Transformers 2

LessonAbout 2 h2 of 2 in Transformers

3-Phase Transformer Bank is an arrangement of three single-phase transformers or one three-phase transformer unit used to step up or step down voltages in three-phase power systems.

1. Delta-Delta Transformer Bank

Phase Voltage Equals Line Voltage: primary and secondary winding voltages match line voltages
Closed Mesh Connection: secondary continues operating if one phase is disconnected
Zero Phase Shift: line voltages on primary and secondary remain in phase

Delta-Delta Voltage and Current Relations

\[ V_1 = V_p \qquad\qquad V_2 = V_s \qquad\qquad a = \dfrac{N_1}{N_2} = \dfrac{V_p}{V_s} = \dfrac{V_1}{V_2} \]

\[ I_1 = \sqrt{3} I_p \qquad\qquad I_2 = \sqrt{3} I_s \qquad\qquad \dfrac{I_p}{I_s} = \dfrac{N_2}{N_1} = \dfrac{1}{a} \]

Where: \(V_1, V_2\) = primary and secondary line voltages (V), \(V_p, V_s\) = primary and secondary phase voltages (V), \(I_1, I_2\) = primary and secondary line currents (A), \(I_p, I_s\) = primary and secondary phase currents (A), \(a\) = turns ratio

Example. A \(150\ \text{kVA}\), \(3\phi\) load is connected to a delta-delta transformer bank with turns ratio \(N_1/N_2 = 1/2\). If the secondary line voltage is \(4160\ \text{V}\), calculate the primary phase current.

Solution.

\[ V_1 = V_2 \left(\dfrac{N_1}{N_2}\right) = 4160(2) = 8320\ \text{V} \implies I_1 = \dfrac{150000}{\sqrt{3}(8320)} = 10.41\ \text{A} \]

\[ I_p = \dfrac{I_1}{\sqrt{3}} = \dfrac{10.41}{\sqrt{3}} = 6.01\ \text{A} \]

Answer: \(6.01\ \text{A}\)

2. Delta-Wye Transformer Bank

Step-Up Application: commonly used at generating stations to elevate distribution voltage
Secondary Neutral Available: neutral wire provides 3-phase 4-wire service
Phase Angle Displacement: secondary line voltage leads primary line voltage by \(30^\circ\)

Delta-Wye Voltage and Current Relations

\[ V_1 = V_p \qquad\qquad V_2 = \sqrt{3} V_s \qquad\qquad \dfrac{V_1}{V_2} = \dfrac{V_p}{\sqrt{3} V_s} = \dfrac{a}{\sqrt{3}} \]

\[ I_1 = \sqrt{3} I_p \qquad\qquad I_2 = I_s \qquad\qquad \dfrac{I_p}{I_s} = \dfrac{1}{a} \]

Example. A delta-wye transformer bank has a turns ratio of \(96\) to \(1\). If the secondary line voltage is \(4160\ \text{V}\), find the primary line voltage.

Solution.

\[ V_s = \dfrac{V_2}{\sqrt{3}} = \dfrac{4160}{\sqrt{3}} = 2401.78\ \text{V} \implies V_1 = V_p = 96(2401.78) = 230571\ \text{V} \approx 230\ \text{kV} \]

Answer: \(230\ \text{kV}\)

3. Wye-Delta Transformer Bank

Step-Down Application: used at distribution substations to reduce high transmission voltage
Stable Secondary Voltage: delta secondary suppresses third-harmonic voltages
Phase Angle Displacement: secondary line voltage lags primary line voltage by \(30^\circ\)

Wye-Delta Voltage and Current Relations

\[ V_1 = \sqrt{3} V_p \qquad\qquad V_2 = V_s \qquad\qquad \dfrac{V_1}{V_2} = \dfrac{\sqrt{3} V_p}{V_s} = \sqrt{3} a \]

\[ I_1 = I_p \qquad\qquad I_2 = \sqrt{3} I_s \qquad\qquad \dfrac{I_p}{I_s} = \dfrac{1}{a} \]

EE Board Exam October 1997

Example. A power transformer rated \(50000\ \text{kVA}\), \(34.5\ \text{kV}/13.8\ \text{kV}\) is connected wye-grounded primary and delta on the secondary. Determine the full load phase current on the secondary side.

Solution.

\[ I_2 = \dfrac{S}{\sqrt{3} V_2} = \dfrac{50000000}{\sqrt{3}(13800)} = 2091.85\ \text{A} \implies I_s = \dfrac{I_2}{\sqrt{3}} = \dfrac{2091.85}{\sqrt{3}} = 1207.73\ \text{A} \approx 1208\ \text{A} \]

Answer: \(1208\ \text{A}\)

4. Wye-Wye Transformer Bank

Economical Insulation: winding insulation rated for phase voltage (\(V_L/\sqrt{3}\))
Neutral Grounding Required: neutral must be grounded to prevent third-harmonic distortion
Zero Phase Shift: line voltages on primary and secondary remain in phase

Wye-Wye Voltage and Current Relations

\[ V_1 = \sqrt{3} V_p \qquad\qquad V_2 = \sqrt{3} V_s \qquad\qquad \dfrac{V_1}{V_2} = \dfrac{V_p}{V_s} = a \]

\[ I_1 = I_p \qquad\qquad I_2 = I_s \qquad\qquad \dfrac{I_1}{I_2} = \dfrac{I_p}{I_s} = \dfrac{1}{a} \]

EE Board Exam October 1997

Example. A power transformer rated \(50000\ \text{kVA}\), \(34.5\ \text{kV}/13.8\ \text{kV}\) is connected wye-wye. What are the line currents at full load?

Solution.

\[ I_1 = \dfrac{S}{\sqrt{3} V_1} = \dfrac{50000000}{\sqrt{3}(34500)} = 836.7\ \text{A} \qquad\qquad I_2 = \dfrac{S}{\sqrt{3} V_2} = \dfrac{50000000}{\sqrt{3}(13800)} = 2092\ \text{A} \]

Answer: \(836.7\ \text{A}\ \text{and}\ 2092\ \text{A}\)

5. Open Delta Transformer Bank

V-V Connection: uses two single-phase transformers to supply a three-phase load
Reduced Bank Capacity: delivers \(57.7\%\) of closed delta bank capacity
Individual Transformer Loading: each transformer operates at \(86.6\%\) of its kVA rating

Open Delta Capacity Formulas

\[ S_v = \sqrt{3} S_{\text{unit}} \]

\[ S_\Delta = 3 S_{\text{unit}} \]

\[ \dfrac{S_v}{S_\Delta} = \dfrac{\sqrt{3}}{3} = 0.577 = 57.7\% \]

Where: \(S_v\) = total kVA capacity of open delta bank (kVA), \(S_\Delta\) = total kVA capacity of closed delta bank (kVA), \(S_{\text{unit}}\) = kVA rating of one single-phase transformer (kVA)

EE Board Exam June 1990

Example. What should be the kVA rating of each two transformers in an open-delta bank when the three-phase balanced load is \(290.4\ \text{kVA}\)?

Solution.

\[ \text{Rating} = \dfrac{S_v}{\sqrt{3}} = \dfrac{290.4}{\sqrt{3}} = 167.66\ \text{kVA} \]

Answer: \(167.66\ \text{kVA}\)

6. Scott or T-Connected Transformer Bank

Phase Transformation: converts 3-phase power to 2-phase power or vice versa
Main Transformer: single-phase unit with \(50\%\) center tap on primary winding
Teaser Transformer: single-phase unit with \(86.6\%\) (\(\sqrt{3}/2\)) tap on primary winding

Scott Connection Voltage and Turns Relations

\[ V_{\text{teaser(pri)}} = \dfrac{\sqrt{3}}{2} V_L = 0.866 V_L \]

\[ \dfrac{N_{1(\text{main})}}{N_2} = \dfrac{V_L}{V_2} = a \qquad\qquad \dfrac{N_{1(\text{teaser})}}{N_2} = \dfrac{0.866 V_L}{V_2} = 0.866 a \]

Where: \(V_L\) = primary 3-phase line voltage (V), \(V_2\) = secondary 2-phase line voltage (V), \(N_{1(\text{main})}, N_{1(\text{teaser})}\) = primary turns of main and teaser transformers, \(N_2\) = secondary turns

Example. A Scott-connected transformer bank converts \(416\ \text{V}\) 3-phase to \(6600\ \text{V}\) 2-phase. If the secondary has \(1500\ \text{turns}\), find the primary turns for the main and teaser transformers.

Solution.

\[ N_{1(\text{main})} = 1500 \left(\dfrac{416}{6600}\right) = 94.5 \approx 95\ \text{turns} \]

\[ N_{1(\text{teaser})} = 1500 \left(\dfrac{0.866 \times 416}{6600}\right) = 81.8 \approx 82\ \text{turns} \]

Answer: \(95\ \text{turns (main)}\ \text{and}\ 82\ \text{turns (teaser)}\)

7. Parallel Operation

Voltage Rating Identity: primary and secondary voltage ratings must be identical
Polarity Requirement: transformers must be connected with correct relative polarity
Equal Impedance Ratio: ratio of equivalent resistance to reactance (\(R_e/X_e\)) should match

7.1 Unequal Turns Ratio

No-Load Circulating Current: flows between windings due to induced voltage difference
Load Division Inequality: transformer with higher secondary emf carries larger load

Parallel Operation with Unequal Turns Ratio

\[ I_1 = \dfrac{(a_2 - a_1)V_2 + (a_2 Z_{e2} I_t)}{a_1 Z_{e1} + a_2 Z_{e2}} \]

\[ I_2 = \dfrac{(a_1 - a_2)V_2 + (a_1 Z_{e1} I_t)}{a_1 Z_{e1} + a_2 Z_{e2}} \]

Where: \(a_1, a_2\) = voltage ratios of transformers 1 and 2, \(V_2\) = terminal load voltage (V), \(Z_{e1}, Z_{e2}\) = equivalent secondary impedances (\(\Omega\)), \(I_t\) = total load current (A), \(I_1, I_2\) = load currents delivered by transformers 1 and 2 (A)

7.2 Equal Turns Ratio (Ohmic Equi Impedance)

Zero Circulating Current: no current flows between transformers at no load
Impedance Inverse Division: shared load current varies inversely as internal impedance

Parallel Load Sharing (Ohmic Impedances)

\[ I_1 = \dfrac{I_t Z_{e2}}{Z_{e1} + Z_{e2}} \qquad\qquad I_2 = \dfrac{I_t Z_{e1}}{Z_{e1} + Z_{e2}} \]

\[ S_1 = \dfrac{S_t Z_{e2}}{Z_{e1} + Z_{e2}} \qquad\qquad S_2 = \dfrac{S_t Z_{e1}}{Z_{e1} + Z_{e2}} \]

Where: \(I_t\) = total load current (A), \(S_t\) = total load apparent power (kVA), \(S_1, S_2\) = apparent power delivered by transformers 1 and 2 (kVA), \(Z_{e1}, Z_{e2}\) = equivalent secondary impedances (\(\Omega\))

EE Board Exam October 1985

Example. Two transformers are connected in parallel to supply a common load of \(175\ \text{kVA}\). Transformer A is rated \(100\ \text{kVA}\) with equivalent impedance of \(9.6\ \Omega\) while Transformer B is rated \(75\ \text{kVA}\) with equivalent impedance of \(12.5\ \Omega\). Find the kVA load of each transformer.

Solution.

\[ S_a = \dfrac{S_t Z_b}{Z_a + Z_b} = \dfrac{175(12.5)}{9.6 + 12.5} = 98.98\ \text{kVA} \approx 99\ \text{kVA} \]

\[ S_b = \dfrac{S_t Z_a}{Z_a + Z_b} = \dfrac{175(9.6)}{9.6 + 12.5} = 76.01\ \text{kVA} \approx 76\ \text{kVA} \]

Answer: \(99\ \text{kVA}\ \text{and}\ 76\ \text{kVA}\)

7.3 Equal Turns Ratio (Per Unit Impedance Different Power Base)

Common Base Conversion: convert all per-unit impedances to a chosen base kVA
Proportional Loading: transformers share load in proportion to kVA ratings if per-unit impedances are equal

Parallel Load Sharing (Per-Unit Impedances)

\[ Z_{\text{pu,new}} = Z_{\text{pu,old}} \times \left(\dfrac{\text{kVA}_{\text{base,new}}}{\text{kVA}_{\text{base,old}}}\right) \]

\[ S_1 = \dfrac{S_t Z_{e2}'}{Z_{e1}' + Z_{e2}'} \qquad\qquad S_2 = \dfrac{S_t Z_{e1}'}{Z_{e1}' + Z_{e2}'} \]

Where: \(Z_{\text{pu,old}}, Z_{\text{pu,new}}\) = original and converted per-unit impedances, \(Z_{e1}', Z_{e2}'\) = per-unit impedances on common base, \(S_t\) = total load (kVA), \(S_1, S_2\) = kVA carried by transformers 1 and 2

EE Board Exam March 1998

Example. A \(125\text{-kVA}\) distribution transformer with \(4\%\) impedance is connected in parallel with another transformer rated \(75\text{-kVA}\) and with \(3\%\) impedance. Both have the same voltage ratio. Neglecting resistance, how much load does each carry when supplying \(140\ \text{kVA}\) total?

Solution.

\[ X_{a2}' = 0.03 \left(\dfrac{125}{75}\right) = 0.05 \implies S_1 = \dfrac{140(0.05)}{0.04 + 0.05} = 77.78\ \text{kVA} \]

\[ S_2 = \dfrac{140(0.04)}{0.04 + 0.05} = 62.22\ \text{kVA} \]

Answer: \(77.78\ \text{kVA}\ \text{and}\ 62.22\ \text{kVA}\)

8. Formula Summary

Formula Name	Formula	Variables
Delta-Delta Voltage Ratio	\(\dfrac{V_1}{V_2} = \dfrac{V_p}{V_s} = a\)	\(V_1, V_2\) = line voltages (V) \(V_p, V_s\) = phase voltages (V) \(a\) = turns ratio
Delta-Delta Current Ratio	\(\dfrac{I_1}{I_2} = \dfrac{\sqrt{3} I_p}{\sqrt{3} I_s} = \dfrac{1}{a}\)	\(I_1, I_2\) = line currents (A) \(I_p, I_s\) = phase currents (A) \(a\) = turns ratio
Delta-Wye Voltage Ratio	\(\dfrac{V_1}{V_2} = \dfrac{V_p}{\sqrt{3} V_s} = \dfrac{a}{\sqrt{3}}\)	\(V_1, V_2\) = line voltages (V) \(V_p, V_s\) = phase voltages (V) \(a\) = turns ratio
Delta-Wye Current Ratio	\(\dfrac{I_1}{I_2} = \dfrac{\sqrt{3} I_p}{I_s} = \dfrac{\sqrt{3}}{a}\)	\(I_1, I_2\) = line currents (A) \(I_p, I_s\) = phase currents (A) \(a\) = turns ratio
Wye-Delta Voltage Ratio	\(\dfrac{V_1}{V_2} = \dfrac{\sqrt{3} V_p}{V_s} = \sqrt{3} a\)	\(V_1, V_2\) = line voltages (V) \(V_p, V_s\) = phase voltages (V) \(a\) = turns ratio
Wye-Delta Current Ratio	\(\dfrac{I_1}{I_2} = \dfrac{I_p}{\sqrt{3} I_s} = \dfrac{1}{\sqrt{3} a}\)	\(I_1, I_2\) = line currents (A) \(I_p, I_s\) = phase currents (A) \(a\) = turns ratio
Wye-Wye Voltage Ratio	\(\dfrac{V_1}{V_2} = \dfrac{\sqrt{3} V_p}{\sqrt{3} V_s} = a\)	\(V_1, V_2\) = line voltages (V) \(V_p, V_s\) = phase voltages (V) \(a\) = turns ratio
Open Delta Capacity	\(S_v = \sqrt{3} S_{\text{unit}}\)	\(S_v\) = open delta capacity (kVA) \(S_{\text{unit}}\) = single transformer rating (kVA)
Open Delta Capacity Ratio	\(\dfrac{S_v}{S_\Delta} = \dfrac{\sqrt{3}}{3} = 0.577\)	\(S_v\) = open delta capacity (kVA) \(S_\Delta\) = closed delta capacity (kVA)
Scott Teaser Voltage	\(V_{\text{teaser(pri)}} = 0.866 V_L\)	\(V_{\text{teaser(pri)}}\) = teaser primary voltage (V) \(V_L\) = 3-phase line voltage (V)
Scott Primary Turns	\(\dfrac{N_{1(\text{teaser})}}{N_2} = 0.866 a\)	\(N_{1(\text{teaser})}\) = teaser primary turns \(N_2\) = secondary turns \(a\) = line turns ratio
Parallel Load Current (Unequal Ratio)	\(I_1 = \dfrac{(a_2 - a_1)V_2 + (a_2 Z_{e2} I_t)}{a_1 Z_{e1} + a_2 Z_{e2}}\)	\(I_1\) = load current of transformer 1 (A) \(a_1, a_2\) = voltage ratios \(Z_{e1}, Z_{e2}\) = impedances (\(\Omega\))
Parallel Load Current (Equal Ratio)	\(I_1 = \dfrac{I_t Z_{e2}}{Z_{e1} + Z_{e2}}\)	\(I_1\) = current of transformer 1 (A) \(I_t\) = total load current (A) \(Z_{e1}, Z_{e2}\) = impedances (\(\Omega\))
Parallel Load Sharing (kVA)	\(S_1 = \dfrac{S_t Z_{e2}}{Z_{e1} + Z_{e2}}\)	\(S_1\) = load on transformer 1 (kVA) \(S_t\) = total load (kVA) \(Z_{e1}, Z_{e2}\) = impedances (\(\Omega\))
Per-Unit Base Change	\(Z_{\text{pu,new}} = Z_{\text{pu,old}} \left(\dfrac{\text{kVA}_{\text{base,new}}}{\text{kVA}_{\text{base,old}}}\right)\)	\(Z_{\text{pu,new}}\) = per-unit impedance on new base \(Z_{\text{pu,old}}\) = per-unit impedance on old base
Parallel Load Sharing (Per-Unit)	\(S_1 = \dfrac{S_t Z_{e2}'}{Z_{e1}' + Z_{e2}'}\)	\(S_1\) = load on transformer 1 (kVA) \(S_t\) = total load (kVA) \(Z_{e1}', Z_{e2}'\) = per-unit impedances on common base