problem with a power transformer 1

[mottt]

I got a power transformer 2500 Kva between a circuit breaker to the grid and a generator (step up transformer) and I want to know typical values of reactive power for diferents states of load, in example...full load...75 % load, the volts are 6000/20000 V, and is Y/Delta....thanks in advance

 

[whycliffrussell]

You need the X/R Ratio of the transformer.

pf = cos(atan(X/R)))

S(Full Load) = 1.7321 * I (amps) * V (l-l)

Power(Real)= S * pf

Power(Reactive)= S * (1- pf)

 

[mottt]

if i have no date could i suppose some value for stimation...

 

[slavag]

Hi.
About 8.5-9.
Regards.
Slava

 

[whycliffrussell]

I have some typical data here for a 2500kVA oil filled transformer at it says:

X/R = 10.67

hope this helps buddy.

 

[slavag]

Whycliffrussell . after re-reading.
Could you please check again your formula.
I don't remember exactly, but seems something not correct:
maybe you meant atan(r/x).
Regards.
Slava

 

[whycliffrussell]

NOPE...that's the correct formula.

 

[slavag]

Sorry, my bad.
Q/P=tanf
Slava

 

[rcw retired EE]

Reactive Power through a step up transformer is affected by the grid voltage, the generator voltage, the generator excitation level and the transformer impedance. The generator excitation could be controlled to maintain 1.0 power factor at the grid intertie, making reactive power flow = zero at all loads. Or it could be over-excited to push maximum reactive power to the grid at all load levels.

The X/R ratio of the transformer does not represent the reactive power flow through the transformer. You can calculate the kW and kVAR losses in the transformer from the X/R ratio, the impedance and the power going through it.

What is the purpose of your question? Maybe we can help with a little more information about your application.

 

[slavag]

Of course!!!!
Not my bad, my stuipid answer!!!. Seems something strange, but I puted attention on formulas and results, not on the Q's and means .
tan, cos, sin!!!!
tanf about 10, of course starnge, it's about 84-86 deg. w/o calculator. cosf near to zero!!!
Isn't important step-up or step down trafo.It's losses and of course reactive losses of trafo are more then active losses and at big trafos are more, more.
And RCWilson remind me also why this Q's.
Motorxplosion. I think once we discussed about it , it's for change PF according to Spain's code.
Slava

 

[waross]

If you are selling VARS and want to know your capacity, use the generators KVA rating or use the maximum current rating to determine the maximum KVA that the generator will produce.
Then, if you simultaneously increase the excitation and reduce the energy input to the prime mover in such a way as to keep the current constant, your output will change from real power to reactive power.
On the drawing board, use the total current as the hypotenuse of a right triangle. Use your load percentage as the base. eg: if total current is 1000 ams and you are running the real power at 75% then use 75% of 1000 amps or 750 amps as the base of the triangle. Use Pythagorus to calculate your reactive component represented by the altitude of the triangle.
Back in the real world, don't forget to check the capability diagram of the generator before implementing proposed changes in VAR production.
Is this what you are asking or am I off course here?

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

 

[slavag]

Hello.
I think we back to Motorexplosion's Q's about PF of generator and this Q connect to his Q about step-up trafo.
For trafo losses aren't important is step-up or step down trafo. All trafos have 4 types of losses:
1. iron active losses
2. iron reactive losses.
3. copper active losses
4. copper reactive losses.
5. iron losses are function of (U/Unom)^2
6. copper losses are function of (I/Inom)^2
where U and I are actual current and voltages.
all losses calculated according to the trafo test sheet.
These losses are calculated for any condition if a several parameters of the
power transformer . The transformer manufacturer provides
this information on the transformer test sheet:
1.Rated total kVA of the trafo.
2.Rated voltage of the trafo.
3.No-load test watts (LWFe) - the active power consumed by the
trafo's core at the rated voltage with no load current (open circuit test).
4.Full-load test watts (LWCu) - the active power consumed by the
transformer's windings at full load current for rated kVA (short circuit test).
5.%Excitation current - ratio of No-load test current (at rated voltage) to full load
current.
6.%Impedance - ratio of Full-load test voltage (at rated current) to rated voltage.

Possible several variation according to IEC or ANSI land for the calculation.
Now with:
1. Data of the trafo.
2. Stability curve of the generator.
you build Excell file with output data of generator and output data of trafo
Pgrid=Pgen-Plosses trafo. ( P losses actually is small value)
Qgrid=Qgen-Qlosses trafo--- import reactive power to grid: this is your variant for day hours: +cosf
Qgrid=Qgen+Qlosses---- export reactive power from grid.this is your variant for night hours: -cosf. absorbe reactive power condition
and variant for big trafos Qlosses=Qgrid +Qgen(-Qaux bus, of course).
If you don't have all data of trafo, give %impedance of trafo ( I assume is 5-7%) and try meas several points P/Q of in/out of trafo, I'm hope possible from this data build some interpolation for the Fe, Cu losses. Or maybe someone have typical data for this size and type of trafo.
Regards.
Slava

 

[slavag]

In case of problem with trafo data sheet , you can use attached document ( DIN 42504 recommendation), very practical too.
Regards.
Slava

 

[slavag]

For some value, see attached catalog, of course isn't your trafo, but for first step, for my pinion is good.
Hope it's help you Motorexposion.
Good Luck.
Slava