# Pascal's Law - Weight of Water Involvement

1. Dec 4, 2011

### joejoekelly1

Pascal’s Law states “pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure ratio (initial difference) remains the same”. My question is, are the two scenarios that I have outlined the same? (see attached).

Scenario 1 is where the right hand cylinder is much higher than the left hand cylinder.

Scenario 2 is where both cylinders are at the same height.

However does the weight of the water affect the system in scenario 1? Because the right hand cylinder is high up in scenario 1, the pressure of the water at the bottom of scenario 1 is greater than scenario 2. Therefore will scenario 1 work as good as scenario 2?

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2. Dec 6, 2011

### bbbeard

Statically, the pressure in a liquid decreases with height. So in Scenario 1 the pressure in the right-hand cylinder will be less than the pressure in the left-hand cylinder, by an amount ΔP=ρgh. The quantity ρg is the weight density of the liquid (about 62.4 lbf/ft3 for water). For water this means that the pressure decreases about 1 atmosphere (14.696 psi) for every 33.9 ft in altitude (or conversely the pressure increases 1 atm for every 33.9 feet in depth). Another way of expressing this is $1/\rho g = 0.016026 ft/psf = 2.3077 ft/psi$.

Now, having said that, whether the gravitational "head" is significant depends on the problem you're working. If you're dealing with a hydraulic lift, for example, with working pressures of roughly 2000 psi, say, then a few psi pressure difference due to the gravitational head may be insignificant. In terms of pressure head (measured in feet or meters), the gravitational head is just the elevation difference between two points, while the static pressure head is ΔP/ρg, so 2000 psi is 2000*2.3077 = 4615 feet. In fact, in a lot of introductory hydrostatics problems it may be assumed that the gravitational head is insignificant.