A wind turbine rotates at 20 rpm. If its power output is 2.0 megawatts, what is the torque produced on the turbine from the wind?
We’re told here that the turbine rotates at 20 revolutions per minute. We’ll call that rev. We’re also told that the turbine produces 2.0 million watts of power. We’ll call that capital 𝑃. We want to know the torque produced on the turbine from the wind. We’ll call that 𝜏. If we draw a sketch of the turbine, we can see that the force of the wind on the blades indeed does rotate the turbine, therefore exerting a torque. To figure out what this torque is, let’s recall the definition of power.
Power 𝑃 is defined as the work done over some amount of time 𝑡. We recall further that work 𝑊 is defined as force times the distance over which that force acts. So in our case, power is equal to force times distance divided by time. Now, if we recall that torque is equal to 𝐹 — the force — times 𝑟 — the distance vector from the rotation point to the applied force.
If we rearrange that equation slightly, we see that the force 𝐹 is equal to the torque 𝜏 divided by 𝑟. And if we look further at 𝑑 over 𝑡 in our equation for power, 𝑑 over 𝑡 distance divided by time is equal to 𝑣 linear velocity, which itself is equal to 𝑟 times the angular velocity 𝜔.
If we substitute in for 𝐹 𝜏 over 𝑟 and for 𝑑 over 𝑡, 𝑟 times 𝜔, then we see that power not only being equal to work divided by time is also equal to torque divided by 𝑟 times 𝑟 times 𝜔. The common factors of 𝑟 cancel out. And we find that power equals torque 𝜏 times 𝜔 the angular velocity. We can rearrange this equation to solve for 𝜏. We find that 𝜏 is equal to 𝑃 power divided by 𝜔 angular velocity.
We’ve been given the power output of the turbine. But recall that the units of 𝜔 are radians per second, whereas we have a rotation speed in units of revolutions per minute. To convert this speed into more helpful units, let’s multiply by one minute divided by 60 seconds, which eliminates the units of minutes. We’ll then multiply by two 𝜋 radians per revolution, which likewise cancels the units for revolutions. Multiplying these three fractions through, we find a resulting value of 𝜔 in units of radians per second of two-thirds 𝜋 radians per second. We’ll call this result 𝜔.
We’re now ready to plug in these values for 𝑃 and 𝜔. The torque 𝜏 is equal to 2.0 times 10 to the sixth watts divided by two 𝜋 over three radians per second. Entering these values on our calculator, we find a result of 9.5 times 10 to the fifth newton metres. This is how much torque the wind exerts on the turbine.