mivey
Senior Member
At least twice. I would have bet more.
Power factor correction experiment
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You're right.
There were previous references.
But I should have thought the simple explanation in post #319 ought to have covered it.
T
T.M.Haja Sahib
Guest
But still your 'explanation' is wrong.See page no 281 & 282 of the reference in your post # 301 below.
http://forums.mikeholt.com/showthread.php?t=140830&page=9
It states that motor torque is directly proportional to the square of voltage in the stable operating portion of motor curve.But you applied that relation in the unstable portion and arrived at the wrong conclusion that motor current is always three times full load motor current in that portion of motor curve.Isn't?
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Really??
Look at equation 6.43.
It gives an expression for developed torque for any value of slip. So, pick a specific value for slip,substitute it in the equation and you can clearly see that torque is proportional to V2.
Equation 6.44 gives a value for torque at s=1. Also proportional to V2.
And s=1 would not generally be considered to be in the stable operating portion of the motor curve.
T
T.M.Haja Sahib
Guest
Well,you are correct.There seems to be one last point not yet considered.That is derating of motor...............
Well, it's been a long time coming and it makes a refreshing change to some of your less than complimentary comments.
Then, by all means, consider it and present the results of your intellectual machinations on this rather weighty matter.
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- Tennessee NEC:2017
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- Semi-Retired Electrician
"Weight" a minute, will the conclusions prove the savings on a power saving contraption?
Stay tuned, and don't let the conclusion "slip" by.
- Location
- Tennessee NEC:2017
- Occupation
- Semi-Retired Electrician
Are you trying to cause trouble Bill?
Roger
Probably! Go ahead and spank me, or get me some work for today!
Let the slip be low
The right way to go
It's good for the motor
And won't cook the rotor
And all will stay cool
When you stick with this rule
Don't make the slip big
It'll burn out the rig
It will go for repair
And cause you despair
The cost will be high
Accountants will sigh
But the cause of your plight
Is a rewinder's delight.
T
T.M.Haja Sahib
Guest
But,that you still hold on to your erroneous idea of '3 times full load motor current at high slip for all induction motors makes me more and more amused and still more determined to relieve you of it.
First see from the equivalent circuit diagram of the motor that
Motor torque =(constant)current squared (I2)*rotor resistance (r')/slip(s).........(1)
Load torque=(constant)(1-s)(1-s)..................................................................(2)
From (1),(2)
Motor Current I=(1-s)*SQRT(S)/SQRT(r').....................................................................(3)
By differentiating (3),
the maximum motor current is found to occur at slip of 0.33.The ratio of this maximum current at this slip to the full load current at slip say 0.05 is 1.75 and not 3.
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kingpb
Senior Member
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- SE USA as far as you can go
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- Engineer, Registered
In the beginning there was some interesting dialogue, but it has gone to the gutter, I vote this thread should be closed..................
I don't believe I made such a sweeping generalisation.
Post #319
Note the words in bold. Particularly typically .
Not as you have stated all induction motors .
If you want to disagree with my points that fine. But please don't attribute comments to me that I have not made.
Your agenda has been very obvious. You're on a hiding to nothing.
I'm tempted to ask dwhat/dwhat......
The speed torque curve tells you what torque the motor develops at all values of speed or slip.
It is not dependent on load torque. Equation (2) is thus irrelevant. It obviously follows that (3) is invalid.
It is what it is.
The actual real motor curves I posted indicate otherwise.
Look again at say, #177. Maximum current is at s=1.
This is normally the case.
Might point out a very simple and obvious error here?
At standstill s=1.
Thus 1-s=0
So
I=(1-s)*SQRT(S)/SQRT(r')
gives
I=0*SQRT(S)/SQRT(r')
i.e I=0
At standstill.
And I'm just sure you know that ain't so, buddy.
readydave8
re member
- Location
- Clarkesville, Georgia
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- electrician
and then it came out of the gutter, I vote for more puns and poetry.
Though your wish is my command
I can't do verses on command
I try my best to make them good
And avoid the downright rude
My lame attempt at wit finds deaf ear
Especially from those not near here
But I'll still l continue to try
I'll post pictures of my supply
I have a good measuring device
The waveforms aren't really nice
Rather really bad
Totally sad
But more than a hunch
Numbers to crunch
The results I have posted
No need to get roasted
But yet I do
By you know who
In the past I shouldn't dwell
I think I played my part quite well
Maybe a response will come my way
Maybe on another day
Watch this space
I do with grace,
T
T.M.Haja Sahib
Guest
No,sir.We have to consider only that value of slip at which motor torque is equal to load torque.Obviously at s=1,the load torque of fan type load is zero;but motor torque # 0.
I want to submit to your kind consideration that the connection of a PFC capacitor across the terminals of a centrifugal pump set may not bring about any reduction in current taken by its motor but actually may increase the motor current,because with slight reduction in voltage,the motor current also drops and increase in voltage causes increase in motor current due to variable nature of its torque requirement.
So we have to consider constant torque type drive such as reciprocating pump set or compressor type pump set used in residence.With such a pump set,a drop in voltage causes a rise in motor current.So connecting a PFC capacitor across its terminals can improve its voltage and thereby bringing down motor current and associated energy saving.What remains now is to calculate that energy saving.
So you accept that your equation (3) is erroneous? It very obviously is, of course. Are you beginning to see the light?
The speed torque curve, such as those I have already presented, gives you what the motor torque is, repeat is, at any speed regardless of how the motor is loaded.
That may seem counter intuitive. But it is what it is.
The load will determine at what point on that curve the motor will settle. But it does not determine what that curve is.
In normal circumstances, that is close to synchronous speed. A slip of 0.33 is not normal circumstances. And not normally the slip at which maximum current occurs. See previous posts.
Tempted to do the Dalek impersonation "does not compute"
The usual reason for fitting PFC is to reduce the current in the supply*. That, in turn, would lead to reduced voltage drop in the supply, all other things being equal so motor terminal voltage would go up. Without rehashing the whole thread, as iWire has pointed out, you can't draw any definitive conclusions about which way the current will go.
Fine. Calculate away.
*The most recent lot of PFC capacitors we installed was to correct the power factor on a couple of 300kW motors. The motors were to be replaced with a different type with lower power factor and the unit transformers supplying them were of insufficient kVA rating to deal with the lower PF.
T
T.M.Haja Sahib
Guest
No.I am afraid that you do not understand.But I do not want to continue our argument here,because it is no more relevant for this thread and if you are pleased,let me start another thread exclusively for this. Do you agree for this?
I'm sure it's pretty clear to all and sundry who it is that does not understand.
To put up an equation that yields motor torque equal to zero at standstill rather displays the depth of that absence of understanding.
The beginning if understanding is the acknowledgement of ignorance.
You failed to take that step.
I don't mean that unkindly.
As you have clearly and repeatedly demonstrated, it simply isn't your field.
Go in peace.
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