Page 79 - Fluid Power Engineering

56 Cha pte r T w o

FIGURE 2B.1 Cylindrical fl uid element fl owing at the pipe center.

FIGURE 2B.2
Laminar fl ow
in pipe line.

speed is constant; it does not change with time. The forces acting on
the fluid element are in equilibrium, thus the shearing forces on the
cylinder are equal to the pressure forces.

τ
2πrdx = dpA = dp πr 2 (2B.1)

or τ= r dp (2B.2)
2 dx

where dp/dx = the pressure gradient. It is a function of x only [≠ fr()].
This equation shows that the shear stress is linearly proportional
to the distance r. For Newtonian fluids, the velocity distribution could
be calculated as follows:

τ = μ du dy (2B.3)
/
−
y = R r then du dy/ =− du dr and τ =− du dr/ (2B.4)
μ
/
τ = r dp = − μ du (2B.5)
2 dx dr

or du =− r dp (2B.6)
dr 2μ dx