AC Distribution PDF

CONTENTS CONTENTS 356 Principles of Power System CHAPTER " A.C. Distribution Introduction Introduction...

CONTENTS CONTENTS 356 Principles of Power System CHAPTER " A.C. Distribution Introduction Introduction I n the beginning of electrical age, electricity was generated, transmitted and distributed as direct current. The principal disadvantage of d.c. system was that voltage level could not readily be changed, except by the use of rotating machinery, which in most cases was too expen- 14.1 A.C. Distribution Calculations sive. With the development of transformer by 14.2 Methods of Solving A.C. Distribution George Westinghouse, a.c. system has become Problems so predominant as to make d.c. system practically extinct in most parts of the world. The present 14.3 3-Phase Unbalanced Loads day large power system has been possible only due to the adoption of a.c. system. 14.4 Four-Wire Star-Connected Unbal- anced Loads Now-a-days, electrical energy is generated, transmitted and distributed in the form of alter- 14.5 Ground Detectors nating current as an economical proposition. The electrical energy produced at the power station is transmitted at very high voltages by 3-phase, 3- wire system to step-down sub-stations for distri- bution. The distribution system consists of two parts viz. primary distribution and secondary dis- tribution. The primary distribution circuit is 3- phase, 3-wire and operates at voltages (3·3 or 6·6 or 11kV) somewhat higher than general utilisation levels. It delivers power to the secondary distri- bution circuit through distribution transformers 356 CONTENTS CONTENTS A.C. Distribution 357 situated near consumers’ localities. Each distribution transformer steps down the voltage to 400 V and power is distributed to ultimate consumers’ by 400/230 V, 3-phase, 4-wire system. In this chap- ter, we shall focus our attention on the various aspects of a.c. distribution. 14.1 A.C. Distribution Calculations A.C. distribution calculations differ from those of d.c. distribution in the following respects : (i) In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system, the voltage drops are due to the combined effects of resistance, inductance and capacitance. (ii) In a d.c. system, additions and subtractions of currents or voltages are done arithmetically but in case of a.c. system, these operations are done vectorially. (iii) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the distributor are generally at different power factors. There are two ways of referring power factor viz (a) It may be referred to supply or receiving end voltage which is regarded as the reference vector. (b) It may be referred to the voltage at the load point itself. There are several ways of solving a.c. distribution problems. However, symbolic notation method has been found to be most convenient for this purpose. In this method, voltages, currents and imped- ances are expressed in complex notation and the calculations are made exactly as in d.c. distribution. 14.2 Methods of Solving A.C. Distribution Pr oblems Problems In a.c. distribution calculations, power factors of various load currents have to be considered since currents in different sections of the distributor will be the vector sum of load currents and not the arithmetic sum. The power factors of load currents may be given (i) w.r.t. receiving or sending end voltage or (ii) w.r.t. to load voltage itself. Each case shall be discussed separately. (i) Power factors referred to receiving end voltage. Consider an a.c. distributor A B with con- centrated loads of I1 and I2 tapped off at points C and B as shown in Fig. 14.1. Taking the receiving end voltage V B as the reference vector, let lagging power factors at C and B be cos φ1 and cos φ2 w.r.t. V B. Let R 1, X 1 and R 2, X 2 be the resistance and reactance of sections A C and CB of the distributor. Impedance of section A C, ZAC = R 1 + j X1 Impedance of section CB, ZCB = R 2 + j X2 Load current at point C, I1 = I1 (cos φ1 − j sin φ1) Load current at point B, I2 = I2 (cos φ2 − j sin φ2) Current in section CB, I CB = I2 = I2 (cos φ2 − j sin φ2) Current in section A C, I AC = I1 + I2 = I1 (cos φ1 − j sin φ1) + I2 (cos φ2 − j sin φ2) Voltage drop in section CB, VCB = I CB ZCB = I2 (cos φ2 − j sin φ2) (R 2 + j X 2) KKKKH KKKKH KH KKH Voltage drop in section A C, VAC ( ) = I AC Z AC = I1 + I 2 Z AC 358 Principles of Power System = [I1(cos φ1 − j sin φ1) + I2 (cos φ2 − j sin φ2)] [R 1 + j X1] Sending end voltage, VA = VB + VCB + VAC Sending end current, I A = I1 + I2 The vector diagram of the a.c. distributor under these conditions is shown in Fig. 14.2. Here, the receiving end voltage V B is taken as the reference vector. As power factors of loads are given w.r.t. V B, therefore, I1 and I2 lag behind V B by φ1 and φ2 respectively. (ii) Power factors referred to respective load voltages. Suppose the power factors of loads in the previous Fig. 14.1 are referred to their respective load voltages. Then φ1 is the phase angle between V C and I1 and φ2 is the phase angle between V B and I2. The vector diagram under these conditions is shown in Fig. 14.3. Voltage drop in section CB = I 2 ZCB = I2 (cos φ2 − j sin φ2) (R 2 + j X2) Voltage at point C = VB + Drop in section CB = V C ∠ α (say) Now I1 = I1 ∠ − φ1 w.r.t. voltage V C ∴ I1 = I1 ∠ − (φ1 − α) w.r.t. voltage V B i.e. I1 = I1 [cos (φ1 − α) − j sin (φ1 − α)] Now I AC = I1 + I2 A.C. Distribution 359 = I1 [cos (φ1 − α) − j sin (φ1 − α)] + I2 (cos φ2 − j sin φ2) Voltage drop in section AC = I AC Z AC ∴ Voltage at point A = VB + Drop in CB + Drop in AC Example 14.1. A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under : (i) 100 A at 0·707 p.f. lagging 200 m from point A (ii) 200 A at 0·8 p.f. lagging 300 m from point A The load resistance and reactance of the distributor is 0·2 Ω and 0·1 Ω per kilometre. Calculate the total voltage drop in the distributor. The load power factors refer to the voltage at the far end. Solution. Fig. 14.4 shows the single line diagram of the distributor. Impedance of distributor/km = (0·2 + j 0·1) Ω Impedance of section AC, Z AC = (0·2 + j 0·1) × 200/1000 = (0·04 + j 0·02) Ω Impedance of section CB, ZCB = (0·2 + j 0·1) × 100/1000 = (0·02 + j 0·01) Ω Taking voltage at the far end B as the reference vector, we have, Load current at point B, I2 = I2 (cos φ2 − j sin φ2) = 200 (0·8 − j 0·6) = (160 − j 120) A Load current at point C, I1 = I1 (cos φ1 − j sin φ1) = 100 (0·707 − j 0·707) = (70·7− j 70·7) A Current in section CB, I CB = I2 = (160 − j 120) A Current in section AC, I AC = I1 + I 2 = (70·7 − j 70·7) + (160 − j 120) = (230·7 − j 190·7) A Voltage drop in section CB, VCB = I CB Z CB = (160 − j 120) (0·02 + j 0·01) = (4·4 − j 0·8) volts Voltage drop in section AC, VAC = I AC Z AC = (230·7 − j 190·7) (0·04 + j 0·02) = (13·04 − j 3·01) volts Voltage drop in the distributor = VAC + VCB = (13·04 − j 3·01) + (4·4 − j 0·8) = (17·44 − j 3·81) volts Magnitude of drop a f a f = 17 ⋅ 44 + 3 ⋅ 81 = 17·85 V 2 2 Example 14.2. A single phase distributor 2 kilometres long supplies a load of 120 A at 0·8 p.f. lagging at its far end and a load of 80 A at 0·9 p.f. lagging at its mid-point. Both power factors are 360 Principles of Power System referred to the voltage at the far end. The resistance and reactance per km (go and return) are 0·05 Ω and 0·1 Ω respectively. If the voltage at the far end is maintained at 230 V, calculate : (i) voltage at the sending end (ii) phase angle between voltages at the two ends. Solution. Fig. 14.5 shows the distributor AB with C as the mid-point Impedance of distributor/km = (0·05 + j 0·1) Ω Impedance of section AC, Z AC = (0·05 + j 0·1) × 1000/1000 = (0·05 + j 0·1) Ω Impedance of section CB, Z CB = (0·05 + j 0·1) × 1000/1000 = (0·05 + j 0·1) Ω Let the voltage VB at point B be taken as the reference vector. Then, VB = 230 + j 0 (i) Load current at point B, I2 = 120 (0·8 − j 0·6) = 96 − j 72 Load current at point C, I1 = 80 (0·9 − j 0·436) = 72 − j 34·88 Current in section CB, ICB = I2 = 96 − j 72 Current in section AC, I AC = I1 + I2 = (72 − j 34·88) + (96 − j 72) = 168 − j 106·88 Drop in section CB, VCB = I CB Z CB = (96 − j 72) (0·05 + j 0·1) = 12 + j 6 Drop in section AC, VAC = I AC Z AC = (168 − j 106·88) (0·05 + j 0·1) = 19·08 + j 11·45 ∴ Sending end voltage, VA = VB + VCB + VAC = (230 + j 0) + (12 + j 6) + (19.08 + j 11.45) = 261.08 + j 17.45 Its magnitude is = a261⋅ 08f + a17 ⋅ 45f 2 2 = 261·67 V (ii) The phase difference θ between VA and VB is given by : 17 ⋅ 45 tan θ = = 0·0668 261⋅ 08 −1 ∴ θ = tan 0·0668 = 3·82 o Example 14.3. A single phase distributor one km long has resistance and reactance per con- ductor of 0·1 Ω and 0·15 Ω respectively. At the far end, the voltage VB = 200 V and the current is 100 A at a p.f. of 0·8 lagging. At the mid-point M of the distributor, a current of 100 A is tapped at a p.f. A.C. Distribution 361 of 0·6 lagging with reference to the voltage VM at the mid-point. Calculate : (i) voltage at mid-point (ii) sending end voltage VA (iii) phase angle between VA and VB Solution. Fig. 14.6 shows the single line diagram of the distributor AB with M as the mid-point. Total impedance of distributor = 2(0·1 + j 0·15) = (0·2 + j 0·3) Ω Impedance of section AM, Z AM = (0·1 + j 0·15) Ω Impedance of section MB, Z MB = (0·1 + j 0·15) Ω Let the voltage VB at point B be taken as the reference vector. Then, VB = 200 + j 0 (i) Load current at point B, I2 = 100 (0·8 − j 0·6) = 80 − j 60 Current in section MB, I MB = I2 = 80 − j 60 Drop in section MB, VMB = I MB Z MB = (80 − j 60) (0·1 + j 0·15) = 17 + j 6 ∴ Voltage at point M, VM = VB + VMB = (200 + j 0) + (17 + j 6) = 217 + j 6 Its magnitude is = a217f + a6f 2 2 = 217·1 V −1 −1 Phase angle between VM and VB, α = tan 6/217 = tan 0·0276 = 1·58 o (ii) The load current I1 has a lagging p.f. of 0·6 w.r.t. VM. It lags behind VM by an angle −1 φ1 = cos 0·6 = 53·13 o ∴ Phase angle between I1 and VB, φ′1 = φ1 − α = 53·13 − 1·58 = 51·55 o o Load current at M, I1 = I1 (cos φ1′ − j sin φ1′) = 100 (cos 51·55º − j sin 51·55º) = 62·2 − j 78·3 Current in section AM, I AM = I1 + I2 = (62·2 − j 78·3) + (80 − j 60) = 142·2 − j 138·3








 Drop in section AM, VAM = I AM Z AM = (142·2 − j 138·3) (0·1 + j 0·15) = 34·96 + j 7·5 Sending end voltage, VA = VM + VAM = (217 + j 6) + (34·96 + j 7·5) 362 Principles of Power System = 251·96 + j 13·5 Its magnitude is = a251⋅ 96f + a13 ⋅ 5f 2 2 = 252·32 V (iii) The phase difference θ between VA and VB is given by : tan θ = 13·5/251·96 = 0·05358 −1 ∴ θ = tan 0·05358 = 3·07 o Hence supply voltage is 252·32 V and leads VB by 3·07º. Example 14.4. A single phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f. lagging respectively ; both expressed with reference to the voltage at A. The total impedance of the three sections AB, BC and CA are (1 + j 1), (1+ j2) and (1 + j3) ohms respectively. Find the total current fed at A and the current in each section. Use Thevenin’s theorem to obtain the results. Solution. Fig. 14.7 (i) shows the ring distributor ABC. Thevenin’s theorem will be used to solve this problem. First, let us find the current in BC. For this purpose, imagine that section BC is removed as shown in Fig. 14.7 (ii). Referring to Fig.14.7 (ii), we have, Current in section AB = 20 (0·8 − j 0.6) = 16 − j 12 Current in section AC = 15 (0·6 − j 0·8) = 9 − j 12 Voltage drop in section AB = (16 − j 12) (1 + j1) = 28 + j 4 Voltage drop in section AC = (9 − j 12) (1 + j 3) = 45 + j 15 Obviously, point B is at higher potential than point C. The p.d. between B and C is Thevenin’s equivalent circuit e.m.f. E0 i.e. Thevenin’s equivalent circuit e.m.f., E0 = p.d. between B and C = (45 + j 15) − (28 + j 4) = 17 + j 11 Thevenin’s equivalent impedance Zo can be found by looking into the network from points B and C. Obviously, Z0 = (1 + j1) + (1 + j 3) = 2 + j4 E0 ∴ Current in BC = Z 0 + Impedance of BC 17 + j11 17 + j11 = = b + 2 j4 + g b+ 1 j2 g 3 + j6 = 2·6 − j 1·53 = 3∠− ∠−30·48º A ∠− Current in AB = (16 − j 12) + (2·6 − j 1·53) A.C. Distribution 363 ∠− = 18·6 − j 13·53 = 23∠− ∠−36·03 A o Current in AC = (9 − j 12) − (2·6 − j 1·53) ∠− = 6·4 − j 10·47 = 12·27∠− ∠−58·56 A o Current fed at A = (16 − j 12) + (9 − j 12) = 25 − j 24 = 34·65∠− − 43·83o A Example 14.5. A 3-phase, 400V distributor AB is loaded as shown in Fig.14.8. The 3-phase load at point C takes 5A per phase at a p.f. of 0·8 lagging. At point B, a 3-phase, 400 V induction motor is connected which has an output of 10 H.P. with an efficiency of 90% and p.f. 0·85 lagging. If voltage at point B is to be maintained at 400 V, what should be the voltage at point A ? The resistance and reactance of the line are 1Ω and 0·5Ω per phase per kilometre respectively. Solution. It is convenient to consider one phase only. Fig.14.8 shows the single line diagram of the distributor. Impedance of the distributor per phase per kilometre = (1 + j 0·5) Ω. Impedance of section AC, Z AC = (1 + j 0·5) × 600/1000 = (0·6 + j 0·3) Ω Impedance of section CB, ZCB = (1 + j 0·5) × 400/1000 = (0·4 + j 0·2) Ω Phase voltage at point B, VB = 400/ 3 = 231 V Let the voltage VB at point B be taken as the reference vector. Then, VB = 231 + j 0 H. P. × 746 Line current at B = 3 × line voltage × p. f. × efficiency 10 × 746 = = 14·08 A 3 × 400 × 0 ⋅ 85 × 0 ⋅ 9 ∴ *Current/phase at B, I2 = 14·08 A Load current at B, I2 = 14·08 (0·85 − j 0·527) = 12 − j 7·4 Load current at C, I1 = 5 (0·8 – j 0·6) = 4 − j 3


 Current in section AC, I AC = I1 + I2 = (4 − j 3) + (12 − j 7·4)


 = 16 − j 10·4 Current in section CB, I CB = I2 = 12 − j 7·4 Voltage drop in CB, V CB = I CB Z CB = (12 − j 7·4) (0·4 + j 0·2) = 6·28 − j 0·56 Voltage drop in AC, V AC = I AC Z AC = (16 − j 10·4) (0·6 + j 0·3) = 12·72 − j 1·44 * In a 3-phase system, if the type of connection is not mentioned, then star connection is understood. 364 Principles of Power System Voltage at A per phase, V A = V B + V CB + V AC = (231 + j 0) + (6·28 − j 0·56) + (12·72 − j 1·44) = 250 − j 2 Magnitude of VA/phase = a250f + a2f 2 2 = 250 V ∴ Line voltage at A = 3 × 250 = 433 V Example 14.6. A 3-phase ring main ABCD fed at A at 11 kV supplies balanced loads of 50 A at 0.8 p.f. lagging at B, 120 A at unity p.f. at C and 70 A at 0·866 lagging at D, the load currents being referred to the supply voltage at A. The impedances of the various sections are : Section AB = (1 + j 0·6) Ω ; Section BC = (1·2 + j 0·9) Ω Section CD = (0·8 + j 0·5) Ω ; Section DA = (3 + j 2) Ω Calculate the currents in various sections and station bus-bar voltages at B, C and D. Solution. Fig.14.9 shows one phase of the ring main. The problem will be solved by Kirchhoff’s laws. Let current in section AB be (x + j y). ∴ Current in section BC, IBC = (x + j y) − 50 (0·8 − j 0·6) = (x − 40) + j (y + 30) Current in section CD, ICD = [(x − 40) + j (y + 30)] − [120 + j 0] = (x − 160) + j (y + 30) Current in section DA, I DA = [(x − 160) + j (y + 30)] − [70 (0·866 − j 0·5)] = (x − 220·6) + j (y + 65) Drop in section AB = I AB Z AB = (x + jy) (1 + j0·6) = (x − 0·6y) + j (0·6x + y) Drop in section BC = I BC Z BC = [(x − 40) + j (y + 30)] [(1·2 + j 0·9)] = (1·2x − 0·9 y − 75) + j (0·9x + 1·2 y) A.C. Distribution 365 Drop in section CD = I CD ZCD = [(x − 160) + j (y + 30)] [(0·8 + j 0·5)] = (0·8x − 0·5y − 143) + j (0·5x + 0·8y − 56) Drop in section DA = I DA Z DA = [(x − 220·6) + j (y + 65)] [(3 + j 2)] = (3x − 2y − 791·8) + j (2x + 3y − 246·2) Applying Kirchhoff’s voltage law to mesh ABCDA, we have, Drop in AB + Drop in BC + Drop in CD + Drop in DA = 0 or [(x − 0·6y) + j (0·6x + y)] + [(1·2x − 0·9y − 75) + j (0·9x + 1·2y)] + [(0·8x − 0·5y − 143) + j (0·5x + 0·8y − 56)] + [(3x − 2y − 791·8) + j (2x + 3y − 246·2)] = 0 or (6x − 4y − 1009·8) + j (4x + 6y − 302·2) = 0 As the real (or active) and imaginary (or reactive) parts have to be separately zero, ∴ 6x − 4y − 1009·8 = 0 and 4x + 6y − 302·2 = 0 Solving for x and y, we have, x = 139·7 A ; y = − 42·8 A Current in section AB = (139·7 − j 42·8) A Current in section BC = (x − 40) + j (y + 30) = (139·7 − 40) + j (− 42·8 + 30) = (99·7 − j 12·8) A Current in section CD = (x − 160) + j (y + 30) = (139·7 − 160) + j (− 42·8 + 30) −20·3 − j 12·8) A = (− Current in section DA = (x − 220·6) + j (y + 65) = (139·7 − 220·6) + j (− 42·8 + 65) −80·9 + j 22·2) A = (− Voltage at supply end A, VA = 11000/ 3 = 6351 V/phase ∴ Voltage at station B, VB = VA − I AB Z AB = (6351 + j 0) − (139·7 − j 42·8) (1 + j 0·6) = (6185·62 − j 41·02) volts/phase Voltage at station C, VC = VB − I BC Z BC = (6185·62 − j 41·02) − (99·7 − j 12·8) (1·2 + j 0·9) = (6054·46 − j 115·39) volts/phase Voltage at station D, VD = V C − I CD ZCD = (6054·46 − j 115·39) − (−20·3 − j 12·8) × (0·8 + j 0·5) = (6064·3 − j 95) volts/phase 366 Principles of Power System TUTORIAL PROBLEMS 1. A single phase distributor AB has a total impedance of (0·1 + j 0·2) ohm. At the far end B, a current of 80 A at 0·8 p.f. lagging and at mid-point C a current of 100 A at 0·6 p.f. lagging are tapped. If the voltage of the far end is maintained at 200 V, determine : (i) Supply end voltage VA (ii) Phase angle between VA and VB The load power factors are w.r.t. the voltage at the far end. [(i) 227·22 V (ii) 2o31′′ ] 2. A single-phase a.c. distributor AB is fed from end A and has a total impedance of (0·2 + j 03) ohm. At the far end, the voltage VB = 240 V and the current is 100 A at a p.f. of 0·8 lagging. At the mid-point M, a current of 100 A is tapped at a p.f. of 0·6 lagging with reference to the voltage VM at the mid-point. o Calculate the supply voltage VA and phase angle between VA and VB. [292 V, 2·6 ] 3. A single phase ring distributor ABC is fed at A. The loads at B and C are 40 A at 0·8 p.f. lagging and 60 A at 0·6 p.f. lagging respectively. Both power factors expressed are referred to the voltage at point A. The total impedance of sections AB, BC and CA are 2 + j1, 2 + j3 and 1 + j2 ohms respectively. Deter- mine the current in each section. [Current in AB = (39·54 − j 25·05) amp ; BC = (7·54 − j 1·05) amp ; CA = (28·46 − j 46·95) amp.] 4. A 3-phase ring distributor ABCD fed at A at 11 kV supplies balanced loads of 40 A at 0·8 p.f. lagging at B, 50 A at 0·707 p.f. lagging at C and 30 A at 0·8 p.f. lagging at D, the load currents being referred to the supply voltage at A. The impedances per phase of the various sections are : Section AB = (1 + j 2) Ω ; Section BC = (2 + j 3) Ω Section CD = (1 + j 1) Ω ; Section DA = (3 + j 4) Ω Calculate the currents in various sections and station bus-bar voltages at B, C and D. [Current in AB = (53·8 − j 46) amp ; BC = (21·8 − j 22) amp. CD = (−−13·55 + j 13·35) amp ; DA = (− −40·55 − j 26·45) amp. VB = (6212·5 j 61·6) volts/phase ; VC = (6103 − j 83) volts/phase − VD = (6129·8 − j 82·8) volts/phase] Phase Sequence Indicator 14.3 3-Phase Unbalanced Loads The 3-phase loads that have the same impedance and power factor in each phase are called balanced loads. The problems on balanced loads can be solved by considering one phase only ; the conditions in the other two phases being similar. However, we may come across a situation when loads are unbalanced i.e. each load phase has different impedance and/or power factor. In that case, current and power in each phase will be different. In practice, we may come across the following unbal- anced loads : A.C. Distribution 367 (i) Four-wire star-connected unbalanced load (ii) Unbalanced ∆-connected load (iii) Unbalanced 3-wire, Y-connected load The 3-phase, 4-wire system is widely used for distribution of electric power in commercial and industrial buildings. The single phase load is connected between any line and neutral wire while a 3-phase load is connected across the three lines. The 3-phase, 4-wire system invariably carries *un- balanced loads. In this chapter, we shall only discuss this type of unbalanced load. 14.4 Four-W Four-W ir -Wir e Star ire -Connected Unbalanced Loads Star-Connected We can obtain this type of load in two ways. First, we may connect a 3-phase, 4-wire unbalanced load to a 3-phase, 4-wire supply as shown in Fig. 14.10. Note that star point N of the supply is connected to the load star point N′. Secondly, we may connect single phase loads between any line and the neutral wire as shown in Fig.14.11. This will also result in a 3-phase, 4-wire **unbalanced load because it is rarely possible that single phase loads on all the three phases have the same magnitude and power factor. Since the load is unbalanced, the line currents will be different in magnitude and displaced from one another by unequal angles. The current in the neutral wire will be the phasor sum of the three line currents i.e. Current in neutral wire, IN = IR + IY + IB...phasor sum The following points may be noted carefully : (i) Since the neutral wire has negligible resistance, supply neutral N and load neutral N′ will be at the same potential. It means that voltage across each impedance is equal to the phase voltage of the supply. However, current in each phase (or line) will be different due to unequal impedances. (ii) The amount of current flowing in the neutral wire will depend upon the magnitudes of line currents and their phasor relations. In most circuits encountered in practice, the neutral current is equal to or smaller than one of the line currents. The exceptions are those circuits having severe unbalance. * No doubt 3-phase loads (e.g. 3-phase motors) connected to this supply are balanced but when we add single phase loads (e.g. lights, fans etc.), the balance is lost. It is because it is rarely possible that single phase loads on all the three phases have the same magnitude and power factor. ** In actual practice, we never have an unbalanced 3-phase, 4-wire load. Most of the 3-phase loads (e.g. 3- phase motors) are 3-phase, 3-wire and are balanced loads. In fact, these are the single phase loads on the 3-phase, 4-wire supply which constitute unbalanced, 4-wire Y-connected load. 368 Principles of Power System Example 14.7. Non-reactive loads of 10 kW, 8 kW and 5 kW are connected between the neutral and the red, yellow and blue phases respectively of a 3-phase, 4-wire system. The line voltage is 400V. Calculate (i) the current in each line and (ii) the current in the neutral wire. Solution. This is a case of unbalanced load so that the line currents (and hence the phase cur- rents) in the three lines will be different. The current in the *neutral wire will be equal to the phasor sum of three line currents as shown in Fig. 14.12. (i) Phase voltage = 400/ 3 = 231 V IR = 10 × 10 /231 = 43·3 A 3 IY = 8 × 10 /231 = 34·6 A 3 IB = 5 × 10 /231 = 21·65 A 3 (ii) The three lines currents are represented by the respective phasors in Fig. 14.13. Note that o the three line currents are of different magnitude but displaced 120 from one another. The current in the neutral wire will be the phasor sum of the three line currents. Resolving the three currents along x-axis and y-axis, we have, = IY cos 30 − IB cos 30 o o Resultant horizontal component = 34·6 × 0·866 − 21·65 × 0·866 = 11·22 A Resultant vertical component = IR − IY cos 60o − IB cos 60o = 43·3 − 34·6 × 0·5 − 21·65 × 0·5 = 15·2 A As shown in Fig. 14.14, current in neutral wire is IN = a11 ⋅ 22f + a15 ⋅ 2f 2 2 = 18·9 A * Had the load been balanced (i.e. each phase having identical load), the current in the neutral wire would have been zero. A.C. Distribution 369 Example 14.8. A 3-phase, 4-wire system supplies power at 400 V and lighting at 230 V. If the lamps is use require 70, 84 and 33 amperes in each of the three lines, what should be the current in the neutral wire ? If a 3-phase motor is now started, taking 200 A from the lines at a p.f. of 0·2 lagging, what should be the total current in each line and the neutral wire ? Find also the total power supplied to the lamps and the motor. Solution. Fig. 14.15 shows the lamp load and motor load on 400 V/230 V, 3-phase, 4-wire sypply. Lamp load alone. If there is lamp load alone, the line currents in phases R,Y and B are 70 A, 84 o A and 33 A respectively. These currents will be 120 apart (assuming phase sequence RYB) as shown in Fig.14.16. = 84 cos 30 − 33 cos 30 = 44·17 A o o Resultant H-component = 70 − 33 cos 60 − 84 cos 60 = 11·5 A o o Resultant V-component ∴ Neutral current, IN = a44 ⋅17f + a11 ⋅ 5f 2 2 = 45·64 A Both lamp load and motor load When motor load is also connected along with lighting load, there will be no change in current in the neutral wire. It is because the motor load is balanced and hence no current will flow in the neutral wire due to this load. ∴ Neutral current, IN = 45·64 A...same as before The current in each line is the phasor sum of the line currents due to lamp load and motor load. Active component of motor current = 200 × cos φm = 200 × 0·2 = 40 A Reactive component of motor current = 200 × sin φm = 200 × 0·98 = 196 A ∴ IR = bsum of active comp.g + breactive comp.g 2 2 = b40 + 70g + a196f = 224·8 A 2 2 IY = b40 + 84g + a196f = 232 A 2 2 IB = b40 + 33g + a196f = 209·15 A 2 2 Power supplied Power supplied to lamps = 230 (70 + 84 + 33) × 1 = 43010 W (Œ cos φL = 1) 370 Principles of Power System Power supplied to motor = 3 VLIL cos φm = 3 × 400 × 200 × 0·2 = 27712 W Example 14.9. The three line leads of a 400/230 V, 3-phase, 4-wire supply are designated as R, Y and B respectively. The fourth wire or neutral wire is designated as N. The phase sequence is RYB. Compute the currents in the four wires when the following loads are connected to this supply : From R to N : 20 kW, unity power factor From Y to N : 28·75 kVA, 0·866 lag From B to N : 28·75 kVA, 0·866 lead If the load from B to N is removed, what will be the value of currents in the four wires ? Solution. Fig. 14.17 shows the circuit diagram whereas Fig.14.18 shows its phasor diagram. −1 o The current IR is in phase with VRN, current IY lags behind its phase voltage VYN by cos 0·866 = 30 −1 o and the current IB leads its phase voltage VBN by cos 0·866 = 30. IR = 20 × 10 /230 = 89·96 A 3 IY = 28·75 × 10 /230 = 125 A 3 IB = 28·75 × 10 /230 = 125 A 3 The current in the neutral wire will be equal to the phasor sum of the three line currents IR, IY and IB. Referring to the phasor diagram in Fig.14.18 and resolving these currents along x-axis and y-axis, we have, = 86·96 − 125 cos 30 − 125 cos 30 o o Resultant X-component = 86·96 − 108·25 − 108·25 = − 129·54 A = 0 + 125 sin 30 − 125 sin 30 = 0 o o Resultant Y-component ∴ Neutral current, IN = a−129 ⋅ 54f + a0f 2 2 = 129·54A When load from B to N removed. When the load from B to N is removed, the various line currents are : IR = 86·96A in phase with VRN ; IY = 125A lagging VYN by 30o ; IB = 0 A The current in the neutral wire is equal to the phasor sum of these three line currents. Resolving the currents along x-axis and y-axis, we have, = 86·96 − 125 cos 30 = 86·96 − 108·25 = − 21·29 A o Resultant X-component Resultant Y-component = 0 − 125 sin 30 = 0 − 125 × 0·5 = − 62·5 A o ∴ Neutral current, IN = b− 21⋅ 29g + b− 62 ⋅ 5g 2 2 = 66·03 A