Thursday, January 16, 2014

More on fuel efficiency and gas consumption

More on fuel efficiency improvements and reductions in gas consumption.

What is the amount of fuel saved per mile driven when the fuel efficiency of a car is improved from 20 mpg to 25 mpg?

How much does one need to improve the fuel efficiency of a 30-mpg vehicle in order to achieve the same reduction in fuel consumption?

Answer:   We know that gas consumption is equal to miles driven divided by mpg.

Gallons = Miles / (Miles/Gallon)

So the gas saved per mile driven from an improvement in fuel efficiency from 20 to 25 miles per gallon is

Reduction in Gas Consumption = (1/20) – (1/25) or 0.05 -0.04 or 0.01 gallon.


Now assume the initial fuel consumption for the car is 30 mpg and the reduction in gas consumption from a one-mile trip is 0.01 gallons

Reduced Gas Consumption = 0.01 = (1/30) – (1/NFE)

Where NFE is the new fuel efficiency.


1/NFE = (1/30) – 0.01 = 0.023333

Take the reciprocal of both sides and get

NFE = 42.857.

We can check our answer by observing that

(1/30) – (1/42.852) is equal to 0.01.

0.033333 – 0.023333 is equal to 0.01.


A Note:  This post shows that a relatively small improvement in fuel efficiency from a very fuel inefficient car can result in large reductions in gasoline. However, returns from improvement in fuel efficiency will eventually diminish.   Consider a 1000-mpg car.    The amount of gas needed for this car to travel 1 mile is 1/1000 or 0.001 gallon.   An increase in the fuel efficiency of this car to 2000 mpg would only result in a 0.0005 reduction in gas consumed for each mile traveled.

Note also we are a long way from 1000 mpg.


This issue was previously looked at more informally in the post below.












No comments:

Post a Comment