Question

The owner of the gas station wants to bury the gasoline so deep that no vacuum pump will be able to extract it. He has hired a general contractor to dig the holes for the tanks. What is the minimum gasoline surface depth h 2 h2 needed to prevent siphoning by any pump

Answer 1

Answer:

The depth of tank so that 13.49 m

Explanation:

As pressure is given as

$P=\rho g h$

here

• P is the pressure which in order to avoid siphoning by any vacuum pump is atmospheric pressure. i.e. P=1.013 x10^5 Pa
• ρ is the density of the gasoline which is calculated from the following equation of specific gravity. Assume the specific gravity of gasoline is 0.766

                                       $\rho=S.G \times \rho_{w}\\\rho=0.766 \times 1000 kg/m^3\\\rho=766 kg/m^3$

• h is the depth which is to be calculated here .

                                        $P=\rho g h\\h=\frac{P}{\rho g}\\h=\frac{1.013 \times 10^5}{766 \times 9.8}\\h=13.49 m$

So the depth of tank so that 13.49 m