Hello Jim,
You can use the Btu's per gallon method. The average gasoline btu's is
about 115,000 btu's per gallon. One kilowatt is 3412 btu's.
Engines vary in efficiency all the way down to 25% and a electric motor may
be as low as 80%. So if all the mechanical efficiency of both rigs are
equal then:
115,000 x 0.25 = 28,750 btus for the engine
28,750 / 3412 = 8.426 kilowatts.
8.326 kw x 0.80 = 6.66 kw for a electric motor
6.66 kw / 746 = 8.9 hp
If you take everything at 100% efficiency then:
115,000 btu's / 3412 = 33.7 kw
33.7 kw / 746 = 45 hp
Lets say that you vehicle is getting 300 watts per mile at 60 mph then a
engine at 100% efficiency would have to get:
33.7 kw / 300 = 112 mpg
Roland
----- Original Message -----
From: "Jim L" <[email protected]>
To: "Electric Vehicle Discussion List" <[email protected]>
Sent: Friday, October 12, 2007 9:49 AM
Subject: [EVDL] kw and mpg conversion?
> Is there a formula for figuring out an approx energy use for a vehicle
> based on its mpg performance?
>
> For example, if a car got 25 mpg on average, what would be the amount of
> kw needed to maintain highway speeds?
>
> If there is no easy formula, is there a range of kw usage that would cover
> most typically used vehicles? i.e. 'X' kwatts for a small aerodynamic
> vehicle to 'Y' kwatts for say a standard 1/2 ton pickup truck?
>
> I realize that this would probably only be estimates.
>
> Any Ideas?
>
> Jim
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>
_______________________________________________
For subscription options, see
http://lists.sjsu.edu/mailman/listinfo/ev
You can use the Btu's per gallon method. The average gasoline btu's is
about 115,000 btu's per gallon. One kilowatt is 3412 btu's.
Engines vary in efficiency all the way down to 25% and a electric motor may
be as low as 80%. So if all the mechanical efficiency of both rigs are
equal then:
115,000 x 0.25 = 28,750 btus for the engine
28,750 / 3412 = 8.426 kilowatts.
8.326 kw x 0.80 = 6.66 kw for a electric motor
6.66 kw / 746 = 8.9 hp
If you take everything at 100% efficiency then:
115,000 btu's / 3412 = 33.7 kw
33.7 kw / 746 = 45 hp
Lets say that you vehicle is getting 300 watts per mile at 60 mph then a
engine at 100% efficiency would have to get:
33.7 kw / 300 = 112 mpg
Roland
----- Original Message -----
From: "Jim L" <[email protected]>
To: "Electric Vehicle Discussion List" <[email protected]>
Sent: Friday, October 12, 2007 9:49 AM
Subject: [EVDL] kw and mpg conversion?
> Is there a formula for figuring out an approx energy use for a vehicle
> based on its mpg performance?
>
> For example, if a car got 25 mpg on average, what would be the amount of
> kw needed to maintain highway speeds?
>
> If there is no easy formula, is there a range of kw usage that would cover
> most typically used vehicles? i.e. 'X' kwatts for a small aerodynamic
> vehicle to 'Y' kwatts for say a standard 1/2 ton pickup truck?
>
> I realize that this would probably only be estimates.
>
> Any Ideas?
>
> Jim
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>
_______________________________________________
For subscription options, see
http://lists.sjsu.edu/mailman/listinfo/ev