Wind power experiment questions

Nov 2015
38
0
Been watching Youtube videos to try and get my head around current and voltage and resistance. Semi-confused. Here is what I understand:

The source produces a voltage, whether there is a load there or not.

But voltage is not power.
A battery provides potential to do work. If there is a high potential difference (PD) between the two terminals of the battery (called voltage) then lots of current can flow.

Power is just a way of assigning a value to the battery by saying P = IV

If battery A is assigned a value of 100 W then this means that the chemicals inside it allow it to provide 20 Amps of current when its current is passing through a LOAD that causes a potential drop of 5 V

If battery B is assigned a value of 100 W then this can mean that the chemicals inside it allow it to provide 50 Amps of current when its current is passing through a LOAD that causes a potential drop of 2 V



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 increase the load.
This seems odd but this is correct.
I'm losing the plot
 
Nov 2015
38
0
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.
So as a after all this discussion I've concluded that I need to place a LOAD in the circuit. But what size of load (wattage of bulb)? How do i work that out.

How do i measure the power output of the turbine?
 
Apr 2015
1,159
305
Somerset, England
I'm losing the plot
I'm not suprised.

If you prefer youtube, then by all means go ahead.

I have been carefully restricting things I introduce to present them in a logical order and not to introduce inappropriate or downright incorrect material.

A battery provides potential to do work. If there is a high potential difference (PD) between the two terminals of the battery (called voltage) then lots of current can flow.

Power is just a way of assigning a value to the battery by saying P = IV

If battery A is assigned a value of 100 W then this means that the chemicals inside it allow it to provide 20 Amps of current when its current is passing through a LOAD that causes a potential drop of 5 V

If battery B is assigned a value of 100 W then this can mean that the chemicals inside it allow it to provide 50 Amps of current when its current is passing through a LOAD that causes a potential drop of 2 V
These misconceptions are most unfortunate and will hinder you until you can dump them.

If you want to put them right tell me what you think a watt is.
 
Nov 2015
38
0
If you prefer youtube, then by all means go ahead.
Was trying to get a quick fix.

These misconceptions are most unfortunate and will hinder you until you can dump them.
If you want to put them right tell me what you think a watt is.
Yes please.


WATT gives a measurement of the rate of consumption of energy.

One WATT is equivalent to one joule of energy consumed in one second.
 
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Apr 2015
1,159
305
Somerset, England
So how do you reconcile this

WATT gives a measurement of the rate of consumption of energy.

One WATT is equivalent to one joule of energy consumed in one second.
with this?

Power is just a way of assigning a value to the battery by saying P = IV
What does it say on the case of a battery about the electrical characteristics?

Here are some facts.

You can flatten a 12 volt car battery in 45 seconds cranking the engine.

It takes 45 minutes to flatten the same battery leaving the headlights on.

After six hours of leaving the doors open and the interior lights on the battery is flat.

What physics can you say about all these facts?
 
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MBW

Apr 2008
668
23
Bedford, England
Experimental practice

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

No matter how carefully an experiment is constructed, there will be some unavoidable variability
(the exact position of the hair dryer, random breezes in the room, mains voltage fluctuations, etc, etc).
Repeating the readings several times allows you to determine the variability of your experiment.
Ideally you should be doing enough repeats to allow basic statistical analysis, to get the average and the standard deviation.
The average is an estimate of what the value "really" is, the standard deviation is an estimate of how good an estimate the average is.
If all the results (for the same experimental conditions) are pretty much identical, then you just need a handful of repeats to prove this.
If the results are more variable, you need enough repeats to give an acceptably well defined average and standard deviation.
Wildly varying results could point to an experimental problem.
However some experiments are unavoidably variable, CERN for example will repeat billions of particle smashups to achieve the required statistically viable result.
 
Nov 2015
38
0
(iii) Explain the number and range of readings that you will take.

No matter how carefully an experiment is constructed, there will be some unavoidable variability
(the exact position of the hair dryer, random breezes in the room, mains voltage fluctuations, etc, etc).
Repeating the readings several times allows you to determine the variability of your experiment.
Ideally you should be doing enough repeats to allow basic statistical analysis, to get the average and the standard deviation.
The average is an estimate of what the value "really" is, the standard deviation is an estimate of how good an estimate the average is.
If all the results (for the same experimental conditions) are pretty much identical, then you just need a handful of repeats to prove this.
If the results are more variable, you need enough repeats to give an acceptably well defined average and standard deviation.
Wildly varying results could point to an experimental problem.
However some experiments are unavoidably variable, CERN for example will repeat billions of particle smashups to achieve the required statistically viable result.
Thanks MBW. Very useful
 
Nov 2015
38
0
You can flatten a 12 volt car battery in 45 seconds cranking the engine.

It takes 45 minutes to flatten the same battery leaving the headlights on.

After six hours of leaving the doors open and the interior lights on the battery is flat.

What physics can you say about all these facts?
Volt = joule per coulomb
Current = coulombs per second
Power = joule per second

The car battery is the LOAD that draws a high amount of current (coulombs per second) from the battery that only provides on a small number of joules per coulomb (12 V or energy per coulomb of charge).

The battery needs a high value of joules per second (energy per second) to operate and it depletes the current in the battery quickly.

Dear Studiot,
My answer may make no sense and you may think I'll never get it BUT I'm happy to work through your instruction. I won't give up if you don't.
Thanks
 
Apr 2018
1
0
Thanks for comment Studiot



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.



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.
nice sharing!!