Voltage Dip During Motor Start
Voltage Dip During Motor Start
(OP)
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
Thanks





RE: Voltage Dip During Motor Start
I believe the NEMA standard for minimum control voltage for contactors is 85%. If voltage dips below that, there is some risk of magnetic contactors dropping out or chattering.
David Castor
www.cvoes.com
RE: Voltage Dip During Motor Start
You should calculate a worst case scenario where all your motors are running, and then the largest motor starts. This would be FLA of all motors + about 6 times FLA of the largest motor. Plug that into an equation and see how much voltage drop you get. I don't know what IEEE requires, but if you stay above 85% you shouldn't have a problem.
RE: Voltage Dip During Motor Start
RE: Voltage Dip During Motor Start
The current of the starting motor may be about 20% PF.
The total current will be much less than the sum of the running and starting currents.
Voltage drop calculations based on the arithmetic sum of the currents will be quite conservative.
The available short circuit current as determined by the impedance of the transformer is valid only for bolted fault conditions. For a rigorous calculation of voltage drop under motor starting conditions, consideration must be given to the phase angle of the current and the X:R ratio of the transformer. Again calculations based on impedance alone may be quite conservative.
Pete, how about crunching some numbers for an example:
Take 6 equal motors running loaded at 80% PF.
Take a similar motor starting at 600% current at 20% PF.
Assume a transformer equal to the KVA load of 10 motors.
Give the transformer an impedance of 6% and an X:R ratio of 6:1
What will be the voltage drop using the simple sum of the currents without regards to phase angles and the PU impedance.
What will be the actual voltage drop considering phase angles and the X:R ratio.
If you take up the challenge I will bribe you with an lps. grin
Bill
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"Why not the best?"
Jimmy Carter
RE: Voltage Dip During Motor Start
I tried to simplify my life by assuming the starting and running currents don't change with voltage (otherwise requires load flow, or more work to solve simultaneous equations).
Results show not much difference.
0.928 using algebraic worst-case method.
0.938 using vectors
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
EE
RE: Voltage Dip During Motor Start
Try this
Regards
Carlos
RE: Voltage Dip During Motor Start
[10% less than motor rated voltage.]This includes the common drop and the individual cable drop in steady state regime.
You have to calculate the voltage drop up to motor terminal for a steady state load and to add the voltage drop up to the last common MCC in the case of the start of the biggest motor [or which presents the biggest current at start].
There is a short way-not so accurate and conservative but efficient-taking the maximum starting current.
If the transformer apparent power is 5 MVA at 4.15 kV the maximum motor power supplied from could be 1/4=1025 HP.
Iftrf=5/sqrt(3)/4.15=0.6956 kA transformer rated current.
Pmot=1025*.746=764.65 kw
Motor rated voltage=4000 V. [as per catalogue].
If the induction motor is squirrel cage normal rotor take it as DOL starting.
Irated=0.764.65/sqrt(3)/4/0.8=0.138 Ka [0.8=pf*eff=0.9*0.89] motor rated current.
Istart=8*0.138=1.1037 kA at full rated voltage.
The total voltage drop up to motor terminals take it as 20%, so you can adjust the Istart at 80% Istart=1.1037*0.8=0.883 kA.
The starting pf of the motor could be 0.2 then the Imotorstart=0.883*(0.2+j*0.98)=0.1766+j*0.865 Ka.
Let's say the transformer load in steady state is 5 MVA pf=0.8.
Itrfsteady=0.6956*(0.8+j*0.6)=0.5565+j*0.4174 kA
Imotorsteady=0.138*(0.9+j*0.436)=0.1242+j*0.06 kA
You can extract the starting motor steady state current in order to find the remaining running load current.
and the remaining load current will be: Itrfremain=0.432+j*0.357 KA.
During the start of the biggest [proposed] motor : Itrfst=Itrfremain+Imotst=0.6086+j*1.222 KA.
Let's calculate the voltage drop through the transformer when this motor will start.
In order to find the transformer resistance let's say the transformer copper losses =40 kw then Rtrf%=40/1000/5*100=0.8%.
The transformer reactance will be sqrt(5.6^2-0.8^2)=5.54%.
Rtrf=0.8*4.16^2/5/100=0.0277 ohm transformer resistance
Xtrf=5.54/100*4.16^2/5=0.19175 ohm transformer reactance.
The voltage drop across the transformer will be:DVolt=sqrt(3)*(Itrfstreal*Rtrf+Itrfstim*Xtrf)=0.4388 kV
The voltage at transformer terminals will be 3.72 kV that means 3.72/4=93% of motor rated voltage.
You have to chose the right cable feeder so that at motor starting terminal the voltage will not be less than 0.8*4=3.2 kV.
and for other running motor not less than 10%[4*.9=3.6kV].
If the cable is in acceptable limits than this could be the maximum power motor you may connect to this transformer.
If not a smaller motor should be taken into consideration.
If the utility supply voltage could be less than rated by 5% and your transformer has not possibility to raise the voltage back,
you have to take it into consideration as a supplementary drop [but only for steady state regime].
RE: Voltage Dip During Motor Start
I learned this on Eng-Tips.
The regulation of a transformer describes the voltage drop of a transformer at full load with a resistive load or mostly resistive load. The PU regulation is less than the PU impedance.
With a short circuit the current is limited by the X and R of the transformer. However, with a resistive load, the X of the total circuit remains the same while the R becomes the sum of the transformer R and the load R.
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.
A simple example to illustrate my point.
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)
Bill
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"Why not the best?"
Jimmy Carter
RE: Voltage Dip During Motor Start
If you have 1 pu voltage on the primary and you load the transformer so that it produces rated current on secondary, the secondary voltage measured at the transformer terminals will be less than 1 pu. It will be equal to Vs_rated*(100-%Z)/100. Are you saying that this is incorrect?
EE
RE: Voltage Dip During Motor Start
First of all the reactive part of all current is negative as the load is inductive, of course.
That does not change the results. But what does change the result is the no-load voltage of the transformer. I took it as rated.
It is the full load transformer voltage [4.15 kV] and the no-load has to be 4.323 kV.
In this case the voltage at the transformer terminals will be 4.323-0.4388=3.884 kV [3.884/4.15=93.5%].
RE: Voltage Dip During Motor Start
Use "Z" for available short circuit current.
Use regulation to describe voltage drop of a load with defined power factor at rated current.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Voltage Dip During Motor Start
IEC standard IEC TR 61000-3-7-2008 looks in detail the issues surrounding voltage dip.
Motors can start with huge voltage dips. The limiting factors are the inertia of the load being started, the thermal limitations of the motor, and the limitations of other connected equipment on the same system.
Last thing you want is to dump a bunch of PLC's and computers every time you start your motor.
Other things to consider are the effects of other motors on the bus. Running motors can provide reverse power for an instant when starting additional motors, just the same way they contribute to fault current. What are all the scenarios that could occur?
Many motors come with thermal and torque curves for 80% motor starting.
Heck, some motor starting contactors have autotransformers for reduced voltage starting.
Throw in soft starters, VFDs, and other possibilities and you have many things to consider.
Really the root question is what constraints do you need to meet, and then with that in mind, start looking at the criteria to shoot for.
Modeling tools like E-Tap and SKM can be helpful to analyze the possibilities.
RE: Voltage Dip During Motor Start
Attached is a load-flow solution using Gauss iteration programmed in Matlab.
The results converge to a secondary voltage of 0.948.
The difference between this load flow solution and the vector solution of Sep 11 23:38 is that the vector solution assumed constant pre-defined current for the starting and running motors. The load flow solution more accurately assumes that the starting current decreased according to the reduced voltage on the secondary and the running current increased according to the reduced voltage on the secondary. Since the starting current (primarily inductive) is more in-phase with the transformer impedance, it dominates the voltage drop, so it is expected the reduced starting current results in slightly higher secondary voltage than the vector method (which assumed constant current)
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
It looks like the consensus is anywhere between 10-20% voltage dip, but I didn't know if there was an industry standard that addressed a short-term voltage dip caused by starting a motor.
RE: Voltage Dip During Motor Start
NEMA MG1 requires successful startin at 90%. It doesn't specifically require starting under 80% voltage conditions, although it may be mentioned as an option.
ANSI C50.41-2000 (Specification for power plant motors) specifies motors must successfuly start their load at 85% rated voltage at motor terminals.
EPRI 1011892 ("Guideline for the Specification of Replacement and Spare AC Squirrel-Cage Induction Motors having Voltage Ratings of 2,300V to 13,200V") specifies large motors must successfully start their load at 80% of rated voltage at motor terminals.
API541-95 just refers to NEMA.
At our plant, the AE specification required all motors to start with 80% voltage.
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
If the starting motors are designed to start at a certain level let's say 80%, I'd think that similarly-designed running motors would not encounter any thermal damage from undervoltage during period of starting another motor. The reasons:
1 - the voltage at running motor may not be quite as low as starting motor if starting motor has long run of cable.
2 - the motor is normally designed for at least one hot restart, which is a lot more severe than just running with reduced voltage.
3 - motors are often run slightly below FLA, at least at our plant.
Still, you might want to check how close you'll get to your time overcurrent trips.
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
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)' ?
RE: Voltage Dip During Motor Start
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).
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)' ?
RE: Voltage Dip During Motor Start
Yours Bill.
Bill
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"Why not the best?"
Jimmy Carter
RE: Voltage Dip During Motor Start
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(2B)+(2B)' ?
RE: Voltage Dip During Motor Start
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.