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Variation of pressure with elevation

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Consider a hypothetical differential cylindrical element of fluid of cross sectional area A and height (z_{2} - z_{1}).

Upward force due to pressure P_{1} on the element = P_{1}A

Downward force due to pressure P_{2} on the element = P_{2}A

Force due to weight of the element = mg = rA(z_{2} - z_{1})g

Equating the upward and downward forces,

P_{1}A = P_{2}A + rA(z_{2} - z_{1})g

P_{2 }- P_{1} = - rg(z_{2} - z_{1})

Thus in any fluid under gravitational acceleration, pressure decreases, with increasing height z in the upward direction.

Equality of pressure at the same level in a static fluid:

Equating the horizontal forces, P_{1}A = P_{2}A (i.e. some of the horizontal forces must be zero)

Equality of pressure at the same level in a continuous body of fluid:

Pressures at the same level will be equal in a continuous body of fluid, even though there is no direct horizontal path between P and Q provided that P and Q are in the same continuous body of fluid.

We know that, P_{R} = P_{S }

P_{R} = P_{P} + rgh à 1

P_{S} = P_{Q} + rgh à 2

From equn.1 and 2, P_{P} = P_{Q}

General equation for the variation of pressure due to gravity from point to point in a static fluid:

Resolving the forces along the axis PQ,

pA - (p + dp)A - rgAds cos(q) = 0

dp = - rgds cos(q)

or in differential form,

dp/ds = - rgcos(q)

In the vertical z direction, q = 0.

Therefore,

dp/dz = -rg

This equation predicts a pressure decrease in the vertically upwards direction at a rate proportional to the local density.

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Last Modified on: 14-Sep-2014

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