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?

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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?

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Answer 1 · 2015-08-31T20:57:38

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

Explanation:

You need to know two things in order to solve this problem, what relative density is and what the formula that establishes a relationship between depth and pressure looks like.

Relative density (you'll sometimes see this being referred to as specific gravity) is simply the ratio between the density of a substance, in your case oil, and the density of a reference substance, which in your case I assume it's water, at specified conditions.

$d = \frac{ρ_{oil}}{ρ_{water}}$ $color(blue)(d = rho_"oil"/rho_"water")$

Most of the time, relative density is compared with water's density at $4^{\circ} C$ $4^@"C"$ , which can be approximated to be $1000 kg/m^{3}$ $"1000 kg/m"""^3$ .

This means that the density of oil will be

$ρ_{oil} = d \cdot ρ_{water}$ $rho_"oil" = d * rho_"water"$

$ρ_{oil} = 0.8 \cdot 1000 kg/m^{3} = 800 kg/m^{3}$ $rho_"oil" = 0.8 * "1000 kg/m"""^3 = "800 kg/m"""^3$

The relationship between depth and pressure is given by the formula

$P = ρ \cdot g \cdot h$ $color(blue)(P = rho * g * h)" "$ , where

$P$ $P$ - the pressure produced at the depth $h$ $h$ ;
$g$ $g$ - the gravitational acceleration;
$ρ$ $rho$ - the density of the liquid.

Rearrange to solve for $h$ $h$ - keep in mind that $\frac{N}{m^{2}}$ $"N"/"m"^2$ is equivalent to $\frac{kg}{m \cdot s^{2}}$ $"kg"/("m" * "s"^2)$ , and don't forget that you have kilonewtons, not Newtons

$h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 800 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("15.3 m")$

To find the depth at which this pressure would be produced in water, simply replace the density of the oil with that of water

$h = P/(rho * g) = (120 * 10""^3color(red)(cancel(color(black)("kg")))/(color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("s"^2)))))/(9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2))) * 1000 color(red)(cancel(color(black)("kg")))/"m"^color(red)(cancel(color(black)(3)))) = color(green)("12.2 m")$

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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?

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