Voltage Dip During Motor Start 2

[cbark]

We have a 4.16 kV MCC fed from a 5 MVA, 5.6% impedance transformer. There are several motors on this MCC and we are looking to add a new motor of undetermined size. I understand the voltage dip requirements for the new motor, but my question is what is the allowable voltage dip on the 4.16 kV bus. NEMA MG1 states that the motors should be able to operate at +/-10% voltage (at rated frequency), but I'm not sure if this applies to momentary dips caused by other motors starting on the bus or just to continuous operating conditions. Is there some standard that dictates the allowed voltage dip on the bus? I would assume that if the other motors on the bus could start at 80% rated voltage, then it would be able to handle a 20% dip while another motor starts. The main problem I have is that I don't have the data sheets on the existing motors. I know that some are rated at 4000 V and some are rated at 4160 V based on the nameplate, but I don't know what they are actually capable of handling.

Thanks

 

[electricpete]

I have a few items to post on load flow:

Returning to peripheral issue of voltage calcs (Bill’s assignment), I did recheck the load flow and found a few errors which are corrected in my new attachment to this thread. I also found oddly I was incorrect in stating my expectation (17 Sep 11 12:24) that the first iteration of the load flow calculation (based on V2=1.0 initial guess) should match the Smath vector solution where currents correspond to full-voltage values. These are different formulations, the Gauss iteration equation has voltage V2 on both sides..... does not represent the same thing as the vector solution.

Attached is revised load flow solution. I have included some checks at the end to convince myself the solution is reasonable (it satisfies everything we know about the problem).

So the results for this problem are now as follows:
load flow solution: V2 = 0.9377 - 0.0260i, |V2| = 0.9381
vector solution(*): V2=0.938-0.0262i, |V2| = 0.938
algebraic solution(*): |V2| = 0.928

* the vector and algebraic solutions assume a value of starting and running current, and do not adjust it based on V2. In contrast the load flow solution varies the starting current in proportion to V2 and the running current in inverse proportion to V2.

I think it is coincidental that the vector and load flow solutions are so close. They are quite different problems as per the asterisked statement above, even though the change in running current and starting current are opposite directions and cancel somewhat.

The vector and algebraic solutions are only 1% apart. That will be the subject of my next post....


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(2B)+(2B)' ?

 

[electricpete]

Pete: Thanks for the work. What I meant was not to compare vectors with algebraic methods but to compare the correct method (vectors or algebraic) with a simple but inaccurate solution using just the sum of the running and starting currents and the PU impedance. Neglecting phase angles and X:R ratios and voltage regulation.
I know you don't have much experience doing it wrong, but that was what I was asking. As the starting current is mostly reactive I expected much more difference.

Don’t underestimate me, I am very good at doing it wrong ;-).
But in this case, I think I have correctly done it wrong. In other words, I think the algebraic solution that I provided (smath file) does match the simplistic style of calculation that you described where we ignore all phase angle. The results may not match your intuition, but I would suggest intuition is not always perfect when you are working vectors in your head.

Attached I have provided a revised/updated Smath file which provides geometrically-developed correction factors which explain the deviation between the algebraic method and the vector method. Specifically, we develop correction factor CF1 based on the error introduced by adding the two current vectors algebraically. It is derived using the law of cosines: C^2=A^2+B^2-2ABcos(theta).
The 2nd correction factor accounts for the fact that we subtract the product |I*Z| from the primary voltage (1.0) algebraically, rather than vectorially. It also is derived from law of cosines. Also, it should be noted the two error types are not independent. The fractional error introduced in the 2nd stage of subtracing |I*Z| from 1 depends on the value |I*Z| that we’re using which depends on CF1. So the approach is to formulate CF2 as a composite error which includes both the effects of subtracting |I*Z| from 1 algebraically and the CF1 error from formulating I algebraically. Hopefully this is clearer in equation form than in the words here.

These geometrically-derived correction factors fully explain the difference between the 2 approaches and validate the complex (real/imaginary) math to all the decimal places available in this program (4).

Assume that running current is at 100% PF and the starting current is at 0% power factor. The resulting current (6 motors x 100% FLA @ 100% PF and one motor at 600% FLA @ 0% PF) will be 70.7% of the simple sum of the currents. (?2/2)

That example is a little more extreme because of large angle difference between the two currents. Moreover, if you look at my Smath, this particular error only shows up in CF1. The effect of CF1 is reduced when we roll it into CF2 as shown studying the following equation in the smath attachment:
CF2 = Cmag / (1 - |B|/CF1)
the importance of CF1 within CF2 is diminished because |B| =0.0673 <<1.

I know the Maple and Matlab computer reports are tough to read. I think (hope) the smath reports like attached are a little easier to read.

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(2B)+(2B)' ?

 

[waross]

Thanks Pete.
Yours Bill.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

No problem. I spent a little more time on it because the results was a little non-intuitive for me also. Attached is vector diagram, which perhaps brings it more into focus. Sorry for beating a dead horse.

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(2B)+(2B)' ?

 

[7anoter4]

In connection with cbark post of September 17 inquiry:
The maximum voltage drop can be appreciated calculated the starting time compared with maximum stall time permitted. Also, the pull-up torque has to be more than the load torque for the same rotor slip- as minimum motor torque [pull-up torque] usually at slip= 0.7-0.8
NEMA MG10 for all 4 type [A,B,C,D] states as pull-up torque for full rated voltage as 65%.Accually the pull-up torque shown in manufacturer catalogue could be more than 100% but the tolerance is -15% then 85% could be.
The starting time may be appreciated using formula: tstart=wk^2*(nsynch)/Tacc/308. [It is better to use rated rpm instead of nsynch but the difference is negligible]. Where wk^2 is the inertia moment of the motor connected with the load [lb*ft^2].
nsynch=synchron rpm Tacc=acceleration torque =motor average torque-load average torque. Taccav=TM-KL*TL.
According to ABB the average motor torque [approx.]TM=0.45*(Tbrk+Tstart).
TL could be the rated [usually less] and the KL=1/3 for fans and centrifugal pumps or 1/2 for piston pump [or other values depends on the load type].
According to NEMA MG10 minimum Tstart=0.7 and Tmax=1.75 and pull-up=0.65 from rated. So Tacc=(0.45*(0.7+1.75)-1/3)*Tfl
Tfl=Trated- for rated power, rated rpm, rated voltage.
Trated=Pmot*5252/rpm Pmot[HP] Trated[lbs.ft].
The motor torque is [approx.] directly proportional with the square of supply voltage.
Let's take an example. From ABB High Voltage NEMA Motors catalogue:
HP=900 rpm= 1783 kV=4.0 pfrun=0.87 eff=0.959 tstart=8 sec tstall=20 sec. rotor inertia=627 lb-ft^2 load inertia=3108
pfstart=0.12. Tfl= 2637 lb-ft Tstart=1.1*Tfl Tbrk=2.6*Tfl. KL=0.64.
Tacc=(0.45*(Tstart+Tbrk)-0.64*Tfl)= 2717
If the voltage drops to 0.8*Ratedvoltage then Tstart+Tbrk will be 0.8^2*( Tstart+Tbrk) rated but the load will stay 0.64
*Tfl[rated]. Then new Tacc=(0.45*0.8^2*( Tstart+Tbrk)-0.64*Tfl)= 1941.2 lb-ft. and the tstart=(627+3108)*1783/1941.2/308=11.3 sec [less than 20 sec stall time].
If you want 2 starts from cold the calculation is more complicated as the actual heating and cooling has to be taken into
consideration.
About the pull-up torque: since the Manufacturer does not mention it we'll take it 0.65*Tfl as minimum per NEMA.
Supposing the load torque will increase from 0 at start to Tfl linearly and at slip=0.7 will be (1-0.7)*Tfl=0.3*Tfl.
[Usually for fans and pumps Tload=(1-slip)^2*Tfl and in this case at 0.7 slip Tload=(1-0.7)^2*Tfl=0.09*Tfl.]
The motor torque will decrease to 0.65*0.8^2=0.417*Tfl that means the motor will overcome the load.