3-phase Power Calculation

[eskim]

Hi,

I am calculating 3-phase power with these datas, all value line-to-line:

Ia= 782.831 @184.8 deg
Ib= 792.247 @306.9 deg
Ic= 763.033 @66.4 deg

Va= 10.973 kV @0
Vb= 11.005 kV @119.9
Vc= 10.990 kV @240.1

3-phase Real Power = ???
Anybody can help me ? Thanks all.

 

[davidbeach]

So, what have you got so far?

 

[waross]

Calculate KVA
Determine power factor from phase angle errors.
Resolve KVA into KVAR and kW.
Start using average I, V, and phase angle errors. If you need more precision this will give you a sanity check number.
I see you have been a site member for six years so this is probably NOT homework.

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

 

[jghrist]

What do you mean "all value line-to-line". The subscripts indicate line-neutral voltages and line currents. Line-to-line means what when applied to the current?

 

[electricpete]

S = Sum (Vk x Ik*)
P = Real({S})
Note that all the Vk need to be referenced to the same point, but it need not be a neutral (it could be one of the phases assuming you have phase to phase voltages available)>

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[electricpete]

Based on magnitude 11kv, I assume that is likely a phase to phase voltage (if it were line to neutral voltage, the system voltage is 17kv?... never heard of it).

In that case you need to know the "polarity" of the phase to phase voltage... is Va = Va-Vb or is it Va-Vc (or something else?)

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[electricpete]

Let's say you have phase-to-phase voltages defined as follows:
Vab = Va-Vb, Vbc = Vb-Vc, Vca =Vc-Va

Choose phase A as reference
Voltage at A is 0
Voltage at B is -Vab
Voltage at C is -(Vab + Vbc)

S = -Vab x Ib* + -(Vab + Vac) x Ic*
We didn't include Ia because it has 0 contribution to the sum based on this voltage reference.

Note "x" represents multiplication here (not cross product).

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[electricpete]

If they are phase to neutral voltages then the calc is much more straightforward of course the neutral can be chosen as the voltage reference and plug directly into S = Sum Vln x I*

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[electricpete]

Note all multiplication and addition are vector operations, and * means conjugate.

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[electricpete]

This is my last post. Go back to the phase to phase case for a correction in bold:

Let's say you have phase-to-phase voltages defined as follows:Vab = Va-Vb, Vbc = Vb-Vc, Vca =Vc-Va
Choose phase A as reference
Voltage at A is 0
Voltage at B is -VabVoltage at
C is -(Vab + Vbc) = -Vac
S = -Vab x Ib* + -Vac x Ic*

This is the type of calc done for a 2 wattmeter method of power measurement

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[electricpete]

Or equivalently:
S = -Vab x Ib* + Vca x Ic*

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[electricpete]

Any more hints as to what those numbers represent.

I notice the Va is roughly 180 degrees from Ia... same for other phases. Could be either:
1 - pure real power flow (power factor = 1.0) under assumption Va = Van and that the positive sense of current was established in a direction opposite real power flow.
2 - 0.86 power factor (30 degree power factor angle) where either Va = Vab or Va = Vac (not sure which one...would have to know whether it is 0.86 leading or lagging and would have to study it a little closer)

How did you measure these? Some PT's in the circuit I imagine. Can you describe the setup?

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[odlanor]

If you measuring were made by three-wattmeter with Point o´Floating of a three-phase Y load , you can use Matlab:
% i= I * exp(-j*teta *180/pi); <==== complex number

for Ia= 782.831 @184.8 deg
=> ia = 782.831 * exp(-j*184.4*180/pi);

for Va= 10.973 kV @0 deg
=> va = 10.973 * exp(-j*10.973*180/pi);
and so on

N = (1/sqrt(3)) * va*ia + vb*ib + vc*ic;
P=real(N)
- 79.21 Watt

 

[eskim]

Thank you all.
Sorry, I am not an active member of this forum I just open this if i have problems.
My calculation showed (-) sign since the angle of current are not in correct direction I think.
I have to correct CT connection for this relay, and will come up eith the result next time ?