In the sidebar of this page you may have noted a block titled “current Flensburg turbine weather”. This article is about the calculations behind the displayed data.  In the sidebar of this page you may have noted a block titled “current Flensburg turbine weather”. This article is about the calculations behind the displayed data. We developed and built our turbine in Flensburg, Germany. However, the turbine is currently not set up outside but resides inside our office on display. With the Flensburg turbine weather we are trying to give a rough estimate of how much energy the turbine would yield if it would be set up under local wind conditions.

For the calculation of the turbine weather we use the WordPress plugin “WP SimpleWeather” that we modified ourselves.

### Wind Speed

We fetch the wind speed from Yahoo weather data in kilometers per hour (kph). This value we transform to meters per second (mps) which is more common in Europe by the formula $v_{\text{wind}} = v_{\text{wind,mps}}=\frac{v_{\text{wind,kph}}}{3.6}$.

Comment: which height estimated?

### Air density

Calculating the air density is important because it is used to calculate the power in the wind. In our lectures, we assume an air density around $1.2 \frac{\text{kg}}{\text{m}^3}$ most of the time. However, the air density varies with the height over sea level, the temperature and the barometric pressure. To get a more exact result for the power in the wind, we calculate the air density from the meteorological quantities we get from the Yahoo weather service which are temperature and barometric pressure. In our script, air density is calculated according to the suggestions from Wikipedia, which claims to have an error in air density calculation of less than 0.2 %: $\rho_{\text{humid air}}=\frac{p_d}{R_d T} + \frac{p_v}{R_v T}$

where: $\rho_{\text{humid air}} =$ Density of the humid air in kg/m^3 $p_d =$ Partial pressure of dry air (Pa) $R_d =$ Specific gas constant for dry air, 287.058 J/(kg · K) $T =$ Temperature (K) $p_v =$ Vapor pressure of water (Pa) $R_v =$ Specific gas constant for water vapor, 461.495 J/(kg · K)

The vapor pressure of water $p_v$ can be found by $p_v = \phi p_{\text{sat}}$

where $\phi$ is the relative humidity we get from the Yahoo weather data. $p_{\text{sat}}$ is the saturation water pressure. We use a formula derived from Shelquist (refer to the Wikipedia article for proper reference) to calculate this value in Pa: $p_{\text{sat}} = 6.1078 \times 10^{\frac{7.5T}{T+273.3}} \cdot \frac{1}{100}$

where $T$ is expressed in degrees Celsius. $P_d$, the partial pressure of dry air, is calculated from the absolute barometric pressure $p$ and the vapor pressure of water $p_v$: $p_d = p - p_v$

With the formulas stated above air density can be calculated from temperature and barometric pressure.

### Power in Wind

We are calculating the power in the wind over the turbine rotor area to find out how much power there is actually available for our turbine to extract. The power in the wind over the rotor area can be calculated by the formula $P_{\text{wind}} = \frac{1}{2}\rho_{\text{humid air}}A_{\text{rotor}}v_{\text{wind}}^3$

where $A_{\text{rotor}}$ is the rotor area $A_{\text{rotor}} = \pi R^2$ and $R$ is the rotor radius. (In the HOLI 300 case $R = 0.8 \text{m}$).

It has to be said that a wind turbine cannot convert all the power that is in the wind into electricity. To easily understand this, one can imagine that the wind turbine decelerates the wind which is going through its rotor as it extracts power. Now, if the wind turbine would extract all the power that is in the wind, the wind right behind the turbine would have a speed of zero, or in other words it would be like a “wall”, not moving at all. In this situation, there is no “space“ for new wind to go through the rotor as it is blocked by the “wall of wind“ behind the rotor. So the wind flow through the rotor can only be decelerated down to a certain value which still allows a lot of power extraction. This limit is associated with the so called “Betz‘s limit“ which basically states that a wind turbine can, in theory, not extract more than 59.3 % of the energy in the wind flow. In practice a value this high is usually never reached. The “power coefficient“ $c_D$ is the fraction of the extracted wind power to the available wind power.

### HOLI 300 Electrical Power

The electrical power output of a turbine depends on a lot more than just the aerodynamic efficiency which is to a part determined by the power coefficient $c_D$. Some more of the energy in the wind gets lost on the way to the electrical mains. In the HOLI 300 some losses occur in the generator and the rectifier.
During the design phase of the HOLI 300, theoretical aerodynamic characteristics of the turbine rotor were determined by the so called “Blade Element Momentum Theory” (BEM), These and the data provided by the generator manufacturer helped to estimate the generator‘s electrical losses.
The following chart shows the calculated values for electrical and aerodynamic power compared to the power which is in the wind (all graphs assuming $\rho_{\text{humid air}} = 1.225$ kg · m³). A discussion of this chart will be made available in the note about the turbine aerodynamics on this website.

The current Flensburg turbine weather displays the data points from the curve for electrical power for the wind speed determined by the Yahoo weather service. It neglects air density variations and dynamic effects, as the aerodynamic characteristics were only calculated for steady state conditions under static wind with the same air density.

### Code

For the implementation in the “WP Simple Weather” plugin the file “jquery.simpleWeather.js” was modified and the code stated below was added. Additionally, some changes were made to “wp-simpleweather.php”. Contact Florian for more information.