## Wind Power generator

The power generated from the aerogenerator is from the kinetic energy of the wind which pass throught it. So, after the wind passed through the aerogenerator, it's kinetic energy decreases. So, will the temperature of the wind drop after passing through the generator since its kinetic energy is lost?
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 Mentor Blog Entries: 9 I would say no to this. The average velocity with respect to the ground will change. The question is how is the microscopic motions which are used to find an average KE of a molecule and therefore the air temperature related to the air velocity wrt the ground. I believe you find that the velocity for the temperature is relative to the neighboring air molecules, so it is velocity that is superimposed on the velocity of air wrt the ground. Thus we can can have a high velocity COLD wind. (If you have ever experienced a winter in Chicago you will now that wind can be cold!) This means that the air molecules are moving slowly among themselves but a large mass of air is moving at a high speed. It is the energy (velocity) wrt the ground which drives an air turbine, not the temperature of the air. So no the energy removed to drive a wind turbine does not NECESSARILY reduce the temperature of the air.
 There is friction at the moving parts of the generator, so there should be heat build up. So, will the temperature of the wind will eventually increase?

## Wind Power generator

From my understanding, most wind generators are not designed to turn rapidly, so there is very little build up of heat.

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 Quote by Integral So no the energy removed to drive a wind turbine does not NECESSARILY reduce the temperature of the air.
Not quite Mr. PF Mentor. The wind generator acts like a turbine rotor. There is an exchange of kinetic energy between fluid and mechanical environment.
In particular, the flow is being accelerated relative to blades. Then pressure decreases as relative velocity to the blade increases. It is possible that such decreasing will be small, because of the constant pressure of the surroundings. The generalized Bernoulli equation for total enthalpy flow shows that total temperature and static temperature will decrease as the flow passes trough blades.

End of first round.
 Recognitions: Gold Member Science Advisor I don't think so. The wind will lose velocity, and this kinetic energy will go to mean places. First, some of the energy (hopefully most of the energy) will be converted to electricity by the generator. Secondly, since no method of power generation is 100% efficient, some energy will be lost to entropy. Most of this energy will be converted into heat due to friction. Friction between the blades and the pair will keep the air directly, while friction within the mechanism of the generator will cause the moving parts to heat up, then radiate this heat out into the wind.

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 Quote by LURCH I don't think so. The wind will lose velocity, and this kinetic energy will go to mean places. First, some of the energy (hopefully most of the energy) will be converted to electricity by the generator. Secondly, since no method of power generation is 100% efficient, some energy will be lost to entropy. Most of this energy will be converted into heat due to friction. Friction between the blades and the pair will keep the air directly, while friction within the mechanism of the generator will cause the moving parts to heat up, then radiate this heat out into the wind.
If you think so, the same reasoning would be employed with turbines. Do you think air behind a blade turbine is hotter due to entropy generation? I hope you don't think that.

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 Quote by Clausius2 If you think so, the same reasoning would be employed with turbines. Do you think air behind a blade turbine is hotter due to entropy generation? I hope you don't think that.
Lurch didn't say the air would heat up, just that there would be heat generated by the interaction of the wind and the turbine - or if you prefer, heat transferred from the wind to the turbine - and then some of it back to the wind.

My take on this is that the turbine does remove some heat from the air (it has to via conservation of energy), but the amount is insignificant.

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 Quote by LURCH First, some of the energy (hopefully most of the energy) will be converted to electricity by the generator.
IIRC, wind power generators have a max efficiency around 35%. I'd have to dig back into my college textbooks to give you the details.

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 Quote by russ_watters Lurch didn't say the air would heat up, just that there would be heat generated by the interaction of the wind and the turbine - or if you prefer, heat transferred from the wind to the turbine - and then some of it back to the wind. My take on this is that the turbine does remove some heat from the air (it has to via conservation of energy), but the amount is insignificant.
The increasing of entropy causes that behind the turbine blades the air is not as cool as it would be if there was not entropy increasing at all (isentropic flow). But, as I stated before, and with generalized (compressible) Bernoulli equation as background of my words, the air is cooled by the turbine blades.

You have said something that has taken me aback. The amount of heat removed by a conventional turbine of air is not insignificant at all.!. In particular the decreasing of total enthalpy is the work per unit mass obtained in the rotating shaft. Do you think this amount is insignificant in a turbine of a nuclear plant?. Surely I have interpreted wrong your words. But in some way you have reason, keep on reading please.

Let's see. Name 1 at blade inlet, and 2 at blade oulet:

$$\tau=h_{t1}-h_{t2}$$; here $$\tau$$ is the work obtained per unit mass and ht is the total enthalpy. This is one of the so-called Euler equations of Turbomachinery. It is similar to energy N-S equation viewed from an inertial frame.

In order to obtain work, ht1>ht2. Thus, the outlet total temperature must be lower than the inlet one. What happens with the static temperature?.

At the inlet: $$\frac{T_{t1}}{T_{1}}=1+\frac{\gamma-1}{2}M_{1}^2$$ where M is #Mach. Moreover: $$\frac{T_{t2}}{T_{2}}=1+\frac{\gamma-1}{2}M_{2}^2$$ at the outlet.

All this stated above is true for isentropic and steady flow. Lately, we will include non-isentropic effects. If Mach# is M<<1, then kinetic energy variations are much smaller than thermal energy itself. So that, $$\tau$$ will be of the order of $$c_{p}(T_{1}-T_{2})$$. Here you can see the reason for my surprising. With isentropic flow, all the work obtained in the shaft is due to static thermal variations at M<<1.

If the flow is not isentropic, the work obtained would be shorten via the isentropic efficiency. So that, the real temperature decreasing must be smaller than that forecasted by isentropic equations. The effect of viscosity on internal dissipation, wall friction, and heat losses is contained in this experimental factor.

I have not said nothing about M>>1. This have not sense, bacause wind generators never have peripheral velocities larger than sound speed. It would cause shock waves at the blade leading edge, and structural bendings. Only high powered rotors, like vapor turbines, have local #Mach supersonic in concrete points.

What happens at M<<<<<1 , I mean, at quasi-incompressible flow?. The Euler equation for isentropic flow is transformed into:

$$\tau=\frac{P_{1}-P_{2}}{\rho}$$. Why? This is because for an incompressible fluid: $$Tds=cdT$$ and $$dh=\frac{dP}{\rho}$$. This is, there is no variation of internal energy at all in an isentropic flowby.

In particular, non isentropic effects would cause a temperature increment. But, as you have said, this increment is very very small. In fact we can obtain an order of this increment:

$$\frac{\delta T}{T}=\frac{\delta P}{P}=\frac{\rho U^2}{P}=\gamma M^2$$

What does it mean?. At very low #Mach, increments of temperature are of the order of the #Mach powered to two.

To sum up, the temperature will remain roughly constant in an hydraulic turbine (hydroelectric power plants), a conventional house fan, compressors at M<<1, and maybe a wind generator at low rotaing speeds.

By contrast, high speed turbines, like nuclear power plant generators, car's turbocompressor, or aircraft's turbines, would cause a severe temperature decreasing.

But in all cases, temperature must be suffer a decreasing.

An example: A 900 KW electric generator connected to some vapor turbine, would need a mass flow of vapor of the order of 10 Kg/s. So that, the decreasing of temperatures will be of the order of:

$$\deltaT=\frac{900.000}{1000*10}=90ºC$$

Best regards for everybody.
 Mentor Blog Entries: 9 If what you say is true then there should be a easily measurable temperature drop through a large wind turbine. Is this the case? So why not set up a large wind turbine up wind of your house to not only to power it, but to cool it on a hot day? Sounds like with a large enough wind turbine we should be able to freeze water in the back draft, why has no one done this?

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 Quote by Clausius2 What does it mean?. At very low #Mach, increments of temperature are of the order of the #Mach powered to two. To sum up, the temperature will remain roughly constant in an hydraulic turbine (hydroelectric power plants), a conventional house fan, compressors at M<<1, and maybe a wind generator at low rotaing speeds. By contrast, high speed turbines, like nuclear power plant generators, car's turbocompressor, or aircraft's turbines, would cause a severe temperature decreasing.
Maybe I'm missing something, but wouldn't the mach number M of a wind turbine necessarily be very low?

The speed of sound at sea level is 761 miles/hour, and if the wind speed driving the turbine is a brisk 20 miles/hour wind, the mach number should be 20/761 = .02

So it seems to me that your own arguments suggest that a wind turbine shouldn't cause much temperature decrease.

Let's look at this another way. 20 mph is about 9 m/s, and the density of air is very roughly around 1kg/m^3 (it depends on temperature). So one cubic meter of air moving at 9 m/s would mass around 1kg and have an energy of 40 joules. Regardless of whether you use a Cp of 1 kilojoule/kg, or a Cv of .7 kilojoule/kg, the maximum possible temperature increase is much less than a degree, somewhere around .05 degrees.

Figures from

http://www.airliquide.com/en/busines...#UnitSelection

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 Quote by Clausius2 Not quite Mr. PF Mentor. The wind generator acts like a turbine rotor. There is an exchange of kinetic energy between fluid and mechanical environment. In particular, the flow is being accelerated relative to blades. Then pressure decreases as relative velocity to the blade increases. It is possible that such decreasing will be small, because of the constant pressure of the surroundings. The generalized Bernoulli equation for total enthalpy flow shows that total temperature and static temperature will decrease as the flow passes trough blades. End of first round.
I'm with Integral. Clausius2, I'm not impressed with your arguments. Integral says temp. won't NECESSARILY decrease, meaning he can imagine mechanical techniques that turn the convective motion of wind into mechanical energy, without changing the air's temperature. Then you say "not quite", and mention a PARTICULAR method in which temperature does decrease. This is obviously not a valid refutation.

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 Let's look at this another way. 20 mph is about 9 m/s, and the density of air is very roughly around 1kg/m^3 (it depends on temperature). So one cubic meter of air moving at 9 m/s would mass around 1kg and have an energy of 40 joules. Regardless of whether you use a Cp of 1 kilojoule/kg, or a Cv of .7 kilojoule/kg, the maximum possible temperature increase is much less than a degree, somewhere around .05 degrees.
This is the sort of temperature change that I would be ok with. Due to mixing behind the blades it would not be measurable.

Does the water temperature change after a water wheel? I think not, if anything it will be increased due to a greater surface area being exposed to the (possiliby) warmer air. This is much more analogous to a wind turbine the any thermal engine.

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 Quote by Clausius2 You have said something that has taken me aback. The amount of heat removed by a conventional turbine of air is not insignificant at all.!. In particular the decreasing of total enthalpy is the work per unit mass obtained in the rotating shaft. Do you think this amount is insignificant in a turbine of a nuclear plant?. Surely I have interpreted wrong your words. But in some way you have reason, keep on reading please.
Well, I don't know what about what I said would have led you to start talking about nuclear reactors. A wind turbine is a wind turbine. Its only a heat engine in the very loosest sense of the term.

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