Internal Laminar Flow
Couette Flow
Consider two parallel plates with infinite span separated by a constant density fluid of height 2b. Each plate is a distance of b away from the centerline. Neglect buoyancy effects and assume the top plate is moving at a constant velocity of U while the bottom plate is stationary. Furthermore, assume the bottom and top plates are isothermal at temperatures of T_s and T_1 respectively. The flow considered here is called parallel flow as all streamlines are parallel. Since all fluid elements move parallel to each other, v=0 and constant density mass conservation is
\frac {\partial u}{\partial x}=-\frac {\partial v}{\partial y}=0
Therefore, the velocity profile u can only be a function of y. Assuming steady flow of a Newtonian fluid, the momentum equation in the x and y directions simplify to
\begin{align*} u\cancel{\frac {\partial u}{\partial x}}+\cancel{v\frac {\partial u}{\partial y}} & =-\frac 1{\rho}\cancel{\frac {\partial p}{\partial x}}+\nu\left(\cancel{\frac {\partial^2u}{\partial x^2}}+\frac {\partial^2u}{\partial y^2}\right) \\ u\cancel{\frac {\partial v}{\partial x}}+\cancel{v\frac {\partial v}{\partial y}} & =-\frac 1{\rho}\frac {\partial p}{\partial v}+\nu\left(\cancel{\frac {\partial^2v}{\partial x^2}}+\cancel{\frac {\partial^2v}{\partial y^2}}\right) \end{align*}
Momentum in the y direction shows there is no pressure gradient in the y direction. There also cannot be a pressure gradient in the x direction as that would violate continuity. If we assumed \frac {\partial p}{\partial x}\neq0, then this would require \frac {\partial^2u}{\partial x^2}\neq0 so u=u(x,y), which violates continuity. The PDE above reduces down to
\frac {\mathrm d^2u}{\mathrm dy^2}=0
subject to the boundary conditions u(-b)=0 and u(b)=U. Integrating twice to determine the velocity profile gives
u(y)=\frac U2\left(1+\frac yb\right) \tag{1}
The energy equation for a constant density fluid is
\rho C_p\left(u\cancel{\frac {\partial T}{\partial x}}+\cancel v\frac {\partial T}{\partial y}\right)=k\frac {\partial^2T}{\partial y^2}+\mu\left(\frac {\partial u}{\partial y}\right)^2