Three phase circuits

Three phase circuits

THREE PHASE CIRCUITS THREE-PHASE ADVANTAGES 1. The horsepower rating of three-phase motors and the kva rating of three-phase transformers are 150% greater than single-phase motors or transformers of similar frame size. 2. The power delivered by a single-phase system pulsates and falls to zero. The three-phase power never falls to zero. The power delivered to the load in a three-phase system is the same at any instant. This produces superior operating characteristics for three-phase motors. 3. A three-phase system needs three conductors; however, each conductor is only 75% the size of the equivalent kva rated single-phase conductors Three-phase power never falls to zero. Three-phase voltages with 120 degrees of phase shift. BASIC PROPERTIES Three-phase systems have either three or four conductors. There are three-phase conductors identified as A, B, and C. The three phases are 120 degrees out of phase with each other (360 divided by 3). There is sometimes a fourth conductor, which is the neutral. The Automobile Society (India) Page 1

STAR-DELTA CONNECTION Used to step down voltage ie end of transmission line Advantages 1. The primary side is star connected. Hence fewer numbers of turns are required. This makes the connection economical 2. The neutral available on the primary can be earthed to avoid distortion. 3. Large unbalanced loads can be handled satisfactory. Disadvantages 1. The secondary voltage is not in phase with the primary. (30 ⁰ phase difference) 2. Hence it is not possible to operate this connection in parallel with star-star or delta-delta connected transformer. POWER IN THREE-PHASE SYSTEMS WITH BALANCED LOAD Power input to a single phase as circuit is VI. Thus, power per phase in 3-phase system is are the rms value of phase voltage and current respectively and. The power factor of the load, being the angle between the phase voltage and the corresponding phase current. Hence, the total power fed to a 3-phase system with balanced load. Three-phase systems are connected wither in star or in delta. (1) The Automobile Society (India) Page 2

POWER IN THREE-PHASE STAR CONNECTED CIRCUIT. In star-connected circuit, Substituting the vale in equation.(1), an expression for power is obtained terms of line voltage and line current. Total power= Total power = POWER IN THREE-PHASE DELTA CONNECTED CIRCUIT In a delta-connected circuit, Substituting these in equation (1) Total power= Total power = Hence in general, it can be stated that the expression for power in a 3-phase system with balanced load remains the same irrespective of the connection of three phases and is given by Total power= RELATIONSHIP B BETWEEN PHASE & LINE CURRENTS AND VOLTAGES Line voltages are voltages between lines either at the generator or at the load. Phase voltages are voltages across the phases. For load, the phases are from line to line. Line currents are current in the line conductors. Phase currents are currents through the phases. PHASORS A phasor is a straight line an arrow marked on one side. The length of this straight line represents the magnitude of the sinusoidal quantity being represented and the arrow represents its direction. Thus phasor representation is similar to the vector representation. However the phasors rotate in the anticlockwise direction as shown in Fig. The length of the phasor represents the rms value of the sinusoidal quantity. Sometimes the length also represents the peak value. The Automobile Society (India) Page 3

Fig. (Phasor representation of sinusoidal quantities) Speed of rotation of the phasor is equal to radians/sec, where = 2 One rotation of the phasor corresponds to one cycle of the alternating waveforms as shown in Fig - (a). Fig. (a): Relation between an alternating quantity and phasor If the length of the phasor is equal to the peak value of the sinusoidal ac quantity, its angular velocity and it rotates in the anticlockwise direction in space than at any given angle its projection on the Y-axis gives you the instantaneous value of the sinusoidal ac quantity at that angle. We can represent two or more sinusoidal simultaneously on the same phasor diagram if and only if their frequencies are same. If their frequencies are not same then they cannot be represented on the same phasor diagram The Automobile Society (India) Page 4

TWO-WATT METER METHOD DESCRIPTION In three-phase currents, all three pairs of the line-to-line voltages have to be measured accurately in order to determine how much power can be consumed by the three-wire load. The two wattmeter method measures the current in two of the three lines and two voltages from the said lines with respect to the third line. From there the total power of the lines, regardless of balance of current, voltage or load variations, can be accurately derived. When using the two wattmeter method, the power of a three-phase circuit drops below zero. This is due to the behavior of the voltage, which leads the current on one wire by 30 and delays the current by 30 in another wire. The correct sum will be obtained when two singlewatt transducers are used in measuring the power they currently have instead of just one. PRINCIPLE Any of the points in a three-phase system can be used as a reference angle when using the two wattmeter method. By means of using a reference point, establishing the measurement of total power by using only two out of three meters is made possible. Many studies show that using this method is only effective on balanced systems, however, whenever a reference point is chosen, it will measure zero, and thus the power obtained from the two remaining meters will continue to be true. LAWS When using the two wattmeter method, it is important to take note of Kirchoff's Laws in order to avoid confusion in power calculation. In his first law, Kirchoff states that when two of three currents are already known, the third current must be equal to the sum of the first two, but carries a different direction. Whereas, the second law states that if two of the three voltages are known, the third should also be equal to the sum of the first two, also also in a different sign or direction. USES The two wattmeter method is used in determining total power derived from the three phase load which is always constant, regardless of variations in phase power as well as angular displacement. Thus, the power output from a three-phase motor is also constant based on the findings using the said measurement. The Automobile Society (India) Page 5

QUESTIONS FOR PRACTICE Q. 1 Two wattmeters connected to measure the power input to 3-Φ circuit indicate 2500w and 500 w respectively. Find the power factor of the circuit (i) When both readings are positive and (ii) When later reading is obtained after reversing the connection to the current coil of one instrument. Ans.: Given: and 500W To find power factor. Case1: Both readings positive and Case 2: { [ ( ) ]} = { [ ( ) ]} ( ) (leading) and { [ ( ) ]} ( ) (Leading) Q. 2 Derive the relation between power in delta and star system. Ans.: Step1: Calculate power in a star system: Power per phase = Total power = 3 = 3 = 3 x. (1) Step 2: total power in a delta circuit: Power per phase Total power P T = = P T = 3. (2) The Automobile Society (India) Page 6

Fig. Compare equations (1) and (2) assuming that the line voltage V L and impendence per phase Z for both the types of loads are identical. Then P T (Delta) = ( ) Hence it is proved that a 3 phase balanced delta load draws 3 times more power than the balance star load. Q. 3 Show that only two Wattmeters are sufficient to measure 3 powers. Also state the advantages of Two Wattmeter Method. Ans: For the balanced load: Assume that the load is balanced and the impedance per phase is inductive impedance so that the phase current lags behind the phase voltage by an angle. Now refer below Fig. The current flowing through the current coil of wattmeter 1 is the phase current I R and the voltage across is pressure coil is line voltage V RB. Hence the reading of wattmeter 1 is given by, W 1 = V RB I R cos (V RB I R ). (1) Similarly the reading of wattmeter 2 is given by, W 2 = V YB I Y cos (V YB I Y ). (2) Now refer to the phasor diagram for a balanced star load as shown in Fig. 2 to obtain the values of (V RB I R ) and (V YB I R ). The phasor diagram of Fig. 2 has been draw for a lagging power factor load. So from Fig. 2, the required angles are as follows: (V RB I R ) = ( ) and (V YB I Y ) = ) Substituting the values in Equation (1) and (2) we get, W 1 = V RB I R cos( ) W 1 = V L I L cos( ) Since V RB and I R are line voltage and line current. Similarly, W 2 = V RB I Y cos( ) W 2 = V L I L cos( ) Hence the sum of W 1 and W 2 is given by, W 1 + W 2 = V L I L ( ) ( ) = V L I L The Automobile Society (India) Page 7

= V L I L = V L I L [ ] W 1 + W 2 = V L I L cos. (3) But V L I L is the total three phase active power. Total three phase active power = W 1 + W 2. (4) Thus we have proved that the sum of the reading of the two wattmeters is equal to the total three phase active power. Advantages of the wattmeter method: 1. We can use it for the balanced as well as unbalanced loads. 2. For the star type loads, it is not necessary to connect the neutral point for connection of the wattmeter. 3. The delta load need not be opened to connect the wattmeters. 4. For the balanced loads, it is possible to measure the power factor along with the power. 5. We need to use only two wattmeters to measure the power in a 3 phase circuit. 6. It is possible to measure the reactive volt amperes for the balanced loads. Fig. : phasor diagram of star connected load Q. 4 A 3-phase RYB system has effective line voltage 173.2 V. Wattmeter s in lines R and Y read 301 W and 1327 W respectively. Find the impedance of the balanced star connected load. Ans: Given: V L = 173.2 V, W 1 = 301 W, W 2 = 1327 W, star load To find: Z ph. Step 1: Calculation of cos and total power: Total power W = W 1 + W 2 = 301 + 1327 = 1628 W { [ ( ) ( ) ]} { [ ( ) ]} ( ) Step 2: Calculation if I L and I ph : V L I L The Automobile Society (India) Page 8

173.2 x I L x 0.6755 I L = 8.034 Amp. I ph = I L = 8.034 Amp. Step 3: Calculation of Z ph : Q. 5 Calculate the total power and readings of two Wattmeters connected to measure power in three phase balanced load, if the reactive power is 15 KVAR, and the load p.f.is 0.8 lagging. Ans: Given: Q = 15 x 10 3 VAR, = 0.8 (lag) Step 1: Calculation of W 1 and W 2 : Reactive power Q = (W 1 W 2 ) = W 1 W 2 W 1 W 2 = 8660.25. (1) Also { [ ( ) ]} 0.8 { [ ( ) ]} 0.75 ( ). (2) 0.75 W 1 + 0.75 W 2 W 1 - W 2 0.982 W 1 2.482 W 2 From Equation (2), W 2 W 2 W 1 = 2.582 W 2. (3) Substituting into Equation (1) to get 2.528 W 2 W 2 = 8.660.25 W 2 = 5669.38 Watt and W 1 = 2.528 x 5669.38 = 14332.2 Watt Step 2: Calculation of total power P = W 1 + W 2 = 14332.2 + 5669.38 Q. 6 Explain how to Watt-meters can be used to measure power and power factor in a three phase balanced star connected load with lagging p.f. Ans: Two wattmeter method: This method used only two wattmeter s to measure the power in 3 phase circuits. It can be used for measuring the power in the star as well as delta connected loads. Below Fig shows the connections for a two wattmeter method for the star connected load. The Automobile Society (India) Page 9

Fig. : Two wattmeter method connections for star type load The current coils of the two wattmeters are connected in any two lines (here R and Y), and the voltage coils are connected between their own current coil terminal and the line without the current coil. (From R to B). The wattmeter connections will remain the same irrespective of the type of load. The total power in a three phase circuit is equal to the algebraic sum of the two wattmeter readings. Total power W = W 1 + W 2 Where W 1 and W 2 are the wattmeter readings. Q. 7 A balanced 3 phase delta connected load draws 10A of line current and 3 kw at 220 V. determine the value of resistance and reactance of each phase of load. Ans: Given: Delta load, I L A, V L = V ph = 220 V, P = 3 k W Step 1: Calculation of (power factor): P = V L I L 3 k W = x 220 x 10 x = 0.787 Step 2: Calculation of I ph : I ph = Step 3: Calculation of Z ph, R ph and X ph : Z ph = 38.10 Ω R ph = Z ph = 38.10 x 0.787 = 29.98 Ω = 0.707 = 45 0 = 0.707 X ph = Z ph = 38.10 x 0.787 = 29.98 Ω Assuming that power factor is a lagging The inductance per phase is given by, X ph = 2 f L ph 29.98 = 2 x 50 L ph L ph = 95.42 mh The Automobile Society (India) Page 10

Q. 8 Each phase of delta connected load consists of a 50 mh inductor in series with a parallel combination of 50Ω resistor and 50 F capacitor. The load is connected to a three phase 550 V, 800 rad/sec ac supply. Find: (i) Phase current (ii) Line current (iii)power drawn (iv) Power factor (v) Reactive power. Ans: Given data: L = 50 x 10-3 H R = 50Ω C = 50 x 10-6 F V L = 550 V = 800 rad/sec Inductive reactance X L = Capacitive reactance X C = = 800 x 50 x 10-3 X L = 40 Ω X C = 25 Ω Admittance of parallel circuit, = 0.02 +j0.04 = (0.045 < 63.43 0 ) S Total impedance of the circuit ( ) ( j X L ) + ( ) ( ) The Automobile Society (India) Page 11

I. For phase current: a. b. II. Line current: a. b. ( ) c. III. Power drawn: Power drawn = 3 x V ph x I ph x a. = 3 x 550 x 24.50 x ( ) b. = 17898.50 W c. = 17.898 k W IV. Power factor: Power factor = ( ) a. ( ) b. = 0.4427 V. Reactive power (Q): a. Q = x V L x I L x b. Q = x 550 x 42,43 x ( ) c. Q = 36.24 k VAR Q. 9 3 220 V, 50 Hz, 11.2 kw Induction Motor has full load efficiency of 88% and 10 draws a line current of 38 Amp, when connected to 3, 220 V supply find the reading on two wattmeter connected in the circuit to measure the input to the motor. Determine also p.f. at which motor is operating. Ans: Given: V L = 220 V, f = 50 Hz, P 0 = 11.2 k W, = 0.88, I L = 38 Amp. To find : W 1, W 2, 1. Find P i = P i = V L I L 12727.3 = x 220 x 38 x 0.8789 (lagging) 2. Find W 1 and W 2 : W 1 + W 2 = 12727.3. (1) W 1 = V L I L ( ) = 220 x 38 ( ) The Automobile Society (India) Page 12

W 1 = 8557.14 W W 2 = V L I L ( ) = 220 x 38 (30 + 28.5) W 2 = 4368.09 W Q. 10 Three inductive coils each with a 15 ohms resistance and 0.03 H of inductance are connected (a) in star and (b) in delta, to 3 phase 400 V, 50 Hz supply. Calculate for each of the case. 1. Phase and line currents. 2. Total power absorbed. Ans: Given: R ph = 15 Ω, L ph = 0.03 H. Part I: Star connected load V L = 400 V, f = 50 Hz 1. Phase and line currents: X LPH = 2 x 50 x 0.03 = 9.42 Ω Z ph = R ph x j X Lph = (15 + j 9.42) Ω = 17.72 < 32.13 0 Ω I ph = I L = 2. Total power absorbed (P T ): P T = V L I L x 400 x 13.03 x (32.13 0 ) P T = 7646.44 Watt Part II: Delta connected load 1. Phase current, I ph = 2. Line current, I L = x I ph = x 22.57 = 39.09 Amp. 3. Total power absorbed, P T = V L I L x 400 x 39.09 x (32.13 0 ) P T = 22939.33 Watt The Automobile Society (India) Page 13