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SMJE 2103 Electrical Power System 3- Ph Power Apparatus Power Apparatus • • • • • Transmission Lines Transformers Circuit Breaker Switchgear Rotating Machine (Generator – Motor) Transformer • • • • • • • • • Introduction to power TX Connection Types and symbol Functionality Application Structure Material Equivalent Circuit Testing 3-Phase Tx Transformer 1. 2. 3. 4. Step up Step down Distribution Tx Special purpose (PT, CT) Transformer Transformer -Connection Types and Symbol Transformer -Connection Types and Symbol Transformer -Connection Types and Symbol Transformer -Connection Types and Symbol Transformer -Connection Types and Symbol Transformer -Operation (Ideal x – Turn Ratio)- v p t vs t Np Ns a i p t 1 is t a Transformer -Operation (Ideal Tx – Power)Pin = Vp Ip cos θp Pout = Vs Is cos θ Pout = Vs Is cos θs Also, Vs = Vp/a and Is = a Ip θp - θ s = θ Pout Vp a aI cos p Pout = Vp Ip cos θ = Pin Also same for power Q and power S Transformer -Operation (Ideal x – Impedance)VL ZL IL VP ZL ' IP VS ZL IS Transformer -Operation (Ideal x – Impedance)aVS VP 2 VS ZL ' a IP IS / a IS ZL’ = a2 ZL Transformer -Operation (Ideal Tx – Analysis)- (a) If the power system is exactly as described above in Figure (a), what will the voltage at the load be? What will the transmission line losses be? (b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step-down transformer is placed at the load end of the line (Figure (b)). What will the load voltage be now? What will the transmission line losses be now? Transformer -Operation (Real Tx - flux)- - Flux generated - The flux from each coil - Average flux level eind d dt N N i i 1 eind d N dt Transformer -Operation (Real Tx – Voltage Across)- 1 vP (t ) dt NP Primary Side: Hence, vP (t ) N P Or rewritten, P M LP dP d d N P M N P LP dt dt dt vP (t ) eP (t ) eLP (t ) Pri or sec voltage due to mutual flux dM eP (t ) N P dt Relationship, Vp and Vs eP (t ) dM eS (t ) NP dt NS eP (t ) N P a eS (t ) N S Transformer -Operation (Real Tx – Magnetization)Magnetization current, iM – current required to produce flux in the core. Core-loss current, ih+e – current required to compensate hysteresis and eddy current losses. iex im ih e Transformer -Application- Transformer -Application- Transformer -Application- Transformer -Structure- Transformer -Structure (Shell Form)- Transformer -Structure (Core Form)- Transformer -Structure- Transformer -Structure- Transformer -Material• Copper conductors • Silicon iron/laminated piece of steel – eddy current • Oil • Insulation materials Transformer -Equivalent Circuit- Transformer -Equivalent Circuit (simplified)- Transformer -Equivalent Circuit (simplified)- Example/Tutorial • The secondary winding of a transformer has a terminal voltage of vs(t) + 282.8 sin 377t V. The turns ratio of the transformer is 100:200 (a = 0.5). If the secondary current of the transformer is is(t) =7.07 sin (377t -36.87o) A, what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are Transformer -Testing- To know the value of the inductance and resistance Transformer -Short circuit test- Z SE VSC I SC Power factor, PF = cos θ = PSC / VSC ISC Z SE The series impedance ZSE (lagging) VSC 0 VSC I SC I SC = Req + jXeq = (RP + a2 RS) + j(XP + a2 XS) Transformer -Open circuit test- Z SE VSC I SC Power factor, PF = cos θ = PSC / VSC ISC Z SE The series impedance ZSE (lagging) VSC 0 VSC I SC I SC = Req + jXeq = (RP + a2 RS) + j(XP + a2 XS) Transformer -Equivalent Circuit (Example)The equivalent circuit impedances of a 20kVA, 8000/240V, 60Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the primary side of the transformer, and the following data were taken: Open circuit test (primary) VOC = 8000 V IOC = 0.214 A POC = 400 W Short circuit test VSC = 489 V ISC = 2.5 A PSC = 240 W Find the impedance of the approximate equivalent circuit referred to the primary side, and sketch the circuit. Example/Tutorial A 1000 VA 230/115 V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below All data given taken from the primary side of the transformer. (a) Find the equivalent circuit of this transformer referred to lowvoltage side of the transformer. (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging. Transformer -Impulse voltage test- Transformer - Voltage Regulation- VR V S ,nlVS , fl VS , fl x 100% Transformer - EfficiencyPout x100% Pin Pout x100% Pout Ploss Types of losses incurred in a transformer: Copper I2R losses Hysteresis losses Eddy current losses VS I S cos x100% PCu Pcore VS I S cos 3-Phase Transformer Transformers for 3-phase circuits can be constructed in two ways: connect 3 single phase transformers Three sets of windings wrapped around a common core The primaries and secondaries of any three-phase transformer can be independently connected in either a wye (Y) or a delta (∆) The important point to note when analyzing any 3-phase transformer is to look at a single transformer. The impedance, voltage regulation, efficiency, and similar calculations for three phase transformers are done on a perphase basis.