How to Calculate Wind Turbine Power Output?

How to calculate wind turbine power output? It’s a simple calculation, which highlights the great potential of these white-spinning machines.

Before we crunch numbers though, let’s go over how they operate.

How do wind turbines work?

When the wind blows past a wind turbine, the blades spin. The blades are capturing the wind’s kinetic energy, and turning it into mechanical energy. Also called, rotational kinetic energy.

In this process, the rotating blades turn a shaft inside of the wind turbine, connected to a gearbox. This then leads to the spinning of the generator’s rotor, which generates electricity.

The below schematic shows the conversion of power step by step.

wind turbine schematic
Wind turbine schematic (Photo Credit: Jalonsom)

While the following schematic shows the guts of a wind turbine:

wind turbine section view schematic
Inside a wind turbine (Photo Credit: U.S. Department of Energy)

Important Note: the power generated by wind turbines depends on the following factors: 

  • Mass of air (density)
  • Speed of air (velocity)
  • Amount of air (volume)
  • The swept area of the wind turbine blades
  • Blade design

Wind turbine power output calculation equations and variables

The following are the equation variables:

m: mass (kg)
v: wind speed (meters/second)
A: rotor swept area (meter^{2})
r: radius (meters)
KE: kinetic energy
P: power
\rho: density (kg/meter^{3})
\frac{dm}{dt}: mass flow rate (kg/second)

swept area of a wind turbine

With our variables defined, we’ll now go over our equations.

Kinetic energy is defined as KE = \dfrac{1}{2}mv^{2}

While power is KE per unit of time, thus P = \dfrac{1}{2}\frac{dm}{dt}v^{2}

Next, fluid mechanics tells us the mass flow rate is \frac{dm}{dt} = \rho \times A \times V

While the change in mass over a unit of time, is seen in the cylinder schematic below. Where the mass flux is the wind flowing through the turbine blades.

We now substitute our mass flow rate equation into our power equation, to get P = \frac{1}{2} \times \rho \times A \times v^{3}

To finalize our equation, we’ll go over the Betz limit variable.

mass flow rate

Betz Limit in wind power extraction

In 1919, Albert Betz, a German physicist, made a discovery on wind turbine efficiency. They can’t convert more than 59.3% of the wind’s kinetic energy into mechanical energy. It doesn’t matter how perfectly designed the wind turbine is either.

This limitation is from generator inefficiencies, drive train friction, and blade design. But also, from wind slowing down the blades. This happens as the turbine extracts energy from the wind. In return, the wind leaving the turbine flows slower than the wind entering the turbine.

For a wind turbine to operate, some wind needs to flow out from the back. If the turbine captures 100% of the wind power, the blades won’t spin. Because there’s no wind to capture energy from.

Think of the wind blockage at the turbine, like traffic on the highway. As cars slow down in front of you, you’ll eventually need to slow down too. No matter how far back you are from the traffic jam.

So, there’s a limit on the amount of air, which can pass through a wind turbine for max power output. This limit we call the Betz Limit.

Power Coefficient, Cp

With the Betz limit explained, let’s look at the Power Coefficient, Cp, below. This is the ratio of extracted power by the wind turbine, to the total available power in the wind source.

C_{p} = \frac{P_{T}}{P_{W}}, where C_{P\:MAX} = 0.59.

The Betz Limit is the maximum possible value of C_{p}. Thus, 16/27 or 0.59.

Our power generation equation now becomes P = \frac{1}{2} \times \rho \times A \times v^{3} \times C_{p}

Important Note: wind turbines cannot operate at this C_{p\:MAX}. Because design requirements for reliability and durability reduce the C_{p}

Also, wind turbines would require perfect wind conditions to maximize power output. In the real world, the C_{P} value falls between 0.25 to 0.45. 

How to calculate wind turbine power output?

To start the calculation, we’ll use the power output from a Vestas V164-8.0 MW. This is an offshore wind turbine and is one of the world’s largest. I haven’t stood next to these monsters, but I’ve done plenty of design work for the V90-3.0 MW.

Below is a picture I snapped of a V90 unit at a project site. Look how small the guy looks standing next to the wind turbine base. Keep in mind, the V164 rotor diameter is 164 meters, while the V90 is 90 meters.

vestas v90 wind turbine

For our calculation data, we’ll use the below table data from the Journal of Physics. The data is for LEANWIND 8 MW, which relates to the V164 unit.

Wind speed (m/s)Power (kW)Cp (Power Coefficient)Thrust (kN)Ct (Thrust Coefficient)
41100.131900.92
56000.372730.85
611400.413810.82
719000.435050.8
829000.446480.78
941550.448000.76
1056300.449450.73
1171500.4210520.67
1278650.359720.52
1379700.288470.39
1480000.237650.3
1580000.187000.24
1680000.156440.19
1780000.136040.16

From the table, we’ll use a wind speed of 14 meters/second for max power output. The following then becomes our input data:

  • V164 blade length: 80 meters
  • Wind speed: 14 meters/second
  • Air density: 1.23 kg/meter^{3}
  • Power coefficient: 0.23

Now first, we calculate the swept area of the turbine blades.

A = \pi r^{2}

The V164 blade length is the radius variable in our equation.

A = \pi\times 80^{2} = 20,106.2 \: meter^{2}

Next, we calculate the generated power from the wind turning the turbine blades.

P = \frac{1}{2} \times \rho \times A \times v^{3} \times C_{p}

P = \frac{1}{2} \times 1.23 \times 20,106.2 \times 14^{3} \times 0.23 = 7.80 MW

While the rated capacity, max power output for the V164, is 8 MW. This is the amount of power the turbine can produce at optimal wind speeds.

Important Note: our calculation is only an estimate of the power output. For a more accurate calculation, we need to consider the following variables:

  • Wind direction
  • The intrusion of water vapor
  • Thermal expansion
  • Equipment corrosion, and wear and tear

Importance of understanding how to calculate wind turbine power output

To learn the relationships between all the power generation variables. Then more accurately, predict wind turbine power output.

In return, this helps smoothly integrate these machines with power grids. Because power grids expect reliable power generation sources. And wind patterns are anything but consistent and overly predictable.

Even more, if you’re in the energy industry, you sell energy before it’s generated. So, the more accurately we can calculate wind turbine power output, the better.

“How to calculate wind turbine power output?” wrap up

These massive white structures look simple as they stand tall in the distance. But their design and integration into power grids are far from simple. Hence the importance of learning how to calculate wind turbine power output. And then, to better understand and appreciate these marvels of engineering.

What are your thoughts on the power output of wind turbines? Do you think the demand for wind power will further increase over the years? 

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4 thoughts on “How to Calculate Wind Turbine Power Output?”

    • Hey Mark; as a quick and dirty calculation:

      Per the article, a nominal generation capacity of 800 MW / 62 turbines = 12.9 MW / turbine

      Thus, a 12.9 MW rated wind turbine will generate 12.9 MWh per hour in peak operating conditions.

      Assuming 15 revolutions/minute (rpm), that’s one revolution every 4 seconds.

      Given there are 3600 seconds in an hour, the turbine will generate 0.11% (or 4/3600) of an hour’s worth of generation in a single revolution. Thus, 14.2 kWh/revolution. Then, using the average U.S. wind capacity factor of 0.35, we can chop this value down to 4.97 kWh/revolution.

      The average daily electricity consumption in Massachusetts is about 20.1 kWh according to a quick Google search. So, we fall well short.

      Reply
  1. What I’m really struggling with in the above analysis is the there is no term for blade area.

    It’s just not possible that say a 2 bladed unit generates the same power as a 4 bladed unit.

    No problem with the limits – it’s just that there is such a huge space between blade tips that trailing tip vortices etc will be long gone such as not to impact the trailing blade.

    After many years trying to optimise marine propellers – I know just how important blade area is – in fact a critIcal variable is DAR or Disk Area Ratio or Blade area / Swept area

    Can someone enlighten me on the trade offs of a scenario where the turbine had say 8 blades mounted on a ring at a significant radius – leave aside the practicality at this stage.

    Interesting to see the very old wind turbines driving pumps in outback Australia with maybe 20 blades mounted only on the outer half diameter.

    I understand the logic of ever larger blades – but – can this provide an opportunity for more blades ?

    Reply
    • Great question and you’re right. The blade area would be included as a variable in a more in-depth analysis.

      Now to preface, the boring answer defaults to costs. The end goal with commercial wind turbines is always to maximize energy while minimizing costs and failures. If you design more blades to increase the power produced, you’ll need to beef up your internal components given the higher failure rate due to the higher RPM in certain case scenarios. Yet, the wind turbine becomes only marginally more efficient. 

      To point out, at lower wind speeds, having more blades will lead to greater rotational speeds. But as wind speeds increase, the blades will stall (they create a wall against the wind), and the turbine speed will slow. So, having more blades you can only operate best in a narrow range of low wind speeds, and poorly outside of the range. But, a lower blade count will allow you to generate more energy since it can operate in a wider range of wind speeds given the distribution of how wind typically blows.

      To dig a bit deeper, the rotor power, P = 2*π*T*n, is proportional to the shaft torque (T) and the rotation frequency (n). The frequency (n) is governed by the tip speed ratio. So, the torque (T) will increase with more blades, because every rotating blade creates “dirty air” (i.e the blade reduces the wind speed for the following blade). Thus, the more blades you have, the greater this “wind shadow” will be. 

      And the generator within the turbine moves let’s say 1,800 RPM to convert the wind’s energy into electricity. So, more blades wouldn’t be conducive, as an electric generator is better with higher speeds, especially when you consider the cost of construction, maintenance, and custom blade designs for a given region (e.g. pitch of the blade). Think about the old mechanically driven windmills used for water pumps as you highlighted in Australia. They need a high starting torque to pump water from underground; thus, why they have so many blades.

      Reply

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