At what depth below the surface of oil, relative density 0.8, will the oil produce a pressure of 120 kN/m2? What depth of water is this equivalent to?

Question

Charles Aeschliman · Accepted Answer

That pressure will be felt at a depth of 15.3 meters.

To solve this problem, you must be aware of two things: relative density and the formula that determines the relationship between pressure and depth. The ratio of a substance's density—in your case, oil—to the density of a reference substance—in your case, water, I assume—at a given condition is known as relative density, or specific gravity as it is sometimes called. #color(blue)(d = rho_"oil"/rho_"water")# Most of the time, relative density is compared with water's density at #4^@"C"#, which can be approximated to be #"1000 kg/m"""^3#. This implies that oil's density will be #rho_"oil" = d * rho_"water"# #rho_"oil" = 0.8 * "1000 kg/m"""^3 = "800 kg/m"""^3# The formula provides the relationship between depth and pressure. #color(blue)(P = rho * g * h)" "#, where #P# - the pressure produced at the depth #h#; #g# - the gravitational acceleration; #rho# - the density of the liquid. Rearrange to solve for #h# - keep in mind that #"N"/"m"^2# is equivalent to #"kg"/("m" * "s"^2)#, and don't forget that you have kilonewtons, not Newtons #h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 800 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("15.3 m")# Just substitute the density of water for that of oil to determine the depth at which this pressure would be generated in water. #h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 1000 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("12.2 m")#

Sebastian Acosta · Answer

undefined To find the depth below the surface of the oil, use the formula:

$$ P = \rho gh $$

Where:
- $ P $ is the pressure (120 kN/m²)
- $ \rho $ is the density of the oil (0.8)
- $ g $ is the acceleration due to gravity (9.81 m/s²)
- $ h $ is the depth below the surface

Rearranging the formula to solve for $ h $, we get:

$$ h = \frac{P}{\rho g} $$

Substituting the given values:

$$ h = \frac{120,000}{0.8 \times 9.81} $$

$$ h \approx 15,306 \, \text{meters} $$

To find the equivalent depth of water, use the same formula but with the density of water (1000 kg/m³):

$$ h_{\text{water}} = \frac{P}{\rho_{\text{water}} g} $$

$$ h_{\text{water}} = \frac{120,000}{1000 \times 9.81} $$

$$ h_{\text{water}} \approx 12.24 \, \text{meters} $$

So, the depth below the surface of oil is approximately 15,306 meters, which is equivalent to a depth of approximately 12.24 meters of water.