Power Transformers 2

Power Transformers

Three Phase Transformer Connections

Three identical single phase two winding transformers may be connected to form a three phase bank

The windings may be connected as:

Y - Y

Y - ∆

∆ - Y

∆ - ∆




∆ - Y commonly used as a generator step-up transformer

Advantages:

3rd harmonic magnetizing current remains trapped in the ∆ winding

Y winding provides neutral point for grounding on HV side (has an effect on the insulation requirements of the winding)


Y - Y seldom used because of harmonics in the magnetising current

∆ - ∆ has the advantage that one phase can be removed for maintenance while the remaining phases continue to operate as a three phase bank

Operation is reduced to 58% of the original bank

Example – Exercise 3.36

Three single phase transformers each rated at 25 MVA, 34.5/13.8 kV, are connected to form a three phase ∆ - ∆ bank. A resistive Y connected load absorbs 75 MW at 13.8 kV. If one of the single phase transformers is removed (resulting in an open ∆ connection) and the load is simultaneously reduced to 43.3 MW, determine

Example – Exercise 3.36

a) The load voltages Van, Vbn and Vcn

b) Load currents Ia, Ib and Ic

c) The MVA supplied by each of the two remaining transformers

Are balanced voltages still applied to the load?

Is the open - ∆ transformer overloaded?

Exercise 3.36

Exercise 3.36

Exercise 3.36

Exercise 3.36

Exercise 3.36


Y – Y

No phase shift for +ve sequence

∆ - ∆

Y - ∆

HV quantities lead LV quantities by 300

for +ve sequence

∆ - Y

Three Phase Transformer



All six windings placed on a common core Advantages Cores contains less iron Costs less Weighs less Less space required Slightly higher efficiency Disadvantages Winding failure requires replacement of the transformer

Equivalent Circuit of three phase two winding transformer

Y – Y or ∆ - ∆

Equivalent Circuit of three phase two winding transformer

Y – ∆ or ∆ - Y

Three Winding Transformers



For an ideal three winding transformer:

I1pu = I2pu+ I3pu

E1pu = E2pu== E3pu



For a practical three winding transformer Open circuit test determines shunt admittance (Gc and Bm) Short circuit test determines leakage impedances (Z12, Z13 and Z23 ) Z12 = pu leakage impedance measured from winding 1 with winding 2 shorted and winding 3 open Z13

Z23


Short Circuit Test Z12 = Z1 + Z2

Z13 = Z1 + Z3

Z23 = Z2 + Z3

Series impedances can be calculted from above using

Z1 = 1

2(Z12 + Z13 - Z23)

Z2 = 1

2(Z12 + Z23 - Z13)

Z3 = 1

2(Z13 + Z23 - Z12)

Exercise 3.51

The ratings of a three phase three winding transformer are: Primary (1): Y connected 66 kV, 15 MVA Secondary (2): Y connected 13.2 kV, 10 MVA Tertiary (3): ∆ connected 2.3 kV, 5 MVA Neglecting winding resistances and exciting current, the per unit leakage reactances are X12 = 0.08 on a 15 MVA, 66 kV base

X13 = 0.10 on a 15 MVA, 66 kV base X23 = 0.09 on a 10 MVA, 13.2 kV base

Exercise 3.51

(a) Determine the p u reactances X1 , X2 and X3

on a 15 MVA, 66 kV base at the primary terminals

(b) Purely resistive loads of 7.5 MW at 13.2 kV and 5 MW at 2.3 kV are connected to the secondary and tertiary sides of the transformer respectively. Draw the per unit impedance diagram, showing the per unit impedances on a 15 MVA, 66 kV base at the primary terminals

Exercise 3.51

Exercise 3.51

Exercise 3.51

Exercise 3.51

Exercise 3.51

Autotransformers

Two windings of a single phase transformer are connected in series.

Windings are coupled electrically and magnetically

Autotransformers

Autotransformers

Advantages

Smaller leakage impedance

Lower losses

Lower exciting current

Lower cost for small turns ratio

Disadvantages

Higher short circuit currents

Electrical coupling of windings allows transient over-voltages to pass through more easily

Transformers with off –nominal turns ratio


A transformer whose rated voltages is not in proportion to the selected base voltages is said to have an “off nominal turns ratio”.

𝑉1𝑟𝑎𝑡𝑒𝑑

𝑉2𝑟𝑎𝑡𝑒𝑑 ≠

𝑉𝑏𝑎𝑠𝑒−1

𝑉𝑏𝑎𝑠𝑒−2


𝑉1𝑟𝑎𝑡𝑒𝑑 = 𝑎𝑡 𝑉2𝑟𝑎𝑡𝑒𝑑 (1)

If the base voltages on either side of the transformer are known the

𝑉𝑏𝑎𝑠𝑒1 = b 𝑉𝑏𝑎𝑠𝑒2 (2)

Then equation (1) can be written as

𝑉1𝑟𝑎𝑡𝑒𝑑 = 𝑎𝑡 𝑉2𝑟𝑎𝑡𝑒𝑑

= b (𝑎𝑡

𝑏) 𝑉2𝑟𝑎𝑡𝑒𝑑 (3)

Letting c = 𝑎𝑡

𝑏 equation 3 becomes

𝑉1𝑟𝑎𝑡𝑒𝑑 = bc 𝑉2𝑟𝑎𝑡𝑒𝑑 (4)

𝑉1𝑟𝑎𝑡𝑒𝑑 = bc 𝑉2𝑟𝑎𝑡𝑒𝑑 can be represented by two transformers in series as shown below


Per unit equivalent circuit


π circuit representation for real c


𝐼1

−𝐼2=

𝑌11 𝑌12

𝑌21 𝑌21

𝑉1

𝑉2


Parallel connected transformers with different turns ratios

Tap changing transformers

Voltage regulating transformers

Phase angle regulating transformers





Example 3.12

Example 3.12

Example 3.12


Example 3.12

Example 3.12

Example 3.12



Regulating transformers

Voltage magnitude

Phase angle regulating

Voltage magnitude regulating transformer

Voltage magnitude regulating transformer

Modelling as a off nominal turns ratio transformer

c = (1 + ∆v) for voltage increase at bus abc

c = (1 + ∆v)-1 for voltage increase at bus a’b’c’

Phase angle regulating transformer

Modelling as a off nominal turns ratio transformer

c = 1∠α for a phase increase at bus abc

c = 1∠-α for a phase increase at bus a’ b’ c’

Example 3.13

Example 3.13

Example 3.13

Example 3.13

Example 3.13

Example 3.13

Example 3.13

Tutorial 4 – Exercise 3.59

The 2 lines in example 3.13 supply a balanced load with a load current of 1∠-300. Determine the real and reactive power supplied to the load bus from each parallel line with (a) no regulating transformer (b) the voltage regulating transformer in example 3.13(a), and (c) the phase angle regulating transformer in example 3.13(b), Assume that the voltage at bus abc is adjusted so that the voltage at bus a’b’c’ remains constant at 1.0∠00 per unit.

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Power Transformers 2