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I am studying about wind turbine and turbine.

I have some questions, but first, I explain some cases as follow:

Low velocity air over an airfoil at a zero-incidence angle does not initiate flow separation. If the air velocity slowly increases, the flow separation starts from the blade trailing edge (TE). If the velocity or the incidence angle increases further, the separation point or stall point moves towards the blade leading edge (LE). If the stall point can be pushed towards the TE, the airfoil performance or the turbine blade performance increases as the attached flow transfers more energy to the blade.

Now, I want to know that how I can claim that efficiency of turbine increase up to given angle of attack?

The efficiency increase with increasing angle of attack while flow separation also increase with increasing angle of attack. I don't understand this case!

With regards to above explanations, if angle of attack increases; separation point or stall point moves towards the blade LE and the turbine blade performance decreases as there is low attached flow. I know that when angle of attack increases, tangential force also increases and for this reason, the efficiency of turbine increases but I want to know about increasing efficiency from the point of flow separation!

How can these two things (increasing efficiency and increasing separation) be linked together?

I am grateful to all who guide me about these cases and if it is possible, please place useful links and papers related to these cases.


In fact, I am studying more about the Ws turbine. In my investigations, the air velocity(inlet velocity or axial velocity) is constant and rotational velocity(r $\omega$) is changed and in other words, rotational speed is decreased in my calculations. this method is performed in the following paper. In this paper, There is the used equation for efficiency.

A comparison of computational...

$\eta = \frac{T \cdot \omega}{\Delta p_\mathrm{t}\cdot Q}$

$T =$ shaft torque

$\omega =$ angular speed

$U = r \cdot \omega =$ rotational speed

$\Delta p_\mathrm{t} =$ total pressure drop across the turbine

$Q = V_\mathrm{inlet} \cdot A$

$A = \pi\left(R_\mathrm{casing}^2-R_\mathrm{hub}^2\right) =$ through flow area

$\frac{V_\mathrm{inlet}}{U} =$ flow coefficient

Also, please see the following figure:

blade profile

In fact, The various values of the flow coefficient were achieved by varying the rotational speed of the rotor at a constant value of the axial velocity, which are similar with those performed by Kim et al. [1,2], Takao et al. 3 and Zahari et al. [4].

1 Kim TH, Setoguchi T, Kinoue Y, Kaneko K. Effects of blade geometry on performance of Wells turbine for wave power conversion. Journal of Thermal Science 2001;10(4):293–300.

2 Kim TH, Setoguchi T, Kaneko K, Raghunathan S. Numerical investigation on the effect of blade sweep on the performance of Wells turbine. Renewable Energy 2002;25:235–48.

3 Takao M, Setoguchi T, Kinoue Y, Kaneko K. Wells turbine with end plates for wave energy conversion. Ocean Engineering 2007;34:1790–5.

[4] Zahari Taha, Sugiyono, Sawada T. A comparison of computational and experimental results of Wells turbine performance for wave energy conversion. Applied Ocean Research 2010;32:83–90.

Now, I am grateful that guide me about these cases and discuss together about these cases.

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You are describing a very interesting engineering problem.

The requirement of a turbine is to convert fluid-energy into mechanical energy over a range conditions. Since the flow around the airfoils is governed by the (elliptic) Navier-Stockes-Equations, an easy answer to your question is not possible in general.

Only in certain edge cases it is possible to simplify the system in such a way that simple answers are possible.

Before I explain what goes into some of those cases, I'd like clarify some words first.

  • Efficiency is usually the ratio between the theoretical maximum and the actually achieved value. The above mentioned equation assumes that the energy extraction is only visible in a total pressure drop, $\Delta p_\mathrm{t}$. For low Mach-Number flows that is a reasonable assumption ($\mathrm{Ma}\ll 0.3$). If higher speeds are achieved a total ethalpy based definition of efficiency is needed.
  • Angle of Attack (AoA) describes the angle between the fluid velocity vector and a somewhat arbitrary line with respect to the airfoil. Since Wells turbines usually feature a symmetric profile the reference line is the symmetry-axis. A change in angle of attack can be achieved by three things:
    • rotating the blades
    • change of rotational speed of the rotor
    • change of windspeed.

As I said there is no easy answer to your question. To delve further into your question, we will tackle it in two steps. First, we have a look on the aerodynamic characteristics of an airfoil. Then, we will look at the turbomachinery part of the problem.

Step 1. Wells Turbines often use symmetric 4 digit NACA profiles. The aerodynamic characteristics of such airfoils can easily be calculated using webtools. The following curves were created using Air Foil Tools. NACA0021-loverd $c_l$ is the lift-coefficient which is defined as: $\frac{F_\mathrm{lift}}{\frac{1}{2}\rho u^2} = c_l$. The drag-coefficient ($c_d$) is defined in equal manner. The angle of attack is defined as alpha. Lift and Drag both result from the aerodynamic forces acting on a profile (see Wikipedia for additional illustrations and explanations).

aeroforces.

There is a lot of information contained in the three curves above. The two messages of interest here are:

  • Within a certain range of AoA the lift changes proportional to the AoA.
  • Within the same range of AoA the drag is considerable lower but increases over-proportional.

Step 2. A Wells Turbine consists (in its original form) a single blade row, which is rotating at a certain speed. The rotor-inflow-angle is a result of the vector addition of the axial inflow velocity and the rotational speed of the turbine rotor. The AoA is the difference of the inflow angle and the blade setting (i.e. angle between the symmetry-axis of the blade and a reference axis). The vector representation of the inflow is called a velocity triangle. This means for every pair of axial-velocity and circumferential-velocity the blade will have a different inflow-angle. And, because of the setting of the turbine blade it will have a different AoA.


When we define a simplified version of $\eta$ as:

$\eta = \frac{\text{WORK}}{\text{WORK }+\text{ LOSS}}$

We can see, that as long as the losses are small and the change with AoA-variation is small (see loss-curves) the change in work will dominate the equation. This results in a higher efficiency despite the increased losses because the work which can be extracted from the fluid increases faster than the loss increases. But: As it can be seen from the $c_l/c_d$-curve this is only true for a certain AoA-range. Outside of this range the separations will create over-proportional more loss.

You are absolutely right to be confused. Because it seems to be strange that despite a larger separation (which is usually results in a loss) the efficiency actually increases. However, this is only true when the separations are small. Therefore correlation between separation-area and efficiency is not a causality.

I would think of this problem in three stages. In the first stage we look at the blade profile. Here the losses increase with higher AoA. In the second stage we also look at the lift, and see that it also increases and increases faster than the losses. In the third stage we see from the definition of the efficiency that the combination of lift and drag mathematically allow for the parallel increase of efficiency and losses if the extracted work increases sufficiently. But this statement is only true for a certain range of AoA. Outside of this range the losses are bigger and grow faster and the efficiency will not increase anymore and even decrease.

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    $\begingroup$ rul30, I wait you that come here until guide me. Thanks $\endgroup$ – user19061 Sep 13 '15 at 11:15
  • $\begingroup$ Wow, @user19061 thanks for the additional information. $\endgroup$ – rul30 Sep 13 '15 at 11:58
  • $\begingroup$ Hi rul30, Can you explain about " over-proportional"? I don't understand this word. Thanks, $\endgroup$ – user19061 Sep 15 '15 at 14:22
  • $\begingroup$ Proportional means (as far as I know) a constant ratio between two things. When something is over-proportional this ratio is getting bigger and bigger. Under-proportional means the ratio gets smaller. $\endgroup$ – rul30 Sep 15 '15 at 20:45

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