C h a p t e r 3 71 CHATE 3 THEE HASE SSTEM (3 Φ ) 3.0 NTODUCTON This chapter is explaining about the three phase system. The learning outcome is student should be able to apply the principles of three phase systems, balanced load, star and delta types of connection to solve electrical circuit correctly. 3.1 THEE HASE SSTEM (3 Φ ) Three-phase system is a common method of alternating current electric power generation, transmission, and distribution. t is a type of poly phase system which has two or more voltage source with fixed phase angle difference. Three phase system is a system that has three source voltage of the same magnitude produced by the three separate coil conductor which space 120 o apart as shown in Figure 3.1. Normally a three phase supply has three (3) live wire or phase conductor of ed phase (), ellow phase () and lue phase (). North ole The emf produced by each coils; 120 o O o 240 o South ole Figure 3.1 : Three hase Conductor Coil coil, E m sin 0 zero, coil, E m sin 120 0 0.866Em volt and coil, E m sin 240 0 equal -0.866 Em volt. Three hase System
C h a p t e r 3 72 The coils were rotating at a constant speed to generate three emf waveform with same peak and frequency as in Figure 3.2. Emf E m E E E 0 O 120 o 240 o 360 O θ Figure 3.2 : Waveform of 3 Φ System Equations for each waveforms ; ed ine oltage, E sinθ. E m o ellow ine oltage, E sin( 120 ) θ (3.1) E m o lue ine oltage, E sin( θ 240 ) E m The above equations can be represent in a form phase as below ; E E E Em o 0 2 Em o 120 2 Em 240 2 o Where the value of E's, E s and E s is a value of E rms ( root mean square). (3.2) Three hase System
C h a p t e r 3 73 3.2 DFFEENT ETWEEN SNGE HASE AND THEE HASE SSTEM Table 3.1: Difference etween Single hase and Three hase Systems SSTEM 1 Φ SSTEM 3 Φ Definition: System with 2 cable connecting from source to load: ife cable () and neutral cable (N). Definition: System with 3 live cabels from supply to load : ed phase(), ellow phase () and lue phase (). Connection : ive Connection: 240 beban 240 415 Neutral N Waveform : Waveform: 240 360 o 0 180 θ 415 0 360 120 o 240 0 θ Three phase system basically used in starting the 3phase machines, supply power to large factories and for transmission and distribution 3 phase voltage. The advantages of 3 phase over single phase systems are: i. More economical due to less construction and maintenance costs. The conductor used is smaller, thus the needs of copper can be reduced Three hase System
C h a p t e r 3 74 ii. Easy to install due to smaller and lighter cable used. Therefore the smaller structures supporters can be used and the distance between the supporters can be relatively long. iii. The characteristics of 3phase equipment have a better operating starting due to its stable and fixed power. iv. Electrical energy supplied is higher than single phase. v. Most large motors used for induction motor 3phase can stand alone without the additional equipment such as capacitors, centrifugal switches or other additional circuits vi. The output roduction and equipment quality of work for 3phase system is better than single phase because of the higher efficiency and power factor. 3.3 THEE HASE OAD CONNECTON The load for three phase may be joined together either in star or delta connection. The load in each phase is usually in balanced which means that the load/impedance at each phase are the same. There are 2 methods of load connection in three phase system: 1. Star or Wai connection 2. Delta connection 3.3.1 Star Connection Star Connection, also known as Wai () connection. This type of connection has fourconductor wires where three for life line (ed(), ellow() and lue()) and one for neutral(n) line as shown in Figure 3.3. Three hase System
C h a p t e r 3 75 (ed ine) oad N (Neutral ine) oad oad (ellow ine) (lue ine) Figure 3.3: STA Connection i. oltage ine voltage ( ) is the voltage measured between two life lines. While, the phase voltage ( ) is the voltage measured between the life line with the neutral line as shown in Figure 3.4. A E N N E E N E N C Figure 3.4: oltage in STA connection Three hase System
C h a p t e r 3 76 E N, E N and E N are called phase voltage ( ). hase difference between phase voltages is equal to the 120. The mathematical relationship between the line voltage to phase voltage is as equation 3.3 below; ine oltage 3 hase oltage 3 ( 3.3 ) ii. Current ine current ( ) is the current that flow in the line, while the phase currents ( ) are defined as current which flow in the phase. (ed line current) (ed phase current) N N (Neutral line current) (lue phase current) (ellow line current) (ellow phase current) (lue line current) Figure 3.5 : Current in STA connection efer to Figure 3.5 above, phase current ( ) which are, and is equal to the line current ( ) which are, and. n the form of mathematical equations, the relationship between line currents and phase currents can be written as equation 3.4. Three hase System
C h a p t e r 3 77 ine current hase current (3.4) Example 3.1 Three uniform impedance each resistance10 Ω and inductance 0.019H, supply 415, 50Hz. Calculate the line current, phase current, phase voltage and line voltage if the three mpedance connected in star. Solution 3.1 ine voltage, 415 (Supply voltage 3 Φ ine voltage) 415 hase voltage, 239. 6 (star connection) 3 3 mpedance, 2 2 X Z +... equation. 1 Where, X 2πf 2π (50)(0.019) 5.97 Ω. Z 2 2 10 + 5.97 11. 65Ω hase current, Z p 239.6 11.65 20.57 A ine current, 20.57 A (star connection) 3.3.2 Delta Connection n the delta ( ) connection, there are only three (3) conductor wire-line, ie red red line(), yellow line () and the blue line (). oint of A, C and AC are the points in the phase. Three hase System
C h a p t e r 3 78 ed line A oad oad ellow line oad C lue line Figure 3.6: DETA Connection i. Current 1 2 3 Figure 3.7: Current in DETA Connection Three hase System
C h a p t e r 3 79 From Figure 3.7 above, current 1, 2 and 3 are the currents of each phase and known as phase current ( ). Meanwhile,, and are the current per line and known as line current ( ). Since the loads are in balance, the current flow through each phase and each line is the same as equation 3.5 and by 120 out of phase with each other. 1 2 3 (3.5) n delta connection, the relationship between line currents and phase currents can be written in the form of mathematical expressions such as equation 3.6. ine Current 3 hase Current 3 (3.6) ii. oltage Figure 3.8: oltage in DETA Connection Three hase System
C h a p t e r 3 80 n the delta connection, the line voltage and phase voltage is the same for both measured between two (2) live wires and it can be expressed in equations 3.7 below. ine oltage hase oltage (3.7) Example 3.2 Three uniform impedance each resistance10 Ω and inductance 0.019H, the supply 415, 50Hz. Calculate the line current, phase current, phase voltage and line voltage if the three impedances connected in delta. Solution 3.2 ine voltage, 415 (Supply voltage 3 Φ ine voltage) hase oltage, 415 (delta connection) mpedance, 2 2 X Z + where, X 2πf 2π (50)(0.019) 5.97 Ω. Z 2 2 10 + 5.97 11. 65Ω hase current, Z p 415 11.65 35.9 A ine current, 3 3 (35.9) 62.18 A (delta connection) 3.4 THEE HASE OWE Three-phase power system is equal to the power that lies within the power of one-phase is the apparent power, the active or real power and reactive power. Three hase System
C h a p t e r 3 81 3.4.1 Apperent power Apparent power is a measure of alternating current power that is computed by multiplying the root-mean-square current by the root-mean-square voltage. The symbol for apparent power is S and the unit is volt-ampere (A) Apperent power in every phase : S (3.8) Total apparent power in the system 3 Φ : S3 Φ 3xS 3 3 (3.9) 3.4.2 eal ower eal power also called the average power or active power. The symbol for real power is and the unit is watt (W). eal power per phase: cosθ (3.10) Total active power in three-phase system: 3 Φ 3x 3 cosθ From equation 3.7 for DETA connection and 3 Three hase System
C h a p t e r 3 82 3 3 3 x x cos 3 Φ θ 3 cosθ 3Φ 3 cosθ (3.11) This is the same formula for the STA connection. 3.4.3 eactive ower eactive power is also called the power of imagination. The symbol for reactive power is and the unit is volt-ampere reactive (A) eactive power per phase: Q sinθ (3.12) Total reactive power in three-phase system: Q3 Φ 3xQ 3 sinθ (3.13) From equation 3.7 for DETA connection and Q 3 sinθ 3 x x cos 3 3Φ θ 3 sinθ 3 Three hase System
C h a p t e r 3 83 Q3Φ 3 sinθ (3.14) This is also the same formula for the STA connection. 3.4.4 ower Triangle diagram All of three power described above can be illustrated using a triangle diagram as shown in Figure 3.9. This triangle is called power triangle. Q S θ Figure 3.9: ower Triangle diagram Example 3.3 ased on Example 3.2, calculate the total apparent power, active power and reactive power Solution 3.3 i). Total apparent power, S3 Φ 3 3 (415)(62.18) 44.7 ka ii). Total active power, 3Φ 3 cosθ 3(415)(62.18)( ) Z 10 44695.1( ) 38.4 kw 11.65 Three hase System
C h a p t e r 3 84 iii). Total reactive power, Q 3Φ 3 sinθ 3(415)(62.18) sin 30.9 o 22.9 ka EFEENCE akshi, U.A and Godse, A., (2007) asic Electrical and Electronics Engineering 1 st Technical ublication une, ndia Edition, OEMS 1. Three balanced load with 10Ω resistance and 0.5H inductor, stars connected supplied with three phase 500v, 50Hz. Calculate: i. ine current ii. Apparent ower iii. eal ower (1.83 < -95.9 A, 1 589 A, 102.2 W) 2. Three impedances of 30Ω resistance and 50Ω inductance reactance supply with 415, 50Hz. The impedances is connected in star connection. Calculate: i. ine voltage ii. hase voltage iii. hase current iv. ine current v. ower factor vi. Apparent ower (415< 0, 239.6< 0, 4.1 < -65.6 A, 4.1 < -65.6 A, 0.514, 2 947 A) 3. y referring to question 2, if the three impedances connected in delta. Calculate (i), (ii), (iii), (iv), (v) and (vi). (415< 0, 415< 0, 7.11 < -65.6 A, 12.32 < -65.6 A, 0.514, 8 855 A) 4. Three balanced load, 10Ω inductance reactance and 12Ω resistance of impedance with star connected was supplied with 400, 50Hz. Calculate: i. ine current ii. ower Factor iii. Apparent power in ka iv. eal ower in kw (14.78 < -44.23 A, 0.768, 10.23 ka, 8.05 kw) Three hase System
C h a p t e r 3 85 5. Calculate the values of (i), (ii), (iii) and (iv) in question 4 for delta connection. (44.35 < -44.23 A, 0.768, 30.73 ka, 23.59 kw) 6. Three balanced load connected in delta with three-phase supply 415, 50Hz. The line current is 15A and 0.8 lagging power factor, calculate: i. hase current, phase voltage and line voltage ii. ower in watt (8.66 < 40.96 A, 415< 0, 415< 0, 8 625W) 7. Three balanced load with 0.15H inductance and 15Ω resistance connected in delta, supply with three-phase 415, 50Hz. Calculate: i. ine oltage ii. hase Current iii. ine Current iv. ower Factor v. eactive power (415 < 0, 8.39 < 80.37 A, 14.54 < 80.37 A, 0.303, 10 451 A) 8. Three balanced load resistance 15Ω, 0.8H inductor and 10µF capacitors for each phase. f the supply voltage three-phase 415, 50Hz is connected in a star. Calculate: i. hase Current ii. ine Current iii. ower Factor iv. eal power (3.49 < -85.97 A, 3.49 < -85.97 A, 0.218, 546.8 W) 9. Three balanced load connected in delta to a three-phase supply, using a total power of 5 kw 415 and take a line current of 12A, specify: i. power factor ii. mpedance of each load iii. iv. Apparent power and reactive power eal power if the loads are star connected. (0.579, 59.88 < 53.34 Ω, 8 625 A, 6 919 A, 5 000W) 10. Three balanced load connected in delta to a three-phase 415 power supply using the power of 5750w and the line current of 10A. i. Draw the load connection ii. Calculate power factor iii. Calculate mpedance of each load (0.667, 71.92 < 53.54 Ω) 11. Three balanced load are connected in delta to a three-phase power supply 400, 50Hz. The line current is 13A and 0.8 power factor leading. Calculate: i. phase current ii. mpedance iii. nductance (7.5 < -40.96 A, 53.33 < 40.96 Ω, 0.1 H) Three hase System
C h a p t e r 3 86 12. Three balance loads with 12Ω resistance and 9Ω inductance reactance, connected in delta to three phase 440v supply. Calculate: i. ine current ii. hase voltage iii. ower factor iv. ka used v. kw used (29.33 < -40.96 A, 440 < 0, 0.8, 12.445 ka, 9.956 kw) 13. Calculate (i), (ii), (iii), (iv) and (v) in question 12 by using star connection. (16.93 < -40.96 A, 254 < 0, 0.8, 12.9 ka, 10.32 kw) Three hase System