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2.5 Exercises
2.5 Exercises
2.5.1. The figure sketches laminar flow between a moving block and a stationary
flat wall. The block moves parallel to the wall at velocity v. There is no pressure
difference in the flow direction in the space between the block and the wall. Reason
that the shear stress τ within the shear layer is constant. Demonstrate that dissipated
work per surface unit and per time unit equals displacement work ( v.τ) exerted by
the object onto the shear flow. Demonstrate that this result remains valid with turbulent flow.
2.5.2. The left figure sketches the velocity profiles in a 2D-channel with fully
developed laminar flow of an incompressible fluid ( ρ = constant, m = constant).
Fully developed flow means that no change of velocity profile occurs in the flow
sense. Derive that the shear stress varies linearly from the wall value τ w to zero in
the channel centre. Demonstrate that the velocity profile is a parabola. Derive that
the energy dissipated per wall surface unit and per time unit equals U m τ w , with
U m being the average velocity within the channel. Note that for attaining these results nothing more is necessary than that the shear stress varies linearly along the
height. So argue that the same result is attained with a turbulent profile (central
figure) ( m t ≠ constant). The right figure sketches the velocity and shear stress profiles with a turbulent flow of a boundary layer over a flat plate at a zero pressure
gradient. The boundary layer grows. So the flow is not fully developed. The shear
stress profile differs little from a linear one. So, the former result is still valid with
a good approximation.
2.5.3. The figure sketches the mixing of two flows with velocities 3/2 v and
1/2 v within a 2D-channel. Both flows occupy half the section. Velocity therefore
is v after complete mixing. Determine the pressure increase by complete mixing.
Determine the energy dissipation by mixing per mass flow rate unit, ignoring
friction.
2.5 Exercises
2.5 Exercises
2.5.1. The figure sketches laminar flow between a moving block and a stationary
flat wall. The block moves parallel to the wall at velocity v. There is no pressure
difference in the flow direction in the space between the block and the wall. Reason
that the shear stress τ within the shear layer is constant. Demonstrate that dissipated
work per surface unit and per time unit equals displacement work ( v.τ) exerted by
the object onto the shear flow. Demonstrate that this result remains valid with turbulent flow.
2.5.2. The left figure sketches the velocity profiles in a 2D-channel with fully
developed laminar flow of an incompressible fluid ( ρ = constant, m = constant).
Fully developed flow means that no change of velocity profile occurs in the flow
sense. Derive that the shear stress varies linearly from the wall value τ w to zero in
the channel centre. Demonstrate that the velocity profile is a parabola. Derive that
the energy dissipated per wall surface unit and per time unit equals U m τ w , with
U m being the average velocity within the channel. Note that for attaining these results nothing more is necessary than that the shear stress varies linearly along the
height. So argue that the same result is attained with a turbulent profile (central
figure) ( m t ≠ constant). The right figure sketches the velocity and shear stress profiles with a turbulent flow of a boundary layer over a flat plate at a zero pressure
gradient. The boundary layer grows. So the flow is not fully developed. The shear
stress profile differs little from a linear one. So, the former result is still valid with
a good approximation.
2.5.3. The figure sketches the mixing of two flows with velocities 3/2 v and
1/2 v within a 2D-channel. Both flows occupy half the section. Velocity therefore
is v after complete mixing. Determine the pressure increase by complete mixing.
Determine the energy dissipation by mixing per mass flow rate unit, ignoring
friction.