Generator Sizing 6

[SilverArc]

Hi Everybody..

I have to size a generator for a mine hoist. The mine hoist is 600 HP that means 600 * 746= 447600 W= 447 .6 KW
For some moments of time, the load even becomes 847 HP during acceleration period of the hoist. But this period is in msecs.

The calculations involved come with a -200 HP power fed back to the motor during hoist going down the shaft.

Now what I understood that, if I size a emergency generator for this application, It should be capable of acting as a sink of -200 HP power which means:

200 * 746= 149 KW now I have to convert this in to KVAR.

The motor is fed with a VFD, some of you who have mine hoist experience can comment on this; Power factor with VFD assumed: .8(Please advise, if this PF is not real when it comes to hoisting application)
149 KW= 186 KVA=1111.75 KVAR

Could you advise me, normally to what percent of its size a generator can act as a sink and source of VARS ?

I heard that to sort out this regenerative power issue with mine hoists: Load banks are used. COuld you advise me, where I can find some information about this.

I have given a best shot of my engineering judgement... An advise from some of you who are experts in this area might give me a better direction.

Thanks

 

[itsmoked]

For starters you will want a generator that is probably at least 2 x motor HP. Possibly more. Someone will be along shortly to confirm this.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[jraef]

The VFD changes everything. You may want to consider hiring someone with specific expertise in this area for a project of that magnitude.

While it is true that the VFD can maintain a displacement PF of around .95, it then adds a distortion PF and can be as bad as .70 under the worst circumstances. In a utility fed system this isn't usually a big issue with regards to source sizing, but in a genset it is. Active front-end drives can be used to reduce that (see below).

By the same token though, you will not necessarily need to consider starting current in the generator sizing because the VFD will be capable of keeping that to a minimum. You may even be able to start that motor with FLA if necessary, although on a mine hoist probably not, more like 150% current for a few seconds, but usually well within the transient capacity of a normal DG set.

Braking is a whole different ball of wax. First, think of the VFD as being an entirely separate power source for the motor; it uses power from the genset as a "raw material" to create the motor power, but other than that there is typically no other direct connection. When braking, the VFD supplies the VARS necessary to allow the motor to regenerate, then takes that back EMF and charges up the DC bus. The "raw material" energy from the genset for this purpose is always less than the motoring energy of course, so you don't need to worry about factoring it in to the genset capacity. From that point, the regenerated energy from the motor needs to go somewhere. In a "standard" VFD, the energy is bled off into a resistor bank as heat. The DC bus voltage is monitored and when it passes a threshold (usually line DC rectification voltage), a separate transistor is switched on to pump the DC energy directly into the resistor. In this mode, the genset is not supplying anything (other than control power) to the VFD. The drawback is that on something like a mine hoist, the resistors will need to be in the circuit for a long time and can be subject to damage if not engineered correctly.

An alternative usually considered for long braking cycles such as this is Line Regenerative Dynamic Braking using an "Active Front-End" VFD. In this mode, the VFD consists of 2 back-to-back inverters, one to supply power to the motor in motoring, another to pump motor regen power back into the line. These VFDs are very expensive (because they involve essentially 2 VFDs) and are highly specialized. You may or may not be able to retrofit an existing VFD to do this. But if you can, then this can also help to take care of the distortion power factor problem mentioned earlier. The drawback is of course, now you WILL need to consider the genset's ability to absorb this power.

So step one (if you insist on pursuing this on your own) will be to determine just HOW that VFD is providing braking. If they are in fact using a resistor bank, no need to worry about the genset sizing AFAIK. The genset will not be sinking any VARs.

JRaef.com
"Engineers like to solve problems. If there are no problems handily available, they will create their own problems." Scott Adams
For the best use of Eng-Tips, please click here -> faq731-376

 

[davidbeach]

If the VFD does the braking through what jraef has called "Line Regenerative Dynamic Braking" you need to worry about the Watts regenerated. If all you have is the generator and the motor, you will have the condition of the generator motoring the engine and that is not a good thing. Best to figure out the maximum regenerated power and have something else, lights are great, on the generator at all times that can absorb the regenerated power without motoring the generator.

 

[peebee]

I think you'd want to have a braking resistor hooked to your VFD.

If you try to pump that energy into your generator -- where will the energy go? It sure won't turn back into fuel.....

Maybe if you had some substantial other loads being served by the generator, you could pump the energy into those loads with a regenerative VFD. But it sounds like this elevator is the only load on your system.

I wouldn't try to pump the energy into the generator without checking with the generator vendor first.

Regarding the gen as a source of VARS -- a typical generator rating is 80% lagging. Note that generators HATE leading power factor, and I guarantee that your generator will NOT like having a leading load pumping VARS into it. Do not let your load become leading under anything over maybe 50% generator kW loading.

 

[waross]

I strongly advise going with a resistor bank to disipate the regenerated energy. With the life safety issues and potential for damage with a mine hoist, you don't want to depend on other loads for an energy sink.
Depending on the mode of operation of the VFD under regenerative conditions, it may overspeed the generator, or it may limit the frequency, in which case it will be ineffective at braking.
In any event, the result will not be good. Either the generator or the mine hoist (or possibly both) will overspeed. Typical control schemes for generators include overspeed trips. You only have to push the generator a few hz. overspeed before the main breaker shunt-trips and the engine shuts down. This circuit may be disabled at risk of voiding the warranty.
In the days before VFDs, a large overhauling load driven by a generator always had a resistor bank connected to the motor to absorb regenerated energy.
respectfully

 

[SilverArc]

Hi waross,

I was surfing this forum for some mine hoisting applications and I came across a post where you mentioned that you have worked on a WaRDLEONARD system of speed control.

I have read about this method of speed control.
I just want to understand one thing here;

If I say: I am using a DC drive for the speed control of a mine hoist DC motor

AND

if I say I am using WARD LEONARD speed control method for DC motor control for mine hoist.
Could you kindly clarify the difference.

Thanks

 

[waross]

As I understand this bit of history, there is more than one definition of a Ward-Leonard system. My first reading was that the original Ward Leonard system was Working backwards) a DC motor with a fully excited field. This was connected permanently to a DC generator. The field current of the DC generator was controlled by reostats.
The systems that I worked on used rotating DC amplifiers called Amplidynes in place of the reostats.
Where the field of the DC generator may consume 1500 watts and require quite large reostats, the amplidyne could supply the 1500 watts with a control current of just 1 or 2 watts.
We called this arrangement an amplidyne drive, but some still called it a Ward-Leonard drive. Later the amplidyne was replaced with a solid state amplifier. In some quarters this was still called an Ward-Leonard drive.
One group considers only the original reostat-generator-motor system to be a Ward-Leonard drive.
Another group refers to any direct connected Motor-generator set with variable generator excitation as a Ward-Leonard drive.
To further the confusion, manufacturers other than Ward-Leonard built amplidyne and solid state exciter drives.
If you were to mention an "Elworthy Drive", I would remember an amplidyne controlled system that others may consider a Ward-Leonard drive.
This is one of those instances where it is best to agree with whoever you are talking to as to what each of you mean by "Ward-Leonard".
A DC drive implies a DC motor. It may have a solid state power supply in place of the generator.
The basic hallmark of a Ward-Leonard drive is a generator directly connected to a DC motor. The motor has full field at all times and the speed and direction are controlled by varying the excitation of the generator.
respectfully

 

[SilverArc]

Thanks Waross. Actually Mechanical Engineers at the last moment changed the AC motor and VFD to other two options which would be DC motor+ Drive & DC motor with Ward-Leonard.

But thanks for your comments though. I would like to add some thing here:
A dileman which I came across today.
I have to size a bus for a switchgear;
60 A: welding receptcle
5KV transformer 600/208/120 v
75 HP motor and a 30 KVAR capacitor bank

The total active current is 141 Amp. 75 HP motor full load is 77 Amp.
So 60+ 4.9(5 KVA transformer)+ 77= 141 A

Now the reactive current is 28.9 A. this is for induction motor capacitor bank connected to spplitter which is feeding all loads and supplied by a main breaker.

Now what is the current that switchgear should be used.
141-28.9 j which is 143 A.

and what current will come through main breaker;
141 A or 143 A.
I am sure it is a small thing, but I could not figure out what I am missing here. I even ran a simulation in ETAP but it did not help me solving my dilemna.

Thanks

 

[mc5w]

If your VFD is of the diode bridge type you would a need a universal harmonic filter such as from MIrus. However, their unversal harmonic filter will not work for regeneration into the line type drives. It will also not work for phase displacement 4 quadrant SCR drives for DC motors.

Very likely you would also need to power water pumps and lighting. For generator of the size that you need you also need a minimum load to stabilize the speed governor and the voltage regulator. This is why when an electrical utility is recovering from a blackout some lucky devils get their power turned back on early. Typically, a black restart capable generating station has some local customers who get their power directly from one or more generator buses. Another reason is that when resynchronizing to an adjacent utility the sychronization is never perfect and having some customer load on the line helps to absorb the bump. You would have the same problem with paralleling generators meaning that some loads need to be connected to the bus after connecting the first generator and before connecting the second generator and so forth.

Since a VFD gradually ramps up power drawn as a motor starts you sould not have any problems with significant voltage and frequency dips when the hoist starts but could have these problems with water pumps and air compressors.

Mike Cole

 

[waross]

Your currents will all be at unknown phase angles, (except for the capacitors).
The current to each of the loads, ie: Motor, Transformer, and Welding machine will be composed of an in phase component and a reactive component.
In order to accurately add the currents you must resolve each current into it'sactive and reactive components. You then use the total of the reactive currents and the total of the active currents to find the actualcurrent.
In practice, it is common to add the load currents and disregard the capacitor current.
I am considering the capacitors as a correction or conditioning device rather than a load.
Adding the currents without regard to phase angle is commonly done when calculating panel loads. The result is conservative.
The capacitors will serve to reduce the current when the motor is running.

For instance, and to make this a little clearer;
Let's resolve the motor current. We will assume that the power Factor is 80%.
The line current is 77 amps.
The in phase current component will be 77 A x .8 PF = 61.6 amps.
From Pythagoris' theorem, the reactive component of the current will be;- sqrt([77x77]-[61.6x61.6])=46.2 amps.
The capacitor current of 28.9 amps is reactive and will subtract from the 46.2 amp reactive component of the motor current.
The reactive current of the motor and capacitors together will be 46.2A - 28.9A = 17.3A
The resulting current of the motor and capacitors together will be sqrt([61.6x61.6]+[17.3x17.3])=64 amps.
This calculation is based on an assumption. (but a reasonable assumption)
An exact solution af the sum of the currents would require that the Power Factor at full load of the motor, transformer, and welding machine be known.

For sizing switchgear, the line currents are added arithmatically. The capacitor current is ignored as long as it is less than the reactive current of a motor on the same bus. In this case, it will be safe to ignore the capacitor current. When the motor is online, the capacitor reduces the motor current. If the capacitors are connected with the motor off-line, the capacitor current will be less than the motor current.
respectfully

 

[SilverArc]

Hi Waross,

We came to the conclusion of using a DC drive(6 pulse or 12 pulse)with a 600 HP 480 V DC motor.

1) 6 Pulse DC Drive:
4160 V switchgear + transformer + DC Drive + DC motor

#A) I am wondering, Does the transformer conf. star/Delta or Delta/Star matters ?

#B)Why these transformers are supposed to have high impedance ?


2) 12 Pulse DC Drive :

# A) We have been advised to have two transformers with star/Delta and Delta/ Star or One primary with a secondary (Star) and tertiary(Delta)
Why this condition exits: with different configuration of sec. and tertiary.

I would appreciate few words to get rid of this dilemna.

Thanks

 

[waross]

Hi SilverArc;
I am flattered by your faith in me and so I will give you the best advice I possibly can.
Wait for skogsgura or jraef or marke or one of a few other drive gurus to post in. They will give you much better advice in this field than I am able to.
respectfully

 

[peebee]

A quick primer on EngTips conventions before I get to your questions:
Star/delta = Yd
Delta/star = Dy

1A) Typically, you would select a Dy transformer. All 3-phase power conversion transformers should have a delta winding to trap zero-sequence currents. Having a wye secondary would permit you to ground the system or more easily detect ground faults if so desired. Your mining application probably prohibits system grounding, but I think ground fault would be required (warning, I know next to nothing about mining applications).

1B) I don't know. I'd need more details on the specific application. How do you define "high impedance", and who told you that it was required? If anything, I'd think you'd want to consider a transformer with a low zero-sequence impedance to help source harmonic current to your rectifier.

2A) The combo Dy and Yd, or single Ddy, is required with a 12-pulse rectifier to convert the 3-phase source into sort of a 6-phase secondary voltage (there will be a 30-degree phase shift between line-to-line voltages on a y secondary and d secondary winding). You need those extra phases (or some sort of phase converter) to make a 12-pulse rectifier work. You should be able to find LOTS more information on this with a bit of Google searching. I assume you understand the fundemental differences between 6 and 12 pulse -- if not, then Mr. Google should be able to help you with that too. Try searching "6 12 pulse rectifier wye delta"

 

[SilverArc]

Thanks Pee bee,


In regards to question 1B)One of the thumb rule for harmonics reduction is a high upstream impedance that is some times, reactors are used in front of drives.

I beleive when it is a DC Drive + DC Motor application

Now if I have a 4160 V switchgear and the motor is 480 V, I have to step it down...
So will that step down transformer will be a K rated transformer.. from Harmonics perspective ?

Thanks

 

[waross]

The combination of wye and delta windings or transformers is used to reduce the ripple.
With one winding, you get 6 pulses every cycle so the pulses are 60 degrees apart whether the winding is wye or delta.
The secondary of a Yd is shifted 30 degrees in relation to the secondary of a Yy. Similar with a Dy and a Dd
By combining a delta winding with a star winding, we get 12 pulses 30 degrees apart.
I think that a Yd and a Dy will give you a phase shift of 30 degrees twice. Depending on the connection they will either cancel or add to 60 degrees. You would be back to 6 pulse either way.
It is my understanding that adding reactance to a circuit increases the "rise time" and so mitigates the higher order harmonics. Designing extra reactance into a drive transformer may save or reduce the cost, labour, and floor space of a stand alone reactor.
respectfully

 

[peebee]

In regards to question 1B)One of the thumb rule for harmonics reduction is a high upstream impedance that is some times, reactors are used in front of drives.

I think you just answered your own question. Beware -- the inclusion of reactors in front of a harmonic load can result in very high voltages on the load side of the reactor. Think of harmonic loads as current sources; when a current source is put in series with a large impedance, a high voltage can result.

Now if I have a 4160 V switchgear and the motor is 480 V, I have to step it down...
So will that step down transformer will be a K rated transformer.. from Harmonics perspective ?

K-rating is only one possible way to address high-harmonic loads. Others include filtering and transformer derating. You need to define the requirements of the upstream system (perhaps based on utility requirements, IEEE guidelines, or physical/electrical system limitations).

 

[SilverArc]

Hi Waross,

Could you advise me, what is this shield thing between Primary and secondary of an Isolation transformer and how this helps.
We are trying to use a step down transformer as an isolation transformer for feeding a DC Drive. I was advised, that a standard step down transformer will not be a good choice.
I would appreciate a word.
Thanks

 

[itsmoked]

SilverArc;

Shielded transformers usually contain a metal shield between the two sides of windings. Remember that there is inductive coupling between the primary and the secondary sides of a transformer,(the whole point of transformers). And.. There is also capacitive coupling between the two sides. The inductive coupling is poor for higher frequencies and with tend to filter them out. But! The capacitive aspect will pass higher frequencies easily. Shielding attempts to physically block this capacitive link by interposing a 'shield' between the two sides, thus blocking the high frequency noise present on either side from crossing over to the other.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[SilverArc]

Thanks Itssmoke.

Could you advise, how to size a reactor?
My current is 1750 A and volatge is 480 V.
Any companies those are into reactor manufacturing.

During emergency, we intend to connect emergency generator directly to DC Drive. No transformer, so for this reactor and to tackle harmonics issues...we will use reactors as we have no isolating transformer for taking care of harmonics.

Thanks