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

 

[dpc]

I would use 90% at the bus unless you want to do a lot more work.

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

http://www.cvoes.com

 

[eeprom]

The problem with voltage drop is usually limited to motor starts. If you have a 10% drop steady state, you are probably overloading the transformer.

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.

 

[RRaghunath]

Generally speaking, 85% at the bus bars and 80% at the motor terminals is the allowable minimum voltages during motor starting.

 

[waross]

The current of the running motors may be about 80% PF.
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
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Hmmm, well attached is a quick attempt using smath free math program.

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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[eeprom]

I understand that the total current would be less than the sum of the starting currents and running currents. I don't consider the low power factor of the starting motor because I would rather be overprotective. But why do you think the X/R of the transformer ratio would affect the magnitude of the voltage drop? The transformer has an impedance of 5.47%. At transformer FLA, the voltage will be 94.53% of no load voltage. This would be true if the X/R ratio was 1, 10, or 40. The only difference that I can see would be how much real power was absorbed by the transformer. Obviously you see something different. What do you see?

EE

 

[petronila]

Hello cbark

Try this

Regards

Carlos

 

[7anoter4]

The voltage shall not drop more than 10% at the terminals of a running motor as per NEMA requirements.
[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 beVolt=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].

 

[waross]

Re X:R ratios:
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
--------------------
"Why not the best?"
Jimmy Carter

 

[eeprom]

Waross,
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

 

[7anoter4]

I have to do some corrections in connection with my above post.
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%].

 

[waross]

eeprom Yes you are correct but "Z" is not just the "Z" of the transformer, it is the "Z" of the transformer plus the "Z" of the load. Voltage drop at full load is described by the regulation of the transformer and is less than the impedance voltage described by the PU impedance of the transformer.
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

 

[gobigorgohome]

The real question is, what other equipment of concern does the voltage dip affect. It's a question of what other stuff on the bus will be affected negatively if the bus voltage drops too much. For instance your point of common coupling with the utility. Utilities don't like to see voltage dips bigger than 6% more than 4 times per day. Less than that if the starting happens multiple times per day or per hour.

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.

 

[electricpete]

Regarding the example problem discussed 13 Sep 11 10:52 and 13 Sep 11 23:38...

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)' ?

 

[electricpete]

The very first solution of the Gauss iteration computes V2 based on the "initial guess", which was V2=1.000. It should have matched the vector method solution (0.938), but it did not (0.949). I think there must be an error in there somewhere. Sorry about that.

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(2B)+(2B)' ?

 

[cbark]

Thanks for the valuable information in all of the posts, but my main question was the allowable voltage dip on the bus. From what I've read on several posts here, it looks like it really depends on what is on the bus and, in this case, there are only motors. I don't have the data sheets on these motors so I don't know what voltage level they are capable of handling. According to NEMA MG1, the minimum requirement is +/-10%, but I didn't know if this actually applied to short-term dips or just for continuous operating conditions.

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.

 

[electricpete]

At what voltage the motors are capable of successfuly accelerating their loads depends on what you specified when you bought them (for large motors like 4kv)

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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[electricpete]

The comments of my preceding post applied to the starting motor.

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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[electricpete]

There are also some transient phenomena like re-acceleration (control valves opening in response to reduced flow, motors reaccelrating when voltage returns) that are usually ignored. Depends how much you want to sharpen your pencil.

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