### Estimating Mileage Change Due to Temperature Change

© Copyright 1994 by Ralph E. Kenyon, Jr.

FEB 25, 1994

It is safe to assume that the variable 'gasoline yield' does not significantly change. And I know that my driver performance does not change from summer to winter months. (I don't "warm up" the vehicle.) Nor do I significantly change the amount of driving, or the proportion of time I spend at different speeds. If anything, I might drive a bit slower, which would tend to increase the mileage.

The main factor influencing mileage is therefore the wind resistance, which itself changes due to air density. Air density itself is changed by temperature and barometric pressure. The barometric pressure may change from day to day, but the average barometric pressure will not change significantly during the year. Fuel consumption is proportional to wind resistance, which is a function of air density and vehicle speed. Since we are averaging over various vehicle speeds, the proportion of which does not change during the year, only the air density is a factor. We may use the universal gas law to estimate the change in density, and therefore the change in mileage, as a function of the change in ambient temperature.

The universal gas law is:

(1) PV = nRT.

Where:

 P = Pressure V = Volume n = number of moles R = Universal Gas Constant T = absolute temperature. (0°F = 459.7°R)

Density is proportional to the number of moles per unit of volume. Dividing both sides of equation (1) by V gives us the equation:

(2) P = (n/V)RT

Equation (2) contains the expression 'n/V', which is the number of moles per unit volume. If we let n/V = cd, where 'c' is a constant of proportionality and 'd' stands for density, we can transform equation (2) into one which contains the variables 'd' for density, and 'T' for temperature. Substituting this in equation (2) yields:

(3) P = cdRT

Since we are interested in the density 'd' we can divide both sides by 'cRT' and get:

(4) P/(cRT)=d or d = (P/cR)/T

'R' and 'c' are both constants, and we reasoned above that the average pressure was essentially constant, so 'P' is also a constant for this equation. That makes the expression 'P/cR' a constant; let us replace the constant value 'P/cR' with a new constant 'K'. That leaves us with the equation:

(5) d = K/T.

This equation says that density is an inverse function of temperature; as the temperature goes up the density goes down and vice versa.

Since power requirements, and therefore gasoline consumption rate (C), is proportional to density, then we may conclude that:

(6) C = K'/T

Also, mileage, (M) goes up as the consumption rate goes down, we can express mileage as

(7) M=K''/C = K''/K'/T =  (K''/K')T = K'''T

Using equation (7) we can compare the expected mileage change from two different temperatures (T1 to T2) simply by dividing equation (7) by itself for the two different temperatures.

 M1 = K'''T1 (8) ==> M1/M2 = T1/T2 M2 = K'''T2

Notice that the constant of proportionality drops out, and we are left with a pure ratio. This ratio can be interpreted as stating that the percentage decrease in the mileage will match the percentage decrease in the temperature measured on an absolute temperature scale.

Remember, the T in this equation is the absolute temperature. If we use the Rankin scale which measure absolute zero at zero, we must add 459.7 to the temperature in Fahrenheit.

(8) (Degrees Rankine = Degrees Fahrenheit plus 459.7.)

Let us compare driving in 90°F weather to driving in -10°F weather.

(9) T1 = 90°F = 90 + 459.7 °R = 549.7°R

(10) T2 = -10°F = -10 + 459.7 °R = 449.7°R

(11) M1/M2 = T2/T1 = 449.7/549.7 = .818 or 81.8%

This means that we can expect, on purely theoretical grounds, that our mileage will drop, due to wind resistance from temperature change alone, at -10°F to 81.8 percent of the mileage we experience at 90°F.

### mileage change as a function of temperature drop

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mpg 98% 96% 94% 92% 90% 88% 87% 85% 84% 82% 81%
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