Therefore it has been seen above that an unbalanced 3-phase system of voltages or currents can be regarded as due to two symmetrical 3-phase systems having opposite phase sequence and a system of zero phase sequence.
The resultant vectors, which we get after combining positive sequence vectors and negative sequence vectors, are shown below:
Va = Va1 + Va2 + Va3 ------- (equation 1)
Vb = Vb1 + Vb2 + Vb3-------- (equation 2)
Vc = Vc1 + Vc2 + Vc3 -------- (equation 3)
In the above equation, the suffix 1 indicates that the vector belongs to the positive –sequence components and a vector with suffix 2 indicates the vector belongs to the negative- sequence components. Also a vector with suffix 0 indicates that the vector belong to the zero-sequence components.
Usually, these vectors are related to each other with the help of the operator e.
The positive sequence vectors have phase sequence a-b-c and hence they are related as:
Va1 = Va1
Vb1 = e2 Va1
Vc1 = e Va1
The negative sequence vector have phase sequence a-c-b and hence the vectors are related as:
Va2 = Va2
Vc2 = e2 Va2
Vb2 = e Va2
The zero sequence vectors are in phase with each other and have the same magnitude, and hence they are related as:
Va3 = Vb3 =Vc3
Therefore substituting the above in equation 1, 2, and 3 we get
Va = Va1 +Va2 +V3
Vb = e2 Va1 + eVa2 +V3
Vc = e Va1 + e2 Va2 +V3
For our convenience, let us write Va1 as V1, Va2 as V2.
Va =V1+ V2 + V3 ----------- (equation4)
Vb = e2V1 + eV2 + V3 ----------- (equation5)
Vc = e V1+ e2V2 +V3 ----------- (equation6)
An unbalanced 3-phase current can also be similarly linked into its symmetrical component.
Ia =I1 + I2 + I3 -------- (equation7)
Ib = e2 I1 + e I2 + I3---------- (equation8)
Ic = e I1+ e2 I2 + I3--------- (equation9)
Estimation of Power Flow or Voltage and Current Flow in Symmetrical Components
Estimation of V1 and I1
From equation 8 and equation 9
We can have
e Ib = e3 I1 + e2 I2 + e I3
e2 Ic = e3 I1 + e4 I2 + e2 I3
Now let us assume that e3 =1 and e4 = e
e Ib= I1 + e2 I2 + e I3---------- (equation 10)
e2 Ic = I1 + e I2 + e2 I3------------(equation 11)
By adding equation 7, equation10, equation11 we get
Ia +e Ib +e2 Ic = I1 +I1 +I2 + I2 + e2 I2 + e I2 + I3 + e I3 +e2 I3
Ia +e Ib +e2 Ic = 2 I1 + I2 ( 1+e+e2) + I3 (1+e+e2)
Ia +e Ib +e2 Ic = 3 I1
I1= 1/3 (Ia +e Ib +e2 Ic)
The same can be applied for V1
V1 = 1/3(Va +e Vb +e2 Vc)
Similarly applying the above methods V2 or I2 and V3 or I3 can be calculated.
V2 = 1/3(Va +e2 Vb + e Vc) and I2 = 1/3(Ia +e2 Ib + e Ic)
V3 = 1/3 (Ia +Ib+Ic) and I3 = 1/3 (Ia +Ib+ Ic)
As we know that power is a product of voltage and current, the power flow can be calculated by using the above relations of vectors and currents.
Power =V1*I1 + V2*I2 +v3*I3
It is clear that the I3 and V3 is one third of the neutral or earth return current or voltage and this will be zero in the unearthed 3-wire electrical system, and also if the vector sum of the current or voltage vectors is zero, then I3 or V3 will be zero.
If there is a flow of positive sequence current components, then there is a flow of negative sequence current, and this is because both are similar, but in case of zero sequence components to flow, then a fourth wire is necessary in the system... so a set of unbalanced 3-phase line to line voltages may be represented by a positive and a negative sequence system of balanced voltages, which are known as symmetrical components.