L-19(NKD)(ET) ((EE)NPTEL) - Module 5 Three-phase AC...

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Unformatted text preview: Module 5 Three-phase AC Circuits Version 2 EE IIT, Kharagpur Lesson 19 Three-phase Delta-Connected Balanced Load Version 2 EE IIT, Kharagpur In the previous (first) lesson of this module, the two types of connections (star and delta), normally used for the three-phase balanced supply in source side, along with the line and phase voltages, are described. Then, for balanced star-connected load, the phase and line currents, along with the expression for total power, are obtained. In this lesson, the phase and line currents for balanced delta-connected load, along with the expression for total power, will be presented. Keywords : line and phase currents, star- and delta-connections, balanced load. After going through this lesson, the students will be able to answer the following questions: 1. How to calculate the currents (line and phase), for the delta-connected balanced load fed from a three-phase balanced system? 2. Also how to find the total power fed to the above balanced load, for the two types of load connections – star and delta? Currents for Circuits with Balanced Load (Delta-connected) V BR I BR V RY I RY V YB I YB 120 ° 120 ° Φ Φ Φ (b) Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced supply (b) Phasor diagram Version 2 EE IIT, Kharagpur A three-phase delta ( )-connected balanced load (Fig. 19.1a) is fed from a balanced three-phase supply. A balanced load means that, the magnitude of the impedance per phase, is same, i.e., Δ BR YB RY p Z Z Z Z = = = , and their angle is also same, as BR YB RY p φ = = = . In other words, if the impedance per phase is given as, p p p p X j R Z + = ∠ , then BR YB RY p R R R R = = = , and also BR Yb RY p X X X X = = = . The magnitude and phase angle of the impedance per phase are: 2 2 p p p X R Z + = , and ( ) p p p R X / tan 1 − = .In this case, the magnitudes of the phase voltages p V are same, as those of the line voltages BR YB RY L V V V V = = = . The phase currents (Fig. 19.1b) are obtained as, p RY RY p RY RY p RY Z V Z V I − ∠ = ∠ ° ∠ = − ∠ ) 120 ( 120 ) 120 ( p YB YB p YB YN p YB Z V Z V I + ° − ∠ = ∠ ° − ∠ = + ° − ∠ ) 120 ( 120 ) 120 ( p BR BR p BR BR p BR Z V Z V I − ° ∠ = ∠ ° + ∠ = − ° ∠ In this case, the phase voltage, is taken as reference. This shows that the phase currents are equal in magnitude, i.e., ( RY V BR YB RY p I I I I = = = ), as the magnitudes of the voltage and load impedance, per phase, are same, with their phase angles displaced from each other in sequence by . The magnitude of the phase currents, is expressed as ° 120 ( ) p p p Z V I / = . The line currents (Fig. 19.1b) are given as ) 30 ( 3 ) 120 ( ) ( p p p p p p BR RY R R I I I I I I θ + ° − ∠ = − ° ∠ − − ∠ = − = − ∠ ) 30 ( p L I + ° − ∠ = ) 150 ( 3 ) ( ) 120 ( p p p p p p RY YB Y Y I I I I I I + ° − ∠ = − ∠ − + ° − ∠ = − = − ∠ ) 150 ( p L I + ° − ∠ = ) 90 ( 3 ) 120 ( ) 120 ( p p p p p p YB BR B B I I I I I I − ° ∠ = + ° − ∠ − − ° ∠ = − = − ∠ ) 90 ( p L I − ° ∠ = The line currents are balanced, as their magnitudes are same and 3 times the magnitudes of the phase currents ( p L I I ⋅ = 3 ), with the phase angles displaced from each other in sequence by...
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L-19(NKD)(ET) ((EE)NPTEL) - Module 5 Three-phase AC...

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