Relative Equilibrium of Liquids - Theory & Concepts

Relative Equilibrium of Liquids

Learning Objectives

Explain why an accelerating liquid can be treated as static in the container reference frame after transients subside.
Determine free-surface slope and pressure distribution during horizontal, vertical, and combined linear acceleration.
Analyze the parabolic free surface produced by rigid-body rotation.
Apply consistent coordinate and sign conventions to open and closed rotating tanks.
Identify spilling, cavitation, and loss-of-contact limits that invalidate the ideal equations.

Analysis of liquids subjected to uniform linear acceleration and rigid-body rotation.

Relative Equilibrium

A liquid is in relative equilibrium when it has no motion relative to its container even though the container may be accelerating or rotating. The liquid moves as a rigid body, so shear deformation is absent and a hydrostatic-type pressure analysis can be performed in the non-inertial container frame.

Effective Gravity

In a frame attached to a container with translational acceleration $\vec{a}_0$ , the liquid behaves as though it is acted on by an effective gravity vector

\vec{g}_{eff} = \vec{g} - \vec{a}_0

The free surface is perpendicular to $\vec{g}_{eff}$ , and pressure increases in the direction of $\vec{g}_{eff}$ . This vector statement is the safest starting point because it prevents sign errors when horizontal and vertical accelerations occur together.

Pressure Gradient in an Accelerating Container

Relates the pressure gradient to effective gravity in the container frame.

\nabla p = \rho \vec{g}_{eff} = \rho(\vec{g}-\vec{a}_0)

Variables

Symbol	Description	Unit
$p$	Fluid pressure	Pa
$\rho$	Fluid density	$kg/m^3$
$\vec{g}$	Gravitational acceleration vector	$m/s^2$
$\vec{a}_0$	Translational acceleration of the container	$m/s^2$

Horizontal Acceleration

Let $x$ be positive in the direction of the container acceleration $a_x$ , and let $z$ be positive upward. For a container accelerating horizontally to the right,

\frac{\partial p}{\partial x}=-\rho a_x, \qquad \frac{\partial p}{\partial z}=-\rho g

Pressure therefore decreases in the direction of acceleration. The free surface rises at the rear and falls at the front.

Free-Surface Slope under Horizontal Acceleration

Determines the slope of the free surface for constant horizontal acceleration.

\frac{dz}{dx}=-\frac{a_x}{g} \qquad\text{or}\qquad \tan\theta=\frac{|a_x|}{g}

Variables

Symbol	Description	Unit
$z$	Free-surface elevation, positive upward	m
$x$	Horizontal coordinate, positive with the acceleration	m
$a_x$	Horizontal container acceleration	$m/s^2$
$g$	Magnitude of gravitational acceleration	$m/s^2$
$\theta$	Magnitude of the free-surface angle from horizontal	degrees or rad

Depth Difference across a Rectangular Tank

For a tank of inside length $L$ measured in the direction of acceleration, the difference in free-surface elevation between the ends is

\Delta h=\frac{|a_x|L}{g}

If the tank is open, compare the raised-end elevation with the available freeboard. Once the predicted surface reaches the rim, spilling occurs and constant-volume formulas must be revised.

Vertical Acceleration

Vertical acceleration does not tilt the free surface, but it changes the apparent specific weight. If the container accelerates upward with magnitude $a_v$ , the effective downward acceleration is $g+a_v$ . If it accelerates downward, the effective downward acceleration is $g-a_v$ .

Pressure under Upward Acceleration

Gauge pressure at depth h when the container accelerates upward.

p=\rho(g+a_v)h

Variables

Symbol	Description	Unit
$p$	Gauge pressure below the free surface	Pa
$\rho$	Fluid density	$kg/m^3$
$g$	Gravitational acceleration	$m/s^2$
$a_v$	Upward container acceleration magnitude	$m/s^2$
$h$	Vertical depth below the free surface	m

Pressure under Downward Acceleration

Gauge pressure at depth h when the container accelerates downward.

p=\rho(g-a_v)h

Variables

Symbol	Description	Unit
$p$	Gauge pressure below the free surface	Pa
$\rho$	Fluid density	$kg/m^3$
$g$	Gravitational acceleration	$m/s^2$
$a_v$	Downward container acceleration magnitude	$m/s^2$
$h$	Vertical depth below the free surface	m

Free Fall and Loss of Contact

At downward acceleration $a_v=g$ , the effective gravity is zero. There is no hydrostatic pressure gradient, and an open liquid is weightless relative to the container. If $a_v>g$ , the assumed liquid contact and ordinary free-surface configuration cannot be maintained without confinement; the simple open-tank equation is no longer applicable.

Combined Horizontal and Vertical Acceleration

For horizontal acceleration $a_x$ and an effective downward acceleration $g_{eff}$ , the free-surface magnitude is

\tan\theta=\frac{|a_x|}{g_{eff}}

Use $g_{eff}=g+a_v$ for upward acceleration and $g_{eff}=g-a_v$ for downward acceleration. The surface falls in the direction of the horizontal acceleration.

Interactive Simulation

Adjust the translational acceleration and angular velocity to observe the free-surface response. The visualization clamps surfaces that would leave the tank, so use the equations to check whether spilling or loss of contact has occurred.

Relative Equilibrium Simulator

Open tank: 2.0 m long, 1.8 m high, initially filled to 1.0 m.

Horizontal acceleration (

a_x

)0.0 m/s²

Vertical acceleration (

a_v

)0.00 m/s²

Positive is upward. At −9.81 m/s² the tank is in free fall.

Effective downward gravity9.81 m/s²

Surface angle0.00°

End-to-end elevation difference0.000 m

Left / right depth1.000 / 1.000 m

The free surface is perpendicular to effective gravity. It falls in the direction of horizontal acceleration, with $\left|dz/dx\right|=|a_x|/(g+a_v)$ .

Rigid-Body Rotation (Forced Vortex)

After a liquid in a cylindrical container reaches constant angular velocity $\omega$ , it rotates as a rigid body. Pressure increases radially outward because a centripetal pressure gradient is required, while pressure still increases downward because of gravity.

Pressure Gradients in Rigid-Body Rotation

Radial and vertical pressure gradients for rotation about a vertical axis.

\frac{\partial p}{\partial r}=\rho\omega^2r, \qquad \frac{\partial p}{\partial z}=-\rho g

Variables

Symbol	Description	Unit
$p$	Fluid pressure	Pa
$r$	Radial distance from the rotation axis	m
$z$	Vertical coordinate, positive upward	m
$\rho$	Fluid density	$kg/m^3$
$\omega$	Angular velocity	rad/s
$g$	Gravitational acceleration	$m/s^2$

Parabolic Free Surface

Elevation of the rotating free surface relative to its vertex.

z-z_0=\frac{\omega^2r^2}{2g}

Variables

Symbol	Description	Unit
$z$	Free-surface elevation at radius r	m
$z_0$	Free-surface elevation at the rotation axis	m
$\omega$	Angular velocity	rad/s
$r$	Radial distance from the rotation axis	m
$g$	Gravitational acceleration	$m/s^2$

Center-to-Wall Elevation Difference

Maximum rise from the vertex to the wall of a cylindrical tank.

\Delta h=\frac{\omega^2R^2}{2g}

Variables

Symbol	Description	Unit
$\Delta h$	Elevation difference between the wall and center	m
$\omega$	Angular velocity	rad/s
$R$	Tank radius	m
$g$	Gravitational acceleration	$m/s^2$

Volume Conservation in an Open Cylindrical Tank

Before spilling, the average liquid level remains unchanged. Because the average value of $r^2$ over a circular plan area is $R^2/2$ , the center drops by $\Delta h/2$ and the wall rises by $\Delta h/2$ relative to the original horizontal level. Check the wall rise against the freeboard and the center drop against the initial depth.

Pressure Field with an Upward Coordinate

Taking the free-surface vertex as $r=0$ , $z=0$ , and using $z$ positive upward, integration gives

p-p_{atm}=\rho\left(\frac{\omega^2r^2}{2}-gz\right)

This expression is zero on the parabolic free surface. At a point below the vertex, $z$ is negative, so the gravity term correctly increases pressure.

Rotating-Liquid Gauge Pressure

Gauge pressure relative to the atmospheric pressure at the free-surface vertex.

p_g=\rho\left(\frac{\omega^2r^2}{2}-gz\right)

Variables

Symbol	Description	Unit
$p_g$	Gauge pressure relative to the vertex atmosphere	Pa
$\rho$	Fluid density	$kg/m^3$
$\omega$	Angular velocity	rad/s
$r$	Radial distance from the axis	m
$z$	Elevation above the vertex, positive upward	m
$g$	Gravitational acceleration	$m/s^2$

Coordinate Convention

If vertical distance $z_d$ is instead measured positively downward from the vertex, the same pressure relation becomes

p_g=\rho\left(\frac{\omega^2r^2}{2}+gz_d\right)

Do not use a minus sign with a downward-positive depth. The previous lesson version mixed these conventions and produced a physically incorrect pressure trend.

Closed Rotating Tanks

A completely filled closed tank has no real free surface, but the same pressure gradients apply. Determine the integration constant from a known pressure at one point. Verify that the minimum absolute pressure remains above the liquid vapor pressure; otherwise cavitation or vapor-pocket formation may occur.

Engineering Applications

Tanker trucks and rail cars: freeboard, pressure distribution, and load transfer during acceleration and braking.
Rotating separators and centrifuges: radial pressure increase and phase separation.
Centrifugal machinery: forced-vortex concepts inside impellers and rotating casings.
Liquid-storage design: spill checks, wall pressure, and uplift or loss-of-contact limits.

Key Takeaways

Relative equilibrium permits a hydrostatic-type analysis in the accelerating container frame.
The governing vector is effective gravity, $\vec{g}_{eff}=\vec{g}-\vec{a}_0$ .
A horizontal acceleration tilts the free surface opposite the acceleration, with $|dz/dx|=|a_x|/g_{eff}$ .
Upward acceleration increases the apparent specific weight; downward acceleration reduces it.
Rigid-body rotation creates a parabolic free surface, $z-z_0=\omega^2r^2/(2g)$ .
Pressure coordinates must be stated explicitly: upward-positive elevation uses a minus gravity term, while downward-positive depth uses a plus gravity term.
Always check the ideal solution against freeboard, tank depth, absolute-pressure, and contact constraints.

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