Maximizing Current Output from a Windmill Generator

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Homework Help Overview

The problem involves a windmill generator and seeks to determine the maximum current output based on the energy lost by the wind as it passes through the windmill. The context includes parameters such as voltage, blade length, wind speeds, and air density.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss energy loss calculations and the relationship between kinetic energy and current output. Questions arise regarding the relevance of blade length, the method to calculate the volume of air, and the implications of not having angular speed provided.

Discussion Status

Some participants have attempted calculations related to kinetic energy and mass flow rate, while others are exploring the assumptions about the cross-sectional area defined by the blades. There is a recognition of the need to clarify how to find the volumetric rate of air passing through the windmill.

Contextual Notes

Participants note the absence of certain information, such as angular speed, and express confusion about the use of blade length in the calculations. There is also mention of a reference answer from a book, which adds to the discussion's complexity.

Hafid Iqbal
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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...
 
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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...
 

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