Question

A very large open tank contains a layer of oil (density 670 kg/m³) floating on top of a layer of water (density 1000 kg/m³) that is 4.2 m high. What should the height of the oil layer be if the gauge pressure at the bottom of the tank is to be 9.5 x 10⁴ Pa?

Answer 1

the height of the oil layer should be approximately
`\(29.38 \, \text{m}\).\\`




To find the height of the oil layer, we can use the hydrostatic pressure formula. The gauge pressure
`(\(P_{\text{gauge}}\))` at the bottom of the tank is related to the density
`(\(\rho\))`, gravitational acceleration (\(g\)), and the height (\(h\)) by the equation:

`\[ P_{\text{gauge}} = \rho \cdot g \cdot h \]We can rearrange this equation to solve for the height (\(h\)):\[ h = \frac{P_{\text{gauge}}}{\rho \cdot g} \]`

Given that the gauge pressure
`(\(P_{\text{gauge}}\))` is
`\(9.5 * 10^4 \, \text{Pa}\)`, the density of the oil
`(\(\rho_{\text{oil}}\))` is
`\(670 \, \text{kg/m}^3\)`, and the density of water
`(\(\rho_{\text{water}}\))` is
`\(1000 \, \text{kg/m}^3\)`, and the acceleration due to gravity (\(g\)) is approximately
`\(9.8 \, \text{m/s}^2\)`, we can substitute these values into the equation:

`\[ h = \frac{9.5 * 10^4 \, \text{Pa}}{(670 \, \text{kg/m}^3 - 1000 \, \text{kg/m}^3) \cdot 9.8 \, \text{m/s}^2} \]`

First, calculate the difference in densities:

`\[ \Delta \rho = \rho_{\text{water}} - \rho_{\text{oil}} = 1000 \, \text{kg/m}^3 - 670 \, \text{kg/m}^3 \]`

Now, substitute into the equation:

`\[ h = \frac{9.5 * 10^4 \, \text{Pa}}{\Delta \rho \cdot g} \`]

Calculate:

`\[ h = \frac{9.5 * 10^4 \, \text{Pa}}{(330 \, \text{kg/m}^3) \cdot 9.8 \, \text{m/s}^2} \]\[ h \approx (9.5 * 10^4)/(3234) \, \text{m} \]\[ h \approx 29.38 \, \text{m} \]`

Therefore, the height of the oil layer should be approximately
`\(29.38 \, \text{m}\).\\`