# Challenge Question on Hydrostatic Pressure! [Challenged Solved]

A large unpressurized tank containing gasoline is hit by a high velocity rock that is kicked up by a lawn mower. The rock creates a 0.5 in hole in the tank 2 ft below the surface of the gasoline (SG=0.8). Just before the rock hits the tank, what is the hydrostatic pressure (psig) at the place where the rock will strike?

Here is the correct answer, which Scott Kolodziej was the first to provide. Congratulations, Scott, on a speedy - and thorough - response!

Scott recently earned his master's in chemical engineering at Carnegie Mellon, where he completed a thesis titled "Global Optimization of the Multiperiod Blending Problem." To learn more about Scott, visit his website: scottkolo.com.

First off, let's assume that this scenario is taking place on Earth, near sea level, and that neither the tank in question nor its contents are not in motion.

Using SI units, the equation for hydrostatic pressure is P = d*g*h, where P is the hydrostatic pressure, d is density, g is the acceleration due to gravity, and h is the vertical height (or head) of liquid above the point being considered. The density of the gasoline is d = 0.8*d_water, so d = 0.8*1000 kg/m^3, or 800 kg/m^3. The acceleration due to gravity is 9.81 m/s^2, and h is 2 ft, or 0.61 m. This leads to an answer of P = (800 kg/m^3)*(9.81 m/s^2)*(0.61m) = 4787 N/m^2, or 4787 Pa. Since 6895 Pa = 1 psig, the final answer would be 4787 Pa / (6895 Pa/psig) = 0.694 psig.

Using AES units, the equation for hydrostatic pressure is P = d*g/g_c*h, where P is the hydrostatic pressure, d is density, g is the acceleration due to gravity, g_c is the conversion factor necessary for converting between lb_m and lb_f, and h is the vertical height (or head) of liquid above the point being considered. The density of the gasoline is d = 0.8*d_water, so d = 0.8*62.4 lb_m/ft^3, or 49.9 lb_m/ft^3. The acceleration due to gravity is 32.2 ft/s^2, and g_c is equal to 32.174 ft*lb_m/(s^2*lb_f). Lastly, h = 2 ft, and remembering that there are 144 in^2 in 1 ft^2, we can finally calculate the answer: P = (49.9 lb_m/ft^3)*(32.2 ft/s^2)/(32.174 ft*lb_m/(s^2*lb_f))*(2ft)*(1 ft^2 / 144 in^2) = 0.694 psig.