lecture 10 outline 10-1 Ideal Transformers Transformers are enormously important. Without them, electrical power would not be as available and widespread as it is today. They allow changes in voltage levels. In particular, voltage can be increased to allow power to be transported at lower current levels. Transformers have two principal components. § A core of high permeability, read ferromagnetic, material which is able to confine a magnetic field § Windings which are wound about the core. An externally supplied AC current on one winding (call the primary winding) produces a magnetic field in the core by Ampere’s law. AC voltages will then be induced by Faraday’s law on other coils (secondary winding) that are wound on the core. The ideal transformer model is based on two assumptions. § The core of an ideal transformer has infinite permeability. The result is that all flux is confined to the core. § There is no power loss in the ideal transformer. The result is that power in must equal the power out.lecture 10 outline 10-2 Transformer action The sinusoidal currents, i1(t) and i2(t), flow in the primary coil and secondary coils, the coils having N1 and N2 turns, respectively. The two currents, working together, produce a sinusoidal magnetic flux, φ(t), in the core. Since the core is assumed to have infinite permeability, all the flux is confined to the core, no flux leaks out. The result is that the same flux links both coils. Sinusoidal voltages are induced in the coils via Faraday’s law. Since they are caused by the same flux, they must be in phase. Their magnitudes are related by the turn’s ratio. φφ1122ddv(t)=N v(t)=Ndtdt v1(t) = V1 cos (ωt + θv) v2(t) = V2 cos (ωt + θv) 111222v(t)NV = = = av(t)NV (a is called the “turns ratio) For an ideal transformer the permeability of the core is infinite and therefore has no reluctance. The magnetic KVL gives N1I1 = N2I2 For an ideal transformer, Pin = Pout (no losses) V1I1 cos(θv-θi1) = V2I2 cos(θv-θi2) Therefore, θi1 = θi2 i1(t) = I1 cos (ωt + θi) i2(t) = I2 cos (ωt + θi)lecture 10 outline 10-3 Phasor relations for the ideal transformer 11122221NN = = NNVIVI The ideal transformer application: impedance reflectionlecture 10 outline 10-4 Example Determine the impedance seen by the source and then determine the primary and secondary currents and voltages. Notice that when the load was referred to the primary side, its value reduced. This is because the load was taken from the high voltage side across to the low voltage side. Impedance will go up when it is moved to a higher voltage and will go down when it is moved to a lower voltage.lecture 10 outline 10-5 EXAMPLE This example demonstrates how transformers reduce the cost of supplying electricity. A single-phase load of 10MW with pf=1 is to be supplied at 2.4kV via a distribution feeder with (0.15 + j0.1)Ω impedance. Electric energy costs 3¢/kWh. Calculate the yearly cost of supplying the line losses if the load is constant and continuous if: a) no transformers are used, and b) two transformers each with a = 10 are used. original system system with step-up and step-down transformerslecture 10 outline 10-6 Energy Costs: %η %VRlecture 10 outline 10-7 Practical Transformer Model A model that accounts for energy storage and energy dissipation is shown below. The model includes an ideal transformer with components added to account for non-ideal behavior of real transformers. • Real transformers are not lossless: R1 and R2 account for energy dissipation in the primary and secondary windings. • In real transformers, the flux is not entirely confined to the core; some flux “leaks out.” X1 and X2 account for the energy stored in this leakage flux. • The core is not lossless. Some energy is lost magnetizing and demagnetizing the core—Rc represents the real power that is dissipated in the core of the transformer due to hysteresis and eddy currents. • Xm accounts for the magnetic energy stored in the core. • Vp and V2 are the primary and secondary voltages that appear at the terminals of the transformer.lecture 10 outline 10-8 The circuit above is known as the “exact equivalent circuit.” A simpler model that still gives accurate results is shown below. The parallel elements, Rc and Xm, are moved in front of R1 and X1 and placed across the supply. This does not introduce much error since Rc and Xm are much larger impedances than R1 and X1 and so appear to be almost open circuits. R2 and X2 are referred to the primary side by impedance reflection. They are added to R1 and X1 since they are in series with them, R = R1 + a2R2 and X = X1 + a2X2. This circuit is called the “approximate equivalent circuit.” It’s the one we’ll work with. Efficiency and Regulation Efficiency and regulation are figures of merit for a transformer; they measure well a transformer performs, how close to ideal it is. For %VR, use Vp and V1. For %η, use Pout = Re{V1I1*] and Pin = Re{VpIp*]lecture 10 outline 10-9 General Analysis Ploss has two components, Pcoil and Pcore § Pcoil varies quadratically with current (goes up with load). § Pcore is almost constant since the voltage Vp doesn’t vary much. Example A 300 kVA, single-phase transformer is rated to transform 2.4 kV to 600 V and has the following equivalent circuit parameters referred to the high voltage side. R = 0.75 Ω, X = 1.5 Ω, Rc = 500 Ω, Xm = 60 Ω Calculate %VR and %η when the transformer supplies rated load at 0.7 lag pf.lecture 10 outline 10-10 Example (cont)lecture 10 outline 10-11 Example Re-work with the load pf corrected to unity (P remains the same). The results are:: § %VR = 2.88% § %η = 92.1% § Pcoil = 7.96 kW § Pcore = 12.2 kWlecture 10 outline 10-12 Experimental determination of equivalent circuit The equivalent circuit parameters can determined from the open-circuit (OC) and short-circuit (SC) tests. Required measurements are Vp, Ip and Pin. OC test The test is performed at rated voltage and no load—the load is an open-circuit (I1 = 0)—R and X can be neglected. Only Rc and Xm load the supply. Voc, Ioc, and Poc are measured once the supply has been adjusted to Vrated. The procedure is: 1. determine θoc