# Effect of Reversed Phase Polarity on Active & Reactive Power of 3-Phase System?

I am trying to investigate the Power/Energy of 3-Phase System under Different Faults.

I have a Three Phase Source and Load, Load is balanced i.e same phase angle and current (Phase-Neutral V = 230 rms, Current = 10 rms) for a power factor of unity (Ideally, typically 0.9-0.95). There is a device (Energy Meter) in between them which is measuring Power/Energy transferred from Source to Load.

Everything below is related to calculations and measurements of that Measuring device. Phase reversal/polarity does not have an effect on source /Load, only the Meter measurements will be affected.

I know about the phase sequence and Power equations of a 3-Phase Balanced System.

(Kindly forgive me if it gives you a headache, I realized after writing this question, how large it became)

I am using +ve Phase Sequence in my calculations.

So the Power Equation is like this;

P_Total = 3 * V_phase * I_phase * cos⁡θ

The Fault Condition is like this: "The Phase_C and Neutral are swapped or Phase C polarity is reversed". i.e. Black Wires goes in 5 and comes out of 6, while yellow wire goes in 7 and comes out of 8 of Meter.

In this condition, Meter shows Active Power of 9.249KW, Reactive Power of 1.89 KVAr, while power factor is still 0.9.

Now I want to measure the voltages, their phasors and ultimately the Power.

This is what I am trying to do:

V_AN will become V_AC because Phase-C is now at Neutral connection.

Here, Positive Phase Sequence is considered for calculations. Then the Phase-A Voltage will be as following,

V_AC= V_A - V_C

∴V_AC= Phase Voltage of A,Don^' t confuse it with Line Voltage of Balanced 3 Phase System

Inserting the values...

V_AC= v_rms<0 - v_rms<-240

V_AC= v_rms (cos⁡0+j sin0) - v_rms (cos⁡240-j sin240)

V_AC= v_rms (1+0)- v_rms (-1/2+j √3/2)

V_AC= v_rms (1+1/2-j √3/2)

V_AC= v_rms (3/2-j √3/2)

V_AC= √3 v_rms (√3/2-j 1/2)

V_AC= √3 v_rms < 30

Similarly,

V_BC= √3 v_rms<-90

V_NC=v_rms <-60

Now it gets tricky. To get Active Power P, i am trying to calculate Apparent POwer S = VI*.

For that, according to simple conjugate and multiplication properties of Complex numbers, Each Phase Voltage and Current will be multiplies after taking conjugate of current.

S_Total= S_Phase-A + S_Phase-B + S_Phase-C

S_Total= (V_A x I_A*) + (V_B x I_B*) + (V_C x I_C*)

S_Total=(V_A<30 × I_A<∅-30)+(V_B<-90 × I_A<∅+90)+(V_C<-60 × I_C <∅+60)

S_Total=[(√3 x v_rms x i_rms) + (√3 x v_rms x i_rms) + (v_rms x i_rms)] <∅

S_Total = 4.464 x v_rms x i_rms <∅^0

So,

Total Active Power,

P_Total = 4.464 x v_rms x i_rms x cos⁡∅

Total Reactive Power,

Q_Total = 4.464 x v_rms x i_rms x sin⁡∅

From Here, assuming, Power Factor =1, Each Phase Voltage = 230 Vrms, and I = 10 A rms,

Power Comes 10.267 KW, instead of 6.9 KW.

• Did i make any mistakes in calculations?

• Why Power Increased?

• It seems like the system is unbalanced now, but why there is no
reactive power, it seems like that reactive power only depends on the power factor which is 1 in this case?

• Please review my calculations and tell me if I am doing something wrong. Because I did measure power through Three Phase meter, it showed the phase voltages as I calculated but overall power was around 6.8KW while my calculation show 10.267KW??

• If I am wrong, what will be right approach to calculate power in this scenario?

Thanks and Best regards..

Power Comes 10.267 KW, instead of 6.9 KW

If the supply and load are balanced then the 3 phase power is 3 x 230 V x 10 A = 6.9 kW irrespective of phase rotation.

Please review my calculations and tell me if I am doing something wrong. Because I did measure power through Three Phase meter, it showed the phase voltages as I calculated but overall power was around 6.8KW while my calculation show 10.267KW??

No, because you have made a mountain out of something that is basically very simple.

If I am wrong, what will be right approach to calculate power in this scenario?

Use common sense and treat the scenario you described in your question as three single phase circuits each having a power of 2300 watts hence, when totalled, becomes 6900 watts.

• Yes, I do know that load consumption will not change, but I am not trying to estimate load power, i do Apologize for confusion, I didn't explain my query correctly, I have added few more details, Commented Oct 18, 2019 at 5:16
• All these calculations I did, are for Meter which is calculating power (in Reverse Polarity of C) that I can't comprehend. Commented Oct 18, 2019 at 5:19