Fluids at Rest — Question 2

Question 2

A density gradient changes the meaning of a depth estimate

Depth $h$ is measured downward from the surface of a stratified liquid. Its density is $\rho(h)=\rho_0(1+h/H)$ for $0\le h\le H$. Surface pressure is $p_0$ and gravity is constant.

1. Derive $p(h)$ by balancing a thin horizontal slice. Evaluate the bottom excess pressure.

2. At what depth is the pressure excess half its bottom value? Compare with $H/2$.

Question 2 - Solution

A density gradient changes the meaning of a depth estimate

(a) Derivation and calculation

For a thin slice of area $A$,$$p(h)A+\rho(h)gA\,dh-p(h+dh)A=0.$$$$\frac{dp}{dh}=\rho_0g(1+h/H).$$$$\boxed{p(h)=p_0+\rho_0g\left(h+\frac{h^2}{2H}\right)},\quad\boxed{p(H)-p_0=\tfrac32\rho_0gH}.$$

(b) Result and interpretation

Let $u=h/H$. Half the bottom pressure excess requires$$u+\tfrac12u^2=\tfrac34\Rightarrow u^2+2u-\tfrac32=0.$$$$\boxed{h/H=-1+\sqrt{5/2}=.581139}.$$The negative root is outside the fluid. More than half the depth is required because the upper liquid contributes less pressure rise per unit depth.

Physical check

At $h=H/2$, the pressure fraction is $(.5+.125)/1.5=5/12<1/2$.