Induction - Squirrel Cage Motor Torque 2

[Wedoca]

knowing Squirrel cage have high inrush current due to low rotor impedence and rotor impedence is equal to

Xrotor + Rrotor and since X = jWL = j 2 pi f L
Rotor impedence is at its max when f reachs the high point and vis versa, and as the (which above I can prove mathmatically)

""Rotor impedence decrease Rotor Current increases and with that Torque decreases ...so Rotor Current goes up Torque goes Down ....Rotor Current goes down Torque goes up .."""

having hard time figure out the above statement (proving it mathmaticly ...) can anyone help?

 

[Skogsgurra]

Yes. Just you wait. Pete will do that easily.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[electricpete]

""Rotor impedence decrease Rotor Current increases and with that Torque decreases ...so Rotor Current goes up Torque goes Down ....Rotor Current goes down Torque goes up .."""

This must be referring to the region where between zero speed (s=1) and breakdown torque, because that is the region where torque and current go in opposite directions as we vary slip.

In addition to your expression for impedance of rotor (Rrotor + j * 2 pi f Lrotor), we have to remember that the voltage induced in the rotor is proportional to slip.

So in the region you referred to (between zero speed and breakdown torque)…
As we increase speed:
[*]Slip decreases
[*]Induced rotor voltage decreases
[*]Rotor current decreases
[*]* Rotor power factor increases by a larger fraction than current decreases, resulting in an increase in rotor mechanical power output
[*]Torque increases.

That is not a proof, just statement of qualitative changes in region of interest.

The only proof I know of would come from the equivalent circuit. You’ll recall in that case, we refer to rotor impedance to the stator which gets rid of the slip-dependent voltage transformation, and changes rotor impedance to Z2 = L2 + R2/s. You can get expressions for current in terms of s, power in terms of s, torque in terms of s.

If you look at 1st and 2nd derivatives of torque with respect to s, you can see how it changes with s, including increasing with speed in the range you discussed. It can be found in any motor textbook. If you want me to cut/paste a bunch of equations, or have a more specific question let us know.


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(2B)+(2B)' ?

 

[Wedoca]

Pete, would you please show me the expressions of current , power , turque in terms of S ( in Zero to Breakdown Torque region )

since you first common , I have went back looking into couple books but yet , I am still having a hard to making connection between them Thanks

 

[Wedoca]

to be more specific , comparing a S-cage motor to a slip ring motor....additional resistance are added to the rotor resistance to increase starting torque ( during take-off) of which will limit the inrush current.. but since Power is equal to torque * internal angular velocity W( when slip equal its MAX @1 which I am assuming W at this time is consider a constant ) with rotor voltage been a fixed vaule at that point of time, power is squared to the size of current ( P=I^2*V)....with that been said ...I dont see how adding external resistance ( reducing Current ) can help increase Torque .......I hope you understand what I am trying to say....

 

[electricpete]

First I will address the original question which I understood to be: why does Torque increase with speed and current decrease with speed in the range of speeds below breakdown torque.

Symbols:
Tbd = Breakdown torque
sTbd = slip at breakdown torque
R1 = stator resistance
R2 = rotor resistance
X1 = stator leakage reactance
X2 = rotor leakage reactance
X = X1 + X2 = total leakage reactance
All parameters referred to stator
w_sync = sync speed = 2*pi*(2*LF)/poles
w_mech = mechanical radian speed = 2*pi*RotorSpeed = (1-s)*w_sync
Pag = airgap power
Pmech = mechanical output power
Tmech = mechanical torque = Pmech / s
Tag = Pag / wsync... “airgap torque”....analogous to mechanical torque expression, and will be proved equal to Tmech

ASSUMPTION: neglect magnetizing branch:
JUSTIFICATION for assumption: In the region of interest s<sTbd, the current in magnetizing branch is much less than current in rotor branch.

Equivalent Circuit #1, neglecting magnetizing branch:
V1---R1---X1---X2---R2/s

ASSUMPTION: neglect R1
JUSTIFICATION: R1 is small compared to the series elements.

New equivalent circuit #2
V1---X-----R2/s

The airgap power entering R2/s element has 2 components: I^2*R2 losses and mechanical output power. To emphasize this distinction, break R2/s into pieces:

Equivalent circuit #3
V1---X-----R2---- R2*(1-s)/s

From equivalent circuit #2:
I = V / sqrt(X^2 + R2^2/s^2)

From equivalent circuit #2:
Pag = I^2*R2/s

From equivalent circuit #3:
Pmech = I^2*R2*(1-s)/s

Prove that Tag = Tmech
Tag / Tmech = [Pag / w_sync] / [Pmech / w_mech]
Tag / Tmech = [<I^2*R2/s> / w_sync] / [<I^2*R2*(1-s)/s> / <s*w_sync>]
Tag / Tmech = [<I^2*R2/s> / w_sync] / [<I^2*R2*(1-s)/s> / <(1-s)*w_sync>]
Tag / Tmech = 1
Tmech = Tag

Tmech = Tag = Pag/wsync
Tmech = (I^2*R2/s) / wsync
Substitute our expression for I:
Tmech = V^2 *R2 / [s*(X^2 + R2^2/s^2) *wsync]
Tmech = V^2 *R2 / [(s*X^2 + R2^2/s) *wsync]
Tmech = V^2 *R2/wsync *[(s*X^2 + R2^2/s)]^(-1)


Find breakdown torque as point where d/ds(Tmech) = 0
d/ds(Tmech)= d/ds{V^2*R2/wsync*u(s)^(-1)}
where u(s) = [(s*X^2 + R2^2/s)]
Using chain rule
d/ds(Tmech)= -V^2*R2/wsync*u(s)^(-2)*du/ds
d/ds(Tmech)= -V^2*R2/wsync*[(s*X^2 + R2^2/s)]^(-2)*[X^2 – R2^2/s^2]
d/ds(Tmech) is 0 where [X^2 – R2^2/s^2] = 0.
d/ds(Tmech) is 0 where s = R2/X
s_Tbd = R2/X

We can do a very easy double-check of this result... airgap torque is maximized when airgap power is maximized, which occurs when the component R2/s matches the source impedance X.... i.e s = R2/X

So, now let’s examine the range of interest:
speed between 0 and breakdown torque
slip between 1 and s_Tbd
R2/X = s_Tbd < s < 1

How does current act for decreasing slip = increasing speed in this region?
I = V / sqrt(X^2 + R2^2/s^2)
I decreases as s decreases ... easy answer.

How does torque act for decreasing slip = increasing speed in this region?
Tmech = V^2 *R2 / [(s*X^2 + R2^2/s) *wsync]
The 2 terms of the sum s*X^2+R2^2/s tend to go opposite direction. But we can see the sum of these terms is dominated by the larger term. The larger term will be s*X^2 for s large and R2^2/s for s small. Our assumption of the region of interest is s>sT_bd=R2/X. This puts us in the large s range where the s*X^2 term dominates the behavior. As speed increases, s decreases, the denominator decreases, the torque increases. This is proof of your original question.

to be more specific , comparing a S-cage motor to a slip ring motor....additional resistance are added to the rotor resistance to increase starting torque ( during take-off) of which will limit the inrush current.. but since Power is equal to torque * internal angular velocity W( when slip equal its MAX @1 which I am assuming W at this time is consider a constant ) with rotor voltage been a fixed vaule at that point of time, power is squared to the size of current ( P=I^2*V)....with that been said ...I dont see how adding external resistance ( reducing Current ) can help increase Torque .......I hope you understand what I am trying to say....

Let’s look at the expression for Tmech
Tmech = V^2 *R2/wsync / [(s*X^2 + R2^2/s)]
When s=1 (starting conditions) we have
Tmech = V^2 *R2/wsync / [(X^2 + R2^2)]
How does it change with R2 when s=1?
The numerator and denominator both tend to increase with R2, which at first glance gives unknown result.

We need to make an assumption that we are varying R2 only over a range where s_Tbd remains less than or equal to 1. s_Tbd = R2/X <= 1.
R2 <=X

Since R2<=X, the change in R2 tends to have only a small effect on the denominator and ceratinly less than a linear effect. In contrast, the numerator has a linear effect when changing R2, so the numerator has bigger effect and it wins. Increasing R2 results in increasing locked rotor Torque under our assumption s_Tbd <=1


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(2B)+(2B)' ?

 

[electricpete]

If there was any jump in the logic that didn't make sense, fell free to ask and I will provide more details.

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(2B)+(2B)' ?

 

[electricpete]

My power and torque expressions represented power and torque from a single phase. Of course the total power and torque would be obtained by multiplying these by 3. Doesn't affect these qualitative conclusions.

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(2B)+(2B)' ?

 

[electricpete]

Correction

Tmech = mechanical torque = Pmech / s

should've been

Tmech = mechanical torque = Pmech / w_mech

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(2B)+(2B)' ?

 

[Wedoca]

Tmech = Tag = Pag/wsyncTmech = (I^2*R2/s) / wsyncSubstitute our expression for I:
Tmech = V^2 *R2 / [s*(X^2 + R2^2/s^2) *wsync]
Tmech = V^2 *R2 / [(s*X^2 + R2^2/s) *wsync]
Tmech = V^2 *R2/wsync *[(s*X^2 + R2^2/s)]^(-1)

Substitude for I,
what was the expression ???

I have difficulty deriving V^2 into the equation

 

[Wedoca]

never mind , I see it now

 

[Wedoca]

correction to my question......

how do you derivate

From equivalent circuit #2:I = V / sqrt(X^2 + R2^2/s^2)

 

[Wedoca]

never mind again.....I got it .....
I just realized how much I have forgotten over the years ....maybe because I was probably never good at it the first place

sorry for the confussion

 

[Wedoca]

for

"airgap torque is maximized when airgap power is maximized, which occurs when the component R2/s matches the source impedance X.... i.e s = R2/X"


if that is the case, then if I take the first derivative of the P_gap with respective to S. I should recieve the same Resistance and Slip relationship as Max Torque(s = R2/X) which you stated that

"its when the component R2/s matches the source impedance X"


Source impedance X, you are referring to the rotor leakage reactance correct??? just need to make sure of that

 

[electricpete]

Questions are good. That shows you are trying to work through it to make sense for yourself. That is worth more than anything I can tell you.

"airgap torque is maximized when airgap power is maximized, which occurs when the component R2/s matches the source impedance X.... i.e s = R2/X"

if that is the case, then if I take the first derivative of the P_gap with respective to S. I should recieve the same Resistance and Slip relationship as Max Torque(s = R2/X) which you stated that

Correct.
Two different approaches that I used above both agree:
(1) - Using derivative of torque with respect to s, we calculated max torque occurs at s = R2/X
(2) - Using max power transfer theorem, we calculate max Pag occurs at s = R2/X. Since Torque = Pag / wsync, this is also where max torque occurs.

Instead of taking derivative of torque with respect to torque as I did in (1), you mentioned taking derivative of Pag with respect to s. This would give the same result as taking derivative of Torque with respect to so, because Pag and mechanical Torque differ by only a constant (wsync).

"its when the component R2/s matches the source impedance X"

Source impedance X, you are referring to the rotor leakage reactance correct??? just need to make sure of that

X would be total leakage reactance X = X1 + X2 as defined in the symbols at the beginning.

The max power transfer theorem tells us that in order to draw the maximum power from a a given source with Thevinin impedance Zth, we need to connect a resistance R =|Zth|.

In our case the element R corresponds to R2/s (because Pag is power transferred to element R2/s) and the source impedance Zth corresponds to X = X1+X2 (according to equivalent circuit #2 above *).

So we maximimize airgap power (and torque) by selecting R2/s equal to X=X1+X2.... and therefore max torque occurs at s = R2/X.

(* equivalent circuit #2 included some simplifications for simplicity. If we used the full equivalent circuit and calculated the Thevinin impedance, we would come up with a slightly different number, but not significantly different.)


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(2B)+(2B)' ?

 

[Wedoca]

Pete, if I understand this correctly.(S= R2/X)which defines the maximium torque and power of which happends when S = 1 when R2 match with X. I do not see that being consistant with the torque VS. Speed graph. because when S = 1 (RPM=0) thats when motor is energized and right before take off. and if you look at the graph , max T (Tdb Breakdown Torque) does not happend at that instant. I am trying connect all the pieces together from math to the graphs. I am seeing gaps , am I missing something here???

another thing is that do you know any computer software that would allow me to plot these equations ? I would like to generate a graph like the attachment is showing.