Hey Dan, I have a HYE-1500 96 volt turbine, they supply you with a charted power curve, But with no voltage points, So in order to program a Aurora Wind inverter, it has to columns. Volts- Power(AMPS). So in order to get the volts and power, is there a way using the given power curve with watts and M/S chart ?
Good question. The problem is that the turbine is always less than 100% efficient, so all of the power in the wind is never fully captured. If you divide the actual output (per the power curve) by the wind power at each speed, you can calculate the efficiency at each speed. There is something called the Betz Law or Betz Limit that tells you the maximum possible turbine efficiency. I believe it's around 60%, but you could Google that to find out. I hope this helps!
Hi Zargham, if you have the raw data (each 10 minutes measured normally), you can use the histogram option from Excel to get the distribution of the wind data according to each specific wind speed, the next step is to stablish the distribution in function of the 8760 hours per year, and the capacity factor.
@@DGCAD not sure if my way works but can you judge it for me ? i have raw data (1hour mean) at 10m as references height and using the Hellman's Power law to calculate the wind speed at the desired hub height. And i have generate it into total of 8670 values of wind speed for a year using Weibull's Distribution and then calculate the capacity factor with the selected wind turbine by using the way taught it this video
These are frequencies (in hours/ year) @ each wind speed bin (m/s), you can calculate this if you have the wind speed data (timeseries data) and then arrange that data with a temporal resolution of 1 hour for the single year and then calculate the frequency the sum of all these frequency should be approx 8760 hours/year, you can also use Python or MATLAB to achieve this.
Thank you for your reply; I appreciate it. These data (time series data) should be the data of wind turbines, or we could use other datasets like ERA5. For example, if I have the data on wind turbines for a year but there are two months of missing data, can I use it or can I find another way? @@m-hamzamalik
@@LilianeUWAJENEZA You can use either of them or both, (local measurments or ERA5), to compare errors, but please check the sources and see how local measurements are different from spatial measurements like ERA5, you also mentioned that you have 2 months missing data which will lead to inaccurate results (you can use methods like interpolation or some other methods like filling data from previous or next year if you have 1+ years of data but still it's an approximation) you can justify it and there is no issue to use ERA5 dataset, please if you are a researcher, also confirm this from your supervisor. If you are ok with spatial measurements, then there is no problem in ERA5.
I'm using the ''Power in the wind'' equation to calculate the Power (kW) for each value of wind speed (m/s). Yet I'm getting much higher values of kW then you are using, Swept area = pi (3.14159) x r (65m blade length)^2 = 13273m2 W = 1/2 x Air Density (1.225kg/m3) x swept area (13273) x wind speed (12m/s)^3 = 14,048,143 kW = 14,048 mW = 14 Yet I know this turbine has a theoretical maximum of 4 mW. So my 14 mW result is highly unrealistic, especially since a wind speed of 12m/s is below average for the location. So I'm making a mistake. Can anyone see what the mistake is?
Hi Ronnet, take in account that there is a mistake in the area formula, also you must consider the power coefficient for the turbine. if you make these changes, you will get the correct value.
How did you get the amount of hrs per year at each wind speed ? do you have a Wind Study or just cheet ? is there a website or link for Lat / Long that shows you the amount of hours at each speed ? for a given location ? Does anyone know the way to answer this. Please share . Thanks
t24money The Location needs to be measured by 80 meter wind masts. Or you could lookup wind atlases and try to do the calculations yourself. Maybe use software called Wasp it is danish software.
That was very useful vedio. However, I am a little bit confused about the difference between the Coefficient of performance (CP)and Capacity factor (CF). I will be so grateful if you can help me in this issue.
Hi, Omar. Coefficient of performance (COP) is used to describe the "efficiency" of a heat pump heating and/or cooling system. It describes how many units of heat you remove (cooling) or add (heating) for each unit of energy you input. E.g. a COP of 3 means that for every one unit of energy input, you remove or add 3 units of heat energy. The units don't matter. (It is not true efficiency because you are not converting the energy into heat, you are using it to move energy around.) Capacity factor refers to any energy generation source, and indicates how much energy you generate in a given period of time (usually a year) divided by the maximum possible energy you could generate with that equipment. E.g. if you have a 1 MW wind turbine, that means that its peak output is 1 MW. Since there are 8760 hours in a year, the max output for a year for that turbine is 8760 MWh. If you actually generate 4380 MWh that year, then your CF would be (4380/8760 =) 50%. This is a high number, by the way, but shows you how it works. Hopefully this helps!
So given the information...is the 28.3% power worth it? Especially when other variables are factored in like fossil fuels used for turbine transportation, human energy used for construction, trees cut down for installation, etc.
Any one can help me, for below mentioned questions? i need answer please Q 1-What is the present value of the payment cost made to decommission the wind farm (in € and answer as an absolute value)? Q 2-What is the present value of all the costs (in € and answer as an absolute value)? Q 3 -What is the levellised annual cost (in €/year and answer as an absolute value)? Q 4-What is the levellised cost of energy (LCoE) in €c/kWh and answer as an absolute value?
Hi Sadbi. It's a very well-known equation. It will be in most energy/power textbooks, and on any number of reputable websites, e.g. www.engineeringtoolbox.com/wind-power-d_1214.html. Hope that helps!
Hey Dan, I have a HYE-1500 96 volt turbine, they supply you with a charted power curve, But with no voltage points, So in order to program a Aurora Wind inverter, it has to columns. Volts- Power(AMPS). So in order to get the volts and power, is there a way using the given power curve with watts and M/S chart ?
Good question. The problem is that the turbine is always less than 100% efficient, so all of the power in the wind is never fully captured. If you divide the actual output (per the power curve) by the wind power at each speed, you can calculate the efficiency at each speed. There is something called the Betz Law or Betz Limit that tells you the maximum possible turbine efficiency. I believe it's around 60%, but you could Google that to find out. I hope this helps!
+Dan Kasper Hi I would like to know how can we calculate the no. of hours/yr for a specific wind speed in excel?
Hi Zargham, if you have the raw data (each 10 minutes measured normally), you can use the histogram option from Excel to get the distribution of the wind data according to each specific wind speed, the next step is to stablish the distribution in function of the 8760 hours per year, and the capacity factor.
@@DGCAD not sure if my way works but can you judge it for me ? i have raw data (1hour mean) at 10m as references height and using the Hellman's Power law to calculate the wind speed at the desired hub height. And i have generate it into total of 8670 values of wind speed for a year using Weibull's Distribution and then calculate the capacity factor with the selected wind turbine by using the way taught it this video
i have the shape and scale parameters for generating total of 8670 values of wind speed in Weibul distribution (leftout information).
Where would you find the wind speed data?
Fantastic lecture.
Thank you for this video; it is very helpful. Could you please tell me how you got the hours of each wind speed? I was just wondering how you got it.
These are frequencies (in hours/ year) @ each wind speed bin (m/s), you can calculate this if you have the wind speed data (timeseries data) and then arrange that data with a temporal resolution of 1 hour for the single year and then calculate the frequency the sum of all these frequency should be approx 8760 hours/year, you can also use Python or MATLAB to achieve this.
Thank you for your reply; I appreciate it. These data (time series data) should be the data of wind turbines, or we could use other datasets like ERA5. For example, if I have the data on wind turbines for a year but there are two months of missing data, can I use it or can I find another way? @@m-hamzamalik
@@LilianeUWAJENEZA You can use either of them or both, (local measurments or ERA5), to compare errors, but please check the sources and see how local measurements are different from spatial measurements like ERA5, you also mentioned that you have 2 months missing data which will lead to inaccurate results (you can use methods like interpolation or some other methods like filling data from previous or next year if you have 1+ years of data but still it's an approximation) you can justify it and there is no issue to use ERA5 dataset, please if you are a researcher, also confirm this from your supervisor. If you are ok with spatial measurements, then there is no problem in ERA5.
Thank you so much@@m-hamzamalik
Thank you so much sir 🙏 for briefly explanation
Wow that was everything.. I can now and try to do something similar for my project.
The gyro wind turbine has smashed Betz limit regards Graham Flowers
What is the power-speed characteristic of wind turbine ?
Great video Dan, thanks.
+Eamonn Lyons You're most welcome, Eamonn. Thanks for the nice comment.
I'm using the ''Power in the wind'' equation to calculate the Power (kW) for each value of wind speed (m/s). Yet I'm getting much higher values of kW then you are using,
Swept area = pi (3.14159) x r (65m blade length)^2 = 13273m2
W = 1/2 x Air Density (1.225kg/m3) x swept area (13273) x wind speed (12m/s)^3 = 14,048,143
kW = 14,048
mW = 14
Yet I know this turbine has a theoretical maximum of 4 mW. So my 14 mW result is highly unrealistic, especially since a wind speed of 12m/s is below average for the location. So I'm making a mistake. Can anyone see what the mistake is?
Hi Ronnet, take in account that there is a mistake in the area formula, also you must consider the power coefficient for the turbine. if you make these changes, you will get the correct value.
How did you get the amount of hrs per year at each wind speed ? do you have a Wind Study or just cheet ? is there a website or link for Lat / Long that shows you the amount of hours at each speed ? for a given location ? Does anyone know the way to answer this. Please share . Thanks
t24money The Location needs to be measured by 80 meter wind masts. Or you could lookup wind atlases and try to do the calculations yourself. Maybe use software called Wasp it is danish software.
That was very useful vedio. However, I am a little bit confused about the difference between the Coefficient of performance (CP)and Capacity factor (CF). I will be so grateful if you can help me in this issue.
Hi, Omar. Coefficient of performance (COP) is used to describe the "efficiency" of a heat pump heating and/or cooling system. It describes how many units of heat you remove (cooling) or add (heating) for each unit of energy you input. E.g. a COP of 3 means that for every one unit of energy input, you remove or add 3 units of heat energy. The units don't matter. (It is not true efficiency because you are not converting the energy into heat, you are using it to move energy around.)
Capacity factor refers to any energy generation source, and indicates how much energy you generate in a given period of time (usually a year) divided by the maximum possible energy you could generate with that equipment. E.g. if you have a 1 MW wind turbine, that means that its peak output is 1 MW. Since there are 8760 hours in a year, the max output for a year for that turbine is 8760 MWh. If you actually generate 4380 MWh that year, then your CF would be (4380/8760 =) 50%. This is a high number, by the way, but shows you how it works.
Hopefully this helps!
you will make it with inplix instructions
So given the information...is the 28.3% power worth it? Especially when other variables are factored in like fossil fuels used for turbine transportation, human energy used for construction, trees cut down for installation, etc.
yes they make like 900k each year per commercial/government run turbine. On average they pay for themselves almost in a year
Any one can help me, for below mentioned questions? i need answer please
Q 1-What is the present value of the payment cost made to decommission the wind farm (in € and answer as an absolute value)?
Q 2-What is the present value of all the costs (in € and answer as an absolute value)?
Q 3 -What is the levellised annual cost (in €/year and answer as an absolute value)?
Q 4-What is the levellised cost of energy (LCoE) in €c/kWh and answer as an absolute value?
I want the reference, Please
Hi Sadbi. It's a very well-known equation. It will be in most energy/power textbooks, and on any number of reputable websites, e.g. www.engineeringtoolbox.com/wind-power-d_1214.html. Hope that helps!
@@dankasper3914 thank you i found it in your comment section
Why it doesnt exceed 2000kW
Do you mean "HOW ?" ... It can use Variable Pitch Blades and/or the turbine turns out of the wind direction
I took great plans from Avasva . It help me a lot with my own solars.
Hi Shanta
Excuse me , what is avasva
Thank you sir !
Thanks!
We have had many years of experience with these devices, calculation is not required as the data gives real figures.
Betz limit has been smashed and debunked regards Graham S Flowers
could i have your whats up ?
just use Avasva Solutions :)))
Sir how to calculate the out put of a permanent magnet rotor axial wind turbine. simple way with formula.
Is this depeds on number of coils sets or magnets