin heat draw overwhelms what it would take to move the car, and the battery would be exhausted long before your destination. On the other hand, If you drove at 120 mph, you would drastically reduce total energy needed to keep the cabin warm for those 30', but the energy cost per mile to move the car at that speed would be very high. I'm sure there is an "ideal" speed at which to make that trip which would net the least total energy consumption from the battery!

It would be very interesting to have a formula to make that calculation based on outside temperature and desired cabin temperature. Naturally there would have to be empirically derived factors to make that work, andI know there are a number of potentially confounding variables to such an equation, but a "ball park" figure would still be useful.

 

[jdw]

Can't really help with the thought problem, but you can see a guesstimate in real time if you enter a trip and bring up the energy panel while driving. You'll get a real time display of the estimated arrival charge at your destination and it will vary as you change speed or adjust the heat. There are a lot variables in the thought experiment (wind, rain, snow, elevation, driving speed, accessory use).

There is a "formula" for driving when using Superchargers - "Don't drive faster than you can charge", meaning there is a point where driving faster extends the charging time and results in a longer trip than driving slower.

 

[electrongeek]

Can't really help with the thought problem, but you can see a guesstimate in real time if you enter a trip and bring up the energy panel while driving. You'll get a real time display of the estimated arrival charge at your destination and it will vary as you change speed or adjust the heat. There are a lot variables in the thought experiment (wind, rain, snow, elevation, driving speed, accessory use).

There is a "formula" for driving when using Superchargers - "Don't drive faster than you can charge", meaning there is a point where driving faster extends the charging time and results in a longer trip than driving slower.

Good point about using the trip energy panel display, it hadn't occurred to me. I haven't played with that enough to know how quickly that updates, but it might be all one would need, thanks.

 

[Rocky_H]

Good point about using the trip energy panel display, it hadn't occurred to me. I haven't played with that enough to know how quickly that updates, but it might be all one would need, thanks.

That display is immensely valuable and updates about a couple of times a minute, so it answers this question directly in real-time. I think that's why some of us were confused at this very theoretical hypothetical question. That display will show you this as you drive.

 

[ewoodrick]

I don't think that it is realistic to say that the heat draws 5k. It will vary based upon the temperature and speed. It should be cycling on and off

 

[electrongeek]

I don't think that it is realistic to say that the heat draws 5k. It will vary based upon the temperature and speed. It should be cycling on and off

It was just a “thought experiment” to illustrate the question. 5 kW was not chosen for a specific reason.

 

[ewoodrick]

Ok, so that answer is about 35 mph. That's the optimal speed for range for a Tesla. Temp will change the range, but that really isn't a player in the calculation.

 

[KenC]

The answer is 42.

 

[electrongeek]

Ok, so that answer is about 35 mph. That's the optimal speed for range for a Tesla. Temp will change the range, but that really isn't a player in the calculation.

You might be right that 35 mph is still the most efficient speed to cover a given distance regardless of whatever load from the cabin heater,, but a pat answer without calculation to back it up is meaningless to me. At very slow speeds over a given distance, cabin heat requirements might predominate total energy consumption. At high speeds, cabin heat becomes far less consequential. At some speed - maybe 35 mph, maybe 42, there will be a nadir in energy consumed for a distance, and that will vary depending on actual kW draw of the cabin heater for that trip.

 

[electrongeek]

Anyone have good figures for Wh/mile for the TM3 at different speeds from 35 mph to 70 mph on level ground? That data would allow a calculation using various cabin heat power draws.

 

[ewoodrick]

You might be right that 35 mph is still the most efficient speed to cover a given distance regardless of whatever load from the cabin heater,, but a pat answer without calculation to back it up is meaningless to me. At very slow speeds over a given distance, cabin heat requirements might predominate total energy consumption. At high speeds, cabin heat becomes far less consequential. At some speed - maybe 35 mph, maybe 42, there will be a nadir in energy consumed for a distance, and that will vary depending on actual kW draw of the cabin heater for that trip.

It comes from a team that went through some calculations and determined that it was the best and now has the record for over 600 miles on a single charge. So I have a little trust in it.

In the Leaf, the manufacturer highly recommends the seat heating instead of cabin heating, as seat heating is dramatically less requirements than cabin heat.

 

[gecko10x]

Couple of thoughts for you:

• Optimal efficiency without heat is @35mph, so that is floor.
• You know the rated efficiency (around 225 wh/mi depending on your model), and can look up what speed that was at.
• Wind resistance effect on efficiency is exponential, but heater use is constant.
• Heater usage has been calculated elsewhere on this forum, but I could not find the reference. However, actual usage will depend on outside temp, so I might just go with an estimate (like 25% hit to efficiency @ rated speed)

Having said that, my guess is that the efficiency hit to going to faster more than overtakes the savings on heat.

 

[electrongeek]

Couple of thoughts for you:

• Optimal efficiency without heat is @35mph, so that is floor.
• You know the rated efficiency (around 225 wh/mi depending on your model), and can look up what speed that was at.
• Wind resistance effect on efficiency is exponential, but heater use is constant.
• Heater usage has been calculated elsewhere on this forum, but I could not find the reference. However, actual usage will depend on outside temp, so I might just go with an estimate (like 25% hit to efficiency @ rated speed)

Having said that, my guess is that the efficiency hit to going to faster more than overtakes the savings on heat.

I have seen a maximum heater core draw of 7 kW quoted in several places. I don't know what kind of outside temperature would require a full 7 kW to maintain comfort in the cabin, but in that extreme instance the 116 Wh/mile draw of the heater core would be a substantial amount compared to the 240 Wh/mile (roughly) of a car traveling at 60 mi/hr. Is a 7kW rate large enough to indicate a speed higher than 35 mph would be more efficient? I don't really know. And if it is, how low a heater core draw would bring you back down to 35 mph as the most efficient speed? Sorry to flog the horse with this stuff, but no one has to read it!

 

[derotam]

So am I missing something...If I drive 60 miles both at 30mph and 60mph, and I have a constant heater draw of lets go with 6kW, that is 12kWh or 200Wh/mile @ 30mph, and 6kWh or 100Wh/mile @ 60mph. So the question then is, what is the difference in driving efficiency between 30 and 60 miles per hour. If the difference is less than 100Wh/mile between 30 and 60mph, then 60mph is better, if less than 100Wh/mile difference then slower is better.

 

[user212_nr]

So some basic algebra

Units
distance in miles
time in hrs elapsed
h = heater kwh average

Speed vs time equivalent equations
speed = distance / time
speed*time = distance
time = distance/speed

Equation for speed/power
Wh/mile = MPH^2 / 30 + MPH * 5/6 + 120
Wh = time*(Wh per mile equation)
Wh = distance/speed*(Wh per mile equation)

Equation for heater
heater kWh = Ht
heater kWh = Ht
heater kWh = H*distance/speed = H*D/MPH

Total equation for watts used (adding above equations together):

D/MPH*(MPH^2 / 30 + MPH * 5/6 + 120) + H*D/MPH

or

(D/x)*(x^2 / 30 + x * 5/6 + 120 + H)

Step 2: Add distance and heater constants

In this case H=5kwh, D=60 miles

D*x/30 + D + (5/6+120)*D/x + H*D/x

2x + 60 + (5/6+120)*60/x + 5*60/x

Step 3: Graph and Find Min

Best Graphing Calculator Online (Easy-to-use & Free)

In this case, min is 61 MPH for 294 kWh.

Now applied to 200 M, the value goes up slightly to 65.64.

For 10 miles, about 39.3 MPH.

This is only as accurate as the math above and the input constants and power/MPH equation, but you can re-apply and fix/improve using the same basic technique.

The heater constant will always be changing, and also it was made up by the OP for example purposes, so these charts attempt to refer to a heater that uses 5,000 W average.

Also, removing the heater from the equation only drops by 10 mph, whereas some say lower is better, so the power input equation is probably wrong.

But this is approximately how the math could be done.

Power equation taken from this user forum post

Table of energy consumption vs. speed | Tesla

 

[ajdelange]

travel time is t = 60/v hours where v is the speed in mph.
Let's use the model for consumption from an earlier post: Wh/mile = v^2 / 30 + V * 5/6 + 120

Thus the energy used in going 60 mi is E = P*60/v + 60*( v^2 / 30 + v * 5/6 + 120) where P is the heater demand in watts. The first derivative of this is

-P*60/v^2 + 60*(2*v/30 +5/6)

Setting this to 0 and cancelling the 60's we get

P/v^2 = 2*v/30 + 5/6
or
P = (2/30)*V^3 +(5/6)*v^2

Ick. It's a cubic! Thus an iterative solution is called for but it's trivial to get one graphically. Just plot P vs 2/30)*V^3 +(5/6)*v^2 and look up the v corresponding to the heater power. It looks like this:

The use of log axes makes it much easier to read. Thus, if we think the heater is going to draw 1 kW (10^3 watts) we read along the 10^3 line until it crosses the curve and then down at that point and see that the optimum speed is 21 mph. IOW, when the heater draws little power, we go slow. If we think the heater is drawing 10 kW we do the same for the 10^4 horizontal line and come down at 50 mph.

Note that the distance driven cancelled out.

Note also that the curve (and the the answers) depends on the Wh/mi model we use. The model we borrowed is monotonic in v. That is not the case with a real vehicle. As we get below about 40 mph the Wh/mi actually goes back up so that the actual optimum speed with low heater power would be about 40 - 45 mpH which is the speed for the maximum efficiency of these cars.

Interesting exercise.

 

[ajdelange]

Some may have noticed that the plot is nearly log-log linear i.e. logP ~ (3/1.3)log V. This means that V ~ P^(1.3/3) = P^(0.43). Thus the optimum speed varies approximately as the square root of the heater power.

 

[StealthP3D]

You're right, this exercise has a lot of variables, the biggest three in no particular order;

1) What was the cabin temperature when the trip started
2) What speed is the fan on
3) Is recirculate on

If there is not an issue with window fogging or icing, you can dramatically lower the consumption of power for cabin heating purposes by using a low fan speed and keeping recirculate "ON". If you are in fogging or icing conditions it may be necessary to turn the fan speed up and/or recirculate off which will dramatically change the consumption

I will say the 5 kW estimate for heating consumption is only accurate during initial heating. Even in *very* cold weather, you are only looking at approximately 1-2 kW to maintain a comfortable cabin (unless you have wasteful climate settings).

I would say in 10 degree F dry weather, starting out with a pre-heated cabin and wearing normal cold weather clothing (like a sweater) most people will get the most range at around 50-60 mph. Do use the seat heaters. They make the cabin feel toasty even if you decide to only heat it to 65 F.

 

[ajdelange]

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.

 

[ewoodrick]

Seems like folks are assuming that the best efficiency is dependent on the battery SoC. I don't think so. I believe that the best efficiency is mainly dependent on the dynamics of the vehicle. So the best efficiency is going to be at the same speed, no matter if full A/C, no A/C, or full heat.