Fluid Mechanics and Boundary Layers

Boundary layer thickness on flat plate

The boundary layer is the thin region adjacent to a solid surface in which viscous effects are important and the velocity rises from zero at the wall to about 99 percent of the free-stream speed U. Its thickness \delta grows with the distance x from the leading edge, and the growth law depends on the flow regime. The regime is set by the local Reynolds number Re_x = Ux/\nu, so you should always evaluate Re_x at the station of interest before choosing a formula. For a laminar layer, the Blasius solution gives \delta = 5.0\,x/\sqrt{Re_x}. For a turbulent layer, the one-seventh power law gives \delta = 0.37\,x/Re_x^{1/5}. Transition on a smooth flat plate typically occurs near Re_x \approx 5\times10^5, so the turbulent expression applies only downstream of that station.

The displacement thickness \delta^* measures the outward shift of the streamlines caused by the slowed fluid near the wall. For the laminar Blasius profile, \delta^* \approx 1.72\,x/\sqrt{Re_x}, so the ratio of the 99 percent thickness to the displacement thickness is \delta/\delta^* \approx 2.9. The shape factor H = \delta^*/\theta \approx 2.59 is also characteristic of the laminar profile. A ratio near 2.9 lies inside the laminar window of roughly 2.5 to 3.4, whereas fuller turbulent profiles give a considerably larger ratio.

For a laminar layer on a long plate, \delta \approx 5L/\sqrt{Re_L}, so the layer is thin compared with the plate length when Re_L is large. Within such a layer u \gg v and \partial u/\partial y \gg \partial u/\partial x, which are the boundary-layer approximations. The layer is also rotational and viscous, because wall shear generates vorticity. It can therefore never be irrotational or inviscid.

Example: water flows at U = 1 m/s over a plate of length L = 0.3 m with \nu = 10^{-6} m^2/s. Then Re_L = 3\times10^5, which is below the transition range, and \delta_L = 5(0.3)/\sqrt{3\times10^5} \approx 2.7 mm.

Flow kinematics of velocity field

A velocity field \vec V = u\hat i + v\hat j + w\hat k is characterised by its divergence and vorticity, which tell you whether the flow is compressible and whether it is rotational. The divergence \nabla\cdot\vec V = \partial u/\partial x + \partial v/\partial y + \partial w/\partial z vanishes for incompressible flow. The z-component of vorticity, \omega_z = \partial v/\partial x - \partial u/\partial y, vanishes for irrotational flow. For example, u = ax and v = -ay have zero divergence and zero vorticity, so the flow is both incompressible and irrotational.

The acceleration of a fluid particle is the material derivative \vec a = \partial\vec V/\partial t + (\vec V\cdot\nabla)\vec V, which consists of a local term and a convective term. In a two-dimensional unsteady field, the components are a_x = \partial u/\partial t + u\,\partial u/\partial x + v\,\partial u/\partial y and a_y = \partial v/\partial t + u\,\partial v/\partial x + v\,\partial v/\partial y. For u = xt and v = 0, the local term gives x and the convective term gives xt^2, so a_x = x + xt^2. Omitting either contribution gives a wrong result.

Vorticity \vec\omega = \nabla\times\vec V is twice the angular velocity \vec\Omega = \tfrac12\nabla\times\vec V of a fluid element, so the ratio of their magnitudes is 2. Solid-body rotation, with u = -\omega_0 y and v = \omega_0 x, illustrates this. The vorticity is 2\omega_0 while the rate of rotation is \omega_0.

Flow past submerged bodies

When a body is fully immersed in a steady, uniform flow, the drag depends on whether the fluid is ideal or real. D'Alembert's paradox states that a closed body in a steady, uniform, inviscid and irrotational (ideal) flow experiences zero drag, which contradicts observation. The paradox is resolved by viscosity. In a real fluid, skin friction and the pressure asymmetry caused by separation give rise to drag, so the zero-drag result holds only for ideal fluids.

Flow separation occurs when an adverse pressure gradient brings the near-wall fluid to rest. The separation point is where the wall shear stress vanishes, so \tau_w = 0 there. Surface devices such as dimples on a golf ball trip the boundary layer into turbulence at a lower Reynolds number. A turbulent layer carries more momentum close to the wall, so it resists the adverse pressure gradient for longer, delays separation, and reduces the size of the wake and the pressure drag.

For a two-dimensional wake, the drag per unit span can be found from a control-volume momentum balance. When the pressure on the control surface is uniform, the drag equals the difference between the inlet and outlet momentum fluxes. In a wake this becomes F_D = \int \rho u(U - u)\,dy, so the drag depends on the full velocity profile through the integral rather than only on its peak velocity deficit.

Along a curved wall, the pressure rises in the streamwise direction on the downstream side of the crest, which gives a positive (adverse) pressure gradient at the wall. In the adverse region the near-wall velocity curves back toward reversed flow, so \partial u/\partial y at the wall is negative there. The leading point where the boundary layer starts is attached rather than separated. A stagnation point is a location where the velocity is zero, which is not the same as a region of reversed near-wall flow.

Bernoulli equation for nozzle flow

Along a streamline in steady, incompressible, frictionless flow, Bernoulli's equation states that p + \tfrac12\rho V^2 + \rho g z is constant. Combined with continuity, A_1V_1 = A_2V_2, it gives the pressure at each section of a nozzle or contraction. As the area narrows, the velocity rises and the pressure falls. For a horizontal nozzle, the inlet pressure must therefore exceed the outlet pressure by \tfrac12\rho(V_2^2 - V_1^2).

For a free jet from a large tank with an open free surface, neglecting losses and taking the free-surface velocity as zero gives Torricelli's result V = \sqrt{2gH}. Here H is the head measured from the free surface down to the centreline of the orifice. The ideal discharge is Q = A\sqrt{2gH}.

In a converging duct, the same two relations are applied between two sections at equal elevation, which gives p_2 = p_1 + \tfrac12\rho(V_1^2 - V_2^2). The downstream pressure is therefore lower than the upstream pressure whenever the velocity increases.

Example: a horizontal nozzle with an inlet area of 0.2 m^2 and an outlet area of 0.02 m^2 carries water with an outlet velocity of 10 m/s. Continuity gives V_1 = 1 m/s, so the gauge pressure at the inlet is \tfrac12(1000)(10^2 - 1^2) = 49\,500 Pa.

Couette flow between parallel plates

In plane Couette flow, one plate is stationary and the other moves parallel to it with velocity U, with no imposed pressure gradient. The steady laminar velocity profile is linear, u(y) = Uy/h, and the shear stress is uniform across the gap, \tau = \mu U/h. The dynamic viscosity is related to the kinematic viscosity by \mu = \rho\nu, which must be used whenever only \nu and \rho are supplied.

The shear force on a surface of area A is F = \tau A. When a plate is pulled with a known force, its velocity follows from V = \tau h/\mu for a linear profile. When a thin plate sits inside the gap, it divides the gap into two layers, and the shear on each face is computed with the thickness of the layer on that side. The total drag is the sum of the two face shears.

Example: in a gap of 2 mm with \mu = 0.001 Pa\cdots, a plate of area 1 m^2 moving at 0.5 m/s experiences \tau = (0.001)(0.5)/(0.002) = 0.25 Pa, so the force is 0.25 N.

Continuity equation

For a fixed control volume \mathcal V bounded by a surface S with outward unit normal \hat n, mass conservation states that the rate of decrease of mass stored inside equals the net mass outflow through the boundary: -\dfrac{\partial}{\partial t}\iiint_{\mathcal V}\rho\,dV = \iint_S \rho\vec V\cdot\hat n\,dS. Equivalently, \dfrac{\partial}{\partial t}\iiint_{\mathcal V}\rho\,dV + \iint_S\rho\vec V\cdot\hat n\,dS = 0. The material derivative D/Dt applies to a fluid system that moves with the flow, so it is not the operator used for a fixed control volume.

The differential form is \partial\rho/\partial t + \nabla\cdot(\rho\vec V) = 0, which reduces to \nabla\cdot\vec V = 0 for incompressible flow. In steady incompressible pipe flow, the volume flow rate is constant, so Q = A_1V_1 = A_2V_2. The velocity ratio is therefore V_2/V_1 = (D_1/D_2)^2. Halving the diameter quadruples the velocity, and the pipe length does not enter the ratio.

Navier-Stokes momentum equation terms

The momentum equation applies Newton's second law to a fluid element. In the form \rho\left[\dfrac{\partial\vec V}{\partial t} + \nabla\!\left(\dfrac{V^2}{2}\right) + (\nabla\times\vec V)\times\vec V\right] = -\nabla(P+\rho gz) + \mu\nabla^2\vec V, the left-hand terms are the local acceleration, the gradient of kinetic energy, and the vortex term. The right-hand terms are the pressure-gravity term and the viscous term. Setting the local acceleration to zero removes the time dependence, which defines steady flow. Euler's equation is obtained by deleting the viscous term while keeping the pressure-gravity term. Deleting the pressure-gravity term instead does not produce Euler's equation.

When the inertial, vortex and viscous terms all vanish, the balance reduces to \nabla(P+\rho gz) = 0, which is the hydrostatic equation. When the local acceleration, the vortex term and the viscous term are all removed, the remaining gradient balance is that of an irrotational flow.

The vector form for incompressible Newtonian flow with constant kinematic viscosity \vartheta is \partial\mathbf u/\partial t + \mathbf u\cdot\nabla\mathbf u = -\nabla p/\rho + \mathbf g + \vartheta\nabla^2\mathbf u. It expresses conservation of linear momentum. The convective term is quadratic in \mathbf u, so the equation is a nonlinear partial differential equation. The viscous term assumes a linear relation between stress and strain rate, so it holds for Newtonian fluids only. Constant density together with \nabla\cdot\mathbf u = 0 restricts the equation to incompressible flow.