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According to the following figure, boundary layer becomes progressively thicker with downstream distance x . Why?
I have an other question as follows:
When turbine blade thickness increases, what dose happen about blade boundary layer thickness? Dose BL thickness increase about blade?
I am grateful that guide me about this question and if it is possible, please give me useful links and papers about these questions.
The greater the length of the flat plate, the more contact area there is with the fluid. Consequently, the fluid near the start of the plate has had less viscous friction forces acting on it than the end of the plate. So, the end of the plate has more fluid slowed down than the front. The width of the boundary layer must necessarily increase with the amount of fluid that had been slowed down by viscous friction forces with the wall. The overall motion of the fluid just pushes more of the slower fluid downstream.
It is very important to mention in your case(given in figure), It not only increasing, but growing like a $parabolic$ profile(dotted line).
Adding to paul,
Imagine boundary layer growth as a momentum diffusion process. (air loosing momentum but not the mass in the boundary layers.) momentum diffusion is attributed to the viscosity of the fluid.
To understand this process, let us take the momentum balance for your problem, It looks like 1D, So,
$ u \frac{\partial u}{\partial x} = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \nu\frac{\partial^2 u}{\partial^2 x}$
we can assume $\nu\frac{\partial^2 u}{\partial^2 y}$ is smaller (imagine you are rubbing your hands one against another, one of them is retarted against another in the rubbing direction.)
also your geometry dont have any curvature. So $- \frac{1}{\rho} \frac{\partial p}{\partial x} = 0$.
now your momentum balance eqn will become,
$ u \frac{\partial u}{\partial x} = \nu\frac{\partial^2 u}{\partial^2 x}$ it is the balance beween convection of your mass by $u$ and diffusion. !!!
It is mathematically interesting equation and restrict yourself for the region $\delta$- boundary layer thickness.
you get good agreement between experiment and this theory, when you assume $\delta = \sqrt{\frac{\nu x}{ U }}$. (this is called similarity solution in boundary layer theory, credited to prandtl). So your profile grows as $\delta \sim x^{\frac{1}{2}}$.
Coming to second question, now you have curvature on the blades. So, $- \frac{1}{\rho} \frac{\partial p}{\partial x} \neq 0$. If you know the velocity profile as a function of $x$ around your turbine blade(one can get through potential flow), then, using bernoulli's equation, convert pressure term interms of velocity and function of $x$. (bernoulli's equation appalicable above the edge of the boudray layer $\sim$ inviscid region)
Once velocity polynomial relation is substituted in the momentum balance equation for pressure, work out the similarity analysis (though complicated)to get your boundary layer profile.
To my knowledge,
Tha boundary layer profile over the $isolated$ turbine blade grows very slowely upto 1/4 th of the chord (favourable pressure), and grows faster after(adverse pressure) that if the boundary layer is atatched till the trailing edge.
(adverse pressure region is pron to separation, where boundary layer theory breaks).
In a cascade turbine you have to find the effect of adjacent blades. still you can proceed with the potential flow theory to get velocity in terms of x co-ordinates.
