11.1 Power Transformers (Equivalent Circuits, Efficiency & Voltage Regulation)

11.1 Power Transformers (Equivalent Circuits, Efficiency & Voltage Regulation)

Executive Overview: Power transformers are the central electromagnetic links in electrical power transmission and distribution networks. On the NCEES PE Power examination, transformer questions evaluate quantitative mastery of turns ratio conversions, per-phase equivalent circuit parameter extraction from standard laboratory tests (Open-Circuit and Short-Circuit tests), full-load voltage regulation under lagging and leading power factors, load-dependent efficiency calculations, autotransformer power magnification, and condition assessment diagnostics including Dissolved Gas Analysis (DGA).

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1. Transformer Construction, Turns Ratio, Polarity & Autotransformers

Turns Ratio and Impedance Referral

An ideal two-winding transformer relates primary and secondary voltages, currents, and impedances through the turns ratio $a$:

$a = \frac{N_1}{N_2} = \frac{V_1}{V_2} = \frac{I_2}{I_1}$

When transferring circuit impedances from one side of the transformer to the other:

• Referring Secondary Impedance to Primary (High-to-Low or Low-to-High): $Z_2' = a^2 Z_2 = \left(\frac{N_1}{N_2}\right)^2 Z_2$
• Referring Primary Impedance to Secondary: $Z_1' = \frac{Z_1}{a^2} = \left(\frac{N_2}{N_1}\right)^2 Z_1$

Primary Side (N1)                     Secondary Side (N2)
   o-------+                             +-------o
           )                             (
   V1     ( N1                         N2 )     V2
           )                             (
   o-------+                             +-------o

Impedance Referral Rule:
  • Move from LV to HV => Multiply impedance by a^2 (Z increases)
  • Move from HV to LV => Divide impedance by a^2 (Z decreases)

Terminal Polarity: Subtractive vs. Additive

Transformer polarity indicates the relative instantaneous direction of induced voltages between high-voltage (HV, labeled $H_1, H_2$) and low-voltage (LV, labeled $X_1, X_2$) terminals:

• Subtractive Polarity: $H_1$ and $X_1$ are adjacent (on the same side). If $H_1$ and $X_1$ are connected together and voltage $V_{H1-H2}$ is applied, the voltage measured across the remaining terminals is $V_{test} = V_{HV} - V_{LV}$.
• Additive Polarity: $H_1$ and $X_1$ are diagonally opposite ($H_1$ adjacent to $X_2$). The test voltage is $V_{test} = V_{HV} + V_{LV}$.

Autotransformers & Power Magnification

An autotransformer connects the primary and secondary windings in series, sharing a common section of winding. Because part of the energy transfers through direct electrical conduction rather than purely magnetic induction, an autotransformer exhibits significantly higher apparent power throughput for a given physical core size.

For a step-down autotransformer with high-voltage $V_H$ and low-voltage $V_X$ ($a_{auto} = V_H / V_X > 1$):

$S_{auto} = S_{two-winding} \times \frac{V_H}{V_H - V_X} = S_{2w} \times \frac{a_{auto}}{a_{auto} - 1}$

The total power throughput decomposes into two components:

1. Transformed (Inductive) Power: Transferred magnetically through the core: $S_{ind} = S_{auto} \left(1 - \frac{1}{a_{auto}}\right) = S_{2w}$
2. Conducted Power: Transferred directly through electrical connection: $S_{cond} = S_{auto} - S_{ind} = S_{auto} \left(\frac{1}{a_{auto}}\right)$

Exam Key Point: When turns ratio $a_{auto}$ is close to 1.0 (e.g., $13.8\text{ kV}$ to $12.47\text{ kV}$, $a = 1.107$), over $90%$ of power is conducted directly. This yields minimal losses, high efficiency ($>99.5%$), and extremely low impedance, but eliminates electrical isolation between primary and secondary circuits.

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2. Per-Phase Equivalent Circuit & Standard Parameter Tests

The complete single-phase equivalent circuit of a real power transformer includes winding resistances ($R_1, R_2$), leakage reactances ($X_1, X_2$), core loss resistance ($R_c$), and magnetizing reactance ($X_m$).

         R1          X1                      R2'         X2'
   o---[####]------[UUUU]---+---------+----[####]------[UUUU]---o
                            |         |
                           [ ] Rc    [ ] jXm
                           [ ]       [ ]
                            |         |
   o------------------------+---------+-------------------------o

In standard power engineering analysis, the shunt excitation branch is moved to the input terminals, and series components are lumped into equivalent values referred to either the primary or secondary:

$R_{eq1} = R_1 + a^2 R_2, \quad X_{eq1} = X_1 + a^2 X_2, \quad Z_{eq1} = R_{eq1} + j X_{eq1}$ $R_{eq2} = \frac{R_1}{a^2} + R_2, \quad X_{eq2} = \frac{X_1}{a^2} + X_2, \quad Z_{eq2} = R_{eq2} + j X_{eq2}$

Open-Circuit (No-Load) Test

• Setup: Performed at rated voltage on the Low-Voltage (LV) winding with the High-Voltage (HV) winding left open-circuit (for safety and instrumentation availability).
• Measurements: Open-circuit voltage $V_{oc} = V_{rated,LV}$, no-load current $I_{oc}$, and no-load power $P_{oc}$.
• Extracted Parameters (Core Excitation Branch referred to LV):

$\text{Core Loss Resistance: } R_{c,LV} = \frac{V_{oc}^2}{P_{oc}}$ $\text{Core Loss Current: } I_c = \frac{P_{oc}}{V_{oc}}$ $\text{Magnetizing Current: } I_m = \sqrt{I_{oc}^2 - I_c^2}$ $\text{Magnetizing Reactance: } X_{m,LV} = \frac{V_{oc}}{I_m} = \frac{V_{oc}}{\sqrt{I_{oc}^2 - (P_{oc}/V_{oc})^2}}$

Short-Circuit Test

• Setup: Performed at rated current on the High-Voltage (HV) winding with the Low-Voltage (LV) winding bolted short-circuited.
• Measurements: Short-circuit voltage $V_{sc}$ (typically $3\text{--}8%$ of rated $V_{HV}$), rated current $I_{sc} = I_{rated,HV}$, and short-circuit power $P_{sc}$.
• Extracted Parameters (Series Winding Impedance referred to HV):

$\text{Equivalent Resistance: } R_{eq,HV} = \frac{P_{sc}}{I_{sc}^2}$ $\text{Equivalent Impedance Magnitude: } Z_{eq,HV} = \frac{V_{sc}}{I_{sc}}$ $\text{Equivalent Leakage Reactance: } X_{eq,HV} = \sqrt{Z_{eq,HV}^2 - R_{eq,HV}^2}$

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3. Voltage Regulation & Phasor Dynamics

Transformer Voltage Regulation ($VR%$) is the percentage change in secondary terminal voltage magnitude from no-load ($V_{NL}$) to full-load ($V_{FL}$) under a constant primary supply voltage:

$VR\% = \frac{|V_{NL}| - |V_{FL}|}{|V_{FL}|} \times 100\% = \frac{|V_1/a| - |V_{2,FL}|}{|V_{2,FL}|} \times 100\%$

Approximate Formula (Kapp Regulation Formula)

Using the equivalent circuit referred to the secondary winding:

$VR\% \approx \frac{I_2 (R_{eq2} \cos \theta_2 \pm X_{eq2} \sin \theta_2)}{V_{2,FL}} \times 100\%$

In per-unit notation:

$VR\% \approx \left[ R_{pu} \cos \theta \pm X_{pu} \sin \theta + \frac{1}{2} (X_{pu} \cos \theta \mp R_{pu} \sin \theta)^2 \right] \times 100\%$

Where the sign convention is:

• $+$ (Plus): Lagging (inductive) load power factor.
• $-$ (Minus): Leading (capacitive) load power factor.

Voltage Regulation Behavior vs Power Factor:
  • Lagging PF (Inductive):    VR% > 0  (V_FL < V_NL => Terminal voltage drops under load)
  • Unity PF (Resistive):       VR% > 0  (Small positive drop due to Req)
  • Leading PF (Capacitive):   VR% can be < 0  (V_FL > V_NL => Terminal voltage rises under load!)

Zero Voltage Regulation Condition: Voltage regulation equals exactly zero when $R_{eq2} \cos \theta = X_{eq2} \sin \theta$, which occurs at a leading power factor angle of $\theta = \arctan(R_{eq}/X_{eq})$.

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4. Efficiency & Condition for Maximum Efficiency

Transformer efficiency $\eta$ is the ratio of active output power to active input power:

$\eta = \frac{P_{out}}{P_{in}} = \frac{P_{out}}{P_{out} + P_{loss}} = \frac{x \, S_{rated} \cos \theta}{x \, S_{rated} \cos \theta + P_{core} + x^2 P_{cu,FL}}$

Where:

• $x = S_{load} / S_{rated}$ is the per-unit load fraction ($0 \le x \le 1.0$).
• $P_{core} = P_{oc}$ is the constant core (no-load) iron loss in Watts.
• $P_{cu,FL} = P_{sc}$ is the full-load series copper loss ($I_{FL}^2 R_{eq}$) in Watts.
• $x^2 P_{cu,FL}$ is the copper loss at load fraction $x$.

Condition for Maximum Efficiency

Differentiating $\eta$ with respect to load fraction $x$ and setting $d\eta / dx = 0$ yields the fundamental maximum efficiency theorem:

$\text{Variable Copper Loss } (x^2 P_{cu,FL}) = \text{Constant Core Loss } (P_{core})$

$x_{\eta,max} = \sqrt{\frac{P_{core}}{P_{cu,FL}}}$

The corresponding load apparent power at maximum efficiency is:

$S_{\eta,max} = S_{rated} \times \sqrt{\frac{P_{core}}{P_{cu,FL}}}$

Efficiency Optimization Curve:
Loss (W)
   ^
   |              /  Total Loss (Pcore + x^2*Pcu)
   |             /   
   |    Pcu ====/==== x^2 * Pcu,FL
   |           / 
   |----------X-------------------- Pcore (constant)
   |         /| 
   |        / | 
   +-------+--+------------------------> Load Fraction x
           0  x_max = sqrt(Pcore/Pcu,FL)

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5. Transformer Testing, Diagnostics & Dissolved Gas Analysis (DGA)

Routine Electrical Field Tests

1. Transformer Turns Ratio (TTR): Verifies the physical turns ratio across all tap positions within $\pm 0.5%$ of nameplate per IEEE C57.12.90. Identifies shorted turns or open windings.
2. Winding DC Resistance Test: Utilizes a 4-wire Kelvin bridge or digital low-resistance ohmmeter (DLRO). Compares winding resistances across phases (must balance within $\pm 2%$) to detect loose internal connections, tap changer contact degradation, or broken conductor strands.
3. Insulation Power Factor / Dissipation Factor ($\tan \delta$): Evaluates dielectric loss in winding insulation and bushings. New liquid-filled transformers must exhibit power factor $< 0.5%$ at $20^\circ\text{C}$. Values between $0.5%$ and $1.0%$ indicate insulation aging or moisture ingress; $> 1.0%$ requires corrective processing.

Dissolved Gas Analysis (DGA per IEEE C57.104)

Mineral insulating oil breaks down thermally and electrically, releasing specific diagnostic hydrocarbon gases. DGA is the single most sensitive method for early internal fault detection.

Exam Diagnostic Ratio Trap: The presence of trace Acetylene ($\text{C}_2\text{H}_2 > 1\text{--}2\text{ ppm}$) indicates active internal arcing and flashover. A $\text{CO}_2/\text{CO}$ ratio dropping below $3.0$ indicates dangerous cellulose paper insulation degradation.

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6. Comprehensive Worked Transformer Calculation

Problem Statement

A single-phase, $500\text{ kVA}$, $13,800\text{ V} : 480\text{ V}$, $60\text{ Hz}$ distribution transformer underwent open-circuit and short-circuit testing:

• Open-Circuit Test (LV Side, HV Open): $V_{oc} = 480\text{ V}$, $I_{oc} = 18.5\text{ A}$, $P_{oc} = 2,200\text{ W}$.
• Short-Circuit Test (HV Side, LV Shorted): $V_{sc} = 650\text{ V}$, $I_{sc} = 36.23\text{ A}$ (rated HV current), $P_{sc} = 4,800\text{ W}$.

Calculate:

1. The equivalent series impedance parameters ($R_{eq,LV}, X_{eq,LV}$) and per-unit impedance $Z_{pu}$.
2. The full-load voltage regulation ($VR%$) at $0.80$ power factor lagging and $0.80$ leading.
3. The full-load efficiency at $0.80$ power factor lagging.
4. The kVA load at which maximum efficiency occurs and the maximum efficiency value at unity power factor.

Step-by-Step Solution

Step 1: Turns Ratio and Base Quantities $a = \frac{13,800\text{ V}}{480\text{ V}} = 28.75$ $I_{rated,LV} = \frac{500,000\text{ VA}}{480\text{ V}} = 1,041.67\text{ A}, \quad I_{rated,HV} = \frac{500,000\text{ VA}}{13,800\text{ V}} = 36.232\text{ A}$ $Z_{base,LV} = \frac{(480\text{ V})^2}{500,000\text{ VA}} = 0.4608\ \Omega$

Step 2: Equivalent Series Impedance from Short-Circuit Test From the SC test on the HV side: $R_{eq,HV} = \frac{P_{sc}}{I_{sc}^2} = \frac{4,800\text{ W}}{(36.232\text{ A})^2} = 3.6565\ \Omega$ $Z_{eq,HV} = \frac{V_{sc}}{I_{sc}} = \frac{650\text{ V}}{36.232\text{ A}} = 17.940\ \Omega$ $X_{eq,HV} = \sqrt{Z_{eq,HV}^2 - R_{eq,HV}^2} = \sqrt{(17.940)^2 - (3.6565)^2} = 17.563\ \Omega$

Referring parameters to the LV side: $R_{eq,LV} = \frac{R_{eq,HV}}{a^2} = \frac{3.6565}{(28.75)^2} = 0.004423\ \Omega = 4.423\text{ m}\Omega$ $X_{eq,LV} = \frac{X_{eq,HV}}{a^2} = \frac{17.563}{(28.75)^2} = 0.021248\ \Omega = 21.248\text{ m}\Omega$ $Z_{eq,LV} = \frac{17.940}{(28.75)^2} = 0.021704\ \Omega = 21.704\text{ m}\Omega$

Per-Unit Series Impedances: $R_{pu} = \frac{R_{eq,LV}}{Z_{base,LV}} = \frac{0.004423\ \Omega}{0.4608\ \Omega} = 0.00960\text{ pu} = 0.960\%$ $X_{pu} = \frac{X_{eq,LV}}{Z_{base,LV}} = \frac{0.021248\ \Omega}{0.4608\ \Omega} = 0.04611\text{ pu} = 4.611\%$ $Z_{pu} = \frac{Z_{eq,LV}}{Z_{base,LV}} = 0.04710\text{ pu} = 4.710\%$

Step 3: Voltage Regulation Calculation

• At $0.80$ Lagging Power Factor ($\cos \theta = 0.80$, $\sin \theta = +0.60$): $VR\% \approx [R_{pu} \cos \theta + X_{pu} \sin \theta] \times 100\%$ $VR\% \approx [0.00960(0.80) + 0.04611(0.60)] \times 100\% = [0.00768 + 0.02767] \times 100\% = 3.535\%$

  Exact Calculation Check: $\mathbf{V}_{NL}' = 480\angle 0^\circ + (1041.67\angle -36.87^\circ)(0.004423 + j0.021248) = 480 + 22.608\angle 41.36^\circ = 496.97 + j14.94 = 497.19\angle 1.72^\circ\text{ V}$ $VR\%_{exact} = \frac{497.19 - 480}{480} \times 100\% = 3.58\%$

• At $0.80$ Leading Power Factor ($\cos \theta = 0.80$, $\sin \theta = -0.60$): $VR\% \approx [R_{pu} \cos \theta - X_{pu} \sin \theta] \times 100\%$ $VR\% \approx [0.00960(0.80) - 0.04611(0.60)] \times 100\% = [0.00768 - 0.02767] \times 100\% = -1.999\% \approx -2.00\%$

Step 4: Full-Load Efficiency at 0.80 Lagging PF $P_{out} = 500\text{ kVA} \times 0.80 = 400\text{ kW}$ $P_{core} = 2,200\text{ W} = 2.20\text{ kW}, \quad P_{cu,FL} = 4,800\text{ W} = 4.80\text{ kW}$ $\eta_{FL} = \frac{400\text{ kW}}{400\text{ kW} + 2.20\text{ kW} + 4.80\text{ kW}} = \frac{400}{407.0} = 0.9828 = 98.28\%$

Step 5: Maximum Efficiency Load and Value $x_{\eta,max} = \sqrt{\frac{P_{core}}{P_{cu,FL}}} = \sqrt{\frac{2,200\text{ W}}{4,800\text{ W}}} = \sqrt{0.45833} = 0.6770$ $S_{\eta,max} = 0.6770 \times 500\text{ kVA} = 338.5\text{ kVA}$

At unity power factor ($\cos \theta = 1.0$): $P_{out} = 338.5\text{ kW}$ $P_{loss} = P_{core} + x^2 P_{cu,FL} = 2.20\text{ kW} + 2.20\text{ kW} = 4.40\text{ kW}$ $\eta_{max} = \frac{338.5\text{ kW}}{338.5\text{ kW} + 4.40\text{ kW}} = \frac{338.5}{342.9} = 0.9872 = 98.72\%$

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7. Common Exam Traps & Strategic Pitfalls

• Test Referral Side Confusion: The Open-Circuit test gives parameters on the test winding side (usually LV); the Short-Circuit test gives series parameters on its test winding side (usually HV). Always verify which side the parameters are referred to before computing voltage drop.
• Leading Power Factor Sign Error: Leading power factors subtract the reactive term in the voltage regulation formula ($R \cos \theta - X \sin \theta$), often producing negative regulation (voltage rise).
• Copper Loss Scaling with Load Squared: $P_{cu}(x) = x^2 P_{cu,FL}$. At half load ($x = 0.5$), copper losses drop to $(0.5)^2 = 0.25$ ($25%$) of full load, NOT $50%$.
• Autotransformer High Fault Current Trap: Because the equivalent impedance of an autotransformer is reduced by $(1 - 1/a)$ compared to a standard two-winding unit, short-circuit fault currents are dramatically higher.