Impedance and p.u. reactance diagram, 5000 VA base

(a) Impedance diagram (no values)

Each generator is shown as an EMF behind its internal impedance. Each transformer is shown as series leakage impedance with a shunt magnetising branch. The line is a series $R + jX$ (short line), and the load is an impedance.

  G1        T1                  T2
 (~)-Zg1-+-Zt1-+--R+jX line--+-Zt2-+-- Load
         |     |             |     |   Z_L
 (~)-Zg2-+    Ym             Ym     |
  G2     |     |             |     |
 ref ----+-----+-------------+-----+---

(b) Selecting the bases

• $S_b = 5000$ VA for the whole system.
• Generator section: $V_b = 250$ V.
• Line section: through T1 (250/800 V), $V_b = 250\times800/250 = 800$ V.
• Load section: through T2 (rated 1000 V on the line side), $V_b = 800\times500/1000 = 400$ V.

Per-unit values

Use $Z_{new} = Z_{old}\,\dfrac{S_{b,new}}{S_{old}}\left(\dfrac{V_{old}}{V_{b,new}}\right)^2$.

$$\begin{aligned} G_1&: j0.2\times\frac{5000}{2000}\times1 = j0.5\\ G_2&: j0.3\times\frac{5000}{2000} = j0.75\\ T_1&: j0.2\times\frac{5000}{4000}\times\left(\frac{250}{250}\right)^2 = j0.25 \end{aligned}$$

Line: the base impedance is

$$Z_b = \frac{800^2}{5000} = 128\ \Omega$$ $$Z_{line} = \frac{40 + j150}{128} = 0.3125 + j1.1719\ \text{p.u.}$$

T2 (rated 1000 V, while the base on its side is 800 V):

$$j0.06\times\frac{5000}{8000}\times\left(\frac{1000}{800}\right)^2 = j0.0586$$

Reactance (per-unit) diagram

  j0.5         j0.25   0.3125+j1.172   j0.0586
 (G1)--+------/\/\---------/\/\--------/\/\----Load
       |                                       |
 (G2)--+ j0.75                                 |
       |                                       |
 ------+------------ reference ----------------+

The magnetising branches are neglected in the reactance diagram. The load impedance is not given, so it is shown as a block. Its p.u. value would be $Z_L/Z_b$ with $Z_b = 400^2/5000 = 32\ \Omega$.