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?

# Answer

*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.

### Short Answer

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.

### Longer Answer

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.

### Longest Answer

Since the rock creates a 0.5 in (diameter) hole, and the question asks what the hydrostatic pressure is at "the place where the rock will strike," we can choose not to treat the area in question as a point. Thus, there will be a small pressure differential between the top and the bottom of where the rock strikes.
Using the previously derived answer of 0.694 psig at the point exactly 2 ft below the top of the tank, we can treat this point as the center of a disk of 0.25 in radius. Since hydrostatic pressure scales linearly with depth (in other words, d and g in P = d*g*h are constant), we can do a simple scaling to determine the pressure at the top of the hole. This yields 0.694 psig/24 inches * 23.75 inches = 0.687 psig at the top of the hole. For the bottom, we can use the same strategy: 0.694 psig/24 inches * 24.25 inches = 0.701 psig. Thus, we can create an equation that describes the hydrostatic pressure at any depth between 23.75 inches and 24.25 inches: (0.694 psig)*(1 + x/(24 inches)), where x can vary in the range [-0.25 in, 0.25 in], where the top of the tank is 0 inches in depth. This yields, as shown above, a range between 0.687 psig and 0.701 psig. Using Bernoulli's equation, we can see that this is pretty small, only enough to make the fluid velocity difference between the top and bottom of the hole when the leak is sprung about 0.1 ft/s.

hydrostatic pressure 31.36 pascal

That is incorrect. The pressure should be only slightly higher than atmospheric conceptually as we are assuming that the tank is at atmospheric pressure since it is an unpressurized tank.