Maximizing Current Output from a Windmill Generator

  • Thread starter Thread starter Hafid Iqbal
  • Start date Start date
AI Thread Summary
To maximize the current output from a windmill generator, the discussion focuses on calculating the energy loss of wind as it passes through the turbine. The wind speed decreases from 12 m/s to 7 m/s, and the air density is given as 1.29 kg/m3. Participants emphasize the importance of determining the volumetric flow rate of air through the windmill's cross-section, which is influenced by the blades' length. The conversation highlights confusion regarding the use of blade length and the calculation of volumetric and mass flow rates. Ultimately, the expected maximum current output is stated to be 77 Amperes, as per the book's answer.
Hafid Iqbal
Messages
15
Reaction score
0
A 120-V generator is run by a windmill that has blades 2.0 m long. The wind, moving at 12 m/s, is slowed to 7.0 m/s after passing the windmill. The density of air is 1.29 kg/m3. If the system has no losses, what is the largest current the generator can produce?
(Hint : How much energy does the wind lose per second?)

Thank you very much...
 
Physics news on Phys.org
What have you tried?

Where are you stuck?

We shouldn't try to help until we have seen that you have tried.
 
SammyS said:
What have you tried?

Where are you stuck?

We shouldn't try to help until we have seen that you have tried.

\Delta \ KE = -\frac{1}{2}.m.(v_2^2-v_1^2)=-\frac{1}{2}.\rho.V.(7^2-12^2)=61.275V (V = Volume of air)

\frac{\tau}{\epsilon} = \frac{N.B.I.A}{N.B.A. \omega} = \frac{I}{ \omega } \ \rightarrow \ \tau= \frac{\epsilon.I}{\omega}=\frac{P}{\omega}

\tau=\frac{\Delta KE}{\omega.t} \ = \ \frac{61.275V}{\omega.t}

And I stuck from that point. Here is my problems :
1. I don't understand what is the use of blades' length
2. How to find volume of air?
3. Is it true angular speed not given?
 
It could be that they want you to assume that the cross sectional area of the airstream is defined by the span of the blades as they rotate.
 
gneill said:
It could be that they want you to assume that the cross sectional area of the airstream is defined by the span of the blades as they rotate.

I still can't get the answer... :(
Anyway, the book says its answer is 77 Ampere
 
Hafid Iqbal said:
I still can't get the answer... :(
Anyway, the book says its answer is 77 Ampere

Start with determining the volumetric rate (m3/s) at which air is passing the windmill cross section. Then turn that into mass rate by multiplying by density. What do you get?
 
gneill said:
Start with determining the volumetric rate (m3/s) at which air is passing the windmill cross section. Then turn that into mass rate by multiplying by density. What do you get?

How to find volumetric rate sir? If I multiply it by density, i wiil get mass rate...
 
Hafid Iqbal said:
How to find volumetric rate sir? If I multiply it by density, i wiil get mass rate...

Volume has units of m3. Cross sectional area has units of m2. Velocity has units of m/s...
 
Back
Top