Cabin Heat / Speed Interplay on kWh/mile

[drtimhill]

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.

Welcome to the exciting world of calculus! The variable here is the speed you drive (lets assume constant speed end-to-end for simplicity) and the derived (computed) value is total energy consumed. You want to solve to find the speed that yields the minimum energy. The calculus is not hard, BUT the relationship between speed and power consumed ("instant" energy, though maths purists will wince at that) is complex, since it involves air resistance, rolling resistance, and motor power consumption. I'm not sure if there is an approximation equation around for this.

 

[Big Earl]

You're telling me that it takes the same amount of power to create a 100F temperature rise as it does to create a 40F temperature rise in a stream of air?

Also, fresh air pressurizes the cabin, forcing conditioned air out of the relief vents in the trunk of the car. Recirculated air stays put.

Try this: once your car is up to temperature, toggle between fresh and recirculate every couple of minutes and watch your power consumption (real time, not Wh/mi displayed in the trip computer). Fresh air can easily use twice as much power as recirculate.

If you'd like, I can make a video demonstrating this with my Fiat 500e, which has a handy gauge that shows HVAC power consumption.

 

[ajdelange]

Honestly, in my view, it seems as if you've posted random information that looks really official.

Well, honestly, in my view, you didn't read the posts very carefully. To begin with OP is interested in a gedanken experiment in order to give himself insight as to whether there is a speed which will give him fewer Wh used for a trip than any other speed.

In No. 15 elijah showed the equation for consumption per mile, showed that you can plot it and find a minimum. To do this one needs a model for traction consumption and he got one from somewhere.

I took the next step to show that his equation can be differentiated, the differential set to 0 and the resulting equation solved for v, the optimum speed. I used his model for traction consumption. I specifically pointed out that the conclusions one can draw from this depend on the suitability of the model and that this particular model was demonstrably flawed because it is monotonic and we know wh/mi goes back up again below 40ish. I mentioned this again in my subsequent posts.

Your equation, without any units or reference to what it is doing just seems random to me.

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.

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.

A total waste of time and space as the math and the plot in no.21 clearly indicate that the model is flawed. And, BTW, it shows that the optimal speed is 0 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.

I don't know where the model came from. You don't know where the model came from. elijah doesn't know where the model came from. Therefore non of us knows how the data was obtained. One suspects they took some data driving at various speeds and fit a second order polynomial to it with the constraint that the polynomial go through 0 at v = 0. There should be a caveat that this polynomial should not be used below the lowest speed at which they actually took data.

And in this case, that's the whole intent.

Even this model shows the OP what he wanted to know and that is whether there is an optimum speed when the heater is on. There is. And we can see from even this inadequate model that speed is only a really relevant factor when heater drain is huge - much bigger than it typically is. Some claim these heaters can draw 6 or more kW. The curves in the two posts show optimum speeds increasing with power in the 35 - 45 mph region as heater power moves above 4 kW thus illustrating quite clearly what OP asked about.

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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.

I posted a couple of graphs both derived from the same math. Thus both graphs and the math tell the same story and contain the same truth.

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

There are two: the power drawn by the heater and the car's actual wH/mi.
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SO, in short, unless you are willing to drive 40 mph, this whole discussion is still moot.

Clearly you have missed the point of the entire discussion. It was to gain insight and in this it has been a great success. I certainly have gained some.

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[ajdelange]

You're telling me that it takes the same amount of power to create a 100F temperature rise as it does to create a 40F temperature rise in a stream of air?

No, I am not telling you that. I am telling you that it takes the same amount of energy to raise 2000 cubic feet of air though a given ∆T whether you do it with a stream of air at 100 ° or one at 140 °F. Enthalpy is conserved. If m1 and m2 be two thermal masses and and ∆T2 two temperature changes then conservation requires m1*∆T1 = m2*∆T2 = the heat transferred. Obviously there are an infinite number of combinations of m1 and ∆T1 which give the same m2*∆T2. You must understand this if you hope to understand the kind of calculations we do in designing/analyzing HVAC systems.

Also, fresh air pressurizes the cabin, forcing conditioned air out of the relief vents in the trunk of the car. Recirculated air stays put.

Well yes, we are aware of that and that's why we advised going to recirculation mode for minimum power consumption.

Try this: once your car is up to temperature, toggle between fresh and recirculate every couple of minutes and watch your power consumption (real time, not Wh/mi displayed in the trip computer).

I was actually sitting in the car earlier checking on the new S/W update and fiddled with this. So I pulled the record which is pretty crude (1 minute averages) and found 1330 W (total for the car including rectifier/inverter inefficiency) and 900 W for incoming and recirculation respectively.

If you'd like, I can make a video demonstrating this with my Fiat 500e, which has a handy gauge that shows HVAC power consumption.

Thank's but no need. I understand how this stuff works pretty well. You might grab a book or find some online stuff on HVAC systems to get you up to speed.

 

[ajdelange]

The calculus is not hard,

That depends on the nature of the consumption power relationship. If that's hard to differentiate - well we can always use machine approximation.

BUT the relationship between speed and power consumed ("instant" energy, though maths purists will wince at that) is complex, since it involves air resistance, rolling resistance, and motor power consumption. I'm not sure if there is an approximation equation around for this.

There is and it has been mentioned at least 5 times in this thread.

 

[mswlogo]

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.

I’ve wondered the same thing. The thing is you don’t have that wide a choice on speed so what does it matter.

Bottom line is try to use recirculate when you can. If not, because of fogging, try using AC (to lower humidity) before resorting to no recirculate. Heating fresh air is expensive.

Auto will often have recirculate off.

The right balance of fan speed contributes to efficiency as well. Too high cost energy for the fan itself and your heating more of the cabin. Too low and the heating coil is not getting as fully utilized. You can also have “cold spots” in the cabin which might make you raise the temperature. I generally like speed around 4. Partly depends on defoggjng requirements too.

 

[ewoodrick]

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In No. 15 elijah showed the equation for consumption per mile, showed that you can plot it and find a minimum. To do this one needs a model for traction consumption and he got one from somewhere.


Clearly you have missed the point of the entire discussion. It was to gain insight and in this it has been a great success. I certainly have gained some.

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So, let's take the calculation and plot it.

So where's the effective dip around the 30's? Since that's missing and this is a discussion about min usage, isn't that an important part of the discussion?
It actually looks as if the data is only valid starting around 40 mph and up. Since the record is about 125 Wh/mi at about 30 mph.

And indeed, looking at the original equation posting, it's from 2014, 2 years before the Model 3 was even announced.

So, given the starting cabin temperature, the thermal characteristics of a car moving through air and varying conditions such as rain, and the outside temperature and the set temperature and ....

A effort can be made, but, the reality is

• The slower the better (down to about 30 mph)
• The minimal speed will increase as the heating requirements increase. (heating use is based on time)

In this case, there's a LOT of SWAGs (wild guesses) and the accuracy is quite questionable.

 

[Big Earl]

No, I am not telling you that. I am telling you that it takes the same amount of energy to raise 2000 cubic feet of air though a given ∆T whether you do it with a stream of air at 100 ° or one at 140 °F. Enthalpy is conserved. If m1 and m2 be two thermal masses and and ∆T2 two temperature changes then conservation requires m1*∆T1 = m2*∆T2 = the heat transferred. Obviously there are an infinite number of combinations of m1 and ∆T1 which give the same m2*∆T2. You must understand this if you hope to understand the kind of calculations we do in designing/analyzing HVAC systems.

Well yes, we are aware of that and that's why we advised going to recirculation mode for minimum power consumption.

I was actually sitting in the car earlier checking on the new S/W update and fiddled with this. So I pulled the record which is pretty crude (1 minute averages) and found 1330 W (total for the car including rectifier/inverter inefficiency) and 900 W for incoming and recirculation respectively.


Thank's but no need. I understand how this stuff works pretty well. You might grab a book or find some online stuff on HVAC systems to get you up to speed.

Respectfully, I think we are talking past one another.

 

[ajdelange]

Again, you miss the point. The purpose of the analysis was to lend insight. OP got that. You don't seem to be able to. Why don't you just leave it at that?

But one more time, Sam: No one ever said the model was current, accurate or good down to low speed. It does seem to be representative at the sort of behaviour of typical cars at higher speed though and given that we see if it will lend us any insight and it does.

 

[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.

Please accept my apology - I misread this post and derailed the conversation with a hasty reply. You are correct and we are on the same page.

 

[ralph142]

No, I am not telling you that. I am telling you that it takes the same amount of energy to raise 2000 cubic feet of air though a given ∆T whether you do it with a stream of air at 100 ° or one at 140 °F. Enthalpy is conserved. If m1 and m2 be two thermal masses and and ∆T2 two temperature changes then conservation requires m1*∆T1 = m2*∆T2 = the heat transferred. Obviously there are an infinite number of combinations of m1 and ∆T1 which give the same m2*∆T2. You must understand this if you hope to understand the kind of calculations we do in designing/analyzing HVAC systems.

Well yes, we are aware of that and that's why we advised going to recirculation mode for minimum power consumption.

I was actually sitting in the car earlier checking on the new S/W update and fiddled with this. So I pulled the record which is pretty crude (1 minute averages) and found 1330 W (total for the car including rectifier/inverter inefficiency) and 900 W for incoming and recirculation respectively.


Thank's but no need. I understand how this stuff works pretty well. You might grab a book or find some online stuff on HVAC systems to get you up to speed.

so if I’m not misinterpreting, and I’ll admit I’m not an hvac engineer, 1300-900 is about space heater consumption, and a far cry from 6 kw. For what it worth running my heater not on recirc, in 35-45 f outside temps, seems to increase my consumption by 30-50 wh per mile on the same commute.

 

[ajdelange]

Please accept my apology - I misread this post and derailed the conversation with a hasty reply. You are correct and we are on the same page.

No apology necessary. Lot's of info goes flying by here and it is very easy to miss something. But thanks.

 

[ajdelange]

so if I’m not misinterpreting, and I’ll admit I’m not an hvac engineer, 1300-900 is about space heater consumption, and a far cry from 6 kw.

Not misinterpreting but don't read too much into those numbers. For starters they were for an X, not a 3, sittting in a not particularly cold garage but on a cold day. They were also power measured at the input to the car and thus include rectifier/inverter loss and thus need to be decreased by at least 10% for that and for whatever else the car might be taking from shore power. Also keep in mind that on a colder day and in motion at high speed where there are convection losses demand for heat will be higher.

The calculated values for heat demand depend on assumptions about the volume of the cabin and about the number of air changes per hour. If air exchange rate is greater or less than what I assumed there will be a change in what the calculation yields. The idea is to explain the principles to the point that interested readers can do their own calculations. We, of course, hope that the numbers presented are representative.

For what it worth running my heater not on recirc, in 35-45 f outside temps, seems to increase my consumption by 30-50 wh per mile on the same commute.[/QUOTE]That would seem reasonable for heater consumption of about a kW and slower (in town) speeds.

 

[RayCanuck]

Loved reading this thread ....