Steam Turbine Failure South Africa

[maersk]

Originally Posted By: Astro14
In the same sentance, you claim that the forces must be equal and must not be equal. Both statements can't be right...


Yo, cap'n! Have you got a tooth against the Normal Force too?

http://en.wikipedia.org/wiki/Normal_force

What, are we dismantling the whole of physics now?

How large would you want the Normal Force to be? Larger than the weight of the bucket of concrete (I assume a 0 weight cable because I'm a noobster so I can) it's keeping up?

Would you like for the steel cable to coil in on itself and hoist the bucket up to the crane hook on its own?

I don't understand your conundrum.

Originally Posted By: Astro14
In fact, while you "feel" the Pseudo force, there is only one force acting - the force that changes your direction...


Well, how about if I used a strong rocket engine to squish you against a rather big asteroid (with lots of mass and, consequently, inertia but still a lot less gravitational acceleration than the moon -> negligible weight).

How fake would you consider the Normal Force stemming from the surface to be then? Really fake, not so fake?

I mean, the Normal Force would only be the result of the asteroid's rather large inertia resisting acceleration by the rocket engine I strapped to your back so you should be fine, right? Inertial forces being fake and all.

Originally Posted By: Astro14
remember, we're describing motion here...not static tension of a hanging bucket.


That's right, drag the goal posts with you as go. Can't be too careful, you know!

Originally Posted By: Astro14
Don't confuse the issue (or yourself) by using the word "force" ambiguously...stick with the force definition from lessons 1 and 2.


Why? Are you formally declaring the Normal Force to be fake as well?

BTW, you still haven't touched on the other inertial forces I mentioned (drag, lift, thrust, etc.). You know, the ones that exact noticeable acceleration or deceleration.

Originally Posted By: Astro14
Newton's second law of motion applies. A force acts on you to continuously change your velocity (direction of motion).


So no force is acting on a styrofoam cup being compressed by the weight of 2000 metres of ocean? Why, just because it isn't going anywhere?

Originally Posted By: Astro14
In the case of the bucket there is no NET force - it is static - it is hanging without acceleration....so we're not talking about motion....F=MA. If A=0, then the other side of the equation (F) must also be zero!


Wow! You've entangled yourself quite nicely there; painted yourself in a corner, as it were. You're basically saying that because I'm not hurtling down to the centre of the earth then gravity doesn't exist. Nice one.

Originally Posted By: Astro14
This is why the definitions and principles matter so much.


Not really. Just like with software development, there are no such things as all-encompassing and always valid definitions and principles.

It's like saying a constitution will never need amendments after its final draft.

 

[Astro14]

I know what the normal force is...do you want to learn this stuff or not?

From a guy who has experienced high-G in multiple-axis motion and the effects of inertial coupling, please spare me the disingenuous squish me against an asteroid example...I know what is the human body feels, including the somatogravic illusions of flight...

Stop picking at each bit I say, as you begin to understand, then I will be able to make it all clear for you...you are not reading to understand, but to debate...and this is not a debate.

The definitions and principles are not a matter of debate.

You let me know if you're ready to start listening and understanding...

 

[Astro14]

I made a mistake in introducing the concept of vector addition too early, so let's make this simple....and let's pick the same reference frame: an observer standing on the ground.

If there is no NET force, then there is no NET acceleration. The hanging bucket is static. It is not moving. So, the solution for the equation F=MA, where A = 0, and M has value must be F = 0.

If I get rid of one of the forces on the bucket, it will accelerate (cut the string, it falls...get rid of gravity and the string goes slack) It only takes one force to make an acceleration happen. Balance the forces, use equivilant forces, and there is no net acceleration because there is no net force, they cancel each other out.

The turbine was rotating. Third lesson, remember? We determined that rotation (circular motion) is ACCELERATED motion. There must be an acceleration, or each of the parts of the turbine would not be constantly changing direction. So, F=MA. A has value. M has value, so F has value (it is not 0).

Just as the string pulling the rock on the hockey rink is only exerting one force, by which the rock accelerates. If there was an equal and opposite force (the pseudo inertial force) on the rock, it would not move...it would be static, like the hanging bucket.

The only thing needed to tear the turbine apart was the centripetal force. The centripetal force is M VSquared / R - so for our turbine, when v increased - the one force increased proportional to V squared...

If I pulled hard enough on that string on our hockey rink rock, I would be able to tear it apart too...that's what is so elegant and fascinating about physics...

Here's an airplane example: If I pull G in an F-14, it follows a curved path (an accelerated path) with respect to the observer on the ground, but also from the pilot's perspective. The only way that it can follow a curved path is because there is a NET centripetal force towards the center of the turn.

This isn't simple...in level flight, lift = gravity. There is no net force to change direction. To pull up into a loop, the pilot increases lift and creates a net force (centripetal) towards the center of the loop. The airplane stays on its curved path. When the airplane is pointing straight up, gravity is not acting to add or subtract from the lift that determines the airplane's path.

BUT, when at the top of the loop, the pilot can ease off the controls and create zero lift...what happens then, when the airplane is inverted and there is no lift? There is Gravity, that is pointed towards the center of the loop, and gravity itself is centripetal and the airplane continues on a curved path...the pilot "feels" zero G...he "feels" weightless...but the airplane is in accelerated motion.

Understand that the radius of the curved path is a function of V squared and force. Solving the above equation and re-arranging the terms, and substituting in F=MA, we are left with R = VSquared / A. So, as A (which we measure in the NET Acceleration, not the total G) increases, radius decreases....this is why fighter pilots pull lots of G - to decrease turn radius (and rate of turn...but we haven't covered that yet...) in a dogfight...

Now, here is where it gets interesting: If you want to fly a constant radius (and the airspeed remains the same, with afterburner...we can assume that part to keep this simple for the noobs...) you need to have a constant NET A....so, at the bottom of the loop, if the pilot "feels" 5 G, there is 4 G NET centripetal acceleration. When the airplane is going straight up or straight down, the pilot "feels" 4 G with a net 4G towards the center. When the airplane is inverted, with net 4G towards the center, the pilot "feels" 3G because there is gravity pulling it towards the center.

The path that is inscribed by the airplane is the result of a constant 4 G centripetal acceleration...but the "feel" changes throughout...

When you look at this turbine - and the VSquared involved and the mass involved, the centripetal force was enormous...that alone was enough to destroy it. There was no pseudo force countering the centripetal force to create tension...it's not required and it doesn't exist...if the opposite force of centripetal force, centrifugal force, existed, then there would be no net force, and no acceleration and no circular motion...

 

[maersk]

Where's your reset button?

 

[Astro14]

Originally Posted By: maersk
Where's your reset button?


I'm a fighter jock...I don't have one...

 

[javacontour]

Is this why round trip tickets are cheaper than one way tickets? Work in a closed loop path is zero.

 

[Astro14]

Touche'

 

[onion]

Still looks like semantics and convention to me.

And that's fine. There's nothing WRONG with the way ya'll are setting up and describing the problem. It's definitely 'better' than showing a centrifugal force. But I can show one. And I have.

Maybe it ain't a "real" force (whatever that means). But it can still break parts. I reckon that's real enough.

 

[Shannow]

It's not, and seeing as it's not, it can't.

 

[Astro14]

Originally Posted By: onion
Still looks like semantics and convention to me.

And that's fine. There's nothing WRONG with the way ya'll are setting up and describing the problem. It's definitely 'better' than showing a centrifugal force. But I can show one. And I have.

Maybe it ain't a "real" force (whatever that means). But it can still break parts. I reckon that's real enough.


Centrifugal force ain't real at all...It can't break parts because it doesn't actually exist...I've been trying to describe what does exist...and what actually DOES break those parts...the truth, the facts are more elegant and subtle than the simple explanation of centrifugal force.

Look, if I were wrong - GPS wouldn't work, among many other things, but physics is applied in everyday technology and life, and most folks simply aren't aware of the complexity of the world around them or how it actually works.

Sorry you don't get it...but remember how, in Galileo's day, people were put on trial for thinking that the Earth went around the sun...

Everyone, I mean everyone, including the Catholic Church, just knew, they KNEW that the sun went around the Earth...they could see it rise every morning and set every night, so they knew that it was real and that they were right...they could see and feel that they were right.

Despite how those crazy people with telescopes explained it...

http://en.wikipedia.org/wiki/Galileo_affair