How much energy would it save to park cars in the shade in summer?

Question

tl;dr: Cars get hot in the sun and it takes a lot of energy to cool them off. But most car trips are short, so the cars may not even get cool during that time. If they were parked in the shade, they would cool off faster, saving energy -- but how much?

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Background

Cars get very hot in the sun. From the study Evaluating the impact of solar radiation on pediatric heat balance within enclosed, hot vehicles (emphasis added):

In direct solar radiation, a greenhouse situation occurs within the vehicle, trapping longwave radiation and heating the vehicle interior to a steady-state with little-to-no airflow. This situation of radiation trapping highlights the importance of solar exposure in vehicle heating and child vulnerability, even on milder days. In as little as five minutes under an ambient temperature of 30°C [86°F], interior car temperatures can rise to 57–68°C [135-154°F].

Using air conditioning (AC) reduces vehicle efficiency. According to the Department of Energy (emphasis added):

Under very hot conditions, AC use can reduce a conventional vehicle's fuel economy by more than 25%, particularly on short trips. The AC's effect on hybrids, plug-in hybrids, and electric vehicles (EVs) can be even larger on a percentage basis.

Most car trips are short. 85% of all car trips in the U.S. are 15 miles or shorter, according to the Department of Energy:

Question

If cars were parked in the shade, how much energy could this save? I'm imagining parking lots at workplaces and shopping centers having canopies installed to shade all vehicles so that they aren't blistering hot at the start of a short trip. If some x% of vehicles in x region were parked in the shade x% of the time, how much gasoline, diesel, or electric energy would this save?

Answer 1

Not very much. The parts of interest to cool are the seats and the air inside the car. If the cabin is three meters long, 1.5 meters wide and 1.5 meters high, then there's 6.75 cubic meters of air. At density 1.2 kg / m3, that's 8.1 kilograms.

Air has 1 kJ/kgC heat capacity. I don't believe that the car gets more than 40 degrees Celsius hotter than the environment. Thus, the cooling needed is 0.324 MJ.

Engines have a marginal efficiency of maybe 35%. Air conditioning compressors generally have an energy efficiency rating of maybe 2.5. Thus, you need 0.324 MJ / 2.5 / 0.35 = 0.370 MJ of fuel to cool down the cabin. Fuel has 34 MJ / liter energy content so that's about 0.01 liters of fuel.

With 0.01 liters of fuel, a car could travel about 160 meters. Compare that to your typical car trip of 10-30 kilometers. That's not even one percent of fuel saving!

Obviously considering the surface layers of the seat insulation too would somewhat increase the mass to be cooled and the needed energy. Note that seats have thermally insulating material so you don't need to cool down the heavy metal seat frames, and not even the whole insulation, but just the very surface of the insulation.

Also consider that if the air is 40 degrees Celsius hotter than outside air, then just blowing in colder outside air without any air conditioning would remove most of the heat. I suspect even when the AC is operating, in the very initial cooling stages, most of the effort is just blowing air in, and not actively cooling it. This would diminish the fuel savings to an even smaller value.