3. Exact parallel flows

The question

Does doubling a channel’s height double its flow? The answer depends on what is held fixed. An exact solution makes that dependence visible before a mesh or solver is involved.

Consider long parallel plates at $y=0$ and $y=H$. Assume a steady, fully developed, incompressible Newtonian flow $\boldsymbol u=(u(y),0)$, constant viscosity, and no body force. The lower wall is fixed and the upper wall moves at speed $U_w$. Let $G=-\mathrm dp/\mathrm dx$ be a constant positive driving pressure gradient.

The advective acceleration vanishes because $u\partial_xu=0$ and the transverse velocity is zero. Momentum reduces to

$$\mu\frac{\mathrm d^2u}{\mathrm dy^2}=-G,\qquad u(0)=0,\quad u(H)=U_w.$$

This is an exact reduction under the stated assumptions, not a low-Reynolds approximation. Entrance regions and disturbances violate the assumed profile.

Integrate twice

The first integration gives $\mu u'=-Gy+C_1$ and the second gives $\mu u=-Gy^2/2+C_1y+C_2$. The lower wall sets $C_2=0$, and the upper wall sets $C_1=\mu U_w/H+GH/2$. Thus

$$u(y)=U_w\frac{y}{H}+\frac{G}{2\mu}y(H-y).$$

The wall-driven part is linear; the pressure-driven part is parabolic. Their sum is possible because the reduced equation is linear.

                    upper wall y = H
Couette: u = Uw      ────────────────→
       u = Uw/2    ────────→
       u = 0      •
                  lower wall y = 0

Poiseuille: u = 0   •                 y = H
          u_max  ────────────────→ y = H/2
          u = 0  •                 y = 0

Arrow length represents streamwise velocity. Both profiles obey no slip at their respective wall velocities.

Flux and shear are different observations

The volume flux per unit span and mean speed are

$$q=\int_0^H u(y)\,\mathrm dy =\frac{U_wH}{2}+\frac{GH^3}{12\mu},\qquad \bar U=\frac qH.$$

The units of $q$ are $\mathrm{m^2/s}$. Multiply by an out-of-plane width to get $\mathrm{m^3/s}$. The shear stress component is

$$\sigma_{xy}=\mu u'(y)=\frac{\mu U_w}{H}+G(H/2-y).$$

For pure Couette flow ($G=0$), shear is constant. For pure Poiseuille flow ($U_w=0$), it changes sign at the centre. Wall tractions also include the wall’s normal orientation: opposite signs of $\sigma_{xy}$ at the two walls do not mean that one wall propels the fluid.

For stationary walls,

$$u_{\max}=\frac{GH^2}{8\mu},\qquad \bar U=\frac{GH^2}{12\mu}=\frac23u_{\max}.$$

This explains the conversion from maximum to mean inlet speed in the cylinder examples. It does not imply that flow around the cylinder remains parallel.

A numerical prediction made on paper

Choose $H=0.01\,\mathrm m$, $\mu=0.001\,\mathrm{Pa\,s}$, $G=1\,\mathrm{Pa/m}$ and stationary walls. Then $u_{\max}=0.0125\,\mathrm{m/s}$, $\bar U=0.008333\ldots\,\mathrm{m/s}$ and $q=8.3333\ldots\times10^{-5}\,\mathrm{m^2/s}$. The lower-wall shear component is $0.005\,\mathrm{Pa}$. These values test different aspects of a future computation: maximum speed, integrated flux and a derivative at a boundary.

At fixed $G$, doubling $H$ multiplies $q$ by eight. At fixed $q$, the required $G$ instead falls by eight. State the controlled quantity before making a scaling prediction.

A smooth transient you can solve

Let $u_s(y)$ denote the steady profile and write $u=u_s+w$. Keep the same wall velocities and pressure gradient. A parallel perturbation satisfies

$$\partial_t w=\nu\partial_{yy}w,\qquad w(0,t)=w(H,t)=0.$$

For $w(y,0)=A\sin(\pi y/H)$, direct substitution gives

$$w(y,t)=A\sin(\pi y/H)\exp\left(-\frac{\nu\pi^2t}{H^2}\right), \qquad \tau=\frac{H^2}{\pi^2\nu}.$$

Its amplitude is reduced by $e^{-1}$ after $\tau$. Higher sine modes decay faster in proportion to the square of their mode number. This is the same mathematical diffusion mechanism as constant-property heat conduction, with momentum diffusivity $\nu$ replacing thermal diffusivity.

These analytic exercises build expectations independent of the cylinder output used later. Carry their flux, shear and time-scale reasoning into that more complicated geometry.

Exercises and failure checks

1. Derive $q$ by integrating the profile, without using the displayed result.
2. For stationary walls, verify that pressure power per unit channel length and span, $Gq$, equals $\int_0^H\mu(u')^2\,\mathrm dy$.
3. Reverse the pressure gradient while retaining a moving upper wall. Determine whether part of the fluid can move opposite to that wall.
4. With $H=0.01\,\mathrm m$ and $\nu=10^{-6}\,\mathrm{m^2/s}$, find $\tau$.
5. A proposed no-slip channel solution has a constant nonzero velocity. Which condition fails even though its divergence is zero?

Check your reasoning. Exercise 2 gives $G^2H^3/(12\mu)$ on both sides. Exercise 4 gives approximately $10.13\,\mathrm s$. A uniform nonzero velocity satisfies mass conservation but violates the stationary wall velocities.

The original teaching materials for MIT’s Couette and Poiseuille lesson provide a complementary treatment of these two experiments.

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