Couette flow with a pressure gradient

(a) Velocity distribution

For steady, fully developed laminar flow, the x-momentum (Navier-Stokes) equation reduces to

$$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}$$

Integrating twice: $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2$.

Boundary conditions: $u = 0$ at $y = 0$, and $u = U$ at $y = h$. Then $C_2 = 0$ and $C_1 = \dfrac{U}{h} - \dfrac{h}{2\mu}\dfrac{dp}{dx}$.

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

The first term is the Couette (linear) part due to the moving plate, and the second is the Poiseuille (parabolic) part due to the pressure gradient.

Discharge per unit width:

$$q = \int_0^h u\,dy = \frac{Uh}{2} - \frac{h^3}{12\mu}\frac{dp}{dx}$$

(b) Effect of pressure gradient

Let $P = -\dfrac{h^2}{2\mu U}\dfrac{dp}{dx}$.

  dp/dx = 0     dp/dx < 0     dp/dx > 0 (large)
      /|           _.|             |
     / |         .'  |            _|
    /  |        /    |          <-'/
   /___|       /_____|         /___|

Shear stress: $\tau = \mu\dfrac{du}{dy} = \mu\dfrac{U}{h} - \left(\dfrac{h}{2} - y\right)\dfrac{dp}{dx}$

At the moving plate ($y = h$): $\tau_h = \dfrac{\mu U}{h} + \dfrac{h}{2}\dfrac{dp}{dx}$

Numerical

$\mu U/h = 0.1 \times 3/0.004 = 75\ \text{Pa}$.

• For $dp/dx = 0$: $\tau_h = 75\ \text{Pa}$.
• For $dp/dx = -10\,000\ \text{Pa/m}$:

$$\tau_h = 75 + \frac{0.004}{2}(-10\,000) = 75 - 20 = 55\ \text{Pa}$$

(The shear stress on the fixed plate would be $75 + 20 = 95\ \text{Pa}$.) The discharge per unit width rises from $Uh/2 = 0.006\ \text{m}^2/\text{s}$ to $0.006 + \dfrac{(0.004)^3 \times 10\,000}{12 \times 0.1} = 0.00653\ \text{m}^2/\text{s}$.

Answer: $\tau_h = 75\ \text{Pa}$ for $dp/dx = 0$, and $55\ \text{Pa}$ for $dp/dx = -10\ \text{kPa/m}$