M091 · Fluid mechanics

Hydrostatics & Hydraulic Press

Does a hydraulic press give out more work as well as more force? No more work comes out than goes in. At the same pressure a bigger piston feels a bigger force, but it moves a correspondingly shorter distance. A piston with ten times the area pushes ten times as hard and moves a tenth as far, so the work is the same.

Move a pressure probe through a cylindrical or flared reservoir and work out the liquid density from the pressures you acquire. Then operate two connected pistons and see how force, stroke and the liquid's gravitational energy trade off.

Probe gauge pressure $\text{Ready}$

Bottom gauge pressure $\text{Ready}$

Probe height above bottom $\text{Ready}$

Liquid volume $\text{Ready}$

Liquid weight $\text{Ready}$

Floor force on liquid $\text{Ready}$

Wall force on liquid $\text{Ready}$

Piston area ratio $\text{Ready}$

Upward output force $\text{Ready}$

Output stroke $\text{Ready}$

Input work $\text{Ready}$

Output work $\text{Ready}$

Fluid gravity energy gain $\text{Ready}$

Work balance residual $\text{Ready}$

Acquired stations $\text{Ready}$

Density from raw data $\text{Ready}$

Fit coordinate residual $\text{Ready}$

Approximation status $\text{Ready}$

Set the geometry, then measure

Does a hydraulic press give out more work as well as more force?

Apparatus: reservoir / press

b

0.00\,

Tank bottom radius

R_0

0.30\,m

Wall flare

f

0.00\,

Liquid height

z_0

1.20\,m

Applied surface gauge pressure

p_s

0.00\,kPa

Input piston radius

r_1

4.00\,cm

Output piston radius

r_2

12.00\,cm

Input force

F_1

150.00\,N

Prescribed input stroke

s_1

0.10\,m

Probe height fraction

z_p/z_{\mathrm{available}}

0.50\,

Reference fluid density

\rho

1000.00\,kg/m^3

Gravity

g

9.81\,m/s^2

Pressure noise bound

\epsilon_p

0.00\,kPa

The liquid has uniform density, gravity is constant, and gauge pressure is measured relative to the surrounding air. The tank is a cylinder or flares out with a radius that grows linearly with height. A nonzero surface pressure is applied by a massless cap fitted to the surface. The side walls supply the rest of the vertical force on the liquid. Cap mass, capillarity, sloshing, seal friction and compressibility are left out.

The two massless pistons start at the same height and are joined by an ideal liquid-filled line. The input piston moves down and the output piston moves up. Pressure changes with height. The input work equals the output work plus the gain in the liquid's gravitational potential energy. A negative output force needs a downward force to balance it, and that case is not recorded as a lifting experiment. The pistons move quasi-statically along a prescribed path, so no acceleration is computed.

An acquisition measures gauge pressure at known depths. A least-squares line with a free intercept, together with the known gravity, gives the density. The reference density is kept out of the fit and the CSV. Moving the probe keeps the readings; changing the apparatus, geometry, fluid, force, stroke or noise clears them. A new reading at the same depth replaces the old one. The noise is bounded and reproducible, so repeated readings are not independent.

The liquid is uniform and incompressible, and every state is in quasi-static balance. Vessel dimensions are real lengths in metres; fixtures are schematic. Sensor readings are simulated. Each piston stroke you set is a balanced state, not a lift in motion.

Start with the essentials

Focus question: Does a hydraulic press give out more work as well as more force?
One-sentence intuition: Depth sets the pressure. Piston area sets the force, and the stroke scales inversely with area.

Core mathematical model

Pressure grows with depth

p(z)=p_s+\rho g(z_s-z)

At a given depth the pressure is the same, whatever the shape of the vessel.

Both pistons sweep the same volume

A_1s_1=A_2s_2

The volume of liquid pushed down by the input piston raises the output piston by the same volume.

Output force when the pistons sit at different heights

F_2=A_2\!\left[\frac{F_1}{A_1}+\rho g(z_1-z_2)\right]

When the piston faces are at different heights, the force gain needs a correction for the liquid column between them.

The missing work raises the liquid

W_1-W_2=\frac{\rho g A_1s_1^2}{2}\!\left(1+\frac{A_1}{A_2}\right)

The two liquid columns start level. As the faces move apart in height, the liquid gains gravitational potential energy, and that is where the work difference goes.

Density from acquired pressure

p_i=\alpha d_i+\beta,\qquad\widehat\rho=\widehat\alpha/g

With known depths and gravity, the fitted slope of pressure against depth gives the density. The intercept takes up the surface pressure.

Common difficulties

Acquire before inferring

Typical misconceptionThe estimate just copies the value on a reference slider.

Better mental modelClear the observations and the estimate disappears. Add noise or change the reference parameter, then acquire again.

Know the approximation

Typical misconceptionIf the apparatus looks smooth, every state it shows must be physically reliable.

Better mental modelCheck the domain status. Extrapolated states stay visible, but you cannot acquire data from them.

Run the experiment

01
Change the vessel
Compare cylindrical and flared presets at equal height.
What to observe: The bottom pressure stays the same; the force from the walls balances the extra weight of liquid.
02
Move and acquire
Drag the gold probe track, then acquire a depth profile.
What to observe: The slope of the fitted pressure profile gives the density.
03
Operate the press
Select the press preset and increase input stroke.
What to observe: The output piston moves less, and part of the input work goes into raising the liquid.
04
Test the head correction
Select the Elevation matters preset and look at the work budget residual.
What to observe: As the piston heights move apart, the force predicted for equal heights drifts away from the actual force.