A single-phase transformer rated at 3000 kVA, 69 kV/4.16 kV, 60 Hz has a total internal impedance Zp of 127 , referred to the primary side. Calculate a. The rated primary and secondary currents b. The voltage regulation from no-load to fullload for a 2000 kW resistive load,

Question

A single-phase transformer rated at 3000 kVA, 69 kV/4.16 kV, 60 Hz has a total internal impedance [latex]Z_{p} ext { of } 127 \Omega[/latex], referred to the primary side.

Calculate
a. The rated primary and secondary currents
b. The voltage regulation from no-load to fullload for a 2000 kW resistive load, knowing that the primary supply voltage is fixed at 69 kV
c. The primary and secondary currents if the secondary is accidentally short-circuited

Accepted Answer

a. Rated primary current

[latex]I_{ np }=S_{ n } / E_{ np }=3000000 / 69000=43.5 A[/latex]

Rated secondary current

[latex]I_{ ns }=S_{ n } / E_{ ns }=3000000 / 4160=721 A[/latex]

b. Because the transformer exceeds 500 kVA, the windings have negligible resistance compared to their leakage reactance; we can therefore write

[latex]Z_{ p }=X_{ p }=127 \Omega[/latex]

Referring to Fig. 26a, the approximate impedance of the 2000 kW load on the secondary side is

[latex]\begin{aligned}Z=E_{ s }^{2} / P &=4160^{2} / 2000000 \&=8.65 \Omega\end{aligned}[/latex]

Load impedance referred to primary side

[latex]a^{2} Z=(69 / 4.16)^{2} imes 8.65=2380[/latex]

Referring to Fig. 26b we have

[latex]\begin{aligned}I_{ p } &=69000 / \sqrt{127^{2}+2380^{2}} \&=28.95 A \a E_{ s } &=\left(a^{2} Z ight) I_{ p }=2380 imes 28.95 \&=68902 V \E_{ s } &=68902 imes(4.16 / 69)=4154 V\end{aligned}[/latex]

Because the primary voltage is held constant at 69 kV, it follows that the secondary voltage at noload is 4160 V.

Voltage regulation is

[latex]\begin{aligned} ext { voltage regulation } &=\frac{E_{ NL }-E_{ FL }}{E_{ FL }} imes 100 \&=\frac{4160-4154}{4154} imes 100 \&=0.14 \%\end{aligned}[/latex]

The voltage regulation is excellent.

c. Referring again to Fig. 26b, if the secondary is accidentally short-circuited, [latex]a E_{ s }=0[/latex] and so

[latex]\begin{aligned}I_{ p } &=E_{ p } / X_{ p }=69000 / 127 \&=543 A\end{aligned}[/latex]

The corresponding current [latex]I_{ s }[/latex] on the secondary side

[latex]\begin{aligned}I_{ s } &= a I_{ p }=(69 / 4.16) imes 543 \&=9006 A\end{aligned}[/latex]

The short-circuit currents in both the primary and secondary windings are 12.5 times greater than the rated values. The I²R losses are, therefore, 12.5² or 156 times greater than normal. The circuit-breaker or fuse protecting the transformer must open immediately to prevent overheating. Very powerful electromagnetic forces are also set up. They, too, are 156 times greater than normal and, unless the windings are firmly braced and supported, they may be damaged or torn apart.

Question 10.5: A single-phase transformer rated at 3000 kVA, 69 kV/4.16 kV...

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