Fluid Mechanics for Mechanical Engineers/Technical Applications

Aerodynamic Drag and Flow Separation
A partially/fully immersed body moving in fluid due to interaction with fluid, will experience a net force. This net force will constitute of different components, a force in the direction of upstream flow velocity (Drag Force) and a force perpendicular to the upstream flow velocity (Lift Force), if the body is assumes as 2 dimensional. For 3 Dimensional bodies, there will be a force perpendicular to the plane of Drag and Lift forces.

While fluid flows over an object, normal pressure forces and tangent shear stresses apply on the object, finding analytically the distribution of this forces on objects is really a difficult work. With the development of computing power, Computational Fluid Dynamics has been a powerful tool to calculate distribution of normal pressure forces and tangent shear stresses.

By integrating the distributions of normal pressure forces and tangent shear stresses in the direction of upstream flow and normal to upstream flow, Drag force and Lift force can be calculated as follow;

$$ \displaystyle dF_x=(p \cos\theta)dA + (\tau_w \sin \theta) dA $$

$$ \displaystyle dF_y=-(p \sin\theta)dA + (\tau_w \cos \theta) dA $$

And by integrating, to obtain total force,

$$ \displaystyle D=\int_{} dF_x=\int_{} p \cos\theta dA +\int_{} \tau_w \sin \theta dA $$

$$ \displaystyle L=\int_{} dF_y=\int_{} -p \sin\theta dA +\int_{} \tau_w \cos \theta dA $$

Obtaining the distribution of the normal pressure forces and tangent shear forces over the shape of the bodies are extremely difficult.

As discussed in Chapter 5, dimensional analysis reduces number and complexity of experimental variables which affect a given physical phenomenon, because of this reason we define to coefficients as coefficient of Drag and coefficient of Lift as;

$$ \displaystyle C_D=\frac{D}{\tfrac{1}{2} \rho\ U^2 A} $$

$$ \displaystyle C_L=\frac{L}{\tfrac{1}{2} \rho\ U^2 A} $$

By knowing the upstream flow properties and these coefficients, the drag and lift forces can be calculated easily. This is while, finding the exact values of $$C_L$$ and $$C_D$$ requires a lot of work in order to define the exact parameters affecting these coefficients. Although, these coefficients are highly dependent on the flow parameters such as Re number and Ma number.

One of the greatest advances if fluid mechanics was done by Ludwig Prandtl (1875-1953). Based on his report, only a thin region on the surface of the body is important since the viscous forces are only important in that region called boundary layer, and outside the flow will be the same as if the fluid was inviscid.

Since the viscosity of the fluid is same everywhere, only the relative importance of the forces matter and viscous forces in this region play a significant role. Based on experiments and computations, the lager the Re number the smaller the region of the flow field in which viscous effects are important.



The velocity gradients within the boundary layer and wake regions are much larger than those in the remainder of the flow field. Since the shear stress is the product of the fluid viscosity and the velocity gradient. It follows that viscous effects are confined to the boundary layer and wake region.

Boundary Layer characteristics
Based on the hypothesis (proved) of Prandtl, if the Reynolds number is large enough, the region in which the viscous stresses are important will be in the small boundary layer region. In the boundary layer, the velocity of the fluid (relative velocity) starts from zero to the upstream fluid velocity. Outer of the boundary layer region, the fluid velocity profile only experiences a small change due to the presence of the body and the velocity gradients are small relatively and it can be assumes that the fluid’s behavior is as if the fluid was inviscid.

If we assume two parallel rectangular fluid elements which one of them travels through the boundary layer and the other one goes from the outer region, the first element will experience distortion which is because of the velocity gradient available in the boundary layer. Also, the particle moving in the boundary layer will start to rotate whilst the latent particle will not experience such phenomena since there is no velocity gradient in that region. The flow is called rotational inside the boundary layer (nonzero vorticity) and irrotational outside of the boundary layer (no vorticity).

By traveling forward in boundary layer, at somewhere fluid elements will experience great distortions due to transition from laminar flow regime to turbulent flow regime. One of the characteristics of the turbulent flow regimes is the complex mixing of fluid particles which occur in ranges from fluid particles to shape characteristic length. In the laminar flow regime, mixing is done only in by the molecular diffusion. The phenomena of transition from laminar flow regime to turbulent flow regime, itself is a highly complex and hard to understand topic. This transition does not occur at a constant location and depends on the surface roughness and instabilities of the upstream flow which may be ranged based on Reynolds number from $$2 \times 10^5$$ to $$3 \times 10^6$$  for a flat plate.

From mathematical and physical point of view, it is not possible to define a sharp edge to separate the boundary layer from the whole flow region and velocity tends to upstream flow velocity as we go further far from of the body in perpendicular direction of the upstream flow velocity. Because of that, boundary layer thickness is defines as a length where u = 0.99U, or in terms of mathematical representation,

$$ \displaystyle y = \delta \quad, \quad u = 0.99U $$

If we assume inviscid flow flowing over the object, to correct the mass flow that is decreased due to the velocity profile change due to viscosity, we must define an extra thickness, such that,

$$ \delta ^* b U = \int\limits_{0}^{\infty} (U-u) b dy $$

Boundary layer momentum thickness
Another thickness which is defined to take into account the momentum flux decrease because of the velocity profile deficit, is called boundary layer momentum thickness;

$$ \int\limits_{}^{} u (U-u) b dy = \rho b\int\limits_{0}^{\infty} u (U-u) b dy \quad \quad \text{momentum flux deficit} $$

$$ \rho b U^2 \Theta = \rho b\int\limits_{0}^{\infty} u (U-u) b dy $$

or

$$ \Theta =\int\limits_{0}^{\infty} \frac{u}{U} (1-\frac{u}{U}) dy \quad \quad \text{Boundary layer momentum thickness} $$

Prandtl/Blasius boundary layer solution
For steady, two-dimensional laminar flows with negligible gravitational effects, Navier-Stokes equations are,

$$ u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}=- \frac{1}{\rho} \frac{\partial p}{\partial x}+\nu(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}) $$

$$ u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=- \frac{1}{\rho} \frac{\partial p}{\partial y}+\nu(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}) $$

and continuity equation as,

$$ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 $$

and boundary conditions are

$$ \displaystyle u = 0, \quad \text{on body surface} $$

$$ \displaystyle u=U, \quad \text{far from the body surface} $$

two assumptions can be made since the thickness of the boundary layer is very small

$$ \displaystyle v \ll u \quad and \quad \frac{\partial}{\partial x} \ll \frac{\partial}{\partial y} $$

implying that the velocity component in the direction normal to the shape is much smaller than the velocity component parallel to body surface and rate of change of any parameter normal to the surface is much bigger than the rate of change of any parameter parallel to the body surface.

By applying these assumptions to the general Navier-Stokes equations, we obtain;

$$ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 $$

$$ u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}=\nu \frac{\partial^2 u}{\partial y^2} $$

There is no pressure gradient terms is seen in the equation. so the pressure gradient normal to the body surface is equal to zero and also pressure gradient in the direction of the upstream velocity is also equal to zero, which implies no change in pressure in the parallel and normal direction to the surface. hence, this equations only show a balance between inertial forces and viscous forces. Solving partial differential equation is not a easy work, so it would better to turn it into a Ordinary differential equation. Blasius turned the PDE into an ODE to solve the equation. It can be shown that the non-dimensional velocity profile can be in the following form,

$$ \frac{u}{U}=g(\frac{y}{\delta}) $$

By using the order of magnitude method (in this method you assign the maximum possible value to parameters to obtain the relative importance of parameters in the equation), it would emerge that,

$$ \delta \sim (\frac {\nu x}{U})^{\frac{1}{2}} $$

Based on this equation, there is a balance between viscous forces and inertial forces. by defining two non-dimensional parameters $$\eta = (\frac {U}{\nu x})^{\frac{1}{2}} y$$ (non-dimensional similarity variable) and $$\psi = (\nu x U)^{\frac{1}{2}} f(\eta)$$ where $$f(\eta)$$ is an unknown function. Since,

$$ u = \frac{\partial \psi}{\partial y} $$

$$ u = - \frac{\partial \psi}{\partial x} $$

if we express velocities in our new defined parameters, we have

$$ u=Uf'(\eta) $$

and

$$ v=(\frac {\nu U}{4x})^{\tfrac{1}{2}}(\eta f^\prime-f) $$

by substituting this parameters in our simplified Navier-Stokes equations, new equation will be stated as

$$ 2f'+ff=0 $$

with the boundary conditions as,

$$ f=f'=0 \quad at \quad\eta = 0\quad and \quad f'\rightarrow 1 \quad as \quad \eta \rightarrow \infty $$

Now, the PDEs are transformed to the Ode which only one variable \eta exist. Although analytical to this problem is also not easy, but this ODE can be solved using the computers. According to the Table 1, numerical solution of the ODE which is called Blasius solution, concludes that $$u = 0.99U$$ where $$\eta = 5.0 $$ and,

$$ \delta=5 \sqrt{\frac{\nu x}{U}} $$

or

$$ \frac{\delta}{x}=\frac{5.0}{\sqrt{Re_x}} $$

$$Re$$ is defined as $$Re = \frac{Ux}{\nu}$$, and displacement and momentum thicknesses are then defined as,

$$ \frac{\delta^*}{x}=\frac{1.721}{\sqrt{Re_x}} $$

$$ \frac{\Theta}{x}=\frac{0.664}{\sqrt{Re_x}} $$

Since by knowing the velocity profile, calculation of the wall shear stress based on $$\tau_w = \mu(\frac{\partial u}{\partial y})_{y=0}$$ will be easy. According to the Blasius solution,

$$ \tau_w =0.332U^{\frac{3}{2}} \sqrt{\frac{\rho \mu}{x}} $$

By moving forward in the $$x$$ direction, the thickness of the boundary layer increases and hence it was expected to decrease the wall shear stress. this behaviour is also seen on the above equation and also wall shear stress varies with $$U^{\frac{3}{2}}$$.

Unless for simple geometries and assumptions to simplify the governing equations, I is impossible to solve the Navier-Stokes equations to find the shear stress distribution and pressure distribution on different shapes. CFD is one of the powerful tools to calculate theses distribution on shapes nowadays. This is while, most of the available data regarding the drag forces on bodies are result of numerous experiments by wind tunnels, wind tunnels and etc.

As described earlier, it is better to express the parameters in a non-dimensional form. So we defined the drag coefficient as:

$$ \displaystyle C_D=\frac{D}{\tfrac{1}{2} \rho\ U^2 A} $$

Which $$C_D$$ itself is a function other parameters such as Reynolds number, Mach number, surface roughness, Froude number and etc.

Characteristics of Drag
As shown in the drag equation, the drag force on a surface is composed of two parts, one is because of shear stresses and the other one is because of the pressure forces.

$$ \displaystyle D=\int_{} dF_x=\int_{} p \cos\theta dA +\int_{} \tau_w \sin \theta dA $$

The friction drag which is the portion of the total drag caused by wall shear stress highly depends on the orientation of the surface. The second part which constitutes the total drag is called Pressure drag. This part is also called form drag which is is strongly dependent on the geometry and form of the body that the pressure acts on. For flat plate, because there was no pressure forces, total drag on surface constituted of shear stress. On the other hand, if the surface was perpendicular to the flow, shear flow would not have any effect on the drag force.

The friction drag on a flat plate with length L and depth of b is calculated as,

$$ D_f=\frac{1}{2} \rho U^2 bl C_{Df} $$

Where the values for $$C_{Df}$$ are calculated based on experiments and shown in figure and Table

As shown on the above figures, surface roughness does not have any effects for laminar flow regime but considerably affects the turbulent flow regime.

Coefficient of drag for some of the popular geometries are shown in figure.



Characteristics of Lift
Another force that any immersed body may experience due to moving in fluid is Lift. The existence and magnitude of the Lift force is highly dependent on the shape of the body and the flow structure. For non-symmetrical objects or the cases in which the flow structure near the object is not symmetrical, Lift force is exerted on the body. Analyzing the Lift force is of much of importance, in some cases scientists and engineers are working to increase it such as in airfoils (to reduce the fuel consumption in airplanes), and in some cases to increase it(in cars to prevent from decreasing the contact area of tires with ground which will decrease traction and cornering ability.).

Based on the equation below the Lift force can be calculated if the pressure distribution and shear stress distribution on the body is known, which is a difficult work.

$$ \displaystyle L=\int_{} dF_y=\int_{} -p \sin\theta dA +\int_{} \tau_w \cos \theta dA $$

generally, the lift force is given by the Lift coefficient as,

$$ \displaystyle C_L=\frac{L}{\tfrac{1}{2} \rho\ U^2 A} $$

which this coefficient generally is calculated using experimental, computational and analytical methods. Lift coefficient as Drag coefficient is a function of different parameters as ,

$$ C_L=f(Re, Fr, Ma, \epsilon /l, Shape, \alpha) $$

Although these parameters all all together important but the most important parameter is the shape of the object, Angle of attack is also another important parameter that affects the coefficient of Lift.

How do airplanes fly?
As mentioned in the previous, an object moving in the fluid experiences two forces if it is considered in 2 dimensions. One of them drag force and second the lift force. By manipulating these forces, humankind has achieved certain ground breaking goals. Airplanes as machines that can flight are one of the main goals. Because of complex 3D shape of the airplane, they experience a 3D force. 2 main forces are Drag and Lift. Here we will discuss the flight process of the airplanes in a simple way.

the airplanes have different types, with different thrust-energy supplied, however their lift process is the same in most of the cases.



here the airplanes which use jet engines for thrust are discussed briefly. Engines take the air inside and at the first step the compress the air to reach a certain value of oxidizer density (the air consists of about 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide, and other gases) and mix it with fuel and then burn the mixture. the products of the combustion have high velocities and high temperatures. these gaseous products pass through the turbine blades at the turbine section of the jet engine and cause the blades to move and hence produce energy as electricity and energy for moving of the compressor. then exhaust gases release into the air.





these exhaust gases have high velocities yet and this difference in momentum between the inlet and the outlet of the jet engine, causes a force to force the airplane in the opposite direction of the inlet air. this force wıll cause the aıplane to move faster and hence the air will pass the wings with airfoil shape faster than before. As mentioned in the previous section, the Drag force and the Lift force are functions of different parameters such as $$Re$$ number and shape of the airfoil. Since the airfoils are non-symmetrical, a Lift force will generate and push the airfoils and hence the whole airplane upward.



You can see an animated video about the airplane's flight here...