How much does failure to compensate for gas pump temperature cost?

Some consumer advocates and politicians are reported to be
that consumers are paying too much for gas because on average it is 5 degrees F warmer than the 60F temperature on which gas pumps are standardized. Apparently, gas pumps in the United States consider a volume of 231 cubic inches to constitute one gallon. The pumps do not compensate for ambient temperature, unlike those here in Canada, which apparently do. The article says that consumers are losing 3-9 cents per gallon. According to this physics site the volumetric coefficient of thermal expansion for gasoline at 20C is 9.5e-4 per degree C, which is 5.2e-4 per degree F, or 2.6e-3 for an increase of 5 degrees F. If gasoline is $3 per gallon, the difference is 7.8e-3 dollars per gallon, that is, about 3/4 of a cent, an order of magnitude less than the 3 to 9 cents per gallon the article gives.

The gas pump standard is set at 60F = 15.6C, but I am hard put to believe that the coefficient of thermal expansion is an order of magnitude higher at 15.6C than at 20C.

I note that the article also mentions that gas pumps in Hawaii are set to consider 234 cubic inches to be a gallon, on the basis of a temperature of 80F. That’s a factor of 0.013 over 20F or 3.2e-3 over 5F, in the same ballpark as the value I computed above.

An additional factor is that the gas is stored underground, where the temperature variation is much less than on the surface. I wonder whether the gas warms sufficiently on its way from the tanks to the customer for it to have even the full theoretical expansion calculated above.

Like most drivers I would love to see lower gas prices, but the effect of temperature on the pumps appears to be quite small, not nearly as large as is being claimed. I wonder where these people get their figures.

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