Wind Turbine Blade Experiment: What is the Optimal Number for Maximum Voltage?

AI Thread Summary
The discussion revolves around an experiment where students measure voltage output from a model wind turbine with varying blade counts. The optimal number of blades for maximum voltage output in the experiment is suggested to be between four to six, while full-sized turbines typically use three blades for efficiency reasons. A key question raised is why the optimal blade count for the experiment differs from the standard used in larger turbines. Possible explanations include cost-effectiveness and the balance between power and voltage output. The conversation highlights the complexity of turbine design principles, which may be challenging for the target age group to fully grasp.
CWatters
Science Advisor
Homework Helper
Gold Member
Messages
10,544
Reaction score
2,324
Not really a homework question but this is probably the best place to post it.

I'm interested in what people think about the following experiment and the wording of one of the follow up questions that came with it. It was set as part of UK GCSE Physics coursework for the 14-15 age group.

1. Homework Statement


The class was instructed to conduct an experiment that involved setting up a desk fan to blow air at a model wind turbine. The model turbine was connected to a small DC motor and that to a voltmeter. They were asked to change the number of blades on the turbine and plot a graph of voltage against number of blades...

GCSE Physics2.jpg


Three related follow up questions were:

3) How many blades do most full-sized modern wind turbines have?
4) What number of blades gives the highest potential difference in the experiment?
5) The number of blades which gives the highest potential difference in this type of experiment is sometimes different from the answer to Question 3. Suggest a reason for this.

Homework Equations



None. (This is a basic physics class not an aerodynamics class!)

The Attempt at a Solution



First two are straightforward.

Q3) 3 blades.
Q4) 6 blades (or perhaps "4 to 6 blades gives the highest potential difference").

My son asked me for advice on how to answer Q5. I can think of possible answers but they aren't expected to know these:

As I recall 3 blades are used for quite technical reasons to do with vibration or resonance but that kind of answer is too advanced for their age range.

Larger turbines are more efficient so need fewer blades? They aren't expected to know that larger blades are more efficient.

More than 3 blades would make the turbine more expensive for little gain? However their experimental data suggests 4 blades would probably be the optimum number.

Large turbines are optimised for maximum power not maximum voltage? (I think).

Is this just a bad question or is there a right answer for this age group that I'm missing?
 
Physics news on Phys.org
CWatters said:
More than 3 blades would make the turbine more expensive for little gain? However their experimental data suggests 4 blades would probably be the optimum number.
That is the one answer I can think of that might be reasonable to expect. And is it not supported by the graph? If we assume that the primary cost of the turbine is the blades, we want a high voltage to blade ratio. It falls off markedly after 3.
 
  • Like
Likes CWatters
  • Like
Likes CWatters
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top