Question 4.10: For case (b) in Fig. 4.12b, flow between parallel plates due...

All parameters can be computed from the basic solution, Eq. (4.134), by mathematical manipulation.

u=-\frac{dp}{dx}\frac{h^2}{2\mu}\left(1-\frac{y^2}{h^2}\right)                        (4.134)

Part (a)

The wall shear follows from the definition of a newtonian fluid, Eq. (4.37):

\tau_{xy}=\tau_{yx}=\mu\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)                      \tau_{xz}=\tau_{zx}=\mu\left(\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}\right)                          \tau_{yz}=\tau_{zy}=\mu\left(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}\right)                   (4.37)

The wall shear has the same magnitude at each wall, but by our sign convention of Fig. 4.3, the upper wall has negative shear stress.

Part (b)

Since the flow is plane, steady, and incompressible, a stream function exists:

Integrating and setting ψ = 0 at the centerline for convenience, we obtain

At the walls, y = ±h and ψ = \pm 2u_{max}h/3, respectively.

Part (c)

In plane flow, there is only a single nonzero vorticity component:

The vorticity is highest at the wall and is positive (counterclockwise) in the upper half and negative (clockwise) in the lower half of the fluid. Viscous flows are typically full of vorticity and are not at all irrotational.

Part (d)

From part (c), the vorticity is finite. Therefore the flow is not irrotational, and the velocity potential does not exist.

Part (e)

The average velocity is defined as V_{av} = Q/A, where Q = ∫ u dA over the cross section. For our particular distribution u(y) from Eq. (4.134), we obtain

In plane Poiseuille flow between parallel plates, the average velocity is two-thirds of the maximum (or centerline) value. This result could also have been obtained from the stream function derived in part (b). From Eq. (4.95),

Q_{1\rightarrow 2}=\int_1^2{(V \cdot n)  dA}=\int_1^2{d \psi}=\psi_2 -\psi_1                          (4.95)

Q_{channel}=\psi_{upper}-\psi_{lower}=\frac{2u_{max}h}{3}-\left(-\frac{2u_{max}h}{3}\right)=\frac{4}{3}u_{max}h  per unit width

whence V_{av} = Q/A_{b=1} = (4u_{max}h/3)/(2h) = 2u_{max}/3, the same result.

This example illustrates a statement made earlier: Knowledge of the velocity vector V [as in Eq. (4.134)] is essentially the solution to a fluid mechanics problem, since all other flow properties can then be calculated.