Fluids at Rest — Question 1

Question 1

The deepest point need not lie below the heaviest layer

An open tank contains $0.40\,\mathrm{m}$ of oil of density $800\,\mathrm{kg/m^3}$ above $0.60\,\mathrm{m}$ of water of density $1000\,\mathrm{kg/m^3}$. Both layers are at rest. Use $g=9.80\,\mathrm{m/s^2}$.

1. Find the gauge pressure at the interface and bottom. Find the depth in pure water giving the same bottom gauge pressure.

2. A thin separator allows the layer order to be reversed without changing thicknesses. Does the bottom pressure change? What physical issue would arise without the separator?

Question 1 - Solution

The deepest point need not lie below the heaviest layer

(a) Derivation and calculation

With depth positive downward, $dp=\rho g\,dh$. Thus$$\boxed{p_I=(800)(9.80)(.40)=3136\,\mathrm{Pa}},$$$$\boxed{p_B=p_I+(1000)(9.80)(.60)=9016\,\mathrm{Pa}}.$$$$\boxed{h_{\mathrm{water}}=\frac{9016}{1000(9.80)}=.920\,\mathrm m}.$$

(b) Result and interpretation

$$p_B=g(\rho_oh_o+\rho_wh_w)=g(\rho_wh_w+\rho_oh_o).$$Reversing the layers preserves the bottom pressure if the separator adds no pressure jump. Without a separator, denser water above lighter oil is gravitationally unstable.

Physical check

The water-equivalent depth is less than the total 1.00 m because the oil is less dense than water.