Oscillations, Waves, and Interactions - GWDG

Sound absorption, sound amplification, and flow control in ducts 89
Herein ɛµ comprises the transport of energy due to viscosity, i. e., diffusion and dissipation,
which will be disregarded, to begin with. Thus the power extracted from the
mean flow (left-hand side of Eq. (5)) is fed into the streamwise flux of kinetic energy
and into the wall-normal energy transport due to the pressure-velocity correlation
(first and second term on the right-hand side, respectively). While the wall-normal
energy transport balances the dissipation of energy in the walls and therefore is irreversible,
the kinetic energy contained in oscillation of the flow can be returned to
the mean flow if the Reynolds shear stress becomes negative. Under this condition
the first term on the right-hand side is negative as well, meaning that the amplitude
of the oscillation decreases in the mean-flow direction, and we will see from Eq. (7)
that the static pressure may increase under the same condition.
3.1.3 Static pressure gradient
The supply or the recovery of mean flow energy in a homogeneous channel cannot
change the kinetic energy of the mean flow when the velocity profile is assumed to
remain constant along the axial coordinate x. Hence the exchange of energy must
be balanced by the work done by the pressure gradient on the flow. For the sake
of simplicity, the following formulae are restricted to the 2D channel (half-width R)
and to the channel with circular cross-section (radius R). The wall shear stress τ w
is constant over the circumference ∂(S) of the channel in both these cases. So the
balance between the forces due to the pressure gradient, due to the wall friction, and
due to the acceleration of the fluid yields
− dp
dx = τ ∂(S) d
w +
S dx
�
S
2 dS
ρU . (6)
S
The axial pressure gradient dp/dx is assumed to be constant over the cross-section
which is a good approximation except for extreme axial alteration of the mean velocity
profile U(r, x). We are particularly interested in the first term on the righthand
side of Eq. (6) which is due to the wall shear stress and which will be denoted
by dp τ /dx. After some analysis the work done by dp τ /dx on the flow, namely
�
S (dp τ /dx)U dS, can be equated with
�
−
S
dp τ
dx UdS = τ wU∂(S)
τ µ≪τ ′′
=
�
0
R
�
τ dU
dy
� S
R dy
� �� �
2D channel
τ µ≪τ ′′
=
�
0
R
�
τ
�
− dU
dr
��
2πrdr
� �� �
circular duct
(7)
wherein
τ(r) = τ ′′ r
(r) + τ µ(r) = τ w
(8)
R
has been used which applies to the circular duct as well as to the 2D channel with
r = R − y. So, according to the middle part of Eq. (8), the right-hand side of Eq. (7)
comprises the power that is fed into the oscillation of the flow according to Eq. (5)
and the power that is dissipated by the mean viscous shear stress τ µ. Furthermore
it turns out that the wall shear stress is a measure of the considered energy transfer.