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Reactance diagram: 300 MVA, 20 kV generator, two motors

Short answer

Using base 300 MVA, Vb=20 kV (gen), 230 kV (line), 13.8 kV (motors). Per-unit reactances: generator j0.2, T1 j0.0857, line (52 Ω) j0.2949, T2 j0.0915, M1 j0.2745, M2 j0.5490.

A 300 MVA, 20 kV three-phase generator has a sub-transient reactance of 20%. The generator supplies two synchronous motors over a 65 kilometre line having transformers at both ends, as shown on the single line diagram. The neutral of one of the motors M1 is grounded while M2 is ungrounded. Rated inputs to the motors are 200 MVA at 13.2 kV and 100 MVA at 13.2 kV for M1 and M2 respectively. For both motors X"d = 20%. The three phase transformer T1 is rated 350 MVA, 230/20 kV with leakage reactance of 10%. Transformer T2 is composed of three single phase transformers, each rated 127/13.2 kV, 100 MVA with leakage reactance of 10%. Series reactance of the transmission line is 0.8 ohm/km. Draw the reactance diagram with all the reactances marked in p.u. Select the generator rating as base in the generator circuit. [Figure: G (Y, neutral grounded through an impedance) → T1 (Δ on generator side, Y grounded on line side) → transmission line TL → T2 (Y grounded on line side, Δ on motor side) → motors M1 (Y, grounded) and M2 (Y, ungrounded).]

In a per-unit reactance diagram every element is expressed on one common MVA base, and the base voltage changes from zone to zone in the ratio of the transformer line-to-line voltages. Resistances, magnetising branches and the grounding impedances are neglected (they carry no current under balanced conditions).

Step 1: Base values in each zone

  • Base MVA (all zones): Sb=300S_b = 300 MVA
  • Generator zone: Vb1=20V_{b1} = 20 kV (given)
  • Line zone (T1 is 20/230 kV): Vb2=20×23020=230V_{b2} = 20 \times \frac{230}{20} = 230 kV
  • T2 is a Y–Δ bank of single-phase units 127/13.2 kV, so its line-to-line ratio is 3×127/13.2=220/13.2\sqrt{3}\times127 / 13.2 = 220/13.2 kV. Motor zone:
Vb3=230×13.2220=13.8 kVV_{b3} = 230 \times \frac{13.2}{220} = 13.8\ \text{kV} Zb,line=Vb22Sb=2302300=176.33 ΩZ_{b,\text{line}} = \frac{V_{b2}^2}{S_b} = \frac{230^2}{300} = 176.33\ \Omega

Step 2: Change of base

Xpu,new=Xpu,old×Sb,newSb,old×(VoldVb,new)2X_{pu,new} = X_{pu,old}\times\frac{S_{b,new}}{S_{b,old}}\times\left(\frac{V_{old}}{V_{b,new}}\right)^2

Step 3: Per-unit reactances

XG=0.20 pu (already on its own rating)XT1=0.10×300350=0.0857 puXline=0.8×65176.33=52176.33=0.2949 puXT2=0.10×300300×(13.213.8)2=0.0915 puXM1=0.20×300200×(13.213.8)2=0.2745 puXM2=0.20×300100×(13.213.8)2=0.5490 pu\begin{aligned} X_G &= 0.20\ \text{pu (already on its own rating)}\\ X_{T1} &= 0.10\times\frac{300}{350} = 0.0857\ \text{pu}\\ X_{line} &= \frac{0.8\times 65}{176.33} = \frac{52}{176.33} = 0.2949\ \text{pu}\\ X_{T2} &= 0.10\times\frac{300}{300}\times\left(\frac{13.2}{13.8}\right)^2 = 0.0915\ \text{pu}\\ X_{M1} &= 0.20\times\frac{300}{200}\times\left(\frac{13.2}{13.8}\right)^2 = 0.2745\ \text{pu}\\ X_{M2} &= 0.20\times\frac{300}{100}\times\left(\frac{13.2}{13.8}\right)^2 = 0.5490\ \text{pu} \end{aligned}

(The three-phase rating of T2 is 3×100=3003\times100 = 300 MVA.)

ElementRating usedX (pu on 300 MVA)
Generator G300 MVA, 20 kV0.2000
T1350 MVA, 20/230 kV0.0857
Line52 Ω, base 176.33 Ω0.2949
T2300 MVA, 220/13.2 kV0.0915
Motor M1200 MVA, 13.2 kV0.2745
Motor M2100 MVA, 13.2 kV0.5490

Step 4: Reactance diagram

     j0.0857   j0.2949   j0.0915
 +---/\/\/-----/\/\/-----/\/\/---+--------+
 |    T1        line       T2    |        |
j0.2                          j0.2745  j0.5490
 |                               |        |
(Eg)                           (Em1)    (Em2)
 |                               |        |
 +-------------------------------+--------+
            reference (neutral) bus

Answer: On 300 MVA base, XG=0.2X_G = 0.2, XT1=0.0857X_{T1} = 0.0857, Xline=0.2949X_{line} = 0.2949, XT2=0.0915X_{T2} = 0.0915, XM1=0.2745X_{M1} = 0.2745 and XM2=0.5490X_{M2} = 0.5490 pu, with base voltages 20 kV, 230 kV and 13.8 kV in the three zones.

✓ Answer checked by Claude Opus 5.5 (high effort)

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