Couette flow

Theory and Concepts
1

can you explain how we derive the solution for couette flow. it seems like we just assume that everything is zero but I don’t know why or when that works

2

This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
To derive the solution for Couette flow, we start with the Navier-Stokes equations and apply several simplifying assumptions based on the geometry and conditions of the flow.

The simplifying assumptions take the form of an ansatz – an educated guess that we check later. The ansatz is only useful if it (a) simplifies the analysis and (b) turns out to be correct. In this case, the ansatz always simplifies the analysis, but it turns out to be correct only some of the time.

Here’s a step-by-step explanation:

  1. Geometry and Flow Conditions:

    • Consider two infinite parallel plates separated by a distance in the y direction.
    • The flow is driven by the motion of one or both plates in the x direction.
    • The flow is steady, unidirectional, and there is no pressure gradient.

  2. Simplifying the Navier-Stokes Equations:

    • The Navier-Stokes equations for incompressible flow are given by:

      \rho\frac{\partial{\vec{u}}}{\partial t} + \rho{\vec{u}}\cdot\nabla{\vec{u}} = -\nabla p + \eta\nabla^2{\vec{u}}

    • For Couette flow, we make the following ansatz:

      • Steady Flow: \frac{\partial{\vec{u}}}{\partial t} = 0.
      • Unidirectional Flow: The velocity vector \vec{u} has only an x component, u(y), and no y or z components.
      • No Pressure Gradient: \nabla p = 0.
      • No Convective Acceleration: Since the flow is unidirectional and steady, \rho{\vec{u}}\cdot\nabla{\vec{u}} = 0.

  3. Resulting Simplified Equation:

    • If the ansatz is correct, the Navier-Stokes equations reduce to:

      \eta \frac{d^2 u}{dy^2} = 0

    • This is a second-order ordinary differential equation indicating that the concavity of the velocity profile is zero, meaning the velocity profile is linear.

  4. Boundary Conditions:

    • To solve this equation, we apply boundary conditions:
      • At y = H, the velocity u = U_H (velocity of the top plate).
      • At y = -H, the velocity u = U_L (velocity of the bottom plate).

  5. Solution:

    • Integrating the differential equation twice gives a linear velocity profile:

      u(y) = C_1 y + C_2

    • Applying the boundary conditions, we solve for the constants C_1 and C_2 to obtain:

      u(y) = \frac{U_H - U_L}{2} \frac{y}{H} + \frac{U_H + U_L}{2}

This is a solution to the Navier–Stokes equations, but nothing we have done shows that it is unique, nor have we shown that the ansatz is correct. We have only shown that if the ansatz is correct, this is the observed solution.

In practice this solution is observed, but only for Reynolds numbers below 500 or so. For Reynolds numbers above 500, the ansatz is not correct, and this solution is not observed. Above Re=500, the flow is turbulent and we cannot find an analytical solution.

This solution (known as laminar Couette flow) shows that the velocity profile is linear between the two plates, with the slope determined by the difference in velocities of the plates and the distance between them.

To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their significance below. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 11.2: Couette Flow Part 2 (HIGHLY RELEVANT)
  2. Chapter 11.2: Couette Flow Part 1 (HIGHLY RELEVANT)
  3. Chapter 11 Fundamentals: Laminar and Turbulent Flow (HIGHLY RELEVANT)
  4. Chapter 11.3: Poiseuille Flow Part 1 (HIGHLY RELEVANT)
  5. Chapter 10 Fundamentals: Solving the boundary layer equations via a similarity transform for the fla… (HIGHLY RELEVANT)

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