NivaarExam PrepOfficial exam papers ↗

22-Elec-A6 Power Systems and Machines · December 2019

Question 3 of 5: Transformer Equivalent Circuit from OC and SC Tests

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Closed-book, 3 hours. Five questions constitute a complete paper and all are of equal value (20 marks each); answer all five. All AC voltages and currents are rms, three-phase voltages are line-to-line, and power is total real power unless noted otherwise.

Reference texts (22-Elec-A6 Power Systems and Machines): S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics, 2nd ed. (Wiley).

Note — Question 2(b). Question 2(b) reads "P required for rated current/voltage" and its assumption is stated as "the core losses are too small to account for", which settles the loss model explicitly.

Question 3: Transformer Equivalent Circuit from OC and SC Tests (20 marks)

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.

Given. A 75 kVA, 2200/220 V transformer with the standard pair of tests: the open-circuit test performed from the LV (220 V) side and the short-circuit test from the HV side.

Given data
Ratings75 kVA, 2200 / 220 V, 60 Hz
Open-circuit test (LV side)$V_{OC}=220\ \text{V},\ I_{OC}=9.6\ \text{A},\ P_{OC}=710\ \text{W}$
Short-circuit test (HV side)$V_{SC}=42\ \text{V},\ I_{SC}=57\ \text{A},\ P_{SC}=1030\ \text{W}$

Find. The shunt (magnetizing) branch $R_c,\ X_m$ and the series branch $R_{eq},\ X_{eq}$, all referred to the HV side, and a sketch of the approximate equivalent circuit.

Approach. The OC test (rated voltage, tiny current) yields the core-loss and magnetizing branch on the LV side; refer it to HV by $a^2$. The SC test (rated current, small voltage) yields the series impedance directly on the HV side. Turns ratio $a = 2200/220 = 10$.

  1. Magnetizing branch from the OC test (LV side). The core-loss resistance and magnetizing reactance are $$R_{c,\text{LV}} = \frac{V_{OC}^2}{P_{OC}} = \frac{220^2}{710} = 68.17\ \Omega.$$ The reactive component uses the OC apparent power $S_{OC}=V_{OC}I_{OC}=2112$ VA: $$Q_{OC} = \sqrt{S_{OC}^2 - P_{OC}^2} = \sqrt{2112^2 - 710^2} = 1989\ \text{var},\quad X_{m,\text{LV}} = \frac{V_{OC}^2}{Q_{OC}} = \frac{48\,400}{1989} = 24.33\ \Omega.$$
  2. Refer the shunt branch to the HV side. With $a = 10$, multiply by $a^2 = 100$: $$R_c = 100\,(68.17) = 6817\ \Omega,\qquad X_m = 100\,(24.33) = 2433\ \Omega.$$
  3. Series branch from the SC test (HV side). The test is already on the HV side, so $$R_{eq} = \frac{P_{SC}}{I_{SC}^2} = \frac{1030}{57^2} = 0.317\ \Omega,\qquad Z_{eq} = \frac{V_{SC}}{I_{SC}} = \frac{42}{57} = 0.737\ \Omega.$$
  4. Series reactance. $$X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2} = \sqrt{0.737^2 - 0.317^2} = 0.665\ \Omega.$$ $$\boxed{R_{eq}=0.317\ \Omega,\ X_{eq}=0.665\ \Omega,\ R_c=6817\ \Omega,\ X_m=2433\ \Omega\ \text{(all referred to HV)}.}$$
  5. Sketch of the approximate equivalent circuit (b). In the approximate model the shunt branch is moved to the input terminals, ahead of the series impedance:
    V_HV (2200 V) R_c = 6817 Ω X_m = 2433 Ω R_eq = 0.317 Ω X_eq = 0.665 Ω V_2' (referred) Approximate equivalent circuit — all quantities referred to the 2200 V (HV) side
    Approximate equivalent circuit referred to the 2200 V (HV) side: magnetizing branch $R_c\parallel X_m$ across the supply, series leakage impedance $R_{eq}+jX_{eq}$ to the load.
Final results — Question 3 (referred to HV)
BranchResistanceReactance
Series (leakage)$R_{eq}=0.317\ \Omega$$X_{eq}=0.665\ \Omega$
Shunt (magnetizing)$R_{c}=6817\ \Omega$$X_{m}=2433\ \Omega$