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Reactance diagram: three generators, two 220 kV lines

Short answer

Base 100 MVA, 11 kV at generators, 220 kV line. All components rated at base MVA and matching voltage, so per-unit reactances equal percent/100: G1 j0.25, G2 j0.20, G3 j0.20, T1 j0.06, T2 j0.07, T3 j0.07, each line j0.10 (parallel combination j0.05).

Draw the reactance diagram of the following figure with its equivalent per unit system. Take base MVA = 100 MVA and base voltage = 11 kV for generator. Compute the reactance diagram. [Figure: G1 → T1 and G2 → T2 feed a common 220 kV bus at the sending side; Line-1 and Line-2 run in parallel from there to a receiving bus, which feeds G3 through T3. Data: G1 100 MVA, 11 kV, X = 25%; G2 100 MVA, 11 kV, X = 20%; G3 100 MVA, 11 kV, X = 20%; T1 100 MVA, 11/220 kV, X = 6%; T2 100 MVA, 11/220 kV, X = 7%; T3 100 MVA, 220/11 kV, X = 7%; Line-1 100 MVA, 220 kV, X = 10%; Line-2 100 MVA, 220 kV, X = 10%.]

Base values

  • Sb=100S_b = 100 MVA, and Vb=11V_b = 11 kV in the generator circuits.
  • Line section (through 11/220 kV transformers): Vb=220V_b = 220 kV, so Zb=2202/100=484 ΩZ_b = 220^2/100 = 484\ \Omega.
  • G3 circuit (through T3, 220/11 kV): Vb=11V_b = 11 kV.

Conversion to the common base

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

Every component is rated at 100 MVA, and its rated voltage equals the base voltage of its section. Both correction factors are therefore 1, and each p.u. value equals its percentage reactance divided by 100.

ElementRatingCalculationp.u. reactance
G1100 MVA, 11 kV, 25 %0.25×1×10.25\times1\times1j0.25
G2100 MVA, 11 kV, 20 %0.20×1×10.20\times1\times1j0.20
G3100 MVA, 11 kV, 20 %0.20×1×10.20\times1\times1j0.20
T1100 MVA, 11/220 kV, 6 %0.06×1×10.06\times1\times1j0.06
T2100 MVA, 11/220 kV, 7 %0.07×1×10.07\times1\times1j0.07
T3100 MVA, 220/11 kV, 7 %0.07×1×10.07\times1\times1j0.07
Line 1100 MVA, 220 kV, 10 %0.10×1×10.10\times1\times1j0.10
Line 2100 MVA, 220 kV, 10 %0.10×1×10.10\times1\times1j0.10

In ohms, each line is 0.10×484=48.4 Ω0.10\times484 = 48.4\ \Omega.

Reactance diagram

 j0.25  j0.06          j0.10 (L1)
+-/\/\--/\/\--+------+--/\/\--+------+ j0.07  j0.20
|             |      |        |      +--/\/\--/\/\--+
(E1)          |      +--/\/\--+                     |
|             |        j0.10 (L2)                  (E3)
|  j0.20 j0.07|                                     |
+-/\/\--/\/\--+                                     |
|                                                   |
(E2)                                                |
+----------------------- reference -----------------+

Simplified values (useful for fault studies)

  • The two lines in parallel: j0.10 ∥ j0.10=j0.05j0.10\,\|\,j0.10 = j0.05.
  • G1–T1 branch: j0.31j0.31. G2–T2 branch: j0.27j0.27.
  • These two branches in parallel: 0.31×0.270.58=j0.1443\dfrac{0.31\times0.27}{0.58} = j0.1443.
  • G3–T3 branch: j0.27j0.27.

✓ Answer checked by Claude Opus 5.5 (high effort)

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