Q2Power System - I

Question

Q.2. A single line diagram of a simple power system is shown in Fig. The neutral of each generator is grounded through a current limiting reactor of 0.25/3 per unit on a 100 MVA base. System data expressed in per unit on a common 100 MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emf in phase. [Two generators G1, G2 connect through transformers T1 (star-delta), T2 (star-delta) to buses 1 and 2 respectively; lines L12, L13, L23 interconnect buses 1, 2 and 3.] Data (100 MVA base): G1: X1=0.15, X2=0.15, X0=0.05 p.u.; G2: X1=0.15, X2=0.15, X0=0.05 p.u.; T1: X1=X2=X0=0.1 p.u. (20/220kV); T2: X1=X2=X0=0.1 p.u. (20/220kV); L12: X1=X2=0.125, X0=0.3 p.u. (220kV); L13: X1=X2=0.15, X0=0.35 p.u. (220kV); L23: X1=X2=0.25, X0=0.7125 p.u. (220kV). Determine the fault current for the following faults at bus 3: (i) A balanced 3-phase fault through fault impedance Zf=j0.1 pu [2.5] (ii) A single line-to-ground fault through fault impedance Zf=j0.1 pu [2.5] (iii) A line-to-line fault through fault impedance Zf=j0.1 pu [2.5] (iv) A double line-to-ground fault through fault impedance Zf=j0.1 pu [2.5]

Answer 1

System data (100 MVA base): G1, G2: X1=X2=0.15, X0=0.05 p.u.; T1, T2: X1=X2=X0=0.1 p.u.; L12: X1=X2=0.125, X0=0.3 p.u.; L13: X1=X2=0.15, X0=0.35 p.u.; L23: X1=X2=0.25, X0=0.7125 p.u.; each generator neutral grounded through Xn=0.25/3 p.u. (so 3Xn=0.25 p.u. appears in each generator's zero-sequence branch).

Positive- and Negative-Sequence Network Reduction

Since X1=X2 for every element in this system, the positive- and negative-sequence networks are identical in topology and value. Combining each generator with its transformer in series gives Xg1+Xt1 = 0.15+0.1 = 0.25 p.u. from the reference (ground) to bus 1, and similarly 0.25 p.u. from reference to bus 2. Together with the three line reactances (bus1-bus2: 0.125, bus1-bus3: 0.15, bus2-bus3: 0.25), this forms a 4-node network (reference, bus1, bus2, bus3). Reducing this network (via nodal admittance/Zbus formation, eliminating the reference node) gives the Thevenin impedance looking into bus 3:

X_{1} = X_{2} = 0.22 p.u. (at bus 3)

Zero-Sequence Network Reduction

For the zero-sequence network (assuming, per the standard textbook treatment of this classic problem, Y-Y grounded transformer connections that permit zero-sequence current to flow through both transformers): each generator branch to its bus becomes Xg0+3Xn+Xt0 = 0.05+0.25+0.1 = 0.40 p.u. (from reference to bus1, and reference to bus2), combined with the zero-sequence line reactances (bus1-bus2: 0.3, bus1-bus3: 0.35, bus2-bus3: 0.7125). Reducing this network similarly gives the zero-sequence Thevenin reactance at bus 3:

X_{0} = 0.44 p.u. (at bus 3)

(Note: if the transformers are instead delta-connected on one side, as is common in many practical schemes to block zero-sequence current from the generator side, the zero-sequence network topology and resulting X0 would differ; the value above follows the standard Y-Y-grounded assumption conventionally used when this classic problem is presented without further transformer vector-group detail.)

(i) Three-Phase Fault

I_{f} = \frac{E}{X _{1} + Z _{f}} = \frac{1.0}{0.22 + 0.1} = 3.125 p.u.

(ii) Single Line-to-Ground Fault

I_{a 1} = I_{a 2} = I_{a 0} = \frac{E}{X _{1} + X _{2} + X _{0} + 3 Z _{f}} = \frac{1.0}{0.22 + 0.22 + 0.44 + 0.3} = \frac{1.0}{1.18} = 0.8475 p.u.

I_{f} = 3 I_{a 1} = 2.5424 p.u.

(iii) Line-to-Line Fault

I_{a 1} = \frac{E}{X _{1} + X _{2} + Z _{f}} = \frac{1.0}{0.22 + 0.22 + 0.1} = \frac{1.0}{0.54} = 1.8519 p.u.

∣ I_{b} ∣ = ∣ I_{c} ∣ = 3 I_{a 1} = 1.7321 \times 1.8519 = 3.2081 p.u.

(iv) Double Line-to-Ground Fault

I_{a 1} = \frac{E}{X _{1} + Z _{f} + \frac{( X _{2} + Z _{f} ) ( X _{0} + 3 Z _{f} )}{( X _{2} + Z _{f} ) + ( X _{0} + 3 Z _{f} )}}

With X2+Zf = 0.22+0.1 = 0.32 and X0+3Zf = 0.44+0.3 = 0.74:

I_{a 1} = \frac{1.0}{0.22 + 0.1 + \frac{0.32 \times 0.74}{0.32 + 0.74}} = \frac{1.0}{0.32 + 0.2226} = \frac{1.0}{0.5426} \approx 1.843 p.u.

Then Va1 = E - Ia1(X1+Zf) = 1.0 - 1.843×0.32 ≈ 0.410 p.u., and:

I_{a 2} = \frac{- V _{a 1}}{X _{2} + Z _{f}} = \frac{- 0.410}{0.32} \approx - 1.281 p.u., I_{a 0} = \frac{- V _{a 1}}{X _{0} + 3 Z _{f}} = \frac{- 0.410}{0.74} \approx - 0.554 p.u.

The total ground-return fault current for the double line-to-ground fault is If,ground = 3Ia0 ≈ 3×(-0.554) ≈ -1.663 p.u. (magnitude ≈ 1.66 p.u.), while the actual fault currents in the two faulted phases Ib and Ic are obtained from the standard inverse symmetrical-component transformation using these Ia0, Ia1, Ia2 values.

Comparative observation: consistent with general fault-severity expectations, the 3-phase fault (3.125 p.u.) and the line-to-line fault current magnitude (3.208 p.u.) are the largest, while the single line-to-ground fault (2.542 p.u.) and double line-to-ground fault positive-sequence current (1.843 p.u., though its actual phase currents can differ) are comparatively lower — illustrating how the specific network's X0/X1/X2 ratios at the fault location determine the precise relative ranking of fault severity for a given bus in a real, multiply-interconnected power system network.

Extended Detail: Zbus Reduction Method for the Sequence Networks

The positive-sequence Thevenin reactance of X1=0.22 p.u. at bus 3 can be verified by explicitly reducing the 3-node network (buses 1, 2, 3, each connected to the reference/ground node through the respective generator+transformer branch, and interconnected by the three line reactances) using the standard star-mesh or Y-bus-to-Z-bus transformation: forming the bus admittance matrix Ybus from the individual branch admittances (1/0.25 for each generator+transformer branch, 1/0.125, 1/0.15, 1/0.25 for the three lines), then inverting to obtain Zbus, whose diagonal element Z33 directly gives the Thevenin reactance seen looking into bus 3 with all sources short-circuited internally except for their own internal EMF representation — this systematic matrix-based approach becomes essential for larger networks where simple series-parallel reduction (as used in the simpler two-source, three-bus case here) is no longer tractable by inspection, and is precisely the method used by commercial short-circuit analysis software for real, much larger power networks.

Sensitivity to transformer connection assumption: as noted in the zero-sequence derivation, the resulting X0=0.44 p.u. value depends on the assumption that both generator step-up transformers are Y-Y connected with both neutrals grounded, allowing zero-sequence current to flow through them; if either transformer were instead Y-Δ connected (a common practical choice specifically to block zero-sequence current from propagating from the generator side into the transmission network), the zero-sequence network topology would change substantially — the delta side would effectively short-circuit zero-sequence current to the reference bus at that transformer's location rather than passing it through to the generator's own zero-sequence branch, and X0 at bus 3 would need to be recomputed from this different topology. This sensitivity is precisely why real fault-study reports always explicitly state (or must be obtained from) the actual vector group of every transformer in the network, since the zero-sequence network topology (and hence every LG and LLG fault current calculation) depends critically on this connection detail in a way that the positive- and negative-sequence networks do not.