Module 5. Three-phase AC Circuits. Version 2 EE IIT, Kharagpur

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1 Module 5 Three-hase AC Circuits ersion EE T, Kharagur

2 esson 9 Three-hase Delta- Connected Balanced oad ersion EE T, Kharagur

3 n the revious (first) lesson of this module, the two tyes of connections (star and delta), normally used for the three-hase balanced suly in source side, along with the line and hase voltages, are described. Then, for balanced star-connected load, the hase and line currents, along with the exression for total ower, are obtained. n this lesson, the hase and line currents for balanced delta-connected load, along with the exression for total ower, will be resented. Keywords: line and hase currents, star- and delta-connections, balanced load. After going through this lesson, the students will be able to answer the following questions:. How to calculate the currents (line and hase), for the delta-connected balanced load fed from a three-hase balanced system?. Also how to find the total ower fed to the above balanced load, for the two tyes of load connections star and delta? Currents for Circuits with Balanced oad (Delta-connected) Φ YB Φ 0 0 Φ YB (b) Fig. 9. (a) Balanced delta-connected load fed from a three-hase balanced suly (b) Phasor diagram ersion EE T, Kharagur

4 A three-hase delta ( Δ )-connected balanced load (Fig. 9.a) is fed from a balanced three-hase suly. A balanced load means that, the magnitude of the imedance er hase, is same, i.e., = = YB =, and their angle is also same, as φ = φ = φyb = φ. n other words, if the imedance er hase is given as, φ = + j X, then = = B =, and also X = X = X Yb = X. The magnitude and hase angle of the imedance er hase are: ( X / ) = + X, and φ = tan.n this case, the magnitudes of the hase voltages are same, as those of the line voltages = = YB =. The hase currents (Fig. 9.b) are obtained as, 0 φ = = φ φ YN 0 YB YB ( 0 + φ ) = = (0 + φ ) YB φ YB + 0 ( 0 φ ) = = (0 φ ) φ n this case, the hase voltage, is taken as reference. This shows that the hase currents are equal in magnitude, i.e., ( YB ), as the magnitudes of the voltage and load imedance, er hase, are same, with their hase angles dislaced from each other in sequence by 0. The magnitude of the hase currents, is exressed as = ( / ). The line currents (Fig. 9.b) are given as θ ( φ ) (0 φ ) = 3 (30 + φ ) 30 + φ ) ( Y θ Y YB ( 0 + φ ) ( φ ) = 3 (50 + φ ) 50 + φ ) B θ ( B ( 90 φ ) YB ( 0 φ ) (0 + φ ) = 3 (90 φ ) The line currents are balanced, as their magnitudes are same and 3 times the magnitudes of the hase currents ( = 3 ), with the hase angles dislaced from each other in sequence by 0. Also to note that the line current, say, lags the corresonding hase current, by 30. f the hase current, is taken as reference, the hase currents are 0 (.0 + j0.0) : YB 0 ( 0.5 j0.866) ; + 0 ( j0.866). The line currents are obtained as ersion EE T, Kharagur

5 0 + 0 {(.0 + j 0.0) ( j 0.866)} (.5 j 0.866) Y YB 0 0 {( 0.5 j 0.866) (.0 + j 0.0)} = (.5 + j 0.866) B {( j 0.866) ( 0.5 j 0.866)} YB j.73) = ( Total Power Consumed in the Circuit (Delta-connected) n the last lesson (No. 8), the equation for the ower consumed in a star-connected balanced circuit fed from a three-hase suly, was resented. The ower consumed er hase, for the delta-connected balanced circuit, is given by ( ) W = cos φ = cos, t has been shown earlier that the magnitudes of the hase and line voltages are same, i.e., =. The magnitude of the hase current is ( / 3 ) times the magnitude of the line current, i.e., ( / 3) is obtained as =. Substituting the two exressions, the total ower consumed ( / 3) cos φ = 3 cos φ W = 3 t may be observed that the hase angle, φ is the angle between the hase voltage, and the hase current,. Also that the exression for the total ower in a threehase balanced circuit is the same, whatever be the tye of connection star or delta. Examle 9. The star-connected load having imedance of ( j 6) er hase is connected in arallel with the delta-connected load having imedance of ( 7 + j 8) er hase (Fig. 9.a), with both the loads being balanced, and fed from a three-hase, 30, balanced suly, with the hase sequence as -Y-B. Find the line current, ower factor, total ower & reactive A, and also total volt-ameres (A). Y Y = (7-j8) N = (-j6) B BB (a) ersion EE T, Kharagur

6 ' = 3. Y B Y ' B YB (b) ' BN ( B ) BN YN N YN ( Y ) YB YB N( ) (c) Fig. 9. (a) Circuit diagram (Examle 9.) (b) Equivalent circuit (delta-connected) (c) Phasor diagram ersion EE T, Kharagur

7 Solution For the balanced star-connected load, the imedance er hase is, = ( j6) = The above load is converted into its equivalent delta. The imedance er hase is, = 3 = 3 ( j6) = (36 j 48) = For the balanced delta-connected load, the imedance er hase is, = (7 + j8) = n the equivalent circuit for the load (Fig. 9.b), the two imedances, & are in arallel. So, the total admittance er hase is, Y = Y + Y = + = = = [( 0.0+ j ) + ( j )] = ( j ) = The total imedance er hase is, = / Y = /( ) = = ( j.98) The hasor diagram is shown in Fig. 9.c. Taking the line voltage, as reference, = 30 0 The other two line voltages are, YB = 30 0 ; = For the equivalent delta-connected load, the line and hase voltages are same. So, the hase current, is, = = = = (8.98 j 0.865) A The two other hase currents are, = ; = YB 8 The magnitude of the line current is 3 times the magnitude of the hase current. So, the line current is = 3 = = A The line current, lags the corresonding hase current, by 30. So, the line current, is = A The other two line currents are, Y = ; B = Also, the hase angle of the total imedance is ositive. So, the ower factor is cos φ = cos 6.04 = lag The total volt-ameres is S = 3 = = ka The total A is also obtained as S = 3 = = ka The total ower is P = 3 cos φ = = kw The total reactive volt-ameres is, Q = 3 sin φ = sin 6.04 = 597. A 5 ersion EE T, Kharagur

8 An alternative method, by converting the delta-connected art into its equivalent star is given, as shown earlier in Ex. 8.. For the balanced star-connected load, the imedance er hase is, = ( j6) = For the balanced delta-connected load, the imedance er hase is, = (7 + j8) = Converting the above load into its equivalent star, the imedance er hase is, = / 3 = (7 + j8) / 3 = (9 + j 6) = n the equivalent circuit for the load, the two imedances, & are in arallel. So, the total admittance er hase is, Y = Y + Y = + = = = [( j 0.04) + ( j 0.058)] = (0.069 j 0.08) = The total imedance er hase is, = / Y = /( ) = = (95+ j 0.976) The hasor diagram is shown in Fig. 8.5c. The magnitude of the hase voltage is, = = / 3 = 30 / 3 = 3. N 8 The line voltage, is taken as reference as given earlier. The corresonding hase voltage, lags by. So, the hase voltage, is = N The hase current, N is, 30 N N = = = N As the total load is taken as star-connected, the line and hase currents are same, in this case. The hase angle of the total imedance is ositive, with is value as φ = The ower factor is cos = lag The total volt-ameres is S = 3 = = ka The remaining stes are not given, as they are same as shown earlier. A N ersion EE T, Kharagur

9 Examle 9. A balanced delta-connected load with imedance er hase of ( 6 j) shown in Fig. 9.3a, is fed from a three-hase, 00 balanced suly with hase sequence as A- B-C. Find the voltages, ab, bc & ca, and show that they (voltages) are balanced. A B X C = C a = 6 A C b X B X c C (a) a 60 A B c 60 Φ Fig. 9.3 (a) Circuit diagram (Examle 9.) (b) Phasor diagram C (b b ersion EE T, Kharagur

10 Solution = ; X = AB 6 C = = j X = 6 j = C For delta-connected load, = = 00 Taking the line or hase voltage AB as reference, the line or hase voltages are, AB = 00 0 ; BC = 00 0 ; CA = The hasor diagram is shown in Fig. 9.3b. The hase current, AB is, = / = = = (8.0 + j 6.0) AB AB ( ) ( ) A The other two hase currents are, = = (.96 j9.98) A BC CA = = ( j 3.98) A The voltage, ab ab ab Bb is, = + = ( j X ) C AB + BC = ( 0) ( ) + (6 0) 83.3 = = ( 7.0 j 96.0) + (9.4 j58.85) = (9.4 j 54.85) = Alternatively, ab = ( j) (8.0 + j 6.0) + 6 (.96 j 9.98) = (9.4 j 54.85) = Similarly, the voltage, is, bc bc Cc bc = + = ( j X ) C BC + CA = ( 0) ( ) + (6 0) = = ( j 4.35) + ( j 6.85) = ( j 48.5) = n the same way, the voltage, ca is obtained as ca = The stes are not shown here. The three voltages, as comuted, are equal in magnitude, and also at hase difference of 0 with each other in sequence. So, the three voltages can be termed as balanced ones. A simle examle (0.3) of a balanced delta-connected load is given in the following lesson The hase and line currents for a delta-connected balanced load, fed from a threehase suly, along with the total ower consumed, are discussed in this lesson. Also some worked out roblems (examles) are resented. n the next lesson, the measurement of ower in three-hase circuits, both balanced and unbalanced, will be described. ersion EE T, Kharagur

11 Problems 9. A balanced load of (9-j6) er hase, connected in delta, is fed from a three hase, 00 suly. Find the line current, ower factor, total ower, reactive A and total A. 9. Three star-connected imedances, = (8-jb) er hase, are connected in arallel with three delta-connected imedances, = (30+j5) er hase, across a three-hase 30 suly. Find the line current, total ower factor, total ower, reactive A, and total A. ersion EE T, Kharagur

12 ist of Figures Fig. 9. (a) Balanced delta-connected load fed from a three-hase balanced suly (b) Phasor diagram Fig. 9. (a) Circuit diagram (Examle 9.) (b) Equivalent circuit (delta-connected) (c) Phasor diagram Fig. 9.3 (a) Circuit diagram (Examle 9.) (b) Phasor diagram ersion EE T, Kharagur