EMT 1143 INTRODUCTION TO ELECTRIC CIRCUITS CHAPTER 7

EMT 114/3 INTRODUCTION TO ELECTRIC CIRCUITS CHAPTER 7: Power System Circuit School of Microelectronic Engineering, Universiti Malaysia

Chapter Outline 7. 1 7. 2 7. 3 7. 4 7. 5 7. 6 What is a Three-Phase Circuit? Balance Three-Phase Voltages Balance Three-Phase Connection Power in a Balanced System Unbalanced Three-Phase Systems Application – Residential Wiring

7. 1 What is a Three-Phase Circuit? Ø It is a system produced by a generator consisting of three sources having the same amplitude and frequency but of phase with each other by 120°. Three sources with 120° out of phase Four wired system

7. 1 What is a Three-Phase Circuit? Advantages: Ø Most of the electric power is generated and distributed in three-phase. Ø The instantaneous power in a three-phase system can be constant. Ø The amount of power, the three-phase system is more economical than the single-phase. Ø In fact, the amount of wire required for a three-phase system is less than that required for an equivalent singlephase system. Ø Households have single phase mains power supplied. Ø This typically in a three wire form, where two 120 V sources with the same phase are connected in series. Ø This allows for appliances to use either 120 V or 240 V

7. 2 Balance Three-Phase Voltages Ø Three phase voltages are typically produced by a three phase AC generator. Ø A three-phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator). A three-phase generator The generated voltages

7. 2 Balance Three-Phase Voltages Ø Ø Ø Three phase voltage sources can be connected the loads by either three or four wire configurations. The three wire configuration is accomplished by a Delta connected source. Two possible configurations: Three-phase voltage sources: (a) Y-connected ; (b) Δ-connected

7. 2 Balance Three-Phase Voltages Ø Ø Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°. A wye connected source is said to be balanced when the sum of the three voltages is zero: Ø This can only happen if: Ø There are two sequences for the phases: Positive sequence Negative sequence

7. 2 Balance Three-Phase Voltages Example 1 Determine the phase sequence of the set of voltages.

7. 2 Balance Three-Phase Voltages Solution: The voltages can be expressed in phasor form as We notice that Van leads Vcn by 120° and Vcn in turn leads Vbn by 120°. Hence, we have an acb sequence.

7. 2 Balance Three-Phase Voltages Ø Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°. They may also be connected in either a Delta or wye configuration. For a balanced wye connected load: Ø For a balanced delta connected load: Ø Ø Two possible three-phase load configurations: (a) a Y-connected load, (b) a Δ-connected load.

7. 3 Balance Three-Phase Connection • Four possible connections 1. Y-Y connection (Y-connected source with a Y-connected load) 2. Y-Δ connection (Y-connected source with a Δ-connected load) 3. Δ-Δ connection 4. Δ-Y connection

7. 3. 1 Balance Y-Y Connection Ø A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load. Ø We will use the positive sequence for this circuit, meaning the voltages are:

7. 3. 1 Balance Y-Y Connection Ø The line to line voltages are: Ø Thus the magnitude of the line voltages VL is: Ø Where:

7. 3. 1 Balance Y-Y Connection Ø If we apply KVL to each phase, we find the line currents are: Ø From this one can see the line currents add up to zero. Ø This shows that the neutral wire has zero voltage and no current. Thus it can be removed without affecting the system. Ø

7. 3. 1 Balance Y-Y Connection Practice Problem 12. 2 Ø A Y-connected balanced three-phase generator with an impedance of 0. 4 + j 0. 3Ω per phase is connected to a Yconnected balanced load with an impedance of 24 + j 19Ω per phase. The line joining the generator and the load has an impedance of 0. 6 + j 0. 7Ω per phase. Assuming a positive sequence for the source voltages and that find: (a) the line voltages, (b) the line currents.

7. 3. 2 Balance Y-Δ Connection Ø A balanced Y-Δ system is a three-phase system with a balanced Y-connected source and a balanced Δ-connected load. Ø Assuming the positive sequence, the phase voltages are: Ø And the line voltages are:

7. 3. 2 Balance Y-Δ Connection Ø The line voltages are equal to the voltages across the load. Ø From this, we can calculate the phase currents: Ø The line currents can be obtained from the phase currents by applying KCL to nodes A, B, and C Ø Since ICA = IAB -240°: Ø Thus:

7. 3. 2 Balance Y-Δ Connection Practice Problem 12. 3 Ø One line voltage of a balanced Y-connected source is. If the source is connected to a Δconnected load of , find the phase and line currents. Assume the abc sequence.

7. 3. 3 Balance Δ -Δ Connection Ø A balanced Δ-Δ system is a three-phase system with a balanced Δ -connected source and a balanced Δ -connected load.

7. 3. 3 Balance Δ -Δ Connection Ø Assuming a positive sequence, the phase voltages are: Ø If line impedances are insignificant, then the impedance voltages are the same as the phase voltages. Ø Hence the phase currents are: Ø By applying KCL at the nodes A, B, and C: Ø The line current is:

7. 3. 3 Balance Δ -Δ Connection Practice Problem 12. 4 Ø A positive-sequence, balanced Δ-connected source supplies a balanced -connected load. If the impedance per phase of the load is 18 + j 12Ω and find and , find IAB and VAB.

7. 3. 4 Balance Δ -Y Connection Ø A balanced Δ-Y system is a three-phase system with a balanced y-connected source and a balanced y-connected load

7. 3. 4 Balance Δ -Y Connection Ø Applying KVL: Ø Thus: Ø Keeping in mind that Ib lags Ia by 120º, we can solve for the line current:

7. 3. 4 Balance Δ -Y Connection Ø Ø Another way to solve this system is to convert both the source and load back to a Wye-Wye system. The equivalent Wye connected source voltages are:

7. 3. 4 Balance Δ -Y Connection Practice Problem 12. 5 Ø In a balanced Δ -Y circuit, = (12 + j 15) Ω. Calculate the line currents. . and ZY

7. 3 Summary

7. 3. 4 Power in a Balanced System Ø We begin by examining the instantaneous power absorbed by the load. Ø The total instantaneous power in the load is: p= 3 Vp. Ip cos θ Ø This equation is true whether the load is Y- or Δconnected. Ø The average power PP per phase for either the Δ connected load or the Y-connected load is P/3 or PP = Vp. Ip cos θ Ø and the reactive power phase is: QP = Vp. Ip sin θ

7. 3. 4 Power in a Balanced System Ø The apparent power phase is: SP = Vp. Ip Ø The complex power phase is: Sp = Pp + j. Qp = Vp Ip* Where Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip , respectively. Ø The total average power is the sum of the average powers in the phases:

7. 3. 4 Power in a Balanced System

7. 3. 4 Power in a Balanced System Practice Problem 12. 5 Ø A Y-connected balanced three-phase generator with an impedance of 0. 4 + j 0. 3Ω per phase is connected to a Yconnected balanced load with an impedance of 24 + j 19Ω per phase. The line joining the generator and the load has an impedance of 0. 6 + j 0. 7Ω per phase. Assuming a positive sequence for the source voltages and that find the complex power at the source and at the load.

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