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Reactance diagram: 30 MVA, 6.6 kV base, three lines

Short answer

Using 30 MVA base, XG1=0.24, XG2=0.30, XM=0.15, XT1=0.10, XT2=0.20 pu; assume T3 is 30 MVA, 115/13.2 kV, X=0.10 pu. Line zone base is 115 kV, Zb=440.83 ohm, so each line reactance in pu is X(ohm)/440.83. Draw the reactance diagram with these series reactances and grounded generator references.

Compute the per unit values taking power base of 30 MVA and 6.6 kV in the circuit of generator G1 and draw the reactance diagram of a power system shown in figure below. The system has the following data: G1: 25 MVA, 6.6 kV, X = 0.2 p.u.; G2: 15 MVA, 6.6 kV, X = 0.15 p.u.; M: 30 MVA, 13.2 kV, X = 0.15 p.u.; T1: 30 MVA, 6.6/115 kV, X = 0.1 p.u.; T2: 15 MVA, 6.6/115 kV, X = 0.1 p.u. [Figure: G1 (Y, grounded) → T1 (Y grounded–Y grounded) → bus; Line-1 from this bus to a junction; Line-3 from the junction to a bus feeding T3 (Y grounded on line side, Δ on motor side) → motor M; Line-2 from the junction down to a bus feeding T2 (grounded windings) → G2 (Y, grounded). No line impedances or T3 rating are printed.]

Base: 30 MVA and 6.6 kV in the G1 circuit. The line impedances and the T3 rating are not printed, so T3 is assumed to be 30 MVA, 115/13.2 kV, X = 0.1 pu, and the line reactances are kept as symbols X1,X2,X3X_1, X_2, X_3 (Ω).

Base voltages

  • G1 zone: 6.6 kV
  • Line zone (T1 6.6/115 kV): 115 kV, Zb=115230=440.83 ΩZ_b = \frac{115^2}{30} = 440.83\ \Omega
  • G2 zone (T2 115/6.6 kV): 6.6 kV
  • Motor zone (T3 115/13.2 kV): 13.2 kV

Per-unit values (30 MVA base)

XG1=0.2×3025=0.24 puXG2=0.15×3015=0.30 puXM=0.15×3030×(13.213.2)2=0.15 puXT1=0.1×3030=0.10 puXT2=0.1×3015=0.20 puXT3=0.10 pu (assumed rating)XLine k=Xk440.83 pu\begin{aligned} X_{G1} &= 0.2\times\frac{30}{25} = 0.24\ \text{pu}\\ X_{G2} &= 0.15\times\frac{30}{15} = 0.30\ \text{pu}\\ X_{M} &= 0.15\times\frac{30}{30}\times\left(\frac{13.2}{13.2}\right)^2 = 0.15\ \text{pu}\\ X_{T1} &= 0.1\times\frac{30}{30} = 0.10\ \text{pu}\\ X_{T2} &= 0.1\times\frac{30}{15} = 0.20\ \text{pu}\\ X_{T3} &= 0.10\ \text{pu (assumed rating)}\\ X_{Line\,k} &= \frac{X_k}{440.83}\ \text{pu} \end{aligned}

For example, a line reactance of 100 Ω would be 100/440.83=0.2268100/440.83 = 0.2268 pu.

ElementX (pu, 30 MVA)
G10.24
G20.30
Motor M0.15
T10.10
T20.20
T3 (assumed)0.10
Lines 1, 2, 3Xk/440.83X_k/440.83

Reactance diagram

     j0.10   jXL1      jXL3    j0.10
 +---/\/\/---/\/\/--+--/\/\/---/\/\/---+
 |    T1     Line1  |  Line3    T3     |
j0.24               |                 j0.15
 |                jXL2 (Line2)         |
(E1)                |                 (Em)
 |                 j0.20 (T2)          |
 |                  |                  |
 |                 j0.30 (G2)          |
 |                  |                  |
 |                 (E2)                |
 +------------------+------------------+
           reference (neutral) bus

Answer: XG1=0.24X_{G1}=0.24, XG2=0.30X_{G2}=0.30, XM=0.15X_M=0.15, XT1=0.10X_{T1}=0.10, XT2=0.20X_{T2}=0.20, XT3=0.10X_{T3}=0.10 pu, and each line =X(Ω)/440.83= X(\Omega)/440.83 pu on the 30 MVA, 115 kV base.

✓ Answer checked by Claude Opus 5.5 (high effort)

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