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

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

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

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

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

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

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

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

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

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