Figure 11 displacement thickness [2]
If a free stream of velocity \(U_{\infty}\) is effectively displaced by\(\delta^{*}\). The loss of the mass flow rate per unit time is given by:-
\(\dot{m}=\rho_{\infty}U_{\infty}\delta^{*}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.1\right)\)
The loss of the mass flow rate per unit time is given by:-
\(d\dot{m}=\rho\left(U_{\infty}-u\right)\text{dy}\)
Then the total mass flow rate per unit time is:
\(\dot{m}=\int_{0}^{\delta}{\rho\left(U_{\infty}-u\right)\text{dy}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.2\right)\)
By equation the above equations, we get
\(\rho_{\infty}U_{\infty}\delta^{*}=\int_{0}^{\delta}{\rho\left(U_{\infty}-u\right)\text{dy}}\)
If the fluid flow is assumed to be incompressible i.e, the density remains constant, and then the above equation will be written as follow;
\(\rho_{\infty}U_{\infty}\delta^{*}=\int_{0}^{\delta}{\rho\left(U_{\infty}-u\right)\text{dy}}\)
\(\delta^{*}=\int_{0}^{\delta}{\left(1-\frac{u}{U_{\infty}}\right)\text{dy}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.3\right)\)
In the fluid mechanics it is recommended to transform the equations into non-dimensional form. In this way;
Let\(\overset{\overline{}}{y}=\frac{y}{\delta}\text{\ \ \ \ \ and\ \ \ \ }\overset{\overline{}}{U}=\frac{u}{U_{\infty}}\)we will get;
\(\delta^{*}=\int_{0}^{1}{\left(1-\frac{u}{U_{\infty}}\right)\text{dδ}\overset{\overline{}}{y}}\)
\(\frac{\delta^{*}}{\delta}=\int_{0}^{1}{\left(1-\overset{\overline{}}{U}\right)d\overset{\overline{}}{y}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.4\right)\)
momentum thickness
Now let us formulate an expression to the momentum thickness;
The loss of momentum flow of the free stream equals to:
\(=\rho_{\infty}U_{\infty}\theta\ U_{\infty}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.5\right)\)
The total loss of momentum is given by;
\(\int_{0}^{\delta}{\rho\left(U_{\infty}-u\right)\text{\ u\ dy}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.6\right)\)
By equating the above two equations, we get
\(\rho_{\infty}U_{\infty}\theta\ U_{\infty}=\int_{0}^{\delta}{\rho\left(U_{\infty}-u\right)\text{\ u\ dy}}\)
For incompressible flow;\(U_{\infty}\theta*U_{\infty}=\int_{0}^{\delta}{\left(U_{\infty}-u\right)\text{\ u\ dy}}\)
\(\theta=\int_{0}^{\delta}{\left(U_{\infty}-u\right)\frac{u}{{U_{\infty}}^{2}}\text{\ dy}}=\int_{0}^{\delta}{\left(1-\frac{u}{U_{\infty}}\right)\frac{u}{U_{\infty}}\text{\ dy}}\)
\(\theta=\int_{0}^{\delta}{\frac{u}{U_{\infty}}\left(1-\frac{u}{U_{\infty}}\right)\text{\ dy}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.7\right)\)
\(\frac{\theta}{\delta}=\int_{0}^{1}{\overset{\overline{}}{U}\left(1-\overset{\overline{}}{U}\right)\text{\ d}\overset{\overline{}}{y}}\)
kinetic energy thickness
Finally, an expression for the kinetic energy thickness will be after derivation something like this;
\(\frac{\delta^{**}}{\delta}=\int_{0}^{1}{\overset{\overline{}}{U}\left(1-{\overset{\overline{}}{U}}^{2}\right)d\overset{\overline{}}{y}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(5.8\right)\)
Momentum equation of hydrodynamics boundary layer over a flat plate
First of all, let us develop the governing equation of the hydrodynamics boundary layer.
The navier-stoke equation in x-direction that derived in section 3;
\(\rho\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\right)=-\frac{\partial P}{\partial x}+\rho g_{x}+\mu\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}}\right)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.34\right)\)
Assumptions
  1. The flow is steady and the fluid is incompressible.
  2. The viscosity of the fluid is constant
  3. The pressure variation in the direction perpendicular to the flow is negligible.
  4. Viscous – shear forces in the y-direction is negligible.
  5. Fluid is continuous both in time and space.
After applying the assumptions mentioned before, we get
\(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu\frac{\partial^{2}u}{\partial y^{2}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(6.1\right)\)
The equation mentioned above is representing the equation of motion of the momentum equation for hydrodynamics boundary layer.
The next pages, two mathematical solutions will be used to solve the momentum equation for hydrodynamics boundary layer. One of the solutions is approximate which is called Von – Karman and the second is the Blasius Exact solution.
  1. Von-Karman Momentum Integral Equation
  2. Derivation of the Momentum Integral Equation
Let us consider control volume of ABCD as in Figure 12 below