Electrical Machines · Transformers

Per-Unit System and Parallel Operation of Transformers

Part 3 · Transformers — normalise every machine to its own rating, then see why equal per-unit impedance is the key to sharing load fairly.

Prof. Mithun Mondal Engineering Devotion Digital Textbook
i Learning Objectives

By the end of this chapter you should be able to:

  • Explain why the per-unit system is used, and calculate base values and per-unit quantities.

  • Convert per-unit impedances to a new base, and show that per-unit values do not depend on which side they are referred to.

  • State the conditions for parallel operation of transformers, and calculate circulating current and load sharing.

Demonstrative Video

Section 1

Why Per-Unit?

A power system contains transformers and machines of very different sizes, ratings and voltage ratios, and their equivalent-circuit parameters in ohms vary over a huge range. Comparing them is much simpler if every parameter is normalised. Percentage values are one form of scaling, but they leave a factor of 100 to carry through the calculations. The per-unit system scales every quantity by a logically chosen base, giving dimensionless values with no extra factors and no dependence on the system of units.

A useful property. Expressed on its own rating, the per-unit impedance of a transformer lies in a narrow band (typically 0.04–0.12) whatever its size. Per-unit values also come out the same whichever side of the transformer they are calculated from.
Section 2

Base Values

Two base quantities are chosen, normally the rated voltage and the rated VA from the nameplate. All the others are derived from them:

\[S_{base} = V_{base}I_{base}, \qquad Z_{base} = \frac{V_{base}}{I_{base}} = \frac{V_{base}^2}{S_{base}}, \qquad Y_{base} = \frac{1}{Z_{base}}\]
\[V_{pu} = \frac{V}{V_{base}}, \qquad I_{pu} = \frac{I}{S_{base}/V_{base}}, \qquad Z_{pu} = Z_{\Omega}\frac{S_{base}}{V_{base}^2}\]

Independence of side. Referring an impedance to the other side divides it by \(a^2\), and the base impedance on that side is also smaller by \(a^2\):

\[Z'_{pu} = \frac{Z_\Omega/a^2}{(V_{base}/a)^2/S_{base}} = Z_{pu}\]
pu
Change of Base
\(Z_{pu,new} = Z_{pu,old}\,\dfrac{S_{base,new}}{S_{base,old}}\left(\dfrac{V_{base,old}}{V_{base,new}}\right)^2\)

When several machines are in one circuit, a common system base is chosen and all impedances are converted to it.

1 Worked Example — Per-unit impedance

Problem. A 50 kVA, 2400/240 V transformer has \(Z_{eq} = 1.5 + j2.5\ \Omega\) referred to the HV side. Find the per-unit impedance, and check it from the LV side.

\[Z_{base,HV} = \frac{2400^2}{50\,000} = 115.2\ \Omega, \qquad Z_{pu} = \frac{1.5 + j2.5}{115.2} = 0.013 + j0.0217\]

From the LV side, \(a = 10\). The impedance is \(0.015 + j0.025\ \Omega\) and \(Z_{base,LV} = 1.152\ \Omega\), which gives the same per-unit value.

Section 3

Parallel Operation

Parallel operation means connecting two or more transformers to the same primary busbars and to a common secondary busbar and load.

Two transformers in parallel
Two transformers A and B supplying a common load.
Why operate in parallel

  • A single large enough unit may not be available.

  • Demand grows, so capacity is added in steps.

  • Reliability: if one unit fails or is under maintenance, the load can still be supplied.

  • Smaller units are easier to transport to site.

Conditions

  • The voltage ratios must be the same.

  • The per-unit impedances on their own ratings should be the same, preferably with the same X/R ratio.

  • The polarity must be the same.

  • For three-phase units, the phase sequence and phase displacement (vector group) must be the same.

With turns ratios \(a_1, a_2\) and impedances \(Z_A, Z_B\) referred to the secondary:

\[I_A + I_B = I_L, \qquad V_L = \frac{V_1}{a_1} - I_AZ_A = \frac{V_1}{a_2} - I_BZ_B\]
Section 4

Unequal Ratios: Circulating Current

Solving the equations shows that each transformer current has two components: its share of the load, and a current circulating around the loop formed by the two secondaries. The circulating current flows even with no load connected:

↻
Circulating Current
\(I_c = \dfrac{E_A - E_B}{Z_A + Z_B} = \dfrac{V_1(a_2 - a_1)}{a_1a_2(Z_A + Z_B)}\)

It increases the copper loss and may overload one transformer, reducing the kVA the bank can deliver. Even a small difference in voltage ratio produces a large \(I_c\), because \(Z_A + Z_B\) is small.

2 Worked Example — Circulating current

Problem. Two transformers have open-circuit secondary voltages of 240 V and 245 V, and each has \(Z = 0.02 + j0.08\ \Omega\) referred to the secondary. Find the circulating current on no load.

\[I_c = \frac{5}{|0.04 + j0.16|} = \frac{5}{0.165} = 30.3\ \text{A}\]
Section 5

Equal Ratios: Load Sharing

With \(a_1 = a_2\) there is no circulating current, and the load divides inversely as the impedances:

\[I_A = \frac{Z_B}{Z_A + Z_B}I_L, \qquad I_B = \frac{Z_A}{Z_A + Z_B}I_L, \qquad \frac{I_A}{I_B} = \frac{Z_B}{Z_A}\]

In terms of complex power, with \(S = VI^*\):

\[S_A = \left(\frac{Z_B^*}{Z_A^* + Z_B^*}\right)S_L, \qquad S_B = \left(\frac{Z_A^*}{Z_A^* + Z_B^*}\right)S_L, \qquad \frac{S_A}{S_B} = \frac{Z_B^*}{Z_A^*}\]
Sharing in proportion to rating. Each transformer delivers rated current at the same time only if \(I_AZ_A = I_BZ_B\) at full load, which means equal per-unit impedances on their own ratings. If the impedance angles also match, the sharing fractions are real and apply directly to kVA magnitudes. Otherwise the units also work at different power factors.
3 Worked Example — Unequal per-unit impedances

Problem. Transformer A (500 kVA, \(Z = 0.01 + j0.05\) pu) and transformer B (250 kVA, \(Z = 0.015 + j0.04\) pu) share a 750 kVA load at 0.8 p.f. lagging. Find the load on each.

On a 500 kVA base, \(Z_B = 2(0.015 + j0.04) = 0.03 + j0.08\) pu.

\[|S_A| = 750\left|\frac{Z_B}{Z_A + Z_B}\right| = 471\ \text{kVA}\ (\text{p.f. } 0.76), \qquad |S_B| = 281\ \text{kVA}\ (\text{p.f. } 0.86)\]

B is overloaded by about 12 % while A carries only 94 % of its rating, because B's per-unit impedance (0.043) is lower than A's (0.051).

Section 6

Summary

  • Per-unit value = actual value ÷ base value. \(Z_{base} = V_{base}^2/S_{base}\).

  • Per-unit impedance is the same on both sides of a transformer.

  • Unequal voltage ratios cause circulating current. Unequal per-unit impedances cause unfair load sharing.

Section 7

Chapter Review

Multiple-Choice Questions

  1. MCQ 1. Transformers in parallel share load in proportion to their ratings if they have equal
    (a) ohmic impedance (b) per-unit impedance on their own ratings (c) kVA (d) copper loss

    Show answer
    (b).

  2. MCQ 2. Circulating current between parallel transformers is caused by unequal
    (a) kVA ratings (b) voltage ratios (c) X/R ratios (d) loads

    Show answer
    (b) voltage ratios.

  3. MCQ 3. If the base kVA is doubled, a per-unit impedance
    (a) halves (b) doubles (c) is unchanged (d) quadruples

    Show answer
    (b) doubles.
Conceptual Questions

  1. Why is a small difference in voltage ratio serious for parallel transformers?

  2. Why must polarity be checked before paralleling?