Transformer Losses versus Frequency, and Voltage along a Feeder: 22-Elec-B7 Power Systems Engineering, December 2014, Q3

Question 3 (Problem 3): Transformer Losses versus Frequency, and Voltage along a Feeder (5 + 10 + 10 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — how frequency affects the several transformer losses. It matters a great deal what is held constant while the frequency changes. The core flux density follows from the applied voltage through $V \approx 4.44\,fNA B_{max}$, so at constant applied voltage $B_{max}\propto 1/f$.

Hysteresis loss obeys the Steinmetz relation $P_{h}=k_{h}fB_{max}^{\,x}$ with $x$ between 1.6 and 2.0. At constant volts per hertz — a machine or transformer fed from a variable-frequency source at constant flux — it rises in direct proportion to frequency. At constant applied voltage the two effects oppose each other and the loss falls roughly as $f^{\,1-x}$, i.e. as $1/f^{0.6}$ to $1/f$.

Eddy-current loss obeys $P_{e}=k_{e}f^{2}B_{max}^{2}t^{2}$, where $t$ is the lamination thickness. At constant volts per hertz it rises as the square of frequency; at constant applied voltage the $B^{2}$ term cancels the $f^{2}$ term exactly and the eddy loss is essentially independent of frequency, depending only on the voltage and the lamination design.

Winding (copper) loss is nominally $I^{2}R$ and so frequency independent, but only at power frequency. Skin effect and proximity effect raise the effective a.c. resistance roughly as $\sqrt{f}$ once the conductor is more than a skin depth thick, which is why high-frequency designs use foil, Litz wire or many strands. The same mechanism, plus increased leakage-flux penetration, raises the stray load loss in the tank and clamping structure faster than in proportion to frequency. The net practical result is the familiar one: a 50 Hz transformer operated at 60 Hz and the same voltage runs at lower flux and lower core loss, whereas a 60 Hz unit operated at 50 Hz and the same voltage saturates and its core loss climbs sharply.

Given. A feeder of 0.17 + j2.2 Ω supplies the high-voltage side of a 400 MVA, 225 kV : 24 kV, 50 Hz, three-phase Y–Δ transformer whose series reactance is 6.08 Ω referred to the high-voltage terminals; the load is 375 MVA at 0.89 power factor leading, at 24 kV line to line.

Given data
Quantity	Symbol	Value
Feeder impedance	Z_f	0.17 + j2.2 Ω
Transformer rating	S_base	400 MVA, 225 kV : 24 kV, 50 Hz, three phase, Y–Δ
Series reactance referred to the high-voltage terminals	X_T	6.08 Ω
Load	S_L	375 MVA at 0.89 power factor leading
Load voltage (line to line, low-voltage side)	V_L	24 kV

Find. The line-to-line voltage at the high-voltage terminals of the transformer, and then at the sending end of the feeder.

Figure 3 — feeder, transformer and load. Everything is referred to the high-voltage side, where the transformer reactance is quoted.

Approach. Normalise both impedances on the transformer rating, express the load as a current phasor, and add the two series drops in turn.

Check — the transformer voltage rating as printed. The source prints the transformer as “22 5kV: 24kV”, with a space where a decimal point or a third digit belongs. It is read here as 225 kV : 24 kV. The reading is settled by the reactance the same sentence gives: 6.08 Ω referred to the high-voltage terminals is 0.0480 p.u. on a 400 MVA, 225 kV base, which is an ordinary value for a large transmission transformer, whereas on a 22.5 kV base the same 6.08 Ω would be 4.8 p.u. — a physically impossible leakage reactance. The 225 kV reading is therefore the only self-consistent one.

Choose a per-unit base and convert the two impedances. Taking 400 MVA and the transformer ratings as the base, $$\begin{gathered}Z_{base,HV}=\dfrac{(225)^{2}}{400}=126.5625\ \Omega \;\Rightarrow\;X_{T}=\dfrac{6.08}{126.5625}=0.048040\ \text{p.u.}, \\ Z_{f}=0.001343 + j0.017383\ \text{p.u.}\end{gathered}$$ Because the transformer ratings are the bases, the low-voltage side of the transformer has its own base of 24 kV and the load voltage is exactly 1.0 p.u.
Express the load as a current phasor. With $V_{L}=1.0\angle 0°$ p.u. and a load of 0.9375 p.u. at 0.89 power factor leading, the current leads the voltage by $\phi=\cos^{-1}(0.89)=27.1268°$: $$\begin{aligned}I&=\dfrac{S}{V_{L}}\angle+\phi=0.9375\angle 27.1268° \\ &=0.834375 + j0.427463\ \text{p.u.}\end{aligned}$$ On the low-voltage side that is 9021.1 A of line current, and 962.3 A on the high-voltage side.
Part (b) — add the transformer reactance drop. Referring the load voltage to the high-voltage side (per unit, the referral is automatic) and adding the series drop, $$\begin{aligned}V_{HV}&=V_{L}+jX_{T}I \\ &=1.0+(-0.020535 + j0.040083) \\ &=0.979465 + j0.040083\ \text{p.u.}\end{aligned}$$ $$\boxed{|V_{HV}|_{LL}=(0.980285)(225)=220.564\ \text{kV}}$$ The drop has a negative real part because the load current leads: the reactance drop $jX_TI$ points partly backwards along the voltage phasor, which is the whole reason a leading load raises the far-end voltage.
Part (c) — add the feeder drop. The same current flows in the feeder, so $$\begin{aligned}V_{S}&=V_{HV}+Z_{f}I \\ &=0.979465 + j0.040083+(-0.006310 + j0.015078) \\ &=0.973155 + j0.055161\ \text{p.u.}\end{aligned}$$ $$\boxed{|V_{S}|_{LL}=(0.974717)(225)=219.311\ \text{kV}}$$ Both voltages come out below the 225 kV nominal even though the load voltage is exactly nominal on its own base. That is the leading power factor at work: a capacitive load makes the sending end the low-voltage end, the reverse of the familiar lagging case.

Problem 3 — results
Quantity	Symbol	Result
Per-unit transformer reactance	X_T	0.048040 p.u.
Per-unit feeder impedance	Z_f	0.001343 + j 0.017383 p.u.
Load current	I	0.9375 p.u. ∠ 27.127° (leading)
(b) high-voltage terminal voltage	\|V_HV\|	0.98028 p.u. = 220.564 kV line to line
(c) feeder sending-end voltage	\|V_S\|	0.97472 p.u. = 219.311 kV line to line

Concept & Theory

Topic. Per-unit voltage-drop calculation through a feeder and a delta–wye transformer, and the frequency dependence of transformer losses.

Key relations. $Z_{base}=V_{base}^{2}/S_{base}$; $V_{HV}=V_L+jX_TI$; $V_S=V_{HV}+Z_fI$; Steinmetz $P_h=k_hfB^{x}$; $P_e=k_ef^{2}B^{2}t^{2}$; $V\approx4.44fNAB_{max}$.

Why this works. Per unit removes the turns ratio from the arithmetic: once the bases are chosen in the transformer’s own ratio, a quantity referred to one side has the same numerical value on the other, so the whole feeder-transformer-load chain becomes a single series circuit. The delta–wye connection introduces a 30° phase shift between the two sides, but because it applies equally to voltage and current it changes no magnitude and no power, which is why it can be ignored in a magnitude calculation and must be restored only when the two sides are compared in phase (for instance in differential-relay settings).

Common pitfall. Forgetting that a leading power factor puts the current angle positive relative to the voltage, so that the reactance drop subtracts from rather than adds to the sending-end magnitude. Candidates who carry the lagging sign convention get a sending-end voltage above nominal and a percentage regulation of the wrong sign.

Source. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press/Wiley, Ch. 5 (transformers, the cantilever equivalent and per-unit normalisation); J. D. Glover, T. J. Overbye and M. S. Sarma, Power System Analysis and Design, 6th ed., Cengage, §3.3–3.5.

Question 3 of 7: Transformer Losses versus Frequency, and Voltage along a Feeder

Question 3 (Problem 3): Transformer Losses versus Frequency, and Voltage along a Feeder (5 + 10 + 10 points)