Question

Although gas station owners lock their tanks at night, a gas station owner in your neighborhood was the victim of theft because his employee had a key to the tank. The owner of the gas station wants to bury the gasoline so deep that no vacuum pump, no matter how powerful, will be able to extract it. He has hired you as a general contractor to dig the holes for the tanks. What is the minimum depth needed for the surface of the gasoline to be unsiphonable? Assume the specific gravity of gasoline is 0.733. (You can ignore the drop rate of the gasoline level inside the tank.) The gas station owner balks at your estimate. He neither has the funds nor approval from the city to dig that deeply. However, you tell him that the best a thief could hope to have at his disposal is a 194-mbar vacuum pump. Anything better would be enormously expensive. Given this information, how deep would the gasoline have to sit below the surface to keep it safe from thieves?

Answer 1

a) The lowest pressure that a vacuum pump could ever hope to attain is 0 bar
The height of a column of fluid (assuming the vapour pressure in the tank is 1 bar) then maximally is
Delta(p) = rho g Delta(h)
Delta(h) = Delta(p) / (rho g)
= 10^5 N/m^2/(733 kg/m^3 * 9.81 m/s^2) = 13.9 meter = 45.6 ft


b) If Delta(p) is not 1 bar but ( 1 - 0.194) bar = 0.806 bar, then the depth needed is reduced by a factor 0.806

So then
D = 0.806 * 45.6 feet = 36.75 feet