Transformer Insulating Oil and the Cantilever Equivalent Circuit: 22-Elec-B7 Power Systems Engineering, May 2013, Q3

Problem 3: Transformer Insulating Oil and the Cantilever Equivalent Circuit (20 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) — what the oil in the tank is for. Mineral insulating oil (or a synthetic ester in modern fire-point-sensitive installations) performs four distinct duties simultaneously, and a complete answer names all four.

Dielectric. The oil is first of all an insulant. Its dielectric strength, of the order of 30 kV across a 2.5 mm gap for new oil to ASTM D877 / CSA test methods, is far higher than that of air, so it permits much smaller clearances between windings and between winding and tank than a dry design of the same rating. Just as importantly it impregnates the cellulose paper and pressboard, displacing air from voids where partial discharge would otherwise start; a paper-oil composite is a far better insulation system than either component alone because the oil fills the pores and the paper subdivides the oil gaps.

Coolant. The oil is the heat-transfer medium. Losses generated in the windings and core are carried by natural or forced convection to the tank walls, radiators or heat exchangers — the ONAN, ONAF, OFAF and ODAF cooling classes are simply descriptions of how the oil and the outside air or water are moved. Oil has roughly 1600 times the volumetric heat capacity of air, which is why an oil-filled unit can be a fraction of the size of a dry-type unit of the same rating.

Arc quenching and protection of the insulation system. The oil suppresses and extinguishes incipient arcing, for instance at a tap-changer contact, by cooling the arc column and by the de-ionising action of the gases evolved. It also excludes oxygen and moisture from the cellulose, which is what actually determines transformer life: paper degrades by hydrolysis and oxidation, and it is the water and oxygen dissolved in the oil that drive both.

Diagnostic medium. Because every incipient fault decomposes oil or paper into characteristic gases, the oil is also the transformer's condition monitor. Dissolved-gas analysis (hydrogen and acetylene for arcing, ethylene for hot spots, carbon oxides for paper degradation), together with routine tests for breakdown voltage, acidity, interfacial tension, water content and furanic compounds, allows a fault to be identified and trended long before it becomes a failure. In Canadian practice these tests and the acceptance limits are covered by the CSA C50 / CAN-CSA-C88 series and by IEEE C57.104 for the gas interpretation.

Check: the “cantilever” equivalent circuit of Figure 1 places the entire excitation branch ($G_c$ in parallel with $-jB_m$) directly across the primary terminals, with $R_{eq}+jX_{eq}$ between that node and the referred secondary. All four parts below are solved on that topology, as the question directs. (The more familiar “exact” T-circuit splits the series impedance either side of the shunt branch; on this transformer the difference in $|V_1|$ is under 0.1 %, but the cantilever model is what makes parts (b) and (d) closed-form rather than iterative.)

[Figure not reproduced: Figure 3.1 — the cantilever equivalent circuit referred to the 2200 V side, redrawn from Figure (1) of the paper. See the official exam paper.]

Given.

Quantity	Value	Meaning
Rating	250 kVA, 2200/220 V, 60 Hz, 1-phase	nameplate
$a$	$2200/220 = 10$	turns ratio (HV : LV)
$R_{eq}$	$0.475\ \Omega$	equivalent series resistance, HV referred
$X_{eq}$	$2.15\ \Omega$	equivalent series reactance, HV referred
$G_c$	$2.5\times10^{-4}\ \text{S}$	core-loss conductance
$B_m$	$4\times10^{-4}\ \text{S}$	magnetising susceptance

Find. (b) $|V_1|$ and $|I_1|$ for a 180 kVA, 0.8 lagging load; (c) $S_1$, primary power factor and voltage regulation at that load; (d) working backwards from a stated primary condition of 125 A at 2200 V and 0.8 lagging, the secondary kVA, the efficiency and the load power factor.

Approach. Refer everything to the 2200 V side once, then walk the cantilever circuit in the direction the data allows: outward from the load in (b) and (c), inward from the primary in (d).

Part (b) — refer the load to the high-voltage side. With $a = 10$, a secondary at rated 220 V refers to $V_2' = a V_2 = 2200\angle 0^\circ$ V, taken as the reference phasor. The referred load current follows from the apparent power, which is invariant under referral: $$I_2'=\frac{S_2}{V_2'}\angle{-\cos^{-1}(0.8)} =\frac{180\,000}{2200}\angle-36.87^\circ =81.8182\,\text{A}\angle-36.87^\circ$$ (the actual secondary current is ten times this, 818.2 A).
Primary voltage. In the cantilever model the whole series impedance lies between the referred secondary and the primary node, and the entire load current flows through it: $$V_1=V_2'+I_2'\left(R_{eq}+jX_{eq}\right) =2200+\left(81.8182\angle-36.87^\circ\right)\left(0.475+j2.15\right)$$ $$\boxed{V_1=2339.58\,\text{V}\angle2.88^\circ\ \ \Rightarrow\ \ |V_1| =2339.6\ \text{V}}$$ The magnitude of the series drop is 180.2 V, about 8.2 % of rated; the voltage regulation computed in part (c) is smaller (6.34 %) because that compares magnitudes rather than the phasor difference. Both are what a 0.8-lagging load does to a unit with an $X/R$ ratio of 4.5.
Excitation current, then primary current. The shunt branch sees the primary voltage in this model, so $$I_\varphi=V_1\left(G_c-jB_m\right) =\left(2339.58\angle2.88^\circ\right)\left(2.5\times10^{-4}-j4\times10^{-4}\right) =1.1036\,\text{A}\angle-55.12^\circ$$ and by Kirchhoff's current law at the primary node $$I_1=I_2'+I_\varphi=81.8182\angle-36.87^\circ+1.1036\angle-55.12^\circ \;\Longrightarrow\;\boxed{|I_1|=82.87\ \text{A}}$$ The excitation current is only 1.10 A, about 1.3 % of the load current, which is why it is so often dropped — but it is what parts (c) and (d) are testing, so it is kept.
Part (c) — apparent power and power factor at the primary. $$S_1=V_1I_1^{*}=193874.3\,\text{VA}\angle39.99^\circ \;\Longrightarrow\;\boxed{S_1=193.87\ \text{kVA}}$$ with $P_1 = 148.55$ kW and $Q_1 = 124.58$ kvar. The primary power-factor angle is $\angle V_1-\angle I_1 = 39.99^\circ$, so $$\boxed{\cos\varphi_1=0.7662\ \text{lagging}}$$ Note it is worse than the 0.8 of the load: the magnetising branch draws 0.91 A of extra lagging current, and the series reactance absorbs a further 14.39 kvar.
Voltage regulation. Regulation compares the secondary voltage at no load with the secondary voltage on load, at constant primary voltage. In the cantilever model no load means no current through $R_{eq}+jX_{eq}$, so the no-load referred secondary voltage is exactly $|V_1|$: $$\text{VR}=\frac{|V_1|-|V_2'|}{|V_2'|}\times100\,\% =\frac{2339.6-2200}{2200}\times100\,\% \;\Longrightarrow\;\boxed{\text{VR}=6.34\,\%}$$ As a by-product, the efficiency at this load is $\eta = P_2/P_1 = 144.0/ 148.5 = 96.94\,\%$.
Part (d) — now work inward from the primary. The data changes: $V_1 = 2200\angle 0^\circ$ V is now the reference and $I_1 = 125\angle-36.87^\circ$ A is given. Because the shunt branch is across the primary in this model, its current is known immediately and does not depend on the load at all: $$I_\varphi=V_1(G_c-jB_m)=2200\left(2.5\times10^{-4}-j4\times10^{-4}\right) =1.0377\,\text{A}\angle-57.99^\circ$$
Referred secondary current and voltage. Subtracting the excitation current from the primary current leaves the load branch, $$I_2'=I_1-I_\varphi=125.0000\angle-36.87^\circ-1.0377\angle-57.99^\circ =124.0326\,\text{A}\angle-36.70^\circ$$ and the series drop then gives the referred secondary voltage $$V_2'=V_1-I_2'\left(R_{eq}+jX_{eq}\right)=2001.39\,\text{V}\angle-5.12^\circ$$ On the real 220 V winding that is 200.1 V at 1240.3 A.
Secondary loading, efficiency and load power factor. $$S_2=V_2'I_2'^{*}=248237.4\,\text{VA}\angle31.58^\circ \;\Longrightarrow\;\boxed{|S_2|=248.24\ \text{kVA}}$$ which is 99.3 % of the 250 kVA nameplate — the transformer is essentially fully loaded, a useful sanity check on the arithmetic. With $P_1 = |V_1||I_1|\cos\varphi_1 = 2200\times125\times0.8 = 220.0$ kW, $$\eta=\frac{P_2}{P_1}=\frac{211.48} {220.0}\times100\,\% \;\Longrightarrow\;\boxed{\eta=96.13\,\%}$$ and the load power factor is the cosine of the angle between $V_2'$ and $I_2'$, $$\cos\varphi_2=\cos\!\left(-5.12^\circ -\left(-36.70^\circ\right)\right) \;\Longrightarrow\;\boxed{\cos\varphi_2=0.8519\ \text{lagging}}$$ The load is better than the 0.80 seen at the primary, for the same reason as in part (c) read backwards: the magnetising branch and the leakage reactance have already taken their share of the reactive power upstream of the load.

Part	Quantity	Result
(b)	Referred load current $I_2'$	$81.818\angle-36.87^\circ$ A
(b)	Primary voltage $\|V_1\|$	2339.6 V
(b)	Primary current $\|I_1\|$	82.87 A
(c)	Primary apparent power $S_1$	193.87 kVA
(c)	Primary power factor	0.7662 lagging
(c)	Voltage regulation	6.34 %
(c)	Efficiency at 180 kVA, 0.8 pf	96.94 %
(d)	Secondary load $\|S_2\|$	248.24 kVA
(d)	Efficiency	96.13 %
(d)	Load power factor	0.8519 lagging

Topic. The cantilever (approximate) equivalent circuit of a single-phase transformer, and the role of transformer oil.

Key relations. $V_1 = V_2' + I_2'(R_{eq}+jX_{eq})$ with the shunt branch at the primary; $I_\varphi = V_1(G_c - jB_m)$; $I_1 = I_2' + I_\varphi$; $\text{VR} = (|V_1| - |V_2'|)/|V_2'|$; $\eta = P_2/P_1$; referral $V' = aV$, $I' = I/a$, $Z' = a^{2}Z$.

Why this works. The exact transformer model puts half the leakage impedance on each side of the magnetising branch, which makes every calculation implicit: the primary voltage depends on the excitation current, which depends on the voltage at the shunt node, which depends on the primary voltage. The cantilever model moves the entire excitation branch to one terminal pair. Because the excitation current is one or two percent of rated, the error introduced is of the same order as the error in the parameters themselves, and every quantity in this question then follows in closed form — forwards in parts (b) and (c), backwards in part (d).

Common pitfall. The commonest error is to apply the shunt branch at the wrong node, i.e. to compute $I_\varphi$ from $V_2'$ rather than $V_1$; in part (d) that mistake propagates into every subsequent answer. The second is to define voltage regulation from $V_1$ against rated secondary rather than against the actual on-load secondary, which changes the number whenever the secondary is off nominal — as it is in part (d).

Source. El-Hawary, Electrical Power Systems: Design and Analysis, transformer chapter (cantilever equivalent circuit); Chapman, Electric Machinery Fundamentals, 5th ed., §2.5–2.8; CSA C50 and IEEE C57.104 for insulating-oil condition assessment.

Question 3 of 7: Transformer Insulating Oil and the Cantilever Equivalent Circuit

Problem 3: Transformer Insulating Oil and the Cantilever Equivalent Circuit (20 points)