alokkumar
- 16
- 2
jbriggs444 said:So far we have ##P=346 Watts## arriving in the air stream. But we are only going to capture some 35 to 38 percent of that. Call this capture efficiency ##e_c=0.38##. We also lose 30% to friction. Let us call that mechanical efficiency ##e_m=0.70##.
You've stated that 100% of the remaining input energy is used to generate electrical energy. But that is a problem. Because if 100% of the remaining energy goes into generating power then there is no energy left to increase the rotation rate of the turbine. That leads to a dead end: RPM = 0.
So let us assume instead that while we are busy spinning up the turbine the generator will not be drawing off any energy. That means that the energy flow rate ##P_\text{rot}## going into increasing the rotation rate of the turbine will be given by:$$P_\text{rot} = P e_c e_m$$
If we substitute in the percentages, we down to 92 Watts going into increasing the rotation rate.
You've already done a good job with the moment of inertia calculation. Let's call the moment of inertial ##I##. Three blades times ##\frac{1}{3}mr^2## for each blade and ##m=4 \text{ kg}## and ##r=1.2 \text{ m}## so we have ##I=5.76 \text{ kg m}^2##
After ##t## seconds at an energy flow rate of ##P_\text{rot}## we will have accumulated rotational kinetic energy:$$KE=\frac{1}{2}I\omega^2=tP_{rot}$$If we solve for ##\omega## that gives:$$\omega=\sqrt{\frac{2tP_\text{rot}}{I}}$$Let us fill in the givens:
##P_\text{rot}##=92 Watts
##t## =10 seconds
##I## = 5.76 ##\text{ kg m}^2##
If I have not screwed up, that comes to
$$\omega = 17 \text{ rad/sec}$$after ten seconds. If I have not screwed up, that is 162 RPM.
Of course, we have also determined that your 1 kw turbine is only gathering 92 watts. And that is before the 85% generator efficiency is factored in. That may be a problem.
@jbriggs444
A quick question. In a practical wind turbine system. The electrical generator is drawing continuous energy out of the turbine system.
So in that case wouldn't the rpm of the overall turbine assemble (including generator housing) will reduce ??
I/2*I*omega*omega = t * P(rot) - Energy drawn by generator ??
So I guess omega (i.e. rotational rpm of overall assembly) will reduce.
While chosing appropriate generator for turbine. Do I need to consider rpm 162RPM( in your example). Or generator chosen should have much lower rpm ? Because actual assembly will rotate with much lower than 162rpm (because electrical generator is drawing the energy out continuously)
Your help on this would be much appreciated!