+0

# Physics question

0
184
9
+78

A windmill is a device for extracting kinetic energy from the moving air. The moving air is used to turn large blades. In modern windmills, the turning blades are used to rotate magnets near loops of wire, which creates electric currents. In this way, kinetic energy in the wind is turned into electric energy and fed into the power grid.

Suppose that when the wind blows at 5.0 m/s the kinetic energy per unit time that blows past a windmill is 0.50 MJ/s This is also known as 0.5 MW The W stands for watts. One watt is one joule per second, so the kinetic energy of the air passing by the windmill per unit time is half a megawatt or half a megajoule per second. Energy per unit time is called power, so the watt is the SI unit for power.

Windmills cannot capture all the kinetic energy from the air passing by them. This is because if they did, the air would come to a stop, and would pile up behind the windmill, preventing any more air from getting through.

The maximum possible efficiency of a windmill is called the Betz limit. At the Betz limit, a windmill can extract 16/27 of the kinetic energy of air passing through it. Even this is just the theoretical maximum efficiency; real windmills are not quite this effective.

Suppose that the wind speed increases to 10.0 m/s If the windmill we considered before can operate at 80% of the Betz limit, how many megawatts of power does it produce?

Jul 18, 2021

#1
+2

From reading, here are the useful facts to solve this question:

1-A windmill is a device for extracting kinetic energy from the moving air.

2- kinetic energy in the wind is turned into electric energy. Denote kinetic energy: $$E_k$$ and electrical energy: $$E_e$$

3-Suppose that when the wind blows at 5.0 m/s the kinetic energy per unit time that blows past a windmill is 0.50 MJ/s

Let the initial speed: $$v_{initial}=5.0 m/s$$, and given the initial kinetic energy, $$E_k=0.50*10^6 J$$
Hint: (Although the text is talking about power and energy, they are the same here due to "1 second" (Since P=E/t -->P=E if t=1))

By the Kinetic Energy equation:

$$E_k=\frac{1}{2}mv^2$$, we can find the mass, m of the air. $$m=\frac{2E_k}{v^2}=\frac{2*0.50*10^6}{(5.0)^2}=40 000 kg$$     (1)

4-Windmills cannot capture all the kinetic energy from the air passing.

From this, we deduce that windmills are not 100% efficient (Thus, we will use the equation of Efficiency), and the text explained this further:

5-The maximum possible efficiency of a windmill is called the Betz limit. Given, the theoretical Betz limit: 16/27 of the kinetic energy of air passing through it.

Now to the question:
Suppose that the wind speed increases to 10.0 m/s  (Denote, $$v_{final}=10.0m/s$$)

If the windmill we considered before can operate at 80% of the Betz limit, how many megawatts of power does it produce?

It is crucial to note from this, that it is the same windmill, thus the following is an accurate assumption in solving: mass of air passing through the windmill blades is constant, as found in (1) -->40000kg.

Applying the efficiency equation: $$Efficiency=\frac{E_e}{E_k}$$

Given, the efficiency is 80% of the betz limit -->$$0.80*\frac{16}{27}=0.47407407$$ is the efficiency.

$$E_k=\frac{1}{2}(40000)(10^2)$$ $$=2.0*10^6 J$$

Thus, $$E_e=0.47407407*2.0*10^6=948148.14 J$$

Recall, this is the energy per second hence the power.

So the question wans the final answer to be in megawatts:

$$948148.14J/s=948148.14W=0.94814814MW$$

Which can be approximated to be: $$0.95MW$$

Jul 20, 2021
#2
+78
+1

Ah, that makes much, much more sense now. My answer was a bit different, sitting at $$\fbox{1.9 MW}$$

PBJcatalinasandwich  Jul 21, 2021
edited by PBJcatalinasandwich  Jul 21, 2021
#3
+2

If you are still confused from anything, you can always ask!

Guest Jul 21, 2021
#4
+78
+1

Thank you!

PBJcatalinasandwich  Jul 21, 2021
#5
+2234
0

Neither of you two dumb-dumbs have the correct answer.

The change in power is proportional to the ratio of the cubed velocities.

$\Delta P = \dfrac {V_2^3}{V_1^3}$

$\Delta P = \dfrac {10^3}{5^3} = 8$

Then

$0.5MW *8 = 4 MW$

This is solvable using kinetic energy equations. But you have two errors in your second equation.

Your second equation has an erroneous value for the mass. Also, you multiplied by the coefficient of efficiency instead of dividing. In this case, you should have done neither.

Dividing gives the actual kinetic energy of the air that’s required to produce this power output.  However, if you do this on the second equation you have to do it on the first equation. The results will be the same if the proportions are the same. Despite its mention in the question, the coefficient is not needed to solve this question. In fact, most of the BS written in the context isn’t needed for the question.

$E_k=\dfrac{1}{2}(40000)(10^2) = 2.0MJ$ – Your result without multiplying by the coefficient of efficiency.

Consider carefully: if you have twice the velocity there will also be twice the mass in the same unit of time.

So...

$E_k=\dfrac{1}{2}(80000)(10^2) = 4.0MJ$

------------

Defining how kenetic energy relates to power:

$KE = \dfrac{1}{2} M * v^2$ the velocity may be expressed as $\left(\dfrac{\dfrac{m}{s}}{s}\right)$

And power as:

$P = \dfrac{KE}{\Delta t} | \;\; t \;\; \text { is in seconds.}$

Then

$P = \dfrac{1}{2}*\left(\dfrac{{\dfrac{\dfrac{m}{s}}{s}}}{s}\right)$

Then

$P = 0.5 M * v^3$

------------

Hint: (Although the text is talking about power and energy, they are the same here due to "1 second" (Since P=E/t -->P=E if t=1))

Power and energy are related and one can be derived from the other, but they are never the same. Just as kilometers per hour is never the same as kilometers. It does not matter that the time unit is one (1); they have different units: for power it’s watts and Joules per second. For energy it’s Joules and watt-seconds.

------------

The general equation for power produced by a wind turbine is

$P = 1/2 \pi * r^2 * v^3 * \rho * \eta \;\;| \;\; \tiny \text {where r is the radius of the turbine blade, v is velocity of the wind,$\rho$is the air density, and$\eta$is the efficiency factor.}$

With this equation you can reconstruct the basic parameters of the wind turbine.

This equation was not included, but a lot of BS filler was.

------------

Both the question and the guest answer look like they could have been written with a quill...

That is, written in a narrative style consistent with the lecture circuit of the nineteenth century.

The great masters of physics and science, such as Michael Faraday and Joseph Henry gave public lectures on various subjects. But for every true master there were dozens of morons, idiots and charlatans, who either did know the subject or couldn’t teach it with any kind of coherent logic.

Here’s an example in written form.

https://web2.0calc.com/questions/what-is-the-linear-value-of-one-knot-at-the-equator#r2

The presented question post is factual, but the most of the facts are worthless as they relate to the question.  Anyone who might have the basic skills needed to answer the question would already know them anyway.

Moving from the mid nineteenth century to the mid twentieth century we might find a radio or teleplay encourage children to pursue a career in a science-related field.

Here’s a hypothetical of such a teleplay:

Tommy: What do you want to do when you grow up, Billy?

Billy: I want to build windmills that catch and stop all the wind. I drew a diagram of one with my quill. I based it on my helicopter hat.

Tommy: If you stop all the wind, what will make the windmills go around?

Billy: I already figured that out. We’ll have bunches of electrical power; so all we need to do is plug in big fans and blow them on the windmills.

Tommy: That is so smart. I never thought of that.

Billy: Well it helps to have some physics education.

Tommy: What’s that?

Billy: I don’t know, but someone told me that.

-------------------

Would anyone care to write a sequel? ...Either in the same era or fifty years later.

GA

Jul 23, 2021
edited by GingerAle  Jul 23, 2021
#6
+1

Oh, that must be the correct answer! Thanks for explaining this idea about mass doubling due to time being considered! (Never thought of it before)!
Sorry guys for my completely wrong answer(guest) above. If u can delete it somehow go ahead as I can't.

Guest Jul 23, 2021
#7
+1

Though a question..:
you referred to "v" as velocity, shouldn't it be speed?

(Because K.E. is scalar not vector, and scalar quantities must be of either: Vector x Vector or scalar x scalar).

Guest Jul 23, 2021
#8
+2234
0

There is no need to delete your answer. I usually leave my mistakes, unless they are a major distraction to the continuity of the post. They are uinique monuments to exceedingly rare events.

Your solution is well thought out and explained; you just have mistakes in the application. The more complex the subject, the easier it is to make mistakes. Physics becomes very complex, very fast.  We learn from our mistakes; others learn from them too. It’s part of educating ourselves.

------

After rereading the question, I realize I made a mistake too, so I’m adding my name to the list of dumb-dumbs.

------

The initial energy of (0.5MW) is what passes through the wind turbine system—the input power, not the output power.  Doubling the wind speed increases the input power to (4MW). The input power is then multiplied by the coefficient of efficiency (4.0MW * 0.47407) giving the output power of 1.9MW.  (This is consistent with the OP’s answer, which I dismissed because there wasn’t any associated work product.)

GA

GingerAle  Jul 23, 2021
#9
+2234
0

you referred to "v" as velocity, shouldn't it be speed?

(Because K.E. is scalar not vector, and scalar quantities must be of either: Vector x Vector or scalar x scalar).

That’s an interesting question.

Most equations use velocity to define kinetic energy. Though I have seen speed used. Kinetic energy is produced when mass is in motion without regard to its direction; however, in most closed systems there is at least an implied direction.

In the case of the wind turbine, the implied direction is at zero (0) degrees relative to the face of blades. If the mass of the wind was vectored at 90 degrees to the face of the blade then none of the kinetic energy would translate to the mass of the blades.

GA

GingerAle  Jul 23, 2021