Dot Physics

First: Car Talk is awesome. I wish I could come up with some class activities that help students become as good at trouble shooting and critical thinking as Tom and Ray are. Anyway, they are quite entertaining.

So, my Dad called and told me he heard a discussion on Car Talk about the effect DC to AC converters and accessories plugged in to it and how they would effect gas mileage. I skimmed through the last Car Talk podcast, but couldn’t find it. He must have heard a re-run on the radio or something (he doesn’t really believe in podcasts). Let me calculate the effect a number of things will have. This will be extremely similar to my calculation of the effect of daytime running lights on fuel efficiency.

Just like the daytime running lights (DRL) calculation, the problem is the conversion of energy from fuel to electrical energy. You need to make some assumptions to do this. I assume (with the symbols I will use for each variable):

  • The energy density of gasoline is 1.21 x 108 Joules/gallon. (dg)
  • A car is 20% efficient at converting this energy to mechanical energy (ee).
  • The alternator is 70% efficient at converting mechanical energy into electrical energy (ea).
  • The DC to AC converter is 70% efficient (eDC).

What am I going to plug in? Suppose it was about 100 Watts total. If this is the case, then the hard work is already completed from the DRL post. So, I am going to take it one step further. I am going to let you input whatever you want by embedding a zoho spreadsheet. There is one big difference between this calculation and the DRL one. For this, I don’t have a specific speed. This means that I can calculate the gallons of gas per second that a device could use. Starting with the power of a device:

i-c3523332ec8040c24a65175bf9a6e4c6-power-1.jpg

So 100 watts worth of stuff would use 100 Joules every second. How much gasoline would this require each second? Well, what is the power produce by consuming a volume of V (in gallons) of gas in 1 second? (including losses due to the alternator, engine etc…):

i-04ab7f8b1ea71102775cba74659ad411-p-gas-2.jpg

With the above expression, I could determine how many gallons per second one would need to run devices with power P. What does this do to the fuel efficiency? Suppose I am going 70 mph. I could determine how far I go in one second and thus get a measure of fuel use in miles per gallon. So, the faster you are going, the less these accessories will reduce your gas mileage. Now, let me put all these calculations in a zoho spreadsheet (so you can change all the variables as you like)

Note: I couldn’t not figure out how to force zoho to show a number in scientific notation. To compensate, I just put the power in with the units. So, this says that 100 watts worth of stuff plugged into a DC to AC converter would use 8.43 x 10-6 gallons per second. If you compare this to a car going 60 mph that gets 20 mpg, that would use 8.35 x 10-4 gallons per second.

Update

I changed this post slightly. The original zoho sheet calculated the loss of efficiency due to the accessories. I need to re-think that.

Comments

  1. #1 Rob
    April 21, 2009

    I haven’t run numbers yet, but I don’t think that can be right. A deeper analysis will (time permitting) follow, but very roughly using figures for my vehicle I get about 23.5 miles per gallon at 55 m.p.h. This uses about 25 horsepower. Power required is proportional (more or less) to speed raised to the third power, so I’d need about 51.5 horsepower to go 70 m.p.h. This is 38,400 watts. Now, fuel used per mile would only go up with the square of velocity so m.p.g. would would go from 23.5 m.p.g. to about 14.5 m.p.g. So, if I need 38,400 watts to go 70 m.p.h. and use 1/14.5 m.p.g. or about 0.069 gallons per mile doing so, then a ratio of (100 watts/38,400 watts)0.069 or .000180 gallons per mile should be about what the 100 watts uses. So, instead of consuming .069 gallons per mile, I’d consume about .06918 g.p.m., reducing my mileage from 14.5 m.p.g. to 14.46 m.p.g. or about 0.04 m.p.g. or 410^-2 m.p.g. less.

    These figures are quite rough, no accounting was made for significant digits, etc., etc., but I still think you are off by many orders of magnitude. I haven’t looked (yet) at the spreadsheet to see if I can see an error because I’m at work, but the 23*10^-8 intuitively felt wrong so I just did the above to see if it made any sense.

  2. #2 Rob
    April 21, 2009

    Hmmm, the formatting on my comment got completely messed up, but hopefully you’ll get the gist of it.

  3. #3 Rhett
    April 21, 2009

    Rob,

    Yes – there is an error. It did seem small. I will try to fix it.