Couette flow is frequently used in undergraduate physics and engineering courses to illustrate
shear-driven fluid motion. A simple configuration corresponds to two infinite, parallel plates separated by a distance h; one plate translates with a constant relative velocity U in its own plane. Neglecting pressure gradients, the
Navier–Stokes equations simplify to :\frac{d^2 u}{d y^2} = 0, where y is the spatial coordinate normal to the plates and u(y) is the velocity field. This equation reflects the assumption that the flow is
unidirectional — that is, only one of the three velocity components (u, v, w) is non-trivial. If the lower plate corresponds to y=0, the boundary conditions are u(0)=0 and u(h)=U. The exact solution :u (y) = U\frac{y}{h} can be found by integrating twice and solving for the constants using the boundary conditions. A notable aspect of the flow is that
shear stress is constant throughout the domain. In particular, the first derivative of the velocity, U/h, is constant. According to
Newton's law of viscosity, the shear stress is the product of this expression and the (constant) fluid
viscosity.
Startup In reality, the Couette solution is not reached instantaneously. The "startup problem" describing the approach to steady state is given by :\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2} subject to the initial condition :u(y,0)=0, \quad 0 and with the same boundary conditions as the steady flow: :u(0,t)=0, \quad u(h,t)=U, \quad t>0. The problem can be made
homogeneous by subtracting the steady solution. Then, applying
separation of variables leads to the solution: :u(y,t)= U \frac{y}{h} - \frac{2U}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} e^{-n^2 \pi^2 \frac{\nu t}{h^2}} \sin \left[n \pi \left(1-\frac{y}{h}\right)\right]. The timescale describing relaxation to steady state is t\sim h^2/\nu, as illustrated in the figure. The time required to reach the steady state depends only on the spacing between the plates h and the
kinematic viscosity of the fluid, but not on U.
Planar flow with pressure gradient A more general Couette flow includes a constant pressure gradient G=-dp/dx=\mathrm{constant} in a direction parallel to the plates. The Navier–Stokes equations are : \frac{d^2 u}{d y^2} =- \frac{G}{\mu}, where \mu is the
dynamic viscosity. Integrating the above equation twice and applying the boundary conditions (same as in the case of Couette flow without pressure gradient) gives :u (y) = \frac{G}{2\mu} y \, (h-y) + U \frac{y}{h}. The pressure gradient can be positive (adverse pressure gradient) or negative (favorable pressure gradient). In the limiting case of stationary plates (U=0), the flow is referred to as
Plane Poiseuille flow, and has a symmetric (with reference to the horizontal mid-plane) parabolic velocity profile.
Compressible flow In
incompressible flow, the velocity profile is linear because the fluid temperature is constant. When the upper and lower walls are maintained at different temperatures, the velocity profile is more complicated. However, it has an exact implicit solution as shown by C. R. Illingworth in 1950. Consider the plane Couette flow with lower wall at rest and the upper wall in motion with constant velocity U. Denote fluid properties at the lower wall with subscript w and properties at the upper wall with subscript \infty. The properties and the pressure at the upper wall are prescribed and taken as reference quantities. Let l be the distance between the two walls. The boundary conditions are :u=0, \ v =0, \ h=h_w=c_{pw} T_w \ \text{at} \ y=0, :u=U, \ v =0, \ h=h_\infty=c_{p\infty} T_\infty, \ p=p_\infty \ \text{at} \ y=l where h is the
specific enthalpy and c_p is the
specific heat.
Conservation of mass and y-momentum requires v=0, \ p=p_\infty everywhere in the flow domain.
Conservation of energy and x-momentum reduce to : \frac{d}{dy} \left(\mu \frac{du}{dy}\right) =0, \quad \Rightarrow \quad \frac{d\tau}{dy}=0, \quad \Rightarrow \quad \tau=\tau_w : \frac{1}{\mathrm{Pr}}\frac{d}{dy} \left(\mu \frac{dh}{dy}\right) + \mu \left(\frac{du}{dy}\right)^2=0. where \tau=\tau_w=\text{constant} is the wall shear stress. The flow does not depend on the
Reynolds number \mathrm{Re}=U l/\nu_\infty, but rather on the
Prandtl number \mathrm{Pr}=\mu_\infty c_{p\infty}/\kappa_\infty and the
Mach number \mathrm{M} = U/c_\infty= U/\sqrt{(\gamma-1)h_\infty}, where \kappa is the
thermal conductivity, c is the
speed of sound and \gamma is the
specific heat ratio. Introduce the non-dimensional variables :\tilde y = \frac{y}{l}, \quad \tilde T = \frac{T}{T_\infty}, \quad \tilde T_w = \frac{T_w}{T_\infty}, \quad \tilde h = \frac{h}{h_\infty}, \quad \tilde h_w= \frac{h_w}{h_\infty}, \quad \tilde u=\frac{u}{U}, \quad \tilde\mu = \frac{\mu}{\mu_\infty}, \quad \tilde\tau_w = \frac{\tau_w}{\mu_\infty U/l} In terms of these quantities, the solutions are :\tilde h = \tilde h_w + \left[\frac{\gamma-1}{2} \mathrm{M}^2 \mathrm{Pr} + (1-\tilde h_w)\right] \tilde u - \frac{\gamma-1}{2} \mathrm{M}^2 \mathrm{Pr} \, \tilde u^2, :\tilde y = \frac{1}{\tilde \tau_w} \int_0^{\tilde u} \tilde \mu \, d\tilde u, \quad \tilde \tau_w = \int_0^1 \tilde \mu \, d\tilde u, \quad q_w = - \frac{1}{\mathrm{Pr}} \tau_w \left(\frac{dh}{du}\right)_w, where q_w is the heat transferred per unit time per unit area from the lower wall. Thus \tilde h, \tilde T, \tilde u, \tilde \mu are implicit functions of y. One can also write the solution in terms of the recovery temperature T_r and recovery enthalpy h_r evaluated at the temperature of an insulated wall i.e., the values of T_w and h_w for which q_w=0. Then the solution is :\frac{q_w}{\tau_w U} = \frac{\tilde T_w-\tilde T_r}{(\gamma-1)\mathrm{M}^2 \mathrm{Pr}}, \quad \tilde T_r =1+ \frac{\gamma-1}{2} \mathrm{M}^2\mathrm{Pr}, :\tilde h = \tilde h_w + (\tilde h_r-\tilde h_w) \tilde u - \frac{\gamma-1}{2}\mathrm{M}^2 \mathrm{Pr} \, \tilde u^2. If the
specific heat is constant, then \tilde h=\tilde T. When \mathrm{M}\rightarrow 0 and T_w=T_\infty, \Rightarrow q_w= 0, then T and \mu are constant everywhere, thus recovering the incompressible Couette flow solution. Otherwise, one must know the full temperature dependence of \tilde \mu(\tilde T). While there is no simple expression for \tilde \mu(\tilde T) that is both accurate and general, there are several approximations for certain materials — see, e.g.,
temperature dependence of viscosity. When \mathrm{M}\rightarrow 0 and q_w\neq 0, the recovery quantities become unity \tilde T_r=1. For air, the values \gamma=1.4, \ \tilde \mu(\tilde T) = \tilde T^{2/3} are commonly used, and the results for this case are shown in the figure. The effects of
dissociation and
ionization (i.e., c_p is not constant) have also been studied; in that case the recovery temperature is reduced by the dissociation of molecules.
Rectangular channel One-dimensional flow u(y) is valid when both plates are infinitely long in the streamwise (x) and spanwise (z) directions. When the spanwise length is finite, the flow becomes two-dimensional and u is a function of both y and z. However, the infinite length in the streamwise direction must be retained in order to ensure the unidirectional nature of the flow. As an example, consider an infinitely long rectangular channel with transverse height h and spanwise width l, subject to the condition that the top wall moves with a constant velocity U. Without an imposed pressure gradient, the Navier–Stokes equations reduce to :\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} =0 with boundary conditions : u(0,z) =0, \quad u(h,z) = U, : u(y,0) =0, \quad u(y,l) = 0. Using
separation of variables, the solution is given by :u(y,z) = \frac{4U}{\pi} \sum_{n=1}^\infty \frac{1}{2n-1} \frac{\sinh (\beta_n y)}{\sinh (\beta_n h)} \sin (\beta_n z), \quad \beta_n = \frac{(2n-1)\pi}{l}. When h/l\ll 1, the planar Couette flow is recovered, as shown in the figure. == Coaxial cylinders ==