The primary, secondary and tertiary windings of a 50-Hz, single-phase, 3-winding transformer are rated as 6.35 kV, 5 MVA; 1.91 kV, 2.5 MVA; 400 V, 2.5 MVA respectively. Three SC tests on this transformer yielded the following results:
(i) Secondary shorted, primary excited: 500 V, 393.7 A
(ii) Tertiary shorted, primary excited: 900 V, 393.7 A
(iii) Tertiary shorted, secondary excited: 231 V, 21 312.1 A
Resistances are to be ignored.
(a) Find the pu values of the equivalent circuit impedances of the transformer on a 5 MVA, rated voltage
base.
(b) Three of these transformers are used in a 15 MVA, Y-Y-\Delta , 3-phase bank to supply 3.3 kV and 400 V auxiliary power circuits in a generating plant. Calculate the pu values of steady-state short-circuit currents and of the voltage at the terminals of the secondary windings for a 3-phase balanced shortcircuit at the tertiary terminals. Use 15 MVA, 3-phase rated voltage base.
Question 3.25: The primary,secondary and tertiary windings of a 50-Hz, sing...
(a) Let us first convert the SC data to pu on 5 MVA base/phase.
For primary, V_{B} = 6.35 kV
I_{B}=\frac{5000}{6.35} = 787.4 A
For secondary, V_{B} =1.91 kV
I_{B}=\frac{5000}{1.91} = 2617.8A
Converting the given test data to pu yields:
| Test No. | Windings involved | V | I |
| 1 | P and S | 0.0787 | 0.5 |
| 2 | P and T | 0.1417 | 0.5 |
| 3 | S and T | 0.1212 | 0.5 |
From tests 1,2 and 3,respectively. 0.0787
X_{12} =\frac{0.0787}{0.5} =0.1574 pu
X_{13} =\frac{0.1417}{0.5} =0.2834
X_{23} =\frac{0.1212}{0.5} =0.2424 pu
From (3.103) – (3.105)
\bar{Z_{1} } =\frac{1}{2} \left(\bar{Z_{12} }+\bar{Z_{13} }-\bar{Z_{23} }\right)
\bar{Z_{3} }=\frac{1}{2} \left(\bar{Z_{13} }+\bar{Z_{23} }-\bar{Z_{12} }\right)
X_{1} =0.5\left(0.1574 + 0.2834 – 0.2424\right) = 0.0992 pu
X_{2} =0.5\left(0.2424 + 0.1574 – 0.2834\right) = 0.05825 pu
X_{3} =0.5\left(0.2834 + 0.2424 – 0.1574\right) = 0.1842 pu
(b) The base line-to-line voltage for the Y-connected primaries is \sqrt{3}\times 6.35 = 11 kV, i.e. the bus voltage is 1 pu.
From Fig. 3.62, for a short-circuit at the terminals of the
tertiary, V_{3} = 0.Then
I_{SC} =\frac{V_{1} }{X_{1} + X_{3} } =\frac{V_{1} }{X_{13} } =\frac{1.00}{0.2834} = 3.53 pu
SC current primary side = 3.53\times 787.4 = 2779.5 A
SC current tertiary side = 3.53 \times \frac{5000\times 1000}{400/ \sqrt{3} } = 76424 A (line current)
Neglecting the voltage drops due to the secondary load current, the secondary terminal voltage is the voltage at the junction point A (Fig. 3.63), i.e
V_{A}=I_{SC} X{3} =3.53\times 0.1842=0.6502 pu
V_{A} (actual)=0.6502\times 1.91 \sqrt{3} =2.15 kV (line-to-line)
