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#1 hadi

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Posted 04 September 2005 - 04:57 AM

How to analysis the transformator that in the attachment file,
I want to know the Short circuit ampere and everything about analysis the transformator

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#2 RalphChristie

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Posted 05 September 2005 - 08:36 AM

Well,

Dyn5
D - Delta primary
y - Star secondary
n - Neutral brought out on secondary side
5 - Secondary lags primary with 150

ONAN
O - Mineral oil or synthetic insulating liquid with a fire point ≤300C
N - Natural thermosiphon flow through cooling equipment and windings
A - Air
N - Natural circulation for external cooling medium

ONAF
O - Mineral oil or synthetic insulating liquid with a fire point ≤300C
N - Natural thermosiphon flow through cooling equipment and windings
A - Air
F - Forced circulation for external coolling medium (fans, pumps)

You have two ratings for your transformer. Under normal conditions 30MVA, and if you have forced external cooling it changes to 35MVA. The rated primary voltage is 20kV and the rated secondary voltage is 6.3kV.
Off-load tapchanger on delta side with two tappositions to either side of the rated tap. Tapchange is 2.5% for every tap.
Temperature rise for topoil is 55C and winding temperature rise by resistance is 60C. This is above ambient temperature, taken normally as 30C average.
Insulation level:
HV windings: power frequency 50kV, lightning impulse 125kV peak
LV windings: power frequency 20kV, lightning impulse 60kV peak

At rated primary voltage (20kV) current is 866A (ONAN) and 1010.4A (ONAF)
At rated secondary voltage (6.3kV) current is 2749.3A (ONAN) and 3207.5A (ONAF)
At rated tap, %Z is 9.07% (30MVA base)


At rated voltage (6.3kV) the short-circuit current on secondary side (ONAN) is 30312.2A (30kA), assuming no source impedance and no motor contributions. (I rated / %Z)
For ONAF rating it would be 35.363.8A (35kA)

I think you will be able to interprete the other info on the nameplate yourself.

#3 RalphChristie

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Posted 05 September 2005 - 08:43 AM

Sorry, should read:

O - Mineral oil or synthetic insulating liquid with a fire point equal or lower than 300C

#4 hadi

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Posted 06 September 2005 - 04:20 AM

Thank you for your information, but I can't define the X/R ratio. How can I define this? SCMVA for 30 MVA base is SCA x V x sqrt 3 is that true?

#5 RalphChristie

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Posted 07 September 2005 - 12:11 PM

The trsf X/R ratio is not on the nameplate, but can be determined from the transformer test sheet or transformer losses, if you've got it.

%R = ((Total watt loss - No-load watt loss) x 100) / transformer rating in volt-amperes

X/R = [(%Z^2 - %R^2)/(%R^2)^0.5

As a last resort you cam assume your X/R ratio. Range of X/R ratio's (from IEEE Std C37.010-1979) shows for a 30MVA rating something between 16 and 50 (average of 33)


Yes, you are correct. (MVA SC = sqrt3 * SCA * V)
At 30MVA base and 6.3kV base MVA SC would be 330MVA

#6 hadi

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Posted 08 September 2005 - 12:37 AM

Dear RalphChristie,

can we determine the X/R ratio without we knowing the result the test transformer?

#7 RalphChristie

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Posted 08 September 2005 - 04:46 AM

No. You can:

1. Obtain the data on the test-report of the transformer,

or

2. Contact the manufacturer and ask for the data,

or

3. Test the transformer yourself to obtain the data,

or

4. As a last resort assume the X/R ratio.

Regards
Ralph

#8 hadi

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Posted 09 September 2005 - 12:25 AM

Dear Ralphchristie,

Actually, If we have got the SC 330 MVA, how long that condition will be happened. is there any rule of this condition base on IEEE standart?

#9 RalphChristie

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Posted 09 September 2005 - 12:21 PM

Just a few points:

Regarding the short-circuit current of 30kA (330MVA)
This is teoretical the highest current (seen from the secondary side of the trsf) your transformer will provide during a 3ph short on the secondary bushings with an infinite bus. Under real conditions your source impedance will also have an effect on the short circuit current, and the actual current will be lower. How much lower depends on your source.

Short Circuit Currents produce very high mechanical stresses in the equipment through which they flow, these stresses being proportional to the square of the currents. Thus, the longer the fault duration, the bigger the damage. You have to remove the faulted part as fast as possible. (A trsf of this size is not manufactured in a week, it can take up to a year's time.) Normally there is a max time in which equipment can handle SCC, say 30kA for 3sec. (just an example) Equipment manufactured a few years ago were much more robust than equipment manufactured today, and normally they will withstand this current longer. Remember, everytime this current flows, you are reducing the life of the transformer.

The time this current will flow depends also on your protective devices. Some kind of devices, like differential relays do not have to coordinate with downstream devices, and can take the faulted part out of the system in a few cycles. Other devices, like IDMT O/C and E/F relays have to coordinate with downstream and upstream devices, and can take longer to remove the faulted part. For more on protective relays you can check the IEEE Red Book.

hope this advice is going to help you

Regards
Ralph

#10 hadi

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Posted 13 September 2005 - 12:27 AM

dear ralph


what the effect X/R ratio transformator in analyse electrical system?

regards
hadi

#11 RalphChristie

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Posted 16 September 2005 - 07:32 AM

I am not sure what you are asking, but the X/R-ratio determines the peak asymmetrical fault current in a power system.

See also:
http://www.ecmweb.co...mplistic_fault/

Regards
Ralph

#12 NAGA

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Posted 19 July 2007 - 03:56 PM

QUOTE(RalphChristie @ Sep 5 2005, 02:36 AM) View Post

Well,
Dyn5
D - Delta primary
y - Star secondary
n - Neutral brought out on secondary side
5 - Secondary lags primary with 150
ONAN
O - Mineral oil or synthetic insulating liquid with a fire point ≤300C
N - Natural thermosiphon flow through cooling equipment and windings
A - Air
N - Natural circulation for external cooling medium
ONAF
O - Mineral oil or synthetic insulating liquid with a fire point ≤300C
N - Natural thermosiphon flow through cooling equipment and windings
A - Air
F - Forced circulation for external coolling medium (fans, pumps)
You have two ratings for your transformer. Under normal conditions 30MVA, and if you have forced external cooling it changes to 35MVA. The rated primary voltage is 20kV and the rated secondary voltage is 6.3kV.
Off-load tapchanger on delta side with two tappositions to either side of the rated tap. Tapchange is 2.5% for every tap.
Temperature rise for topoil is 55C and winding temperature rise by resistance is 60C. This is above ambient temperature, taken normally as 30C average.
Insulation level:
HV windings: power frequency 50kV, lightning impulse 125kV peak
LV windings: power frequency 20kV, lightning impulse 60kV peak
At rated primary voltage (20kV) current is 866A (ONAN) and 1010.4A (ONAF)
At rated secondary voltage (6.3kV) current is 2749.3A (ONAN) and 3207.5A (ONAF)
At rated tap, %Z is 9.07% (30MVA base)
At rated voltage (6.3kV) the short-circuit current on secondary side (ONAN) is 30312.2A (30kA), assuming no source impedance and no motor contributions. (I rated / %Z)
For ONAF rating it would be 35.363.8A (35kA)
I think you will be able to interprete the other info on the nameplate yourself.






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