1. Electro Mechanical System
Loose Couplings
Consider now a perfect core but the windings are loosely coupled
Primary and secondary coils have negligible resistance
The primary is connected to a source Eg
The coil draws no current to drive a mutual flux m1a
The flux produces a counter voltage Ep that equals Eg
The flux produces a voltage E2 on the secondary coil
Under no-load conditions, I2 is zero and no mmf exist to drive any leakage flux
Lecture 07
Electro Mechanical System

2. Electro Mechanical System
Loose Couplings
Now a load Z is connected across the secondary
currents I1 and I2 immediately begin to flow, and are related by: N1 I1 = N2 I2
I2 produces an mmf N2I2 and I1 produces an mmf N1I1 in the opposite direction
mmf N2I2 forms a flux 2, consisting of a mutually coupled flux m2 and a leakage flux f2
mmf N1I1 forms a flux 1 , consisting of a mutually coupled flux m1 and a leakage flux f1
The new mmf’s upset m1a
balance
We combine m1 & m2 into
mutual flux m
E2 = 4.44fN2 m
E1 = 4.44fN1 m
Lecture 07
Electro Mechanical System

3. Electro Mechanical System
Loose Couplings
The new mmf’s upset m1a balance
resolving the modeling conflicts
combine m1 and m2 into a single mutual flux m
Es consists of two parts: E2(N2 m) and Ef2(N2 f2)
They are given by: Ef2 = 4.44fN2 f2 : E2 = 4.44fN2 m
Ep consists of two parts: E1(N1 m) and Ef1(N1 f1)
They are given by: Ef1 = 4.44fN1 f1 : E1 = 4.44fN1 m
Lecture 07
Electro Mechanical System

4. Electro Mechanical System
Leakage Reactance
The four induced voltages can be rearranged in the transformer circuit as shown above
The rearrangement does not change the induced voltages.
Ef2 is a voltage drop across a reactance
Xf2 = Ef2/I2
Ef1 is a voltage drop across a reactance
Xf1 = Ef1/I1
Home Work: Page 201 Example 10-2
Lecture 07
Electro Mechanical System

5. Electro Mechanical System
Equivalent Circuit
Primary and secondary windings are composed of copper or aluminum conductors.
conductors exhibit resistance to the current flow.
the leakage reactance can also be modeled as a series inductance.
Core excitation and losses are modeled as a shunt circuit.
Combining all elements with the ideal transformer forms an equivalent circuit for practical transformers.
Lecture 07
Electro Mechanical System

6. Terminal Markings & Taps
Polarity can be shown by means of dots
H1 & H2 for High voltage and X1 & X2 by low voltage
additive polarity when H1 is diagonally opposite to X1
subtractive polarity when H1 is adjacent to X1
Voltage drops in lines can be regulated by
Adding taps in the distribution transformer
2400V/120V transformer can be connected to a line with voltage never higher than 2000V
Lecture 07
Electro Mechanical System

7. Electro Mechanical System
Losses
As in all machines, a transformer has losses
I2R losses in the primary and secondary windings
Hysteresis losses and eddy-current losses in the core
Stray losses due to primary and secondary leakage fluxes
Losses appear in the form of heat.
Heat produces an increase in temperature.
Drop in efficiency.
Iron losses depend on the mutual flux and hence the applied voltage.
The winding losses depend on the current drawn by the load.
Lecture 07
Electro Mechanical System

8. Electro Mechanical System
Losses
We must set limits to applied voltage and current drawn by the load
These limits determine the nominal voltage Enp and nominal current Inp of transformer winding (primary or secondary)
Power rating is equal to product of the nominal voltage times nominal current.
Result is not in watts because phase angle can have any value.
The result is expressed in voltampre (VA), in kilovoltampre (kVA) or in megavoltampre(MVA).
Lecture 07
Electro Mechanical System

9. Electro Mechanical System
Example
The nameplate of a distribution transformer indicates 250 kVA, 60 Hz, 4160 V primary, 480 V secondary;
Calculate the nominal primary and secondary currents
If we apply 2000V to the 4160V primary, can we still draw 250kVA from the transformer?
a) Nominal current of 4160 V winding is
Inp = nominal S/nominal Ep = Sn/Enp
Inp = (250 x 1000)/4160 = 60A
Nominal current of 480 V winding is
Ins = nominal S/nominal ES = Sn/Ens
Ins = (250 x 1000)/480 = 521A
Lecture 07
Electro Mechanical System

10. Electro Mechanical System
Example
b) If we apply 2000 V to the primary, the flux and iron losses will be lower and core will be cooler. The current should not exceed the nominal value, otherwise the winding will overheat.
S = 2000 V x 60 A = 120 kVA
Lecture 07
Electro Mechanical System