Variation of Pressure with Depth in a Fluid

Consider two points 1 and 2 lie in the ends of fluid prism having a cross-sectional area dA and length L. The difference in elevation between these two points is h as shown in Figure 02 below. The fluid is at rest and its surface is free. The prism is therefore in equilibrium and all forces acting on it sums up to zero.



Note: FFS stands for Free Fluid Surface which refers to fluid surface subject to zero gauge pressure.

The volume of the prism is equal to the length times the base area of the fluid.
$V = L \, dA$

The weight of the fluid prism shown is equal to the product of the unit weight and volume.
$W = \gamma V$

$W = \gamma L \, dA$

Sum up all the forces in x-direction
$\Sigma F_x = 0$

$F_2 = F_1 + W_x$

$F_2 - F_1 =W \sin \theta$

$p_2 \, dA - p_1 \, dA = \gamma L \, dA \sin \theta$

$p_2 - p_1 = \gamma L \sin \theta$

but L sin θ = h, thus

$p_2 - p_1 = \gamma h$

Therefore, in any homogeneous fluid at rest, the difference in pressure between any two points is equal to the product of the unit weight of the fluid and the difference in elevation of the points.

If h = 0 so that points 1 and 2 are on the same horizontal plane, p2 - p1 = 0 or

$p_1 = p_2$

Therefore, in any homogeneous fluid at rest, the pressures at all points along the same horizontal plane are equal.

If point 1 lie on the FFS, the gauge pressure p1 = 0, making p2 - 0 = γh or simply

$p = \gamma h$

This means that the pressure at any depth h below a continuous free fluid surface at rest is equal to the product of the unit weight of fluid and the depth h.

Transmission of Pressure

We can write the equation p2 - p1 = γh into the form

$p_2 = p_1 + \gamma h$

which means that any change in the pressure at point 1 would cause an equal change of pressure at point 2. In other words, a pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid.

Pressure Head

The equation p = γh may be written into the form

$h = \dfrac{p}{\gamma}$

where h or its equivalent p/γ is in hydraulics called the pressure head. Pressure head is the height of column of homogeneous fluid of unit weight γ that will produce an intensity of pressure p.

To convert pressure head of liquid A to equivalent pressure head of liquid B

$h_B = \dfrac{s_A}{s_B}h_A = \dfrac{\rho_A}{\rho_B}h_A = \dfrac{\gamma_A}{\gamma_B}h_A$


To convert pressure head of any liquid to equivalent pressure head of water

$h_{water} = s_{liquid} \times h_{liquid}$

s = specific gravity
γ = unit weight
ρ = density

Properties of Water

s = 1.0
γ = 9.81 kN/m3 (or 62.4 lb/ft3 in English system)
ρ = 1000 kg/m3 (or 1.94 slugs/ft3 in English system)