*(Note: Fixed 2 typos)*

Assuming an example of a Chrysler 413-3 truck engine spec of:

Fuel consumption, brake specific: .45pounds of gasoline/ horsepower/ hour of run time

Running time: 1 hour at constant speed

Max hp developed at most effecient RPM: 65 road load horsepower at 2175 engine RPM and 65mph road speed

You will burn:

.45*65=29.25 pounds of fuel per hour

...or:

*29.25*/6.39 (pounds of fuel per gallon)=4.577 gallons of fuel per hour

...and get:

65/4.577=14.20 miles per gallon

Now, of that 65hp road load, your fuel burn rate of 29.25 pounds of fuel per hour means you are releasing (125,000 BTU/ pound of gasoline or 36,633.883771528 Watts

*of*energy - this figure assumes 91RON at 60 degrees F):

4.577*125,000= 572,125 BTU/hour or 167,561.11 Watts of energy

....while you are only developing 65hp or 48,470.49 Watts of energy. This means you are wasting:

167561.11-48,470.49=119,090.62 Watts of energy or, said another way:

119090.62/167561.11=.7107*100 (to get percent)=

*71.01 percent of the fuel is wasted through engine friction losses, heat rejection to the cooling system, and exhaust heat.*
**Edited by Bob_Sheaves, December 1, 2007 at 01:49 am.**