Transformer Basics
A Transformer May Look Like This
Any transformer consists of the following three basic parts in it. • Primary coil • Secondary coil • Transformer (Magnetic) core
1. Primary coil The primary coil is the coil to which the source is connected. It may be the high voltage side or low voltage side of the transformer. An alternating flux is produced in the primary coil.
2. Secondary coil The output is taken from the secondary coil. The alternating flux produced in the primary coil passes through the core and links with there coil and hence emf is induced in this coil. 3. Magnetic core The flux produced in the primary passes through this magnetic core. It is made up of laminated soft iron core. It provides support to the coil and also provides a low reluctance path for the flux.
Construction of Power and Distribution Transformers Construction • Core Type High voltage • Shell Type Less leakage flux
Type of Cooling • Ventilated Dry-Type Transformers They are cooled by natural air convection.
•
Gas-Filled Dry-Type Transformers Cooled with nitrogen or other gases
•
Liquid-Immersed Transformers Hermetically sealed tanks with insulated liquid (mineral oil, silicone oil)
Transformer Types In a step-down transformer, the secondary voltage is less than the primary voltage and Nsec < Npri
In a step-up transformer, the secondary voltage is greater than the primary voltage and Nsec > Npri
Step Up Transformer
In an isolation transformer, the secondary voltage is equal to the primary voltage and Nsec = Npri https://www.youtube.com/watch?v=GMePE7NZcxw
Principle of Transformer Action Faraday’s Law d d e1 N1 e2 N 2 dt dt • The same flux (mutual flux) exists in both coils. • Flux is generated by i1 (right-hand rule). • The induced emf e1 and e2 are generated to oppose the buildup of flux in its window. • i2 is generated by the induced emf e2.
d 0 dt
In DC, the induced emfs are transients, in steady state:
e1 e2 0
AC Assuming: • Core permeability constant • No leakage flux
E p 4.44 N p f max
Ep Es
Es 4.44 N s f max
Np Ns
Transformer
VP - is the Primary Voltage VS - is the Secondary Voltage NP - is the Number of Primary Windings NS - is the Number of Secondary Windings Φ (phi) - is the Flux Linkage
Example A ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 120 V, 60Hz source. What is the effective voltage across the secondary terminals?
Solution:
𝑁2
𝐸2
→ 2250
𝐸2
𝑁1 = 𝐸1 90 = 120 → 𝐸2 25 = 120 → 𝐸2 = 25 × 120 = 3000 𝑉
Example An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A. What is the effective value of the primary current?
Solution:
𝑁2
𝐼1
→ 2250
𝐼1
𝑁1 = 𝐼2 90 = 2 → 𝐼1 25 = 2 → 𝐼1 = 25 × 2 = 50 𝐴
Single Phase Transformer information: HV 115 kV LV 7200 V Capacity 16,670 KVA primary
Primary amps 16,670,000 ÷ 115,000 = 145 amps Secondary amps 16,670,000 ÷ 7,200 = 2315 amps Turns Ratio 115,000 ÷7200 = 15.97/1 Current ratio Check 2315÷145 = 15.97/1
secondary
Calculating single phase transformer MVA Voltage X Current 115,000 X 145 amps = 16,675,000= 16.67MVA 7,200 X 2315 amps = 16,668,000 = 16.67 MVA
Transformer efficiency The efficiency of a transformer is the ratio of power delivered to the load (Pout) to the power delivered to the primary (Pin). That is 𝑃𝑜𝑢𝑡 ƞ= × 100% 𝑃𝑖𝑛
𝑃𝑜𝑢𝑡 =
𝑉𝑆 2
152 𝑃𝑖𝑛 = 𝑉𝑝 𝐼 = 120 × 20 × 10−3 = 2.4 = = 2.25 𝑅 100 𝑃𝑜𝑢𝑡 ƞ= × 100% = 2.25 2.4 × 100% = 93.75% ≅ 94% 𝑃𝑖𝑛
In-Rush current • When connecting an AC source to a transformer circuit the current called the transient component or in-rush current • Although the in-rush component to a transformer decays rapidly, dropping to the normal no-load current within 5 to 10 cycles, it may exceed 25 times the full-load rating during the first halfcycle. • This high in-rush must be taken into consideration when selecting fuses and/or circuit breakers. • The magnitude of the in-rush depends on the magnitude and phase angle of the voltage wave at the instant the switch is closed.
Transformer Polarity • Transform polarity refers to the relative phase relationship of transformer leads • On power transformers the terminals are designated by the symbols H1, and H2 for the high-voltage (HV) winding and by X1, and X2 for the low-voltage (LV) winding. • By convention, H1 and X1 have the same polarity. • The transformer has either additive or subtractive polarity. • A transformer is said to have additive polarity when terminal H1 is diagonally opposite terminal X1 • A transformer has subtractive polarity when terminal H1 is adjacent to terminal X1
Three-Phase Systems • Three-phase power is preferred over single-phase power for several important reasons:
– Three-phase motors, generators, and transformers are simpler, cheaper, and more efficient – Three-phase transmission lines can deliver more power for a given weight and cost – The voltage regulation of 3-phase transmission lines is inherently better
Three Phase Transformers • A three-phase transformer can be built by constructing a threephase transformer on a common magnetic structure or can be built by suitably connecting a bank of similar three singlephase transformers • Three-phase transformers use much less material than three single-phase transformers for the same three-phase power and voltage ratings. • Three-phase transformers have all three phases wound on a single magnetic core or in shell-type and core-type construction. • The principal disadvantage of a three-phase transformer, compared with its three-transformer counterpart, is that failure of one phase puts the entire transformer out of service
Balanced Three-Phase System • Three-phase transformers are required to step up or step down voltages in the various stages of power transmission. • The primary and secondary windings may be connected in either wye (Y) or delta (Δ) configurations. • There are therefore four possible connections for a three-phase transformer: » Y -Δ » Δ -Y
» Δ-Δ » Y-Y
Three Phase connections of three Single Phase Transformers
Connections Wye connected windings: Have a common connection point Have two voltages available “L-L & L-N” Has only one current : Line current
Connections Delta connected windings: Have no common connection point Have only one voltage available “L-L” Delta can be closed “connected” in more than one way Has two currents Line current Phase or winding current
Three-phase Voltage and Current Connection
Phase Voltage
Line Voltage
Phase Current
Line Current
Star
VP = VL ÷ √3
VL = √3 × VP
IP = I L
IL = IP
Delta
VP = VL
VL = VP
IP = IL ÷ √3
IL = √3 × IP
𝑃=
3𝐼𝐿 𝑉𝐿
or
𝑃 = 3𝐼𝑃𝐻 𝑉𝑃𝐻
𝑡𝑢𝑟𝑛 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑃𝐻 𝑝𝑟𝑖𝑚𝑎𝑟𝑦/𝑉𝑃𝐻 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦
HS Line 230 kV LS Line 23 kV Transformer 3 Ø 230/13280/23000 Gnd. Wye 25 MVA
230 kV B
A
C
H2
H1
W1
H3
W2
W3
W5
W4
W6
N X1
X2
X3
23 kV a
b
c
HS Line 115 kV LS Line 12 kV Transformer: 3 Ø 115/12000/6928 Gnd. Wye 40MVA
Voltage on HS 115 kV. Only one voltage because it is delta B
A
H2
H1
Voltage on LS 12,000 & 6,928 because it is wye connected
C
W1
H3
W3
W2
16.6/1 Transformer Ratio 16.6/1
W4
W6
W5
N X1
a
X2
X3
b
c
Transformer Solutions • Information: Δ-Υ 230/23kV 25 MVA transformer 230 kV (Given)
•
What is the voltage on High voltage line?
•
What is the voltage on low voltage line?
•
What is the voltage on low voltage phase?
13,280 kV ( V 𝐿 − 𝐿 ÷ √3)
•
What is the voltage across W1, W2, W3?
230 kV (Same as the line for Δ winding)
•
What is the voltage across W4, W5, W6?
13,280 kV ( L-N voltage)
•
What is the current in line A, B, C ?
•
What is the current in W1, W2, W3?
•
What is the current in W4, W5, W6?
23 kV (Given)
25 𝑀𝑉𝐴 ÷ 230,000 ÷ 3 = 62.75 25 𝑀𝑉𝐴 ÷ 230,000 ÷ √3 ÷ √3 = 36.23 25 MVA / 23 kV / √3 = 627.56 amps
Transformer Solutions •
What is the current in line a, b, c?
•
The line current ÷ √3 = Winding Current ; On the delta side of the transformer 62.76 ÷ √3 = 36.23
Same as winding W4, W5, W6= 627.556 amps
Calculating three phase transformer MVA Voltage X Current X three windings 230,000 X 36.3 amps = 8,349,000 x 3 (windings)= 25,047,000 = 25 MVA 13,280 X 627.57 amps = 8,334,129 X 3 (windings)= 25,002, 388 = 25MVA
Transformer Ratio 3 Phase 230/23 kV 25 MVA Transformer is 230/ 23 kV Ratio is NOT this ratio but is the ratio of the voltages across the windings. W1 230 kV, W4 13,280 Ratio is 230,000 / 13,280 = 17.32/1
A
B
C
H2
H1
H3
W 1
W2
W4
W5
W3
W6
N
Note: W4 current/ W1 current = 627.56/36.3 = 17.3 627.56 amps X 13,280 volts = 8.3 MVA 36.3 amps X 230,000 volts = 8.3 MVA 8.3 MVA X 3 (windings) =25 MVA
X1 a
X2
X3
b
c
How Are They Connected?
H
primary
H
H primary
secondary
H
H
H primary
secondary
H
H
H
secondary
H
H
H
A
B
H1
H2
W1
W2
H3
W3
W5
W4
C
W6
N X1
a
X2
b
X3
c
Both transformers are Δ-Υ but notice the Δ windings are not closed the same
A
B
H1
H2
W 1
W 2
N
X1
A H1
B
C
H2
H3
W 1
N
W 2
W 3
W 5
W 4 X1
a
a
W 6 X2
X3
b
c
H3
W 3
W 5
W 4
C
W 6 X2
X3
b
c
Transformer Information For Practice Calculations • • • • • •
3Ø, Δ-Υ 115/12 kV 25MVA 3Ø, Δ-Υ 115/23 kV 40 MVA 3Ø, Δ- Δ 23/ 13.8 kV 5 MVA 3Ø Υ- Υ 4160/ 2400 5 MVA 3Ø, Δ- Δ 115/ 13.8 kV 10 MVA 1Ø Υ- Δ 230/22 kV 350 MVA
Delta and Wye 3-phase circuits : Electronics Worksheet
