Hello all, this is my first post he

[Genosphere]

Hello all, this is my first post here so be gentle. Currently enrolled in an Electrical Engineering program and finally finishing up next year. I have 7 years of avionic and airframe experience in the aerospace industry and more hours around automotive junk. I have worked plenty with coding, CAD design, and PCB board design. I'm not a idiot, but some times I need to warm my brain up like an old vacuum tube.

Anyway,
I'm looking for some direction on how to go about the following project. I am suppose to find two motors(just data not buy them) from online to use as references, which I am suppose to determine which of the two is best suited at the end of report(and why). I am not sure how to go about this, I am not looking for an exact answer per say, but rather some direction on how this should be presented with formulas and data. I have done a Design Report in another class, but not sure if that will fit this professor's "criteria."

I was not given any sort of references, examples, or rubric so I am at a bit of a loss on how its should be laid out, organized, and how the math would/should be expressed. So any help to get me started and guide me along the process would be great.

So what I know right now is by motor starting he's talking about how the coils or stators are wired. This article shows the main methods from what I see. 4160 @ 800 is specific and a very real world size motor rating. An PF(power factor) is going to play a critical role here as it usually does.

Again, I'm not looking for an actual answer, but guidance and some explanation to help me learn how one goes about figuring out a "real world" problem like this. Learning how to figure this out is the goal here.

Thanks,
Geno

p.s.(This class has been all over the place and an absolute nightmare in terms of organization and consistency, which is a shame since I was looking forward to electrical power, so this info from the email below is all I have to go off of.)

~~~~The Project Info~~~~~

You are the utility with an end customer who has a 4160V motor that they need to start. You must list advantages and disadvantages of each type of motor starting. Determine the best solution and give detailed design for the motor starting. The customer will be at the end of the line 12.47kV with 1000 Amperes of fault current available.

The customer has an 800 hp motor, LRC = 6 Times FLA, FLA = 1kVA=1HP. Must calculate FLA of the motor and give recommendations on how the motor should be started with detailed design. You must calculate the transformer rating as well.

Include the following: System problem, Motor starting types with advantages and disadvantages of each, Conclusion for each and detailed recommendations with design.

 

[zeusfaber]

DANGER!

Read the question again. This is not a question about comparing two motors. This is a question about comparing two different types of control gear for a motor that has already been specified.

A.

 

[electricpete]

From your link:

> This huge current at the starting of a motor can damage the motor windings and also this current can cause a large voltage drop in the line.
> These voltage spikes may affect the other appliances connected to the same line. Therefore, a starter is necessary to limit this starting current to avoid damage to the motor as well as to other adjoining equipment.

I take minor exception to the word "spike". It is reduction in voltage to other equipment that is the concern.

I'll mention there is an eng-tips regular who runs a great site on motor controls. Mark Empson aka MarkE

Here is his summary of induction motor starting methods:

http://www.lmphotonics.com/m_start.htm

You'll have to build a model of your system to predict some voltages and currents during starting.

The specification of the fault current and source voltage allows you to build a Thevenin model for your supply system upstream of the transformer.

The transformer impedance will depend on the size of the transformer. Bigger transformers tend to have lower impedances (in absolute terms that is... the p.u. impedances are somewhat constant accross transformer size).

I assume you can calculate the starting impedance of the motor for DOL start (assume something like 0.2 starting p.f.). For each of the starting methods the combined impedance of the motor / starter will be higher than that, also possibly shifted in phase angle.

What exactly is the acceptance criteria would depend on the situation and I'm not very familiar with those. It might involve a max current drawn at the transformr primary during start or a minimum voltage at the transformer primary during start (so you don't affect other customers served by the same line that feeds the transformer). Or if there are other loads fed from the same transformer then you might be concerned about the minimum voltage downstream of the transformer during start (beyond its effects on motor).

There will be some assumptions required along the way.

Bill (waross) knows these types of calculations inside out. He and others might have some more comments.

=====================================
(2B)+(2B)' ?

 

[Sparweb]

Welcome to Eng-Tips,
True enough that many members here can answer your question (not me) because it is a part of the basics that you should learn. I would expect that your course material should describe doing so since the load-to-power-supply interface is a fundamental question is almost all electrical circuits. Possibly your professor is asleep at the wheel.

Teach yourself what you need to know from real books you can hold in your hand. Books are not used for getting good marks in classes. Books are used for your entire career to build and expand your understanding. The subject you need to know is well covered by many electrical engineering textbooks. If you had, one handy an evening's study would have given you the answer. This is just one that I have, and is good. Others may be just as good, or even better. Used copies are cheap:
Rotating Electric Machinery and Transformer Technology, by Donald V. Richardson

http://www.sparweb.ca

 

[waross]

I suggest losing your motor starting link.

> This huge current at the starting of a motor can damage the motor windings and also this current can cause a large voltage drop in the line.
> These voltage spikes may affect the other appliances connected to the same line. Therefore, a starter is necessary to limit this starting current to avoid damage to the motor as well as to other adjoining equipment.

This statement does not apply to over 90% of the three phase inductions motors in North America which are started DOL.
The author is struggling up the first peak of the Dunning-Kruger curve.
There is some oversimplification and outright wrong statements that may mislead you in the future.
For a good overview of motors try the Cowern papers.
Cowewrn Papers
A few tips;
Transformer; Review regulation versus PU impedance voltage.
The motor starting current is quite reactive. It will be more accurate to use percent impedance for motor starting when the transformer is closely sized to the motor. The result will be conservative and may be closer to reality than calculations based on transformer regulation. (It may be well for your to verify this and add a note to your submitted work.)
When the transformer is closely sized to the motor, the transformer will act a a primary impedance and soften the starting current somewhat.
What to look for in the Cowern Papers;
NEMA Locked Rotor Code, pages 8 and 9. The locked rotor KVA/HP ranges for different motors. pages 8 and 9
Motor Design Letters Pages 1, 9 and 10
The National Electrical Manufacturer’s Association
(NEMA) has defined four standard motor designs
using the letters A, B, C and D. These letters refer to
the shape of the motors’ torque and inrush current vs.
speed curves. Design B is the most popular motor. It
has a relatively high starting torque with reasonable
starting currents.
PF is very low when starting and when running unloaded.
Reactive current is highest at locked rotor, lowest running unloaded and increases slightly as the load increases.

Your given information simplifies things quite a bit.
Take some time to absorb this and then post back with your work.
Start with DOL starting. Suggestion: Start with a transformer sized for 125% of the motor full load current.
Go online and select a suitable transformer. You may select a standard transformer and a high efficiency transformer, and compare the results.
Once you get the voltage sag for the two transformers we will walk you trough the various reduced current starting methods.
Locked rotor KVA may be at about 50%. The voltage sag may be more than that calculated based on the transformer regulation and less than a calculation based on the %impedance voltage. Use %Imp V with a note that the result will be conservative.

Extra information:
I suggest that this motor may be suitable as a real world example of the information available for a large motor.
When you get out in the real world you may be finding information such as this on your own.
ABB/Baldor 800 HP motor

Home Catalog

Print
EM50804L-2340
800HP, 1791RPM, 3PH, 60HZ, 5012, TEFC, F1

Product Information Packet PDF
Ship Weight 6,440.000 LB
UPC 781568735060
Product has been discontinued

Catalog Number EM50804L-2340
Enclosure TEFC
Frame 5012
Frame Material Iron
Frequency 60.00 Hz
Output @ Frequency 800.000 HP @ 60 HZ
Phase 3
Synchronous Speed @ Frequency 1800 RPM @ 60 HZ
Voltage @ Frequency 4000.0 V @ 60 HZ
2300.0 V @ 60 HZ
Agency Approvals CCSA US
Ambient Temperature 40 °C
Bearing Grease Type Polyrex EM (-20F +300F)
Current @ Voltage 102.000 A @ 4000.0 V
178.000 A @ 2300.0 V
Design Code -
Duty Rating CONT
Efficiency @ 100% Load 95.8 %

High Voltage Full Load Amps 102.0 A
Insulation Class F
Inverter Code Not Inverter
KVA Code F
Motor Lead Termination Flying Leads
Motor Lead Quantity/Wire Size 6 @ 2 AWG
Number of Poles 4
Power Factor 87
Product Family General Purpose
Pulley End Bearing Type Ball
Service Factor 1.15
Shaft Diameter 4.125 IN
Shaft Rotation Reversible
Speed 1791 rpm
Speed Code Single Speed
Motor Standards NEMA
Starting Method Direct on line
Thermal Device - Bearing None
Thermal Device - Winding RTD Only

Note, The standard motor voltages are about 95.83 of the standard supply voltages. (115V/120V)
4000 Volt rated motors are commonly used on 4160 Volt supply systems.
ABB lists 13 2300V/4000V Volt motors and one 2300V/4160V motor.
Assignment. Do a root three comparison on those voltage pairs and post your comments.


Disclaimer: I am willing to help you understand and learn.
I will try to point you in the right direction.
I will help you find your own answers.
I will not do your homework for you.

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

 

[electricpete]

The original problem as stated is wide open, but I think Bill gave you great tips about how to select the transformer and the motor which will help bring the design into focus.

I think everything we talked about so far falls in the category of reduced voltage starting methods. At the risk of complicating the problem, I'll mention that in the real world some additional options for motor starting that would be on the table would be: electronic soft starters (another reduced voltage starting method) and vfd's (a whole category of their own) and of course Direct On Line starting (sort of the opposite of reduced voltage starting). I'm not sure what your professor had in mind but you obviously have to draw the line somewhere on how many options you evaluate so it might be fair to ignore electronic soft starter and vfd (or ask your prof)....

Direct On Line start is probably worth reviewing/discussing, if nothing else as a starting point to contrast to the reduced voltage starting options. DOL start may well end up creating an unacceptable voltage drop on the line feeding the transformer, which is usually the motivation for the more complicated/expensive reduced voltage starting methods to begin with.

> So what I know right now is by motor starting he's talking about how the coils or stators are wired.

That's not true in general for motor starting methods. It is true for wye-delta start which is relatively common reduced voltage starting method. It is also true for part winding start which is relatively rare reduced voltage starting methhod (rare because it requires a motor special built for that type of starting). But there are a number of other reduced voltage starting methods discussed in the links which don't involve changing how the motor is wired (for example adding an impedance in series with the motor).


=====================================
(2B)+(2B)' ?

 

[waross]

So what I know right now is by motor starting he's talking about how the coils or stators are wired.

While I am sure that the author is able to find examples to support this assertion, as a general statement it is very misleading.
There are too many millions of motors that this does not apply to.
There are two main reasons for reduced current starting:
To reduce voltage sag on the supply system.
Starting high inertia loads where the acceleration time may cause unacceptable heating of the motor.
Search Eng-Tips for:
Star delta starting (Wye delta starting).
Primary resistance starting. (Primary impedance starting).
Part winding starting.
Auto-transformer starting.
Electronic soft starting.
Variable Frequency Drive (VFD) starting.


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

 

[electricpete]

True enough that many members here can answer your question (not me) because it is a part of the basics that you should learn. I would expect that your course material should describe doing so since the load-to-power-supply interface is a fundamental question is almost all electrical circuits. Possibly your professor is asleep at the wheel.

That's a baffling response to me. This is not a simple textbook question with one answer. It seems to be a broad exploratory assignment... the kind that builds an appreciation of the depth of the subject.

=====================================
(2B)+(2B)' ?

 

[LionelHutz]

Starting high inertia loads where the acceleration time may cause unacceptable heating of the motor.

Now we're giving bad information to the OP. Reduced current starting generally causes more heating than DOL. Not much more if you're careful with the current limit, but still more.

 

[electricpete]

I’ll throw in my two cents about whether reduced voltage starting improves thermal performance or hurts thermal performance: it can be either.

[ul]
[li]If you neglect deep bar effect, you can mathematically derive that the total heat energy produced by rotor I^2*R losses over the entire course of an UNLOADED start is equivalent to the final kinetic energy of the rotating system, regardless of the voltage level applied during start (no change with reduced voltage). You could also reach the same conclusion with a thought experiment: imagine how the unloaded start looks while putting yourself in the reference frame of the rotating stator field; now the stator appears to you to be equivalent to stationary permanent magnet and the rotor is a conductor cage that at the beginning appears to you to be rotating at sync speed; you have a dc braking scenario in your ref frame, so the rotor slows down to zero speed (in your reference frame); where did the initial kinietc energy go to?; since the magnets aren't moving they have done no work in your reference frame; we neglect friction losses; we conclude the kinetic energy must have been dissipated in rotor I^2*R losses. The conclusion for the rotor total heat energy is unchanged regardless of your reference frame (it is the same rotor current flowing through same rotor resistance either way causing same I^2*R... there are unrelated energy differences in these two frames... when viewed from the normal stationary reference frame energy input would be required to keep those magnets rotating which constitutes power delivered to the stator from the power system but is not relevant to the rotor heating conclusion). Total heat energy from stator I^2*R losses are proportoinal to total heat energy from rotor I^2*r losses (could be computed by multiplying total heat energy in rotor times the ratio of stator resistance to rotor resistances). Attached is an ugly nasty math proof (still neglecting deep bar effects) that total rotor I^2*R energy during unloaded start equals kinetic energy, which does not rely on the thought experiment above and illustrates some of my other points below.[/li]
[li]When you add torque load during start, that’s when the reduced voltage can increase the total I^2*R heat energy (in the limit where reduced voltage causes reduced torque to the point of stall, we can see the start lasts forever and the total heat energy from rotor I^2R goes infinite assuming no trip).[/li]
[li]But if we start unloaded (or very little load), then reduced voltage causes no significant increased total heat energy during start. In that case, the reduced voltage can provide an thermal advantage because that same total heat energy is produced over a longer starting time. If it happens to be a high inertia load as Bill mentioned, then we are talking about very long start durations and the cooling that occurs during that long starting period can be significant and result in lower max temperature for reduced voltage start (unloaded).[/li]
[li]So I agree with both you guys. Maybe Bill's scenario could clarify it applies to high inertia load with relatively low torque load during start which I'm sure he assumed. Lionel's point is also valid that some motors that start their load ok DOL might damage themselves if you try to start with reduced voltage. [/li]
[/ul]
I’m sorry for dragging this in a direction that won’t be helpful to the op.

=====================================
(2B)+(2B)' ?

 

[Genosphere]

The reason for the confusion, aside from my initial remarks about the class being a mess, is that the professor initially stated he wanted us to find two motors online and compare them for the given problem at that time. The next day in the email they sent, it does appear he changed the assignment almost completely by giving us some motor parameters.

Regardless, I need to read more on all of this since It doesn't seem these starting methods and their intricacies were discussed at any great length.

~~~~~~~~~~~~~~~~~

So, from what I understand right now, Star-Delta starting is commonly used due to low cost, space requirement, and reduced starting current, but it offers no adjustable starting characteristics and a soft stop is not possible including additional cost in setup.

Soft starting require less maintenance and offer various starting and stopping selections, heat generation only occurs during short periods of starting and stopping. You can also couple these with VFDs, which I have personally worked with so I am at least familiar with them. Though from what I read, controlling speed consistently doesn't seem to be one of its strong points.

Well, since our goal is to only start the motor, would this soft starting method with a VFD be my best choice?

As for the transformer, ill need a stepdown transformer to drop the 12k to 4k.

[LRC=6*FLA; FLA=1kVA=1HP] 800HP=800kVA=FLA; 6*(800kVA)=4800kVA

Transformer 125% rating = 1000kVA, Primary=12.47
Primary Full-Load Current: 46.3A
Secondary Full-Load Current: 138.79A
Turns Ratio: 2.998
Transformer Type: Three Phase Step Down Transformer

Are the proper formulas I should consider using?

Motor:
Motor Full Load Current = (kW x 1000)/(1.732 x Volt(L-L) x PF
Motor Locked Rotor Current(LRC) = Multiplier x FLA(Full Load Current)
Motor In-Rush kVA at Start (Irsm) = Volt x LRC x FLA x 1.732/1000

Transformer:
FLA = kVA/(1.732 x Volt)
Short circuit current at TC secondary (Isc) = Transformer full load current / Impedance
Maximum kVA of TC at rated Short circuit current (Q1) = (Volt x Isc x 1.732) / 1000
Voltage drop at transformer secondary due to Motor Inrush (Vd) = (Irsm) / Q1 (%)
Voltage drop at Transformer secondary is 10% which is within permissible limit.


Does this would like I am on the right path???

Thanks for al the input so far.

By the way, thanks for the links to the book waross, looks like a good read.

 

[waross]

Now that you have done some of your own work, some comments:

Star delta starting;
The transition from star to delta is problematic.
The motor will be generating a back EMF when the star connection is opened.
An asynchronous induction motor will be running slightly below synchronous speed so that the frequency of the back EMF will be slightly lower than the grid frequency.
When the motor is reconnected in delta, the back EMF may be out of sync with the grid.
A very large transient will be developed.
When the change is made from star to delta there is an inherent phase shift of 30 degrees, either plus or minus.
each phase winding may be considered to have a start end and a finish end.
The delta connection may be formed by connecting together either the start ends or the finish ends.
The switching transient may often be much less with one connection than with the other connection.
The inrush may also be reduced by varying the timing of the star/delta transition.
If the load at startup is not consistent then mitigation efforts will be less successful.
Star delta starting is falling out of favour.
Search Eng-Tips.com for star delta starting, star delta starters.

Soft starting:
Soft starting works reasonably well.
The SCRs are triggered into conduction partway through each cycle.
Search Eng-Tips.com for soft starting, soft starters.

VFDs;

You can also couple these with VFDs

Maybe not.
The first stage of a VFD rectifies the grid current to DC.
Putting a soft start ahead of that will not work well.
A VFD generates a variable frequency and voltage via Pulse Width Modulation.
The control algorithms maintain the proper Volts per Hertz ratio to avoid magnetically saturating the motor at low frequencies.
A motor driving a load such as a centrifugal pump or a centrifugal fan, the motor may be started with a starting ramp that actually draws less than rated full load current until the very last of the acceleration curve.
Staring currents of 150% are quite common.
The VFD also has the advantage of variable speed.
If money is no object, the VFD is the best solution.
The VFD often has the highest losses, so the cost of ownership includes the ongoing cost of the losses as well as the initial purchase and installation.

Transformer;

Transformer 125% rating = 1000kVA, Primary=12.47
Primary Full-Load Current: 46.3A
Secondary Full-Load Current: 138.79A
Turns Ratio: 2.998
Transformer Type: Three Phase Step Down Transformer

Done. Use this transformer.

The voltage sag is interesting.
At full load current and normal loads the voltage sag is described by the regulation expressed as a percentage of rated voltage.
The regulation is stated at a stated power factor. (80%?) You can verify this figure.
Short circuit current is calculated from the percent impedance voltage. With a short circuit, only the transformer windings are in the circuit so the power factor is determined by the X/R ratio of the transformer. Usually a very low power factor.

For you to research:
Some transformers rate the voltage at full load and some rate the voltage at no load.
Find out how your selected transformer is rated.
Your formula for voltage dip at the motor may be based on:
1> 6 times starting current x 800/1000 = 4.8 times transformer full load current
4.8 times transformer rated current times % regulation = % voltage sag.
The actual sag will be more than this due to the low power factor of the starting motor. (About 50%)
or
4.8 times transformer rated current times % impedance voltage = voltage sag.
The actual sag will be less than this due to the lower power factor of of a shorted transformer. (Much less than 50%)

2> If the transformer is rated at full load, You will be allowed one time the transformer rated current before the voltage sag is calculated.
You do this one and we will check it for you.

Not mentioned;
Auto-transformer starting,
Primary resistance or impedance starting,
Part winding starting.
Do some quick Googling and search this site.
Show us that you have done some work and we will add some comments.
Please don't cut and paste anything from Eng-Tips.
Read it and when you understand it, use your own words.
Yours
Bill
(It's late and I'm too tired to check for silly typos.)

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

 

[Genosphere]

Thanks for all the info so far. Lot of info above im going to have to get ready by tomorrow, so im not sure I will be able to get back to you on my findings until after the hand in. Either way, Ill touch back to get more info and see where I may have gone wrong or if I'm actually headed in the right direction.

A motor driving a load such as a centrifugal pump or a centrifugal fan, the motor may be started with a starting ramp that actually draws less than rated full load current until the very last of the acceleration curve.

By this, you are referring to something such as "Low Frequency Torque Boost" that I have used in setting up VFDs here at work, yes? From my understanding, It applies a small % of EXTRA voltage to compensate for voltage drop to help jump the motor at start. I don't want to digress to much, but wanted to make sure my understanding was correct at least on one thing. haha

So, transformer calculations seem to be right based on what you said, great.

Now for the starting method selected, this is where it might get a little harry again, since all have to go off of is the initial instructions mentioned, I don't know if "Money is no object" is justified; though since most people/companies usually want to cut costs where they can one would assume the soft start/VFD method might be to pricey. For the sake of simplicity and lack of specifics from instructor, I think I would go with the VFD option due to:

{Soft starting works reasonably well.}vs{The transition from star to delta is problematic.}

Ill check back throughout the day, but for now Ill go in the direction of the soft start and pretend I have government funding.