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#1

# LOOP TEST OF TURBOGENERATOR

I would like to calcolate the load current of a loop test of a turbogenerator. I know the characteristic of the machine and the magnetization curve of iron. Has anyone solved this problem?
01-13-2014 11:20 PM
Top #2
Hello Flavio, normally one uses flux densities of 1,4-1,5 Tesla for the loop test. please look the required ampere turns from the magnetization curve of the iron. Then you have to calculate the average circumference of the stator yoke, it is the average of stator lamination outer diameter and diameter of slot bottom. then you multiply these two.. For example, for 1,5T you need 2000 Amperes per metre , stator outer diameter is 1,5 meters and slot bottom diameter is 1,1 meter, you have 1,3 meters times pi is ca. 4,1 meter. 2000 x 4,1 = 8200 Ampere-turns, that is with four windings you need 2050 Ampere and so forth.
01-14-2014 01:56 AM
Top #3
Thank you for your answer, but I do the more complex problem!!!
In my case:
- i need 1000 Amperes per metre (1.3 T)
- outer diamter is 1.5 m
- diamter of slot bottom 1 m
- eight slot 0,17 m

I calculated an average magnetic length of about 4 m, so about 4000 Amperes-turns.
If I put 4 turns at the center of the magnetic core I should have 100 amps for 1.3 T

But, i have available a supply voltage of the coils of 400 V (constant!!!!!).
How I do I know the number of loops to be put in to have 1.3 T? (my tension is not variable!!!!)

This is a typical problem when I do the test site.

Thanks Harri
01-14-2014 04:11 AM
Top #4
hello Flavio,

A comment: for 4000 Ameper-turns and 4 truns you should have 1000 Amps, not 100, (maybe there was a typo). Anyway, you can calculate the voltage per turn with the standrad formulae:
voltage per turn = 4,44 * frequency * flux, where flux = flux density (here 1,3 Tesla) * cross sectional area.
area = yoke heigh (here (1,5 m - 1m)/2 * (total packet length - total length of cooling slots).

then you divide the voltage available (here 400 V) with the voltage per turn and you get the number of turns. The voltage per turn is also the voltage for the serch coil in the opposite side to check the corrct flux density.
01-14-2014 06:51 AM
Top #5
Correct 1000 A!!!
Congratulations for professional answers and thanks for removing my doubts!
01-14-2014 09:30 AM
Top #6
Excellent discussion!
01-14-2014 11:51 AM
Top #7
Hello Flavio
You have a problem with just 400V!
Like Harri said, you first need to compute the volts/turn.
For core length = 4m, core back depth = 0.5m, flux = 1.3Tpk, this is 2*pi*freq*flux*core_cross_section_area/1.414 = 578Vrms at 50Hz.
Thus the maximum flux you can ever achieve is 0.9T. You will need a higher voltage supply, and of course all the safety implications of this.

If for example you used a 6600V supply, you could use 11 turns. The mean core back circumference (magnetic length) is 2*pi*1.25 = 7.9m, thus at 1000A/m peak gives 1000*7.9/11/1.414 = 508Arms supply current. (At high flux levels the crest factor of the current is rather "peaky" so the rms value will in fact be more like 1/2 rather than the formal 1/1.414.)

The lamination joints in larger machines tend to make the mmf needed at higher flux larger than that predicted from just electrical steel datasheets. Note the steel datasheet magnetising field values are normally given as peak and for example Cogent M310-50A steel is shown as only needing 251A/m peak (=178A/m rms) at 1.3T. The guide figures in IEEE std 62.2 are however realistic for low loss steel, as is the 2000A/m @1.5T that Harri suggests and your 1000A@1.3T (both assumed to be peak currents).

You could consider a low flux electromagnetic stator core test instead, such as the EL CID. This just uses 4% flux and can be run off a 16A mains socket. No safety issues!

good luck
01-14-2014 02:48 PM
Top #8
Thank you David for your corrrection! I forgot the division by square root of two = conversion to RMS value