Hydraulic Press: Include Height Differences

Apply hydrostatic pressure changes without misusing Pascal's principle.

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For static liquid of density rho, pressure rises with depth: delta p = rho g h. A pressure change imposed on an enclosed fluid is transmitted, but absolute or gauge pressures at different heights need not be equal. Use gauge pressure and ignore piston weight and friction. Use diagrams and calculations only; never build or operate a press for this activity.

Worked example

With rho = 800 kg/m^3 and g = 10 m/s^2, a point 0.5 m lower has 4,000 Pa more pressure. From 30,000 Pa at the upper face, the lower face has 34,000 Pa.

A lower point is half a meter below an upper point in liquid of density 800 kilograms per cubic meter. With g equal to 10 meters per second squared, gauge pressure rises from 30 to 34 kilopascals.

Would you believe it?

Two points in still liquid can have different pressures if they are at different depths. Transmitting a pressure change does not make all pressure readings identical.

OpenStax: pressure and depth

Question 1 Calculate the pressure increase 800 x 10 x 0.5. 400 Pa 40,000 Pa 4,000 Pa

Question 2 What is lower-face gauge pressure for the example? 30,000 Pa 34,000 Pa 26,000 Pa

Question 3 What net fluid force acts on lower area 0.002 m^2? 68 N 60 N 8 N

Question 4 If an extra 2,000 Pa is imposed, with heights fixed, what is the upper pressure? 34,000 Pa 28,000 Pa 32,000 Pa

Question 5 What is the lower pressure after that same change? 34,000 Pa 36,000 Pa 32,000 Pa

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