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Old 04-02-2011, 11:02 PM   #1 (permalink)
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So does it take almost 4000 lb of lead to= 1 gal. of gas?

I've got several automotive lead acid batteries that actually state the A/h rating on the sticker.

So I started weighing them and calculate the Wh/lb. They are all pretty close to about 18 Wh/lb.

According to what I can find,energy in a gallon of gasoline is 36.6 Kw/hr.

36600(W/hr) / 18(W/hr) = 2033 lb. Let's just say 2000 pounds. But that's calculating with the 20 hour discharge A/hr rating of the battery!

So unless we want to drive around at 0.5 mile/hr in super granny gear for 20 hours to justify the 20 hour discharge rating of the batteries,I think it's about half of the Amp/hour rating is what is an appropriate capacity for a one hour discharge? Is that about right?

In that case can we say that it takes about 4000 pounds of lead acid batteries to eaqual the energy available in a gallon of gas?

Just an interesting comparison if I did it correctly.

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Old 04-02-2011, 11:26 PM   #2 (permalink)
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I think you are forgetting the efficiency difference. Gasoline engines are maybe 30% efficient, while an electric motor should be at least double that.
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Old 04-03-2011, 12:08 AM   #3 (permalink)
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I think you are forgetting the efficiency difference. Gasoline engines are maybe 30% efficient, while an electric motor should be at least double that.
Well I meant that question purely from the available energy aspect of the source. Let me rephrase.

Is it fair to display the inefficiency of a gas car by saying it takes about 4000lb of lead acid to equal a gallon of gas under normal driving conditions?

I know that due to the efficiency of EV's it does not take that much lead to go 30 miles for example but that's kind of taking away from the EVs.

This way if we tell the average joe that we carry a battery pack with the energy equivalent of 1/3 of a gallon of gas and can still go 40 miles ( I'm just making these numbers up ) maybe they get to think about it!?
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Old 04-03-2011, 03:07 AM   #4 (permalink)
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If we are talking about heat energy then yes.
But if you are talking about doing psychical work then a gallon of gas is closer to 800-1000 pounds of battery power.
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Old 04-03-2011, 04:05 AM   #5 (permalink)
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I did a generator test yesterday where 8 fluid ounces of gasoline in a gas engine was turned into an approximately constant 1.6kW of electrical power for 15 minutes out of a DC motor (and was just burned up in humungus resistors).
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Old 04-03-2011, 06:32 AM   #6 (permalink)
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Well if you figure it 3dplane's way and adjust for efficiency its about 40KWH.
Pretty close to MPHomes test. So 2000 lbs would be close. I never did the math like that. Nice perspective.

But hang on 3dplane, soon someone will pop in to explain to you about equivilent math and MPGe*

It's sort of like Afirmative Action for EVs.........
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Old 04-03-2011, 10:25 AM   #7 (permalink)
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If we are talking about heat energy then yes.
But if you are talking about doing psychical work then a gallon of gas is closer to 800-1000 pounds of battery power.
I'm not sure what kind of energy I'm talking about. Energy that is used out of the source ( gasoline and batteries) turned into motive force regardless of how much wasted heat is generated in the process.

I've heard that 800lb of lead acid to a gallon of gas analogy but how is that fair to the EV?

18 W/hr X 800 (pounds) = 14.4 Kw/hr pack. A gallon of gas=36.6 Kw/hr
so that battery pack is less than .4 gallon of gas and again that is at the
20 hour discharge rate!
So accounting for Peukert by discharging it in an hour and not 20 hours makes it .2 gallon of gas equivalent? Am I doing this right?

So if that EV can go 20 miles on this pack (.2 gallon equivalent) that would be a 100 mpg comparison.
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Old 04-03-2011, 10:40 AM   #8 (permalink)
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Quote:
Originally Posted by MPaulHolmes View Post
I did a generator test yesterday where 8 fluid ounces of gasoline in a gas engine was turned into an approximately constant 1.6kW of electrical power for 15 minutes out of a DC motor (and was just burned up in humungus resistors).
Hey Paul!

Math is not my strong side but a quick google search told me that there is 128 fl oz. in a gallon so 128/8=16

36.6/16= 2.28 Kw/hr worth of gas you turned into .4 Kw/hr (1.6 Kw for 15 min=.4 kw/hr).

That would be about 18% eff? That is actually not bad for a gasser turning a DC motor. Must be a pretty decent DC motor you are using!
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Old 04-03-2011, 01:14 PM   #9 (permalink)
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If you're just comparing chemical energy, you need to also add in the weight of the oxygen from the air that's going into the chemical reaction, which IIRC is about 4:1 by weight. So if you had to carry around your own oxygen instead of getting from the air, you'd need (I think) about 35 lbs/gal.
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Old 04-03-2011, 09:32 PM   #10 (permalink)
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Quote:
Originally Posted by jamesqf View Post
If you're just comparing chemical energy, you need to also add in the weight of the oxygen from the air that's going into the chemical reaction, which IIRC is about 4:1 by weight. So if you had to carry around your own oxygen instead of getting from the air, you'd need (I think) about 35 lbs/gal.
Wow,
Obviously missed expecting that when I typed post #6

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