Fluids at Rest — Question 8

Question 8

Free fall removes the hydrostatic gradient

A sealed liquid container accelerates vertically upward at $a_z$, with upward positive. Liquid of density $\rho$ is at rest relative to it after transients. A top pressure reference is maintained at $p_t$. Two points are separated vertically by $H$, the lower below the upper.

1. Derive $p_b-p_t$. Evaluate for $\rho=1000\,\mathrm{kg/m^3}$, $H=0.50\,\mathrm{m}$, $g=9.80\,\mathrm{m/s^2}$ and $a_z=+2.0\,\mathrm{m/s^2}$.

2. What does the ideal model predict for $a_z=-g$? Does uniform pressure mean zero pressure?

Question 8 - Solution

Free fall removes the hydrostatic gradient

(a) Derivation and calculation

With upward $z$ and container acceleration $a_z$,$$0=-\frac{dp}{dz}-\rho g-\rho a_z\Rightarrow\frac{dp}{dz}=-\rho(g+a_z).$$$$\boxed{p_b-p_t=\rho(g+a_z)H}.$$$$\boxed{p_b-p_t=(1000)(9.80+2.0)(.50)=5900\,\mathrm{Pa}}.$$

(b) Result and interpretation

In free fall,$$a_z=-g\Rightarrow g_{\mathrm{eff}}=0\Rightarrow\nabla p=0.$$$$\boxed{p_b=p_t}.$$The pressure is uniform, not necessarily zero: its absolute value is fixed by the maintained boundary reference.

Physical check

Upward acceleration increases the pressure difference; the expression reduces to ordinary hydrostatics when $a_z=0$.