Using the AC or opening the windows

[Backstageff]

Homework Statement

I've been going crazy over solving this problem properly. We were suppose to design a simple problem and then present it to the class, and i decided to solve this very question - is it more energy efficient to use the AC or open the windows on the highway?

the CdA for the car I am using (a 2007 Toyota Prius) is 0.67
i assumed air density was constant at 1.29
i am comparing a car traveling at different highway velocities (90 - 140 km/h)
The mass of the car is 1500 +1 person = 1565
When all 4 windows are open, drag increases by 20% (SAE http://www.liveleak.com/view?i=b9e_1214963046)
The AC increases fuel consumption from between 5% (low settings) to 25% (extreme external heat)
All other factors, excluding AC or open windows, are exactly the same in both trials
Cars do not stop for 100km and travel at uniform velocity. The trial begins after they have reached said speed.
There are 34.6M in 1L of gasoline (sorry Americans)
And the car consumes 4.9L /100km

Homework Equations

Drag equation
Power to overcome drag

The Attempt at a Solution

Using this model i have already arrived at a solution. When traveling at 105km/h the AC and force to overcome drag is approximately 46.94MJ. With just the windows open, the consumption is 44.12MJ. I realize that this model is painfully simple, but i would really like to know why these numbers are not entirely accurate for a real life situation. In a real situation, why is it more fuel efficient to drive with the AC on in the highway? What forces dramatically affect the trial with the windows open that the AC trial does not suffer from? I thought it had something to do with increasing RPM to overcome drag.. but everything in the car relies on RPM for power, so I am doubtful


Just Windows

D=1/2*ρ*v2*A*Cd=1/2*0.67m2*1.29kg/m3*(27.17m/s2)= 367.7N

Fnet= 367.7N * 1.20

W=(f * D)= (441.24N*100000m) / 10^6 J = 44.12 MJ

Just AC

D=1/2*ρ*v2*A*Cd= 1/2 * 0.67m2* 1.29kg/m3* (27.17m/s^2)= 367.7N
Fnet= 367.7N
W=(f * D)= (367.7N*100000m) / 10^6 J = 36.77 MJ
AC usage of 6% = 4.9 * 0.06 = (0.294 * 34.6) = 10.17 MJ

Total Fuel Consumption with AC = 46.94 MJ

 

[JaWiB]

It seems strange to me that AC power consumption changes with the fuel consumption. It seems like it should be relatively constant for a constant setting (unless your car has climate control and you can set the a desired temperature). That is, I would think more fuel would go to keeping the car moving at higher speeds than would go to keeping the inside cool...

 

[jdwood983]

It seems strange to me that AC power consumption changes with the fuel consumption. It seems like it should be relatively constant for a constant setting (unless your car has climate control and you can set the a desired temperature). That is, I would think more fuel would go to keeping the car moving at higher speeds than would go to keeping the inside cool...

It's actually the other way around. Your fuel consumption changes with the AC power. This makes sense since the engine (and thus gas) is what powers the AC! At lower fan speeds, the engine doesn't need to expend that much power for it, but at higher fan speeds the engine needs to expend a lot more power to run the fans.

Backstageff, the math seems a little awkward when calculating the effect of the AC power. Turning the AC on for 6% of the power yields a 30% increase in fuel consumption? That seems really high.

 

[JaWiB]

It's actually the other way around. Your fuel consumption changes with the AC power. This makes sense since the engine (and thus gas) is what powers the AC! At lower fan speeds, the engine doesn't need to expend that much power for it, but at higher fan speeds the engine needs to expend a lot more power to run the fans.

I understand that. My issue is with where he gets the 6% figure. It seemed from his post that the AC consumption was 5% to 25% depending on outside temperature, but that means that the AC uses 5% of the fuel regardless of whether the car is going 25mph or 70mph, which does not make sense to me.

 

[Backstageff]

It's actually the other way around. Your fuel consumption changes with the AC power. This makes sense since the engine (and thus gas) is what powers the AC! At lower fan speeds, the engine doesn't need to expend that much power for it, but at higher fan speeds the engine needs to expend a lot more power to run the fans.

Backstageff, the math seems a little awkward when calculating the effect of the AC power. Turning the AC on for 6% of the power yields a 30% increase in fuel consumption? That seems really high.

Yeah the math is what i was wondering about. I understand that the fuel consumption changes with the AC power, which is demonstrated by the change in % fuel consumption that the AC uses, sorry if i was confusing, i guess i wasn't really thinking when i was typing.

Thanks for the response, and yeah i thought it was a bit weird too, but using the model i started with, i think that the answer is at least accurate. The reason that i am so unsure, however, is that the SAE did a similar experiment and found that the speed i listed is about the same speed where it is more energy efficient to use the AC instead of opening the windows. However, they were testing V8 cars, and I am cheating and using a hybrid.

I was thinking that in a real life situation, that the duration of the test would somehow factor into the answer@JaWib
The 6% figure was for how much energy the AC uses at a low setting as a percentage of total fuel consumption. I would have rather calculated the AC usage by getting finding out how much power the AC requires, but that value is impossible to find on the internet

 

[jdwood983]

Homework Statement

the CdA for the car I am using (a 2007 Toyota Prius) is 0.67
i assumed air density was constant at 1.29
i am comparing a car traveling at different highway velocities (90 - 140 km/h)
The mass of the car is 1500 +1 person = 1565
When all 4 windows are open, drag increases by 20% (SAE http://www.liveleak.com/view?i=b9e_1214963046)
The AC increases fuel consumption from between 5% (low settings) to 25% (extreme external heat)
All other factors, excluding AC or open windows, are exactly the same in both trials
Cars do not stop for 100km and travel at uniform velocity. The trial begins after they have reached said speed.
There are 34.6M in 1L of gasoline (sorry Americans)
And the car consumes 4.9L /100km

...

Just AC

D=1/2*ρ*v2*A*Cd= 1/2 * 0.67m2* 1.29kg/m3* (27.17m/s^2)= 367.7N
Fnet= 367.7N
W=(f * D)= (367.7N*100000m) / 10^6 J = 36.77 MJ
AC usage of 6% = 4.9 * 0.06 = (0.294 * 34.6) = 10.17 MJ

Total Fuel Consumption with AC = 46.94 MJ

Looking back at this, in particular the bold part:

$$\frac{4.9 L}{100km}\cdot\frac{6}{100}\cdot\frac{34.6M}{1L}=10.2MJ??$$

So one mole per km is equal to a MegaJoule??

 

[JaWiB]

@JaWib
The 6% figure was for how much energy the AC uses at a low setting as a percentage of total fuel consumption. I would have rather calculated the AC usage by getting finding out how much power the AC requires, but that value is impossible to find on the internet

I'm not sure I'm getting my point across...I think that AC power should NOT be a percentage of total fuel consumption. For instance, at 25mph you are using a different amount of fuel to keep the car going than at 70mph, but you can still have the AC on the same setting.

Since at lower speeds air resistance is likely less of a factor, keeping the windows open should not significantly impact fuel economy. At higher speeds, AC should take the same energy to run (unless I'm missing something?), but opening windows will have a larger impact on fuel economy.

 

[Backstageff]

Looking back at this, in particular the bold part:

$$\frac{4.9 L}{100km}\cdot\frac{6}{100}\cdot\frac{34.6M}{1L}=10.2MJ??$$

So one mole per km is equal to a MegaJoule??

The closest estimate i could find for AC fuel consumption was listed as a % of the total fuel consumption. Using the listed fuel consumption for 100km on the highway, from the manufacturer i just calculated the extra percentage that it would take to run the car + the AC. I got that .294 (i believe) extra liters and then i multiplied that by how much energy is in one liter of gasoline.

I'm not sure I'm getting my point across...I think that AC power should NOT be a percentage of total fuel consumption. For instance, at 25mph you are using a different amount of fuel to keep the car going than at 70mph, but you can still have the AC on the same setting.

Since at lower speeds air resistance is likely less of a factor, keeping the windows open should not significantly impact fuel economy. At higher speeds, AC should take the same energy to run (unless I'm missing something?), but opening windows will have a larger impact on fuel economy.

I would have rather not done it this way, but i couldn't find AC usage as KWh since for some reason, no car manufacturers list how much power it takes to run just the AC. The % fuel consumption was the only thing i could find, and it was all over the place, so i thought that it would be accurate to a certain extent

The main reason I am having problem with this is a lack of reliable car specs Oo
Thanks for the help guys

 

[jdwood983]

The closest estimate i could find for AC fuel consumption was listed as a % of the total fuel consumption. Using the listed fuel consumption for 100km on the highway, from the manufacturer i just calculated the extra percentage that it would take to run the car + the AC. I got that .294 (i believe) extra liters and then i multiplied that by how much energy is in one liter of gasoline.

Okay...It turns out that you forgot the J in the 34.6MJ/L. But even with that, you are still trying to compare MJ/km to MJ. You'd have to multiply your 10.2 by the distance (100km) and get something far more ludicrous.

I am 100% confident that your source of error is that one line I highlighted in my last post. That seems very fishy to me that turning the AC on for an extra 6% increase in fuel consumption leads to a 30% increase in energy. If this were actually true, no car manufacturer would install an AC into the car because it'd be way too inefficient.
If I had to guess, I'd say that only a 6% increase in total energy by turning on the AC would be moderately accurate. This would increase the energy with AC to be 38.98 MJ instead of 36.77 MJ (or rather 46.94 MJ).