Wind Power B/C

soyuppy
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Re: Wind Power B/C

Postby soyuppy » March 19th, 2017, 1:23 pm

Here's a PSA to all of you with low-pitch designs. Make sure that the pitch is adjustable. Our blade was designed for an average fan, and we medaled at every competition. Then at regionals their high speed was lower than our low speed, and our blade didn't move. We couldn't change anything because the pitch was fixed. Keep that in mind
was it even a 20" box fan? This is probably the one event where regardless of how well your blade design it, the outcome rest on the equipment used by the proctor. Similar situation happen in N.Cal last year. A lot of good blade failed to move/spin because of the quality of the fan and the min distance from fan to blade is set to 5 inches instead 5 cm

0ddrenaline
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Re: Wind Power B/C

Postby 0ddrenaline » March 19th, 2017, 4:19 pm

Here's a PSA to all of you with low-pitch designs. Make sure that the pitch is adjustable. Our blade was designed for an average fan, and we medaled at every competition. Then at regionals their high speed was lower than our low speed, and our blade didn't move. We couldn't change anything because the pitch was fixed. Keep that in mind
was it even a 20" box fan? This is probably the one event where regardless of how well your blade design it, the outcome rest on the equipment used by the proctor. Similar situation happen in N.Cal last year. A lot of good blade failed to move/spin because of the quality of the fan and the min distance from fan to blade is set to 5 inches instead 5 cm
Yeah, it was a proper box fan. I can't be frustrated with the proctors, because it was completely legal within the rules. I just should have been prepared for an exceptionally weak fan.

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Re: Wind Power B/C

Postby chalker » March 20th, 2017, 7:02 am

Are there any rules against touching the blade during the testing period? Yesterday we went to the state competition, and oddly the proctor told us we could not touch the blade once it had been put on the stand. Because our blade needs manual help to start spinning, it never moved.

I looked at the rules again, and competition rule f. says, "teams are allowed to adjust, modify, start and stop the blade assembly rotation and re-position the blade assembly or support stand during the measurement period.". The only time you can't touch the blade is in rule g. where it says, "the team must not touch or reposition the blade assembly or supporting stand during the measurement period".

So I am confused where and how the proctor came up with "you can't touch the blade once it is on the stand".

Thanks,

Lisa
corretion: what I meant was - competition rule f. says, "teams are allowed to adjust, modify, start and stop the blade assembly rotation and re-position the blade assembly or support stand during the testing period.".
Yep, you're correct. One of the things I emphasize time and time again is that there are 300+ tournaments all over the country each year, which results in more than 13,000 event supervisors! The vast majority of them are volunteers who are donating their time and don't necessarily have specialize expertise. As such, mistakes will be made. This is why we have an official appeals/arbitration process in place, which I encourage coaches to utilize. It also helps to bring a printed version of the rules to each event, so that you can show them in writing what you are concerned about, instead of it being a 'he said-she said' type situation.

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chalker
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Re: Wind Power B/C

Postby chalker » March 20th, 2017, 7:04 am

Here's a PSA to all of you with low-pitch designs. Make sure that the pitch is adjustable. Our blade was designed for an average fan, and we medaled at every competition. Then at regionals their high speed was lower than our low speed, and our blade didn't move. We couldn't change anything because the pitch was fixed. Keep that in mind
was it even a 20" box fan? This is probably the one event where regardless of how well your blade design it, the outcome rest on the equipment used by the proctor. Similar situation happen in N.Cal last year. A lot of good blade failed to move/spin because of the quality of the fan and the min distance from fan to blade is set to 5 inches instead 5 cm
Yeah, it was a proper box fan. I can't be frustrated with the proctors, because it was completely legal within the rules. I just should have been prepared for an exceptionally weak fan.
One of the things far too many competitors do is the OVER-FIT their solution by optimizing to a particular setup. We deliberately write the rules to include some ambiguity in the testing setup in order to test the ability of competitors to adapt to varying situations. This is exactly what 'real-world' scientists and engineers have to do day in and out.

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National Event Supervisor
National Physical Sciences Rules Committee Chair

heiber
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Re: Wind Power B/C

Postby heiber » March 21st, 2017, 11:36 am

My team and I are having a very hard time understanding how to solve for calculating power loss along a line. I think the terms / equations are similar to Ohms Law but not quite the same and that is mixing us up. Here is an example question for an older test:

A power plant in Chicago is generating 900 MW on 700 KV. Assuming the line is 100 kilometers long, with a resistance along the line of 0.2 ohms
A. What is the current flowing along the line?
Using Ohms Law, current is I=P/V 900 MW / 700 KV = 1286 amps

B. How much power is lost in the lines?
This is where we get confused. The answer key says find the voltage drop = current (amps) * resistance (ohms) 1286 * 0.2 = 257.2 kV. Then find power lost = amps * kV 1286 amps * 257.2 kV = 330.76 MW
We don't understand these equations and how does the line distance factor in?

C. What is the percentage of the power lost?
This should be easy taking the answer to B / 900 MW.

Can someone please help explain how to solve these kinds of problems or point us to resources to learn from? Thank you for the help.

science_nerd_foreves
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Re: Wind Power B/C

Postby science_nerd_foreves » March 21st, 2017, 7:14 pm

My team and I are having a very hard time understanding how to solve for calculating power loss along a line. I think the terms / equations are similar to Ohms Law but not quite the same and that is mixing us up. Here is an example question for an older test:

A power plant in Chicago is generating 900 MW on 700 KV. Assuming the line is 100 kilometers long, with a resistance along the line of 0.2 ohms
A. What is the current flowing along the line?
Using Ohms Law, current is I=P/V 900 MW / 700 KV = 1286 amps

B. How much power is lost in the lines?
This is where we get confused. The answer key says find the voltage drop = current (amps) * resistance (ohms) 1286 * 0.2 = 257.2 kV. Then find power lost = amps * kV 1286 amps * 257.2 kV = 330.76 MW
We don't understand these equations and how does the line distance factor in?

C. What is the percentage of the power lost?
This should be easy taking the answer to B / 900 MW.

Can someone please help explain how to solve these kinds of problems or point us to resources to learn from? Thank you for the help.
For part B, you are not looking at the units. You should do 1286 * 0.2 = 257.2 kV (like you did). But then, 1286 amps * 257.2 kV = 330,759 kW. The reason that the correct answer is in MW, is because the values that started the problem were given in MW. To convert your answer, know that 1 MW is 1000 kW. So it is 330.76 MW.
For part C, you are correct, it is 330.76MW / 900MW = 36.75%.

I once had a teacher that took off points if we did not carry the units throughout the problem. We could only cross off the unit if they matched. This habit stuck with me, and it has really helped me, so try that.

I hope this helps

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Re: Wind Power B/C

Postby science_nerd_foreves » March 21st, 2017, 7:18 pm

I know that scores tend to change based on the fans used, but does anyone have a base line estimate of what i should aim for in terms of scoring?
I honestly do not know what to tell you, other than to trust your gut. I had a turbine that worked really well for my set up at home and reached over a volt. At regionals, with the same turbine, it did not even reach 300mV. I spent the rest of the test freaking out, but then got within the top 3, so I did well, even though it seemed as though I did really badly.

heiber
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Re: Wind Power B/C

Postby heiber » March 22nd, 2017, 5:38 am

My team and I are having a very hard time understanding how to solve for calculating power loss along a line. I think the terms / equations are similar to Ohms Law but not quite the same and that is mixing us up. Here is an example question for an older test:

A power plant in Chicago is generating 900 MW on 700 KV. Assuming the line is 100 kilometers long, with a resistance along the line of 0.2 ohms
A. What is the current flowing along the line?
Using Ohms Law, current is I=P/V 900 MW / 700 KV = 1286 amps

B. How much power is lost in the lines?
This is where we get confused. The answer key says find the voltage drop = current (amps) * resistance (ohms) 1286 * 0.2 = 257.2 kV. Then find power lost = amps * kV 1286 amps * 257.2 kV = 330.76 MW
We don't understand these equations and how does the line distance factor in?

C. What is the percentage of the power lost?
This should be easy taking the answer to B / 900 MW.

Can someone please help explain how to solve these kinds of problems or point us to resources to learn from? Thank you for the help.
For part B, you are not looking at the units. You should do 1286 * 0.2 = 257.2 kV (like you did). But then, 1286 amps * 257.2 kV = 330,759 kW. The reason that the correct answer is in MW, is because the values that started the problem were given in MW. To convert your answer, know that 1 MW is 1000 kW. So it is 330.76 MW.
For part C, you are correct, it is 330.76MW / 900MW = 36.75%.

I once had a teacher that took off points if we did not carry the units throughout the problem. We could only cross off the unit if they matched. This habit stuck with me, and it has really helped me, so try that.

I hope this helps
Thanks but this does not answer my question. I understand the units of measure changes. It is the overall question itself. Intuitively it seems that power loss should factor in to distance but we don't see that included in any of the above answers - even though it is included in the question. What if the line was 50 KM long - how would that change things? The easy way is just to memorize the following equations, but I want my team to understand the concepts as well and I can't explain it to them.
Voltage Drop = Current * Resistance
Power Loss = Current * Voltage Loss

Thanks for the help
(your poor suffering Industrial Engineer who should have apparently taken more Electrical Engineering classes)

kenniky
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Re: Wind Power B/C

Postby kenniky » March 22nd, 2017, 7:24 am

My team and I are having a very hard time understanding how to solve for calculating power loss along a line. I think the terms / equations are similar to Ohms Law but not quite the same and that is mixing us up. Here is an example question for an older test:

A power plant in Chicago is generating 900 MW on 700 KV. Assuming the line is 100 kilometers long, with a resistance along the line of 0.2 ohms
A. What is the current flowing along the line?
Using Ohms Law, current is I=P/V 900 MW / 700 KV = 1286 amps

B. How much power is lost in the lines?
This is where we get confused. The answer key says find the voltage drop = current (amps) * resistance (ohms) 1286 * 0.2 = 257.2 kV. Then find power lost = amps * kV 1286 amps * 257.2 kV = 330.76 MW
We don't understand these equations and how does the line distance factor in?

C. What is the percentage of the power lost?
This should be easy taking the answer to B / 900 MW.

Can someone please help explain how to solve these kinds of problems or point us to resources to learn from? Thank you for the help.
For part B, you are not looking at the units. You should do 1286 * 0.2 = 257.2 kV (like you did). But then, 1286 amps * 257.2 kV = 330,759 kW. The reason that the correct answer is in MW, is because the values that started the problem were given in MW. To convert your answer, know that 1 MW is 1000 kW. So it is 330.76 MW.
For part C, you are correct, it is 330.76MW / 900MW = 36.75%.

I once had a teacher that took off points if we did not carry the units throughout the problem. We could only cross off the unit if they matched. This habit stuck with me, and it has really helped me, so try that.

I hope this helps
Thanks but this does not answer my question. I understand the units of measure changes. It is the overall question itself. Intuitively it seems that power loss should factor in to distance but we don't see that included in any of the above answers - even though it is included in the question. What if the line was 50 KM long - how would that change things? The easy way is just to memorize the following equations, but I want my team to understand the concepts as well and I can't explain it to them.
Voltage Drop = Current * Resistance
Power Loss = Current * Voltage Loss

Thanks for the help
(your poor suffering Industrial Engineer who should have apparently taken more Electrical Engineering classes)
Resistance changes with length yeah but I think you don't need length in this problem. Resistivity I believe measures resistance per length, but the resistance is given to you so I think the length can safely be ignored.
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Re: Wind Power B/C

Postby heiber » March 22nd, 2017, 8:30 am

[Resistance changes with length yeah but I think you don't need length in this problem. Resistivity I believe measures resistance per length, but the resistance is given to you so I think the length can safely be ignored.[/quote]

OK. That helps. We've been trying to find equations and information that include distance. So the distance is included in the resistance calculation. Here is another example from the same test.

21. If a power plant is generating 100 MW, running on a 20 KV line (20 kilometers long), loses 10 MW of power after running along the line, how much resistance does the line have? (2 pts) Their answer:
1) Find Amps: P/V 100MW / 20KV= 5000A
2) Find Ohms: V/I= 20KV / 5MA= 4 ohms

My questions:
From Ohms Law, Resistance = Power / (Current * Current) or 100 MW / (20 KV * 20 KV) = 250 which is different. Why can't you use this equation and why is the answer different from above. All of the equations appear to be coming from the electrical formula wheel so would think it would work out either way.
Again - have information such as power loss of 10 MW and length of 20 kilometers that does not appear to even be used.

Thanks again.


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