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Old Nov 15th 2015, 09:21 AM   #1
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Wind power experiment questions

Hello I'm working through questions in a book and the task is :

Plan an investigation to evaluate wind power as an energy source with equipment:

model wind turbine
multimeter to measure the voltage generated
anemometer to measure wind speed
hair dryer to generate wind power, set on cold
metre rule to measure distance.

Questions & My Answers:

(i) Explain the aim of your investigation.

Aim of investigation is to find out how increasing or decreasing wind speed changes the power output of the wind turbine. Hence decide if a light wind (hair dryer at a low setting) in an average town would be a good alternative fuel OR if only high winds (on a hill on in the sea) would generate adequate power.

(ii) Explain what you will measure.

The turbine is effectively a battery so I would find out the voltage output of the turbine by connecting the voltmeter in parallel with it + connect an ammeter in series to find the current supplied. With voltage and current know I will know the power generated by the turbine from the formula P=IV.

I would measure the distance of hairdryer from turbine. The shorter the distance the more wind power (if i can't just control the wind with a button on the side of the dryer). So I will get an idea of the minimum wind speed required to generate a current from the turbine.

(iii) Explain the number and range of readings that you will take.

Repeat 3 times (DON'T KNOW WHY - just sounds right).
Measure the wind speed with the anemometer.

(iv) Explain the independent variable.

This is the wind speed or the distance of the hairdryer from the turbine propellors.

(v) Explain the dependent variable.

The voltage output from the turbine.

(vi) Explain the control variables.

Size of the propeller blades.

(vii) Explain how you will make your experiment a fair test.

Repeat the experiment 3 times ???
Use the same sized blades on the turbine for each change of wind speed.


viii) Draw out a results table that you would use in your investigation.

Table used in investigation would include list of voltages produced, current flowing, distance of hairdryer from propellers.

(ix) Write an evaluation identifying aspects of your experiment where modifications are possible.

Include a light bulb in a series circuit (running from the generator to the bulb and back to generator)to see how the bulb lit up with different winds.



Please can someone comment on my answers and advise what I've got wrong and where to improve. Thank you. This is not a homework.
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Old Nov 15th 2015, 03:46 PM   #2
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1) What about wind direction?

2) You need an electrical load connected to the generator output to measure power.
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Old Nov 16th 2015, 03:27 AM   #3
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Thanks for comment Studiot

Originally Posted by studiot View Post
1) What about wind direction?
I'm assuming the hairdryer is pointed towards the turbine propellers but I think (after reading your comment) that I should make this obvious in the write-up.

Originally Posted by studiot View Post
1) What about wind direction?

2) You need an electrical load connected to the generator output to measure power.
I'm glad you mentioned 'load'. I've heard this word before but never knew what it meant. Just did a bit of research and here's what I think:

The turbine is effectively a battery so I would connect the output (+ve terminal) to the -ve terminal of the turbine. Then connect a voltmeter in parallel with the wires (to measure the voltage generated) + connect an ammeter in series to find the current supplied.

BUT having a wire leading from +ve to -ve terminal without an electrical device (THIS IS THE LOAD???) (light bulb, beeper, motor, etc.) would lead to a high rate of charge flow and will cause a short circuit.

With charge flowing rapidly between terminals (without the LOAD in-between) means the rate at which energy would be consumed would be high, and would heat the wires excessively (and they may melt) and drain the battery of its energy quickly.

Okay so with the BULB in the circuit I can measure voltage and current supplied by the turbine. But isn't that an under-estimate of the power generated by the turbine because some of the power is used by the bulb?

How do I know what load (bulb wattage??) is high enough to place i the circuit so that it doesn't burn out? By LOAD do you mean we need something that offers RESISTANCE to the flow of current in the circuit?



With voltage and current know I will know the power generated by the turbine from the formula P=IV.
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Old Nov 16th 2015, 04:47 AM   #4
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Originally Posted by studiot
1) What about wind direction?
I'm assuming the hairdryer is pointed towards the turbine propellers but I think (after reading your comment) that I should make this obvious in the write-up.
Wind direction changes (and even reverses) all the time in most places so this is very important for wind driven installations.

I'm glad you mentioned 'load'. I've heard this word before but never knew what it meant. Just did a bit of research and here's what I think:
Some basic facts about electrical circuits and installations you need to know.

a) The purpose of an electrical source such as a battery or generator is to provide electrical energy for use by some load.
Some sources are safe when 'not on load' (no load is connected) some are not, that is why some wind turbines are locked so they do not turn in very high winds.

b) The purpose of the load such as a light bulb (good choice), electric heater or electric motor, is to use (convert) the electric energy to some other form of energy such as mechanical, light or heat.

c) I can't stress this point enough.

Unless the demand from the load is greater than the source can supply the current and power drawn is determined by the load, not by the source.
If the demand is too great the load determines the demand up to the level where supply is limited by the source's ability to supply.

Your same mains wall socket outlet supplies a 3kw space heater and the charger to your laptop correctly because of this.
The load only draws what it needs.
The source cannot impose a certain current and voltage on the load.
It can impose one or the other, but not both.

So rethink your experiment in the light of this information.

But well done you are thinking and learning.



One further question to investigate.

Does your turbine generator output alternating voltage/current or direct voltage/current (AC or DC) ?

Last edited by studiot; Nov 16th 2015 at 04:53 AM.
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Old Nov 16th 2015, 05:17 AM   #5
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Thanks Studiot

Originally Posted by studiot View Post
a) The purpose of an electrical source such as a battery or generator is to provide electrical energy for use by some load.
Thanks. Now I know a LOAD is required in a circuit to use the current supplied by the source. BUT WHY? Why can the current just flow around and around the circuit? Why will the wires heat up?

If my turbine say produces 100 Watts of power and if my LOAD is a low 'wattage' bulb (2 W) (I think this means a bulb of low resistance) then with all that unused power (98 W) will the wires still heat up and melt / short circuit?
How do you determine the minimal LOAD that you must place into a circuit for safe use?




Originally Posted by studiot View Post

c) Unless the demand from the load is greater than the source can supply the current and power drawn is determined by the load, not by the source.
If the demand is too great the load determines the demand up to the level where supply is limited by the source's ability to supply.
Then the load will be running ineffectively e.g the bulb will not be as bright as it should be or the mobile phone will not charge up so fast??




Originally Posted by studiot View Post

Your same mains wall socket outlet supplies a 3kw space heater and the charger to your laptop correctly because of this.
The load only draws what it needs.
The source cannot impose a certain current and voltage on the load.
It can impose one or the other, but not both.
So a 240V mains socket supplies a 3kW heater and a laptop. What prevents the laptop taking in excessive current and melting? Is it the transformer fitted to the plug? Is the size of transformer calculated by the manufacturer?



Originally Posted by studiot View Post

Does your turbine generator output alternating voltage/current or direct voltage/current (AC or DC) ?
This I haven't got a clue about. Don't know why something generates AC or DC ..... I though it just happened to be AC or DC
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Last edited by Hero; Nov 16th 2015 at 05:53 AM.
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Old Nov 16th 2015, 05:43 AM   #6
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This I haven't got a clue about. Don't know why something generates AC or DC ..... I though it just happened to be AC or DC
Well yes your generator will happen to be AC or DC.
DC generators are called dynamos
AC generators are called alternators or just generators.

You need to know the difference because you have different voltmeters and ammeters for AC and DC.
Or you use different settings on meters that can measure both.

Isn't this is a good learning exercise?

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Old Nov 16th 2015, 05:50 AM   #7
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Originally Posted by studiot View Post
Isn't this is a good learning exercise?
It is indeed. I've just gone and but BIG QUESTION MARKS in the parts where i'm still confused. Hope you can help me further.
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Old Nov 16th 2015, 05:56 AM   #8
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It absolutely fundamental that you understand this statement, which was why I made it bold.

the current and power drawn is determined by the load, not by the source.
It is also the commonest misconception amongst beginners to think that the source produces any power whatsover in the absence of a load.

It does not, it produces a big fat zero.

If you short circuit it that is just connect wires across the teminals you will likely dmage or destroy the generator.

If you leave it open circuit that is no connection across the terminals it is working on no-load condition and produces no power.
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Old Nov 16th 2015, 06:04 AM   #9
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Originally Posted by studiot View Post
It absolutely fundamental that you understand this statement, which was why I made it bold.
"CURRENT & POWER AND DRAWN iS DETERMINED BY THE LOAD, NOT BY THE SOURCE"; Studiot 2015

From this statement I understand that if I connected a low wattage bulb (say 2 W) directly to the 240 V mains supply with a wire. Then the bulb will draw in current and glow BUT will not explode/blow out with all that current from the mains supply. I do not need a transformer connected to the plug to reduce the current entering the bulb.

I sense I've made many mistakes in my statement but don't know where exactly
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Last edited by Hero; Nov 16th 2015 at 06:07 AM.
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Old Nov 16th 2015, 06:19 AM   #10
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The source produces a voltage, whether there is a load there or not.

But voltage is not power.

If you connect a load current will flow and the source will attempt to maintain its voltage across the load.

As you reduce the resistance of the load you incease the load.
This seems odd but this is correct.

When you increase the load by reducing its resistance, one of two things will happen.

Initially more and more current will be drawn as the resistance decreases and the source maintains the supplied voltage.

Thus more and more power is drawn as the load increases (greater curent tims constant voltage(.

Then

1) Either the current will be so great it destroys or damages the load

or

2) The source cannot supply any more current and its voltage starts to fall. Prolonged running at the maximum current may or may not damage the source.
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