A 100-MVA 230/115-kV Δ-Y three-phase power transformer has a per-unit resistance of 0.015 pu and a per-unit reactance of 0.06 pu. The excitation branch elements are RC =100 pu Xm =20 pu (a) If this transformer supplies a load of 80 MVA at 0.8 PF lagging, draw the phasor diagram of one phase of the

Question

A 100-MVA 230/115-kV [latex] \Delta [/latex]-Y three-phase power transformer has a per-unit resistance of 0.015 pu and a per-unit reactance of 0.06 pu. The excitation branch elements are [latex] { R }_{ C } [/latex] = 100 pu and [latex] { X }_{ M } [/latex] = 20 pu

(a) If this transformer supplies a load of 80 MVA at 0.8 PF lagging, draw the phasor diagram of one phase of the transformer.

(b) What is the voltage regulation of the transformer bank under these conditions?

(c) Sketch the equivalent circuit referred to the low-voltage side of one phase of this transformer. Calculate all the transformer impedances referred to the low-voltage side.

(d) Determine the losses in the transformer and the efficiency of the transformer under the conditions of part (b).

Accepted Answer

(a) The transformer supplies a load of 80 MVA at 0.8 PF lagging. Therefore, the secondary line current of the transformer is

[latex]I_{L S}=\frac{S}{\sqrt{3} V_{L S}}=\frac{80,000,000 \mathrm{\ VA}}{\sqrt{3}(115,000 \mathrm{~V})}=402 \mathrm{~A}[/latex]

The base apparent power is [latex]S_{ ext {base }}=100 \mathrm{\ MVA}[/latex] , and the base line voltage on the secondary side is [latex]V_{L S, ext { base }}=115 \mathrm{\ kV}[/latex] , so the base value of the secondary line current is

[latex]I_{L S, ext { base }}=\frac{S_{ ext {base }}}{\sqrt{3} V_{L S, ext { base }}}=\frac{100,000,000 \mathrm{\ VA}}{\sqrt{3}(115,000 \mathrm{~V})}=502 \mathrm{~A}[/latex]

so the per-unit secondary current is

[latex]\mathbf{I}_{L S, \mathrm{pu}}=\frac{I_{L S}}{I_{L S, \mathrm{pu}}}=\frac{402 \mathrm{~A}}{502 \mathrm{~A}} \angle \cos ^{-1}(0.8)=0.8 \angle-36.87^{\circ}[/latex]

The per-unit phasor diagram is shown below:

(b) The per-unit primary voltage of this transformer is

[latex]\mathbf{V}_P=\mathbf{V}_S+\mathbf{I} Z_{\mathrm{EQ}}=1.0 \angle 0^{\circ}+\left(0.8 \angle-36.87^{\circ} ight)(0.015+j 0.06)=1.039 \angle 1.7^{\circ}[/latex]

and the voltage regulation is

[latex]\mathrm{VR}=\frac{1.039-1.0}{1.0} imes 100 \%=3.9 \%[/latex]

(c) The secondary side of this transfer is Y-connected, so the base phase voltage of the low voltage (secondary) side of this transformer is:

[latex]V_{\phi S, ext { base }}=\frac{V_{L S, ext { base }}}{\sqrt{3}}=\frac{115 \mathrm{\ kV}}{\sqrt{3}}=66.4 \mathrm{\ kV}[/latex]

The base impedance of the transformer referred to the low-voltage side is:

[latex]Z_{ ext {base }}=\frac{3 V_{\phi S, ext { base }}^2}{S_{ ext {base }}}=\frac{3(66.4 \mathrm{\ kV})^2}{100 \mathrm{\ MVA}}=133 \ \Omega[/latex]

Each per-unit impedance is converted to actual ohms referred to the low-voltage side by multiplying it by this base impedance. The resulting equivalent circuit is shown below:

[latex]\begin{aligned}& R_{\mathrm{EQ}, S}=(0.015)(133 \ \Omega)=2.00 \ \Omega \& X_{\mathrm{EQ}, S}=(0.06)(133 \ \Omega)=7.98 \ \Omega \& R_C=(100)(133 \ \Omega)=13.3 \ \mathrm{k} \Omega \& X_M=(20)(133 \ \Omega)=2.66 \ \mathrm{k} \Omega\end{aligned}[/latex]

(d) The per-unit losses in the series resistance are

[latex]P_{E Q}=I^2 R_{E Q}=(0.8)^2(0.015)=0.0096 \mathrm{\ pu}[/latex]

and the actual losses in the series resistance are:

[latex]P_{E Q}=S_{ ext {base }} P_{E Q, \mathrm{pu}}=(0.0096 \mathrm{\ pu})(100 \mathrm{\ MVA})=0.96 \mathrm{\ MW}[/latex]

The per-unit losses in the excitation branch are:

[latex]P_{E X, \mathrm{pu}}=\frac{V^2}{R_{E X}}=\frac{(1.039)^2}{100}=0.0108 \mathrm{\ pu}[/latex]

and the actual losses in the excitation branch are:

[latex]P_{E X}=S_{ ext {base }} P_{E X, \mathrm{pu}}=(0.0108 \mathrm{\ pu})(100 \mathrm{\ MVA})=1.08 \mathrm{\ MW}[/latex]

The per-unit power supplied to the load

[latex]P_{ ext {load.pu }}=\frac{P_{ ext {load }}}{S_{ ext {base }}}=\frac{80 \mathrm{\ MW}}{100 \mathrm{\ MVA}}=0.80 \mathrm{\ pu}[/latex]

Therefore, the transformer’s efficiency is

[latex]\eta=\frac{P_{ ext {OUT }}}{P_{\mathrm{IN}}} imes 100 \%=\frac{0.80}{0.80+0.0096+0.0108} imes 100 \%=97.5 \%[/latex]

Question 2.11: A 100-MVA 230/115-kV Δ-Y three-phase power transformer has a...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions