In this section, attention is focused on the discussion of a fluid at rest or moving in a way such that there is no presence of shearing stresses. The variations of pressure with direction and depth in a fluid will be examined.

    Pressure at a Point

Force Balance on a
Triangular Element



Consider a small triangular wedge of fluid taken from an arbitrary location within a fluid mass to determine the variation of pressure with direction. With the absence of shear stresses, the only forces acting on the fluid element are the normal and gravitational forces. Note that the forces in the x direction are not shown for clarity. According to Newton’s second law, the forces in the y and z directions can be summed as

     py ΔxΔz - ps ΔxΔs sinθ = ρ (ΔxΔyΔz/2) ay

     pz ΔxΔy - ps ΔxΔs cosθ - ρ (ΔxΔyΔz/2) g
          = ρ (ΔxΔyΔz/2) az
Since Δz = Δs sinθ and Δy = Δs cosθ, the above equations reduce to

     py - ps = ρ(Δy/2) ay

     pz - ps = ρ (Δz/2) (g + az)

By shrinking the fluid element to a point, i.e., Δx, Δy, and Δz approach zero, it can be seen that

     py = pz = ps

These results are known as Pascal’s law, which states that the pressure at a point in a static fluid is independent of direction. In other words, pressure is a scalar for fluids.

    Pressure Field



Force Balance on a
Rectangular Element


Next, consider a small rectangular fluid element taken from an arbitrary location within a fluid mass. Let the pressure at the center of the element be p. Then, use the first order Taylor series expansion to determine the forces at the faces of the fluid element (forces in the x-direction are not shown for clarity). For the force at the x+ face:


Similarly, the forces at other faces can be written as:


By assuming the only body force acting on the fluid element is due to gravity (i.e., the weight of the fluid element) and the absence of shearing forces (i.e., inviscid assumption), then applying Newton’s second law in the x-direction and yields



Similarly, according to Newton's second law, the force balances in the y- and z- directions are given by



The above equations can be simplified to obtain


These equations can be recast into vector notation:


where is the gradient operator and is the acceleration vector of the fluid element.

    Pressure Variation in a Static Fluid

Derivation of Hydrostatic
Pressure Distribution


For a fluid at rest (i.e., = 0), we have
or in scalar form

Hence, it can be concluded that p varies only in the z-direction and is not a function of x and y, giving

      dp/dz = -ρg

The above equation determines how pressure changes with elevation for a fluid at rest. Integration can be performed to obtain the changes in pressure once the variation of gravity and fluid density with elevation is known. For most practical engineering applications, the variation of g is negligible over the change in height. This gives,


p1 - p2 = ρg (z2 - z1)

Hydrostatic Pressure Distribution
in a Static Fluid

Thus, pressure increases with depth in a linear fashion. This type of pressure distribution is called hydrostatic. We can also rearrange the above equation to yield an expression for the pressure head as follows:

     h = (p1 - p2) / ρg

Physically, the pressure head represents the height of a homogeneous fluid column required to produce a pressure difference of (p1 - p2). The term ρg is often written as γ (specific weight) in civil engineering applications.

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