dq transformation for 3 phase systems

Thread Starter

almotions

Joined Feb 6, 2009
46
Hi,
recently,i'm doing some experiment to convert from 3 phase balanced input to dq components to calculate the active power.

As you can see from the attached diagram,i've included the wattmeter to measure the active power drawn by the load(R=20 ohms) which shows 605Watts at steady state. But when i use the dq transformation and calculate the active power based on the formula P=VdId +VqIq
which is shown on the first graph ,P=V23*V21+V24*V22 ,it gives a value of 400W at steady state.Is there anything wrong with my assumptions or diagram?
Another question. for dq transformation, why the direct and quadrature component remain at a constant value despite of change in theta with time (theta=wt) Another question. for dq transformation, why the direct and quadrature component remain at a constant value despite of change in theta with time (theta=wt) .Thanks

Regards
alvin
 

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Hi!

I have recently read about the dq transformation.

The reason why the dq transformation gives DC values, despite the changing ωt, is because the reference frame of the dq transformation moves as the three phase vectors move. So when you look at the positions of the dq vectors, relative to the phase vectors, their relative position is always the same. Wikipedia has a good artice about this (search for "Dqo transformation"), and the picture at the article end should help in the understanding of vector relative positions.

Hope this helps
 

steveb

Joined Jul 3, 2008
2,431
Hi,
recently,i'm doing some experiment to convert from 3 phase balanced input to dq components to calculate the active power.

As you can see from the attached diagram,i've included the wattmeter to measure the active power drawn by the load(R=20 ohms) which shows 605Watts at steady state. But when i use the dq transformation and calculate the active power based on the formula P=VdId +VqIq
which is shown on the first graph ,P=V23*V21+V24*V22 ,it gives a value of 400W at steady state.Is there anything wrong with my assumptions or diagram?
You need to be very careful with dq-transformations (Park's transformations) because there are many variants and you need to carefully identify the meaning of the d-q voltages, and the correct power formula. It's hard to be sure what your issue is unless you post the exact equations you are are using to generate the dq voltages and currents. The schematic does not make this clear.

However, it looks like you have it mostly correct and whatever scale factor is incorrect can just be put into your final formula most likely.
 

Thread Starter

almotions

Joined Feb 6, 2009
46
Hi all,
can anybody please give me a SIMULINK example of how to calculate the active and reactive power from a three phase supply supplying a load? Please help! I've tried everything but the values still comes out wrong.
 

1223spy

Joined May 17, 2010
1
Well it is quit General formula but for specific circuit check the current flow.

P=SQRT(3)*V(rms)*I(rms)cos(theta)
Q=SQRT(3)*V(rms)*I(rms)sin(theta)

S=P+iQ

I hope this will work for you
 
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