calculate voltage 1

[pumpo]

We are using an outdated DA program to record test data. our system calculates start current from the rms measured by a few yokagawa's. the start current is calculated
Irms/Vrms X Vrated= Istart
to really mess things up upon motor start we are seeing a 36% Vdrop
A 500Kw motor being powered By a cummins 2Mw gen through a 1-4 trans. any assiatnce would be great.

 

[jbartos]

Question: What is the calculated start current required for?

 

[pumpo]

the start current is for the customer, we manufacture and test large cryogenic pumps. and the customer wants every piece of data available.

 

[jbartos]

Suggestion: Incidentally, the equation is actually calculating the starting current meaningless way since the ratio Vrated/Vrms produces the result for Istart bigger for Vrms < Vrated and smaller for Vrms > Vrated. The motor starting current should be measured and provided what it is per specific application, without any scaling up or down. If the current is scaled up, the motor protection may be selected unsafely to the motor. If the current is scaled down, the motor protection may be selected too sensitive, thus causing a nuisance tripping. Sometimes, this is done for &quot;good business&quot; purposes.

 

[pumpo]

The problem is the motors are unrated.since the rotor and stator are the only two items produced by the manufacturer all else is made in Japan by Parent company, they only supply an estimated start current, thus we need to measure or calculate those numbers. currently we are using an outdate DA system to record test data and are moving to and all digital DeltaV system this year.

 

[enginesrus]

Calculate voltage????
What does the generator say on its data plate???
What is the current drawn by the motor???
Need at least 2 things to calculate voltage.

Like the power and the current.

Is the trans a step up or down?

 

[jbartos]

Comment on the previous Pumpo posting:
If the motor starting current is known, the rated terminal voltage is unknown and output HPs are unknown, the motor has to be properly analyzed for all its parameters, not only for the start current.

 

[pumpo]

the generator is a 2Mw at 460V so it should be at 4347Amperes. corretct??, the Transformer is a step-up 4-1 the generators or generator in this case are matched to the transformer for the correct high side volatge. the software says that we are pulling 564Irms and 1494Vrms.
But we are experiencing a 36% voltage drop at start. And as i understand at motor start if there is a Vd of 36% then the I will be proporitionally decreased.

 

[pumpo]

Sorry I forgot to list the motor V 2300 is the V rating of the motor. At 500Kw the I rms should be 216 approx. is there any refrences or Recomended books that i can use to bone up on this. I am a computer guy who was asked to look into this and could use all the help I can get. Thank you to all who reply.

 

[alehman]

2MW @ 460V is 2510A at 1.0 power factor. I get 125A for a 500kW motor at 2300V, 1.0 power factor. If this is an induction motor, the power factor will be about 0.9, so the running current would be about 140A. I think you're forgetting to divide by sqrt(3).

Calculating the starting current accurately is not a simple task. There are many factors to consider including the generator winding impedances, transformer impedance, genset voltage regulator and speed control performance, motor winding data and mechanical load.

Assuming full voltage starting of a &quot;typical&quot; 500kW motor and 2MW genset, my Caterpillar motor starting program says you should expect roughly 34% instantaneous voltage dip. The transient reactance of the Cat unit is 36%. So 34% instantaneous voltage drop would indicate roughly 100% of rated current, or 3100A (rated current at nameplate kVA).

 

[pumpo]

We tried to run a resistive load bank on the pump and it made a less than 1% diffrence @ 500Kw pre-load. A local electrical engineer told us that on a setup with two 2Mw gensets starting a 6600V 1100Kw motor with the same load bank would give a 28% Vd. So I assumed that we should see at leat that same Vd % change. Also I was told that at motor start Astart/Ameasured = Vstart/Vmeasured is this corresct. Thank you for all your assistance I am a computer guy not a elelctrical guy.

 

[pumpo]

One more thing if you could. If I collect the information nessecary whatare the calcs necessary to achive an accurate #, is there a site you would recomend or a book title out there that would assit me. If these tpes of situations come up again. or just something that will give me a good feel for three phase motors and electrical calcs. any help would be appreciated

 

[alehman]

My motor amp numbers neglected to account for efficiency. They should be divided by the motor efficiency.

I'm confused about the load bank. Explain in more detail what you are doing with it. &quot;Astart/Ameasured = Vstart/Vmeasured&quot; doesn't make sense if &quot;Ameasured and &quot;Vmeasured&quot; are measured while running. During starting the current is much higher than normal while the voltage is lower than normal.

I think you would be wise to invest in the services of a qualified engineer if the results of this analysis are important.

 

[pumpo]

The load bank was setup to pre load the gen set so that when we switch the load over to the pump the generator would already be off idle and help to reduce Vd from 36%
or at least that was tge idea from our Test engineers.
I was tols that that is a start only equation and if our system was reading correctly we should be able to run that cal and have it come out right. Now our voltage is Rms not start voltage. Our system which I have no idea why does not measure start V or I, it calculates the I from the Rms Values. That did not seem right to me either. to get the voltage drop at start they are taking Irms/Vrms X 2300V (Rated V) = Vd although probably over simplified is that correct at the motor.

 

[alehman]

Pre-loading may help a little with votlage regulator performance, but I would be surprised if it made a big difference in the starting voltage dip you see. Most of the dip is due to internal impedance of the generator which is not effected by load.

I still confused about how you're measuring voltage dip. When you refer to &quot;RMS&quot; quantities, do you mean these are values for the unloaded condition? RMS means &quot;root mean square&quot;. Running and starting values are normally expressed as RMS.

Very simply,
Vdip = Istart * Z'd
where Istart is the starting current of the load and Z'd is the direct axis transient impedance of the generator and all quantities are complex. Z'd should be available from the generator nameplate. It may be shown as X'd and R.

 

[jbartos]

Comment on pumpo (Computer) Feb 1, 2004 marked ///\\the start current is for the customer, we manufacture and test large cryogenic pumps.
///Please, who is the motor-pump integrator?\\ and the customer wants every piece of data available.
///It appears that you, as a pump manufacturer, would be responsible for pump data only. The motor-pump set integrator would be responsible for the motor electrical data and its starting requirements via the generator power supply. The generators usually have some leeway in the voltage dip control when it comes to the starting motor load. There are also industry standards covering this, e.g. IEEE Std 141-1993 Section 3.11.6 Effect of Motor Starting on Generators.\\\