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### Re: Wind Power B/C

Posted: April 16th, 2017, 3:21 pm
Regarding transmission lines, what type of information should you know about the voltages, resistance, power, etc.?
Does anyone know a reliable source to find energy production percentages?(like what percentage hydroelectric or wind accommodates for the world energy production)

### Re: Wind Power B/C

Posted: April 17th, 2017, 7:30 am
I'm a person who has significant trouble with all the math and formulas involved with wind power (maybe because i can't figure out which formulas that can even be used in this event) so can someone walk me through how to solve these problems:

1. A transmission line is providing 800 MW at 900 Amps. Calculate the operating voltage, in kilovolts, to 2 significant digits.

2. The transmission line in problem 1 has a total resistance 6.25 Ω. What is the power lost, expressed as a percent of the total 800 MW?

3. You mount a turbine 10 feet about the ground. Your neighbor puts up a 90 foot tower where the wind speed is 25% higher. The power available at the 90 foot tower is how many times larger than the power at the 10 foot tower?

### Re: Wind Power B/C

Posted: April 17th, 2017, 8:16 am
I'm a person who has significant trouble with all the math and formulas involved with wind power (maybe because i can't figure out which formulas that can even be used in this event) so can someone walk me through how to solve these problems:

1. A transmission line is providing 800 MW at 900 Amps. Calculate the operating voltage, in kilovolts, to 2 significant digits.

2. The transmission line in problem 1 has a total resistance 6.25 Ω. What is the power lost, expressed as a percent of the total 800 MW?

3. You mount a turbine 10 feet about the ground. Your neighbor puts up a 90 foot tower where the wind speed is 25% higher. The power available at the 90 foot tower is how many times larger than the power at the 10 foot tower?
1. Megawatt is a unit of power, amperes are a unit of current. Power = Voltage * Current, so 800,000,000 Watts = (voltage) * 900 amps. Voltage (in volts) equals 800,000,000 divided by 900 = 89,000 volts, which is 89 kilovolts.

#2 involves combining Ohm's law (current = voltage / resistance) and power=voltage*current. Essentially just plug it in, power = voltage^2 / resistance; this gives the "power" (as heat) produced by the resistor (the transmission wire). Strangely, calculating this out gives me 1,267,000,000 watts, which is actually a larger amount than the 800 MW. Dividing the two will get you 158.42%. I'm not entirely certain whether this is a valid answer since I haven't done these problems in a while; someone else can confirm these two.

3. I have no idea how to do this one, I suspect it relates to some formulas specific to the event (the rest of the stuff is general electromagnetism).

(someone please correct me if I'm wrong, it's been well over a year since I've done this stuff)

### Re: Wind Power B/C

Posted: April 17th, 2017, 8:47 am
I'm a person who has significant trouble with all the math and formulas involved with wind power (maybe because i can't figure out which formulas that can even be used in this event) so can someone walk me through how to solve these problems:

1. A transmission line is providing 800 MW at 900 Amps. Calculate the operating voltage, in kilovolts, to 2 significant digits.

2. The transmission line in problem 1 has a total resistance 6.25 Ω. What is the power lost, expressed as a percent of the total 800 MW?

3. You mount a turbine 10 feet about the ground. Your neighbor puts up a 90 foot tower where the wind speed is 25% higher. The power available at the 90 foot tower is how many times larger than the power at the 10 foot tower?
1. Megawatt is a unit of power, amperes are a unit of current. Power = Voltage * Current, so 800,000,000 Watts = (voltage) * 900 amps. Voltage (in volts) equals 800,000,000 divided by 900 = 89,000 volts, which is 89 kilovolts.

#2 involves combining Ohm's law (current = voltage / resistance) and power=voltage*current. Essentially just plug it in, power = voltage^2 / resistance; this gives the "power" (as heat) produced by the resistor (the transmission wire). Strangely, calculating this out gives me 1,267,000,000 watts, which is actually a larger amount than the 800 MW. Dividing the two will get you 158.42%. I'm not entirely certain whether this is a valid answer since I haven't done these problems in a while; someone else can confirm these two.

3. I have no idea how to do this one, I suspect it relates to some formulas specific to the event (the rest of the stuff is general electromagnetism).

(someone please correct me if I'm wrong, it's been well over a year since I've done this stuff)
Number 1 seems correct to me

For #2, first you want to find the voltage drop over the line caused by the resistance in the line. So using Ohm's law for Vdrop = current * resistance, you get 5625 volts. From there, you multiply the voltage drop by the current again (since Power Lost = Vdrop * current) to find the power lost. This should leave you with 5.06 MW. Divide that by the power supplied on the line to get the percentage of power lost, from which I got 0.632%.

For #3, you would use the equation for calculating power available from a wind turbine:
Power = 0.5 * air density * swept area * (wind speed)^3 * power coefficient
(Note that swept area is found by the equation: Swept area = (blade length)^2 * pi)
From there, you could put in a test value for wind speed (say 100 for the 10 foot tower) for the equation and use that to find out how much more power the 90 foot tower would generate.
I ended up with 1.95 times larger.

Please correct me if I'm wrong - I'm not that great at math in Wind Power either. And tell me if my sigfigs are wrong - I can't do sigfigs. ### Re: Wind Power B/C

Posted: April 17th, 2017, 9:13 am
I'm a person who has significant trouble with all the math and formulas involved with wind power (maybe because i can't figure out which formulas that can even be used in this event) so can someone walk me through how to solve these problems:

1. A transmission line is providing 800 MW at 900 Amps. Calculate the operating voltage, in kilovolts, to 2 significant digits.

2. The transmission line in problem 1 has a total resistance 6.25 Ω. What is the power lost, expressed as a percent of the total 800 MW?

3. You mount a turbine 10 feet about the ground. Your neighbor puts up a 90 foot tower where the wind speed is 25% higher. The power available at the 90 foot tower is how many times larger than the power at the 10 foot tower?
1. Megawatt is a unit of power, amperes are a unit of current. Power = Voltage * Current, so 800,000,000 Watts = (voltage) * 900 amps. Voltage (in volts) equals 800,000,000 divided by 900 = 89,000 volts, which is 89 kilovolts.

#2 involves combining Ohm's law (current = voltage / resistance) and power=voltage*current. Essentially just plug it in, power = voltage^2 / resistance; this gives the "power" (as heat) produced by the resistor (the transmission wire). Strangely, calculating this out gives me 1,267,000,000 watts, which is actually a larger amount than the 800 MW. Dividing the two will get you 158.42%. I'm not entirely certain whether this is a valid answer since I haven't done these problems in a while; someone else can confirm these two.

3. I have no idea how to do this one, I suspect it relates to some formulas specific to the event (the rest of the stuff is general electromagnetism).

(someone please correct me if I'm wrong, it's been well over a year since I've done this stuff)
Number 1 seems correct to me

For #2, first you want to find the voltage drop over the line caused by the resistance in the line. So using Ohm's law for Vdrop = current * resistance, you get 5625 volts. From there, you multiply the voltage drop by the current again (since Power Lost = Vdrop * current) to find the power lost. This should leave you with 5.06 MW. Divide that by the power supplied on the line to get the percentage of power lost, from which I got 0.632%.

For #3, you would use the equation for calculating power available from a wind turbine:
Power = 0.5 * air density * swept area * (wind speed)^3 * power coefficient
(Note that swept area is found by the equation: Swept area = (blade length)^2 * pi)
From there, you could put in a test value for wind speed (say 100 for the 10 foot tower) for the equation and use that to find out how much more power the 90 foot tower would generate.
I ended up with 1.95 times larger.

Please correct me if I'm wrong - I'm not that great at math in Wind Power either. And tell me if my sigfigs are wrong - I can't do sigfigs. On the answer key for the test, the answer for number two says "0.5% or 0.005 (4 MW)", but your answer seems to be making more sense, so which should I go with?

### Re: Wind Power B/C

Posted: April 17th, 2017, 12:31 pm
Quick question regarding the "functional modification": if my partner and I take fan blades from a PC case fun, but we attach the blades to our own setup (like our own thing that holds the blades), is that allowed by the rules?

### Re: Wind Power B/C

Posted: April 17th, 2017, 12:39 pm
Quick question regarding the "functional modification": if my partner and I take fan blades from a PC case fun, but we attach the blades to our own setup (like our own thing that holds the blades), is that allowed by the rules?
I would encourage you to try and design your own. I promise that you can make a better turbine using very simple materials.

### Re: Wind Power B/C

Posted: April 17th, 2017, 12:44 pm
Quick question regarding the "functional modification": if my partner and I take fan blades from a PC case fun, but we attach the blades to our own setup (like our own thing that holds the blades), is that allowed by the rules?
I would encourage you to try and design your own. I promise that you can make a better turbine using very simple materials.
We are building our own but we're making multiple designs. I'm wondering this specific one is legal. I think it is based off of others I've seen but just want to make sure.

### Re: Wind Power B/C

Posted: April 17th, 2017, 1:36 pm
I'm a person who has significant trouble with all the math and formulas involved with wind power (maybe because i can't figure out which formulas that can even be used in this event) so can someone walk me through how to solve these problems:

1. A transmission line is providing 800 MW at 900 Amps. Calculate the operating voltage, in kilovolts, to 2 significant digits.

2. The transmission line in problem 1 has a total resistance 6.25 Ω. What is the power lost, expressed as a percent of the total 800 MW?

3. You mount a turbine 10 feet about the ground. Your neighbor puts up a 90 foot tower where the wind speed is 25% higher. The power available at the 90 foot tower is how many times larger than the power at the 10 foot tower?
1. Megawatt is a unit of power, amperes are a unit of current. Power = Voltage * Current, so 800,000,000 Watts = (voltage) * 900 amps. Voltage (in volts) equals 800,000,000 divided by 900 = 89,000 volts, which is 89 kilovolts.

#2 involves combining Ohm's law (current = voltage / resistance) and power=voltage*current. Essentially just plug it in, power = voltage^2 / resistance; this gives the "power" (as heat) produced by the resistor (the transmission wire). Strangely, calculating this out gives me 1,267,000,000 watts, which is actually a larger amount than the 800 MW. Dividing the two will get you 158.42%. I'm not entirely certain whether this is a valid answer since I haven't done these problems in a while; someone else can confirm these two.

3. I have no idea how to do this one, I suspect it relates to some formulas specific to the event (the rest of the stuff is general electromagnetism).

(someone please correct me if I'm wrong, it's been well over a year since I've done this stuff)
Number 1 seems correct to me

For #2, first you want to find the voltage drop over the line caused by the resistance in the line. So using Ohm's law for Vdrop = current * resistance, you get 5625 volts. From there, you multiply the voltage drop by the current again (since Power Lost = Vdrop * current) to find the power lost. This should leave you with 5.06 MW. Divide that by the power supplied on the line to get the percentage of power lost, from which I got 0.632%.

For #3, you would use the equation for calculating power available from a wind turbine:
Power = 0.5 * air density * swept area * (wind speed)^3 * power coefficient
(Note that swept area is found by the equation: Swept area = (blade length)^2 * pi)
From there, you could put in a test value for wind speed (say 100 for the 10 foot tower) for the equation and use that to find out how much more power the 90 foot tower would generate.
I ended up with 1.95 times larger.

Please correct me if I'm wrong - I'm not that great at math in Wind Power either. And tell me if my sigfigs are wrong - I can't do sigfigs. I got the same answers for all of these.

### Re: Wind Power B/C

Posted: April 17th, 2017, 1:59 pm
1. Megawatt is a unit of power, amperes are a unit of current. Power = Voltage * Current, so 800,000,000 Watts = (voltage) * 900 amps. Voltage (in volts) equals 800,000,000 divided by 900 = 89,000 volts, which is 89 kilovolts.

#2 involves combining Ohm's law (current = voltage / resistance) and power=voltage*current. Essentially just plug it in, power = voltage^2 / resistance; this gives the "power" (as heat) produced by the resistor (the transmission wire). Strangely, calculating this out gives me 1,267,000,000 watts, which is actually a larger amount than the 800 MW. Dividing the two will get you 158.42%. I'm not entirely certain whether this is a valid answer since I haven't done these problems in a while; someone else can confirm these two.

3. I have no idea how to do this one, I suspect it relates to some formulas specific to the event (the rest of the stuff is general electromagnetism).

(someone please correct me if I'm wrong, it's been well over a year since I've done this stuff)
Number 1 seems correct to me

For #2, first you want to find the voltage drop over the line caused by the resistance in the line. So using Ohm's law for Vdrop = current * resistance, you get 5625 volts. From there, you multiply the voltage drop by the current again (since Power Lost = Vdrop * current) to find the power lost. This should leave you with 5.06 MW. Divide that by the power supplied on the line to get the percentage of power lost, from which I got 0.632%.

For #3, you would use the equation for calculating power available from a wind turbine:
Power = 0.5 * air density * swept area * (wind speed)^3 * power coefficient
(Note that swept area is found by the equation: Swept area = (blade length)^2 * pi)
From there, you could put in a test value for wind speed (say 100 for the 10 foot tower) for the equation and use that to find out how much more power the 90 foot tower would generate.
I ended up with 1.95 times larger.

Please correct me if I'm wrong - I'm not that great at math in Wind Power either. And tell me if my sigfigs are wrong - I can't do sigfigs. I got the same answers for all of these.
Yup that's what I thought because the answers that you gave me made sense with the explanations and everything. Thanks for the help!