Three Phase Transformer Connections MCQ Quiz in தமிழ் - Objective Question with Answer for Three Phase Transformer Connections - இலவச PDF ஐப் பதிவிறக்கவும்

We known that diameter of the circumscribing is same as diagonal of the square.

Area of square core = l × b

\(l = b = \frac{{0.5}}{{\sqrt 2 }}\)

\(\therefore Area = {\left( {\frac{{0.5}}{{\sqrt 2 }}} \right)^2} = 0.125\;{m^2}\)

∴ Net cross section area = 0.125 × 0.92 = 0.115

E.M.F/Turn = 4.44 fBA = 4.44 × 50 × 1.3 × 0.115 = 33.19

Phase turn ratio \(= \frac{{11000}}{{\sqrt 3 }}:440\)

3- phase to 6 - phase conversion:

• In certain applications like thyristors and rectifiers six-phase supply is required. Therefore it becomes necessary to convert three-phase ac supply into six-phase.
• By using three identical single-phase transformers suitably interconnected this can be achieved.
• The primary winding is connected in delta whereas its secondary winding is split into two halves.
• A greater number of phases also results in the reduction of harmonics in the alternating current.
• Thus conversion from three-phase to six-phase can be obtained by having two similar secondary windings for each of the primaries of the three-phase transformer. 
• There are several methods of connecting transformers for obtaining a 3 - to - 6 phase transformation. A simple arrangement is known as a 6 phase star connection is below in figure

• In this connection, the primary is connected in a delta configuration and a 3-phase supply is given to the primary.
• The secondaries are provided with center taps, which are connected together to form the neutral of the 6 - phase side.
• Now we got six phases as shown in the above figure and the six-phase voltages are V_an, V_dn, V_cn, V_fn, V_en, and V_bn.
• And the phaser diagram of six-phase voltages is shown below

Figure: 3 - phase voltage phasor diagram           figure: 6 - phase voltage phaser diagram

Star-Delta transformer:

• Consider a 3-phase star-delta transformer with primary side Y connected (star connected) and secondary side with delta connected as shown below

• The polarity markings are indicated on each phase.
• The dots on the winding indicate the terminals which are positive at the same time on the undotted terminals.
• Phase on star side – A, B, C
• Phase on delta side – a, b, c
• The labeling is indicated to the diagram corresponding to +90° connections in which the positive sequences on the delta side are lead by 90° corresponding to the star side.
• Thus the line current flows through the phase and A.
• If we label delta as b → a, c → b, and a → c. There we will get standard Yd1, –30° connection
• If polarities on the delta side are also reversed, we get standard Yd11, 30° connection as shown in the figure given below.
• Thus we can get a phase difference of 30° between input and output of three-phase transformer by using a star-delta connection.

Important Points:

The phase shift in different connections of transformer are given below.

Open-Delta Connection:

• An open delta connection transformer uses two single-phase transformers to provide a 3-phase supply to the load.
• An open-delta connection system is also called a V-V system.
• The Open-delta connection system is usually only used in emergency conditions, as its efficiency is low when compared to the delta-delta (Closed delta) system.
• The total load carried by an open delta system is 57.7% of the closed delta system.

The capacity of open-delta system = √3 × Rating of one transformer

= 0.577 × Rating of the closed delta system

Calculation:

Given,

Rating of one transformer = 150 kVA

The capacity of open delta system = √3 × Rating of one transformer

= √3 × 150 = 259.8 kVA

Therefore, the maximum three-phase load that they can carry is 259.8 kVA

Explanation:

Given figure can be redrawn as,

With primary star (γ) side,

Phase voltage (v_a) = \(\rm \frac{line voltage (V_{ab})}{\sqrt{3}}=\frac{230}{\sqrt{3}}\)

At secondary delta (Δ) side,

Phase voltage (V_A) = line voltage (V_AB) = 115 V

Transformation ratio (k) = \(\frac{V_A}{V_a}=\frac{115}{\frac{230}{\sqrt{3}}}=0.87\)

Phase current (I_A) on Δ side is,

\(I_A=\frac{I_s\rm(line)}{\sqrt{3}}=\frac{100}{\sqrt{3}}=57.7\)

∴ I_s lags I_A by 30°

So, I_A = 57.7 ∠30°

Now, \(k=\frac{I_p}{I_A}\)

⇒ I_P = k I_A = 0.87 × 57.7∠30°

⇒ = 50.199 ∠30°

Hence, the correct answer is (A).

Delta – star connection is most economical for step-up applications(transmission side) and also the secondary side (Y) connection provides a stable neutral point for 1-ϕ loads in the distribution network (3-ϕ, 4 wire).

∴ Δ - Y connection is popular for both the step-up (transmission side, high voltage) and step-down(distribution side, low voltage) applications.

• Delta-star connected distribution transformers are widely used in low power distribution for 3 phase 4 wire supply.
• The primary winding connected in delta providing a three-wire balanced load to the utility company
• While the secondary winding connected in the star connection, to provide the required 4th-wire neutral or earth connection.

Important Points

Star – delta connection is the most uneconomical among all the transformer connections because, for the same voltage rating, it requires more number of turns per phase. The economical use of this connection is connecting (Y) on HV side and (Δ) on LV side (i.e.Step-down application in transmission). But it can't be used in the distribution network due to the absence of a stable neutral point on secondary.

Star – star connection is used for high voltage and low current rating (small kVA) transformers.

Delta – delta connection is used for low voltage and high current rating (large kVA) transformers.

Star to Delta transformer:

A star to delta transformer has a primary side as star connected and secondary is delta connected

Circuit diagram of the star to delta transformer:

Where,

I_R I_Y I_B are line currents

I_RY I_YB I_BR are phase currents

V_R V_Y V_B are line voltages

V_RY V_YB V_BR are phase voltages

S = √3 VL IL 

Where,

IL is the line current

I_P is the phase current

S is the apparent power

IL = S / √3 VL

Calculation:

Given,

Apparent power S = 230 kVA

Secondary line voltage V_L = 2300 V

S = √3 V_L I_L 

I_L is the line current

I_L = S / √3 V_L

= (230 × 10³) / (√3 × 230)

= 100 / √3 A

Phase current IP = IL / √3

= 100 / 3 = 33.33 A

∴ Its rated secondary current/phase is 33.33 A

Points to remember:

Differences between star and delta connection

Phase voltage on primary \(\left( {{V_{p1}}} \right) = \frac{{{V_{L1}}}}{{√ 3 }} = \frac{{11 × {{10}^3}}}{{√ 3 }}V\)

Phase current on primary (I_p1) = I_L1 = 20 A

Per phase turns ratio = 12

Phase voltage on secondary \(\left( {{V_{p2}}} \right) = \frac{{11 × {{10}^3}}}{{12 × √ 3 }} = 529.23\;V\)

Phase current on secondary (I_p2) = 20 × 12 = 240 A

Output kVA = 3 V_p2 I_p2 × 10^-3

= 3 × 529.23 × 240 × 10^-3

= 381.05 kVA

Alternate solution:

In a transformer, the power & frequency remains constant from primary to secondary

So, output power = primary side power i/p

= √3 × V_L × I_L

= √3 × 11 kV × 20 = 381.05 kVA

The correct answer is option "3".

Concept:-

Parallel operation of Transformer:

When two transformers operate in parallel then the input and output of the two transformers must be the same. Suppose two transformers A and B are operated in parallel then the primary of transformer A is star connected then the primary of the second transformer B should be star connected.

The transformer windings can be connected in various ways that produce magnitudes and phase displacements in the secondary voltage with respect to the primary voltage.

All the transformer connections can be classified into distinct vector groups.

Group 1: Zero phase displacement (Yy0, Dd0, Dz0)
Group 2: 180° phase displacement (Yy6, Dd6, Dz6)
Group 3: - 30° phase displacement (Yd1, Dy1, Yz1)
Group 4: + 30° phase displacement (Yd11, Dy11, Yz11)

1. In order to have zero relative phase displacement of secondary-side line voltages, the transformers belonging to the same phase angle (the only magnitude not sign) to be connected in parallel.
2. The transformers of groups 1 and 2 can only be paralleled with transformers of their own group.
3. The transformers of groups 3 and 4 can be paralleled by reversing the phase sequence of one of them.

Transformers with Yd11 and Dy1 connections can be paralleled.

Δ-Y and Δ-Y belong to the same vector group.

Additional Information

Condition for parallel operation of transformers:

For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.

Various conditions that must be fulfilled for the successful parallel operation of transformers:

1. Same voltage Ratio & Turns Ratio (both primary and secondary Voltage Rating is same).

2. Same Percentage Impedance and X/R ratio.

3. Identical Position of Tap changer.

4. Same KVA ratings.

5. Same Phase angle shift (vector group is same).

6. Same Frequency rating.

7. Same Polarity.

8. Same Phase sequence.

Some of these conditions are convenient and some are mandatory.\

Important Points The convenient conditions are the Same voltage Ratio & Turn Ratio, Same Percentage Impedance, Same KVA Rating, and Same Position of Tap changer.

The mandatory conditions are the Same Phase Angle Shift, Same Polarity, Same Phase Sequence, and Same Frequency.

When the convenient conditions are not met paralleled operation is possible but not optimal.