# Power measurement in balanced 3 phase circuits and power factor improvement. 1 Power in Single Phase Circuits. Experiment no 1

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1 Experiment no 1 Power measurement in balanced 3 phase circuits and power factor improvement 1 Power in Single Phase Circuits Let v = m cos(ωt) = cos(ωt) is the voltage applied to a R-L circuit and i = I m cos(ωt θ) = I cos(ωt θ) is the current flowing in it ( and I are the rms values of the voltage and current respectively). The power at any instant of time is p = vi = m I m cos(ωt) cos(ωt θ) p = mi m cos θ + mi m cos(ωt θ) p = mi m cos θ + mi m cos(ωt) cos θ + mi m sin θ sin (ωt) careful study of the above equation reveals the following characteristics: The average value of power = P = m Im cos θ = I cos θ Watts. In C circuits, when we talk of power rating, we are usually referring to the average power. For example, a 100 W bulb implies that average power output of the bulb is 100 W. The frequency of instantaneous power is twice the frequency of the voltage and current. Note that instantaneous power may be negative for a portion of each cycle, even if the load is passive. In a passive network, negative power at a given instant implies that energy stored in inductors or capacitors is being given back to the supply at that instant. The fact that the instantaneous power varies with time explains why some single phase motor driven appliances such as refrigerators, which are fed from single phase supply experience vibration, and hence require resilient mountings to prevent excessive vibration of the equipment. If the load is purely resistive, the voltage and current are in phase, which means that θ = 0. The above equation reduces to p = P + P cos(ωt). In this case instantaneous real power can never be negative, i.e., power cannot be given back to the supply from a purely resistive network. If the circuit is purely reactive ( either capacitive or inductive, θ = ±90 o ), the expression for instantaneous power is p = mim sin(ωt). The average power is zero. Therefore in a purely reactive circuit there is no average power dissipation. Energy is drawn from the supply for a part of a cycle is stored, and given back in another part of a cycle. However, since there is a current flow, the supply system must be designed to source such currents. Therefore it is useful to define a quantity called reactive power as follows: Q = mi m sin θ = I sin θ Dimensionally P and Q are same. However, in order to distinguish between average (also called real power ) and reactive power, we use the term r for reactive power. Q is positive for inductive load and negative for capacitive load. In other words inductors demand or absorb reactive power while capacitors furnish or deliver the reactive power. For purely resistive circuits, Q=0. 1

3 Three Phase System The generation, transmission, distribution and utilization of large blocks of electric power are accomplished by means of three-phase circuits. balanced three phase system consists of three sinusoidal voltages of identical amplitudes and frequency but are out of phase with each other by 10 o. In discussing three-phase circuits, standard practice is to refer the three phase as,, and C ( or R,Y, ). Furthermore, the -phase ( or R-phase) is almost always used as the reference phase. ecause the phase voltages are out of phase by 10 o, two possible phase relationships can exist between the phase ( or R phase) voltage and the and C phase voltages ( Y and ). One possibility is for the phase to lag the phase by 10 o ( phase C will be lagging phase by 10 o, or lead phase by 10 o ). In this case the phase relationship is known as C (or RY) phase sequence. The only other possibility is for the C phase to lag phase ( or phase to lead phase by 10 o ). This phase relationship is known as C sequence. In phasor notation, two possible sets of balanced phase voltages are: a = 0; b = 10 o ; c = 40 o ; and a = 0; c = 10 o ; b = 40 o = +10 o. n important characteristic of balanced three phase voltages is that the sum of the voltages is zero ( a + b + c = 0). Three phase voltage sources can be either Y ( star) or delta connected. These connections are shown in Fig. (a) and (b) respectively. Let an, bn and cn are the phase voltages ( the voltage between line and neutral). Now there are 4 wires - three for three lines and fourth one for the neutral. ll the single phase loads are connected between any one of the lines and neutral. The relationship between the line-line voltage ( ab, bc and ca ) and phase voltage can be derived as follows: Referring to Fig.(a), the line-line voltages for C sequence are: ab = an bn = 0 10 o = 3 30 o Similarly, bc = 3 90 o ca = 3 10 o In three phase system the load is said to be balanced if the magnitude and phase of the load impedance connected to each phase is the same. For the Fig. (b), the phase current ( the current in each phase of the load) is I = an an = 0 = θ I θ I = bn = I (10 + θ) I C = cn bn = 10 θ cn = 40 θ = I (40 + θ) Line current is the current in each phase of the line. In star connected system, phase currents & line currents are identical. The complete vector diagram for a three phase system feeding a star connected load is shown in fig.3(a). For delta connected system shown in Fig. (b), line to line voltage is the same as phase voltage. However, the relationship between the phase currents and line currents can be determined as follows: The phase currents are given by: I ab = ab ab = 0 = θ I θ I bc = bc = I (10 + θ) I ca = ca bc = 10 θ ca = 40 θ 3 = I (40 + θ)

4 We can write the line currents in terms of the phase currents by direct application of KCL: I = I ab I ca = I θ I (40 + θ) = 3I (30 + θ) I = I bc I ab = 3I (150 + θ) & I C = I ca I bc = 3I (70 + θ) The complete vector diagram for a three phase system feeding a delta connected load is shown in fig.3(b)..1 Power in three phase circuits: In a balanced three phase system, the magnitude of each line-to-neutral voltage (also line-line) is the same, as is the magnitude of each phase current. Therefore, the expression for total power in terms of phase current and phase voltage is given by: P = 3 I cos θ Watt, θ = (I& ) In-terms of line voltage & line current the above expression becomes: P = 3 L I L cos θ Watt, θ = (I & ) In the above expression and I are the rms values of the phase voltage and phase current respectively. We can also calculate reactive power (Q) and complex power (S). For balanced load, the expressions for Q and S are given by: Q = 3 L I L sin θ r, θ = ( I & ), S = 3 L I L ll the above equations are valid for both the connections (Y or Delta).. Measurement of Power in Three phase circuit Real power in balanced (or unbalanced ) 3 phase circuits can be measured by using two wattmeters connected in any two lines of a three-phase three wire system. In Fig.4(a) the current coils are connected in lines and C, with the voltage coil reference connections in line. The current flowing through the current coil of the wattmeter connected in line is I and that in line C is I C. Similarly, the voltage applied to the voltage coils is ab and cb (and not bc ) respectively. Their readings will be: W = ab I cos ( I & ab ) and W = cb I C cos ( I C & cb ) From Fig.3 (a or b) angle between ab and I is (30 + θ) and between cb and I c is (30 θ). Therefore the wattmeter readings will be: W 1 = 3 I cos(30 + θ) W = 3 I C cos(30 θ) W 1 + W = 3 I cos θ = 3 L I L cos θ Watt Note that if θ is greater than 60 o, W 1 will be negative. In actual power measurement it is possible to encounter such a situation when one of the wattmeters has negative reading. The point to remember is that the total power in three phase circuit is obtained by adding the two wattmeter readings, taking the signs of the readings into account. It may be worthwhile to identify the following cases: If θ is zero (load is purely resistive), W 1 = W and both are positive. For θ = 60 o lag, W 1 = 0. Similarly for θ = 60 o lead, W = 0. Note that you will be using a Power analyzer to measure power. It is also possible to measure other quantities such as voltage, current, Q, S, power factor etc. using this equipment. 4

8 C 0 N I bn I b IC n a an cn c 0 10 C 40 I I b IC Iab Ibc a Ica c Fig. (a) Star Star System Fig. (b) Delta Delta System Figure : Schematic configuration of star delta supply and load system a I a an ca cn I c 30 o ab = = 3 an = bn N cn I b 30 o I a b c I b Ic bn bc = C Fig. 3(a) ector Diagram for a 3 phase system supplying a balanced star connected load Ia I b I c b Iab I bc a I ca c I b ca I bc 30 o I ca 30 o I c 30 o I ab ab bc Ia Fig. 3(b) ector Diagram for a 3 phase system supplying a balanced delta connected load 8

9 utotransformer 415, 3 phase, 50Hz C N W1 5/500 M L C Power nalyzer C a b c 3 Phase Y or Connected Load Fig. 4(a) Circuit Diagram to measure power consumed by 3 phase Load using Wattmeter method Y & a b c a b c Fig. 4(b) 3 Phase Load connection: Star & Delta Ic= Iy Lumped Transmission Line Impedance Ix s I = Ix jiy L R L Ic= Iy L C Iy Θ Ix I L L (a) Schematic Diagram (b) Phasor Diagram Figure :5 Power Factor Improvement 9

10 utotransformer W1 5/500 5 M L C 3 Phase Lagging Load Power nalyzer (Induction m/c) S 1 S S 3 10µF 10µF 10µF 10µF 10µF 10µF C C 10µF 10µF 10µF Delta connected capacitor bank 1 Delta connected Delta connected capacitor bank capacitor bank 3 C Fig.6 Circuit diagram for Power factor improvement C 415, 3 phase, 50Hz N 10

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