Fluids at Rest — Question 7

Question 7

An accelerating tank has a tilted free surface

A tank accelerates horizontally to the right with constant $a=2.0\,\mathrm{m/s^2}$. After transients, its liquid is at rest relative to the tank. Let $x$ point right and $z$ upward; use $g=10\,\mathrm{m/s^2}$. The tank length is $L=1.2\,\mathrm{m}$. Ignore waves and assume no spilling or exposed bottom.

1. Derive the pressure gradients in the tank frame and find the right-minus-left free-surface height difference.

2. For $\rho=1000\,\mathrm{kg/m^3}$, find the right-minus-left pressure difference between two submerged points at the same height separated by $L$.

Question 7 - Solution

An accelerating tank has a tilted free surface

(a) Derivation and calculation

The tank-frame effective body acceleration is $-a\hat{\mathbf i}-g\hat{\mathbf k}$. Therefore$$\boxed{\frac{\partial p}{\partial x}=-\rho a},\qquad\boxed{\frac{\partial p}{\partial z}=-\rho g}.$$$$dp=0\ \text{on the surface}\Rightarrow a\,dx+g\,dz=0.$$$$\boxed{z_R-z_L=-aL/g=-2.0(1.2)/10=-.240\,\mathrm m}.$$

(b) Result and interpretation

At a fixed submerged elevation,$$\boxed{p_R-p_L=\int_0^L(-\rho a)\,dx=-1000(2.0)(1.2)=-2400\,\mathrm{Pa}}.$$Equal height does not imply equal pressure when the effective body force has a horizontal component.

Physical check

The higher pressure on the left supplies the rightward force needed to accelerate the liquid in the laboratory frame.