russ_watters said:
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.
