Sequence Networks of Synchronous Generator & Fault Calculations of a single Synchronous Generator

Sequence Networks of Synchronous Generator:

Consider an unloaded synchronous generator which generates balanced emf Ea,Eb, Ec in phase. Let impedances of each winding are Z. The generators neutral terminal is grounded through an impedances Zn.

Let unbalanced short-circuit fault occurs at the terminal of the generator then unbalance current Ia, Ib,Ic flows in phase. Now their symmetrical components are:

Ia=Ia0+Ia1+Ia2

Ib=Ib0+Ib1+Ib2

Ia=Ic0+Ic1+Ic2

Now, paths for each sequence in generator and its equivalent sequence networks are:

The terminal voltage for +ve,-ve and zero sequence network can be written as:

Where from fig of zero sequence network

Zo=3Zn+Zgo

Fault calculations of a single synchronous generator:

Line to ground faults:

Consider an unloaded synchronous generator which generates balanced emf Ea, Eb, Ec in phase a, b and c respectively. Let impedances of each winding are Zs. The generators neutral terminal is grounded through impedances Zn. The initially generator is unloaded. Let fault occurs at phase a as in fig. then unbalance current Ia, Ib, Ifc flows in phase.

Then the conditions are:

Va=IaZf, Ib =0, Ifc =0

Now writing these unbalanced phase current in terms of their symmetrical components in matrix form.

Since, Ib = 0, Ic = 0.Now solving the matrix, we get:

............1

Now phase a voltage in term of symmetrical components:

..........2

Where:

.........3

From eq.1, 2 and 3

..........4

Or,

The fault current is

From the above equations, we can determine the sequence network and combination for a single line to ground fault as in the figure below.

Since all symmetric components of current are equal they are in series.

Line to Line Faults:

Let the fault occurs due to impedance Zf between phase b and c. Initially, the generator is on no-load. The conditions for this circuit are:

Ib= -Ic

Ia =0

Vb –Vc=Zf Ib

Now writing these unbalanced phase current in terms of their symmetrical components in matrix form for the above condition .

Solving the matrix, we get

From above ,

Also,

Or,

Now, substituting Ib

Since (a-a­²)( a­²-a) =3 ,we get

The phase current are

The fault current is

From the above equations, we can determine the sequence network and combination for the line to line fault as in the figure below.

Here zero sequence current is 0 so its sequence network is not involved whereas positive and negative sequence series are in series but the sequence current is opposite to each other.

Double Line to Ground Fault:

Let the fault occurs due to impedance Zf between phase b and c to ground as in the figure below. Initially, the generator is on no-load. The conditions for this circuit are:

....1

Writing the symmetrical components of voltages:

....2

Since Vb=Vc

....3

Substituting symmetrical components of current above condition

...4

Using eq 2and 3 in eq 4 ,we get

......5

Substituting symmetrical component of voltage and solving for zero sequence current, we get....6

Also, from eq 3 and 5 and writing its sequence component of voltage

........7

From Eq. 6, 7 and 1, we get;

....8

Equation 6,7 and 8 can be represented by connecting positive sequence impedance in series with parallel with negative and zero sequence network as shown in the equivalent circuit of the figure below. Finally, fault current is obtained from