Fundamentals of Electrical Drivess - Chapter 5 ppt

Chapter 5 CONCEPT OF REAL AND REACTIVE POWER 5.1 Introduction In this chapter the meaning of ‘real’ and ’reactive’ power are explored for sinusoidal systems. Initially, single phase (so-called two wire) circuits are discussed as to gain an understanding of the energy flow within a circuit config- uration that is representative for electrical machines. We will then extend this analysis to three-phase (three wire) circuits. In the final part of this chapter a set of tutorials is introduced to reinforce the concepts discussed. 5.2 Power in single phase systems The concept of power is introduced with the aid of figure 5.1, which shows a load in the form of an inductance L, resistance R and voltage source u e in series connection. The voltage source u e is generally known as the induced voltage in electrical machines. The circuit configuration as described above is representative for electrical machines hence its use here. A current source i (t) is connected to this network. The reason for using a sinusoidal supply current source insteadof a supply voltage sourceis to simplifythe mathematical analysis. Application of Kirchhoff’s voltage laws to this circuit shows that the Figure 5.1. R-L-u e load connected to current source 122 FUNDAMENTALS OF ELECTRICAL DRIVES voltage across the current source can be written as u = u R + u L + u e (5.1) where u R ,u L ,u e represent the instantaneous voltages across the resistance, inductance and voltage source u e respectively. If we multiply equation (5.1) with the instantaneous current produced by the current source, the so-called power balance equation appears as shown in equation (5.2). ui  p in = iu R  p R + iu L  p L + iu e  p e (5.2) The instantaneous power, which is a physical quantity, is simply the product of the instantaneous voltage and instantaneous current. If this value is positive then power flows into the circuit element or internal voltage source u e . With reference to equation (5.2) the term p in refers to the power supplied to the network. For the resistance and inductance the instantaneous power is given as p R and p L respectively. The same definition in terms of energy/power flow also applies to the voltage source u e . For this circuit element energy can for example be converted to mechanical energy. It is noted that for a resistance, energy is dissipated which means that it is given as heat to the environment. Note that energy (Ws=joule) is defined as ∆W e =  t 0 p (τ ) dτ, i.e. in a time-diagram the area underneath the respective power function and the horizontal time line. It is instructive to discuss these concepts with the aid of an example where we assume a sinusoidal current time function of the form i (t)= ˆ i cos ωt (5.3) The ‘steady-state’ voltage across the current source will be of the form u (t)=ˆu cos (ωt + ρ) (5.4) In addition we will assume that the voltage across the induced voltage source u e can be written as u e (t)=ˆe cos (ωt + η) (5.5) It is convenient at this stage to introduce a phasor representation of the variables u (t) ,i(t) according to the approach discussed in section 2.5. The variables according to equations (5.3), (5.4) may also be written as i (t)=    ˆ i  i e jωt    (5.6a) u (t)=    ˆue jρ  u e jωt    (5.6b) Concept of Real and Reactive Power 123 Figure 5.2. Phasor diagram of phasors i ,u An observation of figure 5.2 learns that we can also represent the voltage phasor in terms of the vector sum of the two phasors u re =ˆu cos ρ and u im = jˆu sin ρ, i.e. u = u re + u im . The phasor u re is aligned (in phase) with the current phasor, the other u im is orthogonal, i.e. at right angles to i. If the nature of the circuit is ‘inductive’ (as shown) the angle ρ will be greater than zero. Alternatively, circuits which exhibit a negative ρ are referred to as being ‘capacitive’. This definition ties in which the fact that the phasor diagram of for example an inductance, corresponds to the case ρ = 1 2 π rad, whereas for a capacitance the angle is equal to ρ = − 1 2 π rad. The voltage phasor components u re ,u im can also be converted to variables as functionoftime (byusing the transformationsgiven in equation(5.6)) namely u re (t)= ˆu cos ρ cos ωt (5.7a) u im (t)=−ˆu sin ρ sin ωt (5.7b) Note that expression (5.7) may also be written in space vector form as u (t)=u re (t)+ju im (t) (5.8) The introduction of phasor components u re ,u im allows us to rewrite the input power equation p in = iuas p in = iu re  p re + iu im  p im (5.9) which shows that the instantaneous input power expression is now defined in terms of two components which with the aid of equations (5.3), (5.7) can also be written as p re =ˆu ˆ i cos ρ cos 2 ωt (5.10a) p im = −ˆu ˆ i sin ρ cos ωt sin ωt (5.10b) 124 FUNDAMENTALS OF ELECTRICAL DRIVES Equation 5.10 can also be rewritten in the form given below p re = ˆu ˆ i 2 cos ρ    P (1 + cos 2ωt) (5.11a) p im = ˆu ˆ i 2 sin ρ    Q (−sin 2ωt) (5.11b) An analysis of equation (5.11a) learns that there is a time-independent term, known as the ‘real power P in ‘watts’ which represents the average power level P of the function p re . In other words, the average power level corresponds to the total amount of energy supplied to the circuit for one cycle period T (the area enclosed by the power function p re and time line) divided by T . Note that the average power level is also present in the function p in given that the average value of the variable p im is zero. The so-called ‘reactive’ power value Q expressed in VA is tied to the energy flow associated with expression (5.11b). Note that this power expression is based on the use of the voltage phasor component, which is at right angles to the current phasor (see figure 5.2). An observation of expression (5.11b) learns that the average power level is zero. The amplitude of the power function p im is known as the reactive power value Q. The real and reactive power value of the circuit may be written as P = UI cos ρ W (5.12a) Q = UI sin ρ VA (5.12b) where U =ˆu/ √ 2,I= ˆ i/ √ 2 are the respective RMS values of the volt- age/current waveforms of the source connected to the R-L-u e circuit. The term cos ρ is referred to as the power factor. It is at this stage helpful to consider a simple numerical example where we assume the current to be of the form i =cosωt,whereω = 100 π rad/s. The circuit elements are chosen purposely as to arrive at a voltage across the cur- rent source which is of the form u =2cos(ωt + ρ), with ρ = π/3. Hence, the circuit is ‘inductive’ given that the voltage/time function leads the cur- rent/time function. For this example the values of P and Q are according to equation (5.12), equal to P =0.5WandQ =0.866VA respectively. The input power versus time plot together with its components p re ,p im are shown in figure 5.3 for one 20ms cycle of operation. An observation of figure 5.3 shows that the energy flow is towards the circuit for some parts of the cycle (shown in ‘green’) and back to the current source for other parts (shown in ‘red’). There is however, an average energy flow (the difference between the ‘green’ and Concept of Real and Reactive Power 125 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 −0.5 0 0.5 1 1.5 time (s) p in VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 −0.5 0 0.5 1 1.5 time (s) p re VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 −0.5 0 0.5 1 1.5 time (s) p im VA P=0.5 P=0.5 Q=0.866 Figure 5.3. Power plot for R-L-u e circuit, supply side ‘red’ areas) and the power associated with this net energy is known as the ‘real’ power P given in watts as was discussed above. Also shown in figure 5.3 are the two components p re ,p im of the input power function. An observation of this example confirms that the energy linked with the power function p re is precisely the energy supplied to the circuit over one period T =20ms. The amount of energy supplied (shaded ‘green’ in the power function p re ) is equal to P times the cycle time T. The energy linked to the power waveform p im represents the energy which is temporarily stored in the circuit (in either the inductive, capacitive elements or the internal voltage source). This energy oscillates between circuit and supply source (as shown by the ‘green’ and ‘red’ areas which identify the energy direction). The reactive power value Q is equal to the amplitude of the waveform p im . Note that the value of Q can be positive or negative, in both cases its value remains linked to the amplitude of the energy fluctuations. It is interesting to consider thechanges tofigure 5.3for thecase ρ =0. Under thesecircumstances the reactive power level Q is zero, hence there are no energy fluctuations linked to this term. The energy level still fluctuates, but the energy flow is unidirec- tional, i.e. from supply to circuit. On the other hand, if we choose ρ = 1 2 π the 126 FUNDAMENTALS OF ELECTRICAL DRIVES real power P will be zero in which case the average amount of energy trans- ferred from supply to the circuit is zero. The energy under these circumstances is stored and recovered from the circuit, i.e. for parts of the cycle it flows from source to the circuit and for the other (equal amount) it flows in the opposite direction. At thispoint wehave considered the power and energy situationwhen viewed from outside the circuit, i.e. from the source connected to the circuit. We will now consider the energy flow within the circuit itself. Firstly, we examine theresistive componentwhere thevoltage across this ele- ment is given as u R = iR, which can also be linked to its phasor representation namely u R (t)=    ˆ iR  u R e jωt    (5.13) The phasor u R is in phase with the current phasor i as shown in figure 5.4. The Figure 5.4. Phasor diagram for R-L-u e circuit, compo- nent side instantaneous power linked to this component is equal to p R = i 2 R, which after substitution of equation (5.3) can (after some manipulation) be written as p R (t)= ˆ i 2 R 2 (1 + cos (2ωt)) (5.14) An important observation from equation (5.14) is that the power p R is greater than or equal to zero. Furthermore, the real power P R (which is defined as the average value of p R (t)) is equal to P R = I 2 R,whereI = ˆ i/ √ 2 is equal to the RMS value of the current. We will now consider the inductance L of the circuit. The voltage across this element is given as u L = L di dt . Substitution of the current expression (5.3) leads to u L = − ˆ iωLsin ωt, which can also be written in terms of the phasor Concept of Real and Reactive Power 127 u L linked to this function namely u L (t)=      j ˆ iωL  u L e jωt      (5.15) Equation (5.15) shows that the voltage phasor u L is orthogonal (at right an- gles) to the current phasor as shown in figure 5.4. The power linked with this component is given as p L = u L i, which can also be written as p L (t)=− ωL ˆ i 2 2 sin (2ωt) (5.16) The first observation to be made from equation (5.16) is that its average value is zero, which is to be expected given that an inductance cannot dissipate energy. Hence energy taken into the device is stored and must (at a later stage) be recovered. The peak value of p L represents the reactive power Q L of this component and is equal to Q = I 2 ωL. Note that a similar analysis of this type can also be done for the capacitor in which case the reactive power is taken to be negative and of the form I 2 1 ωC . Finally, the circuit component u e is considered, which is of the form given be equation (5.5). The phasor e linked to this voltage function is of the form u e (t)=    ˆee jη  e e jωt    (5.17) The phasore shown infigure5.4 (withη = −π/4)can bedefinedin termsof two phasors e re ,e im (also given in figure 5.4 according to the approach outlined for the voltage phasor (see equation (5.7)). The corresponding voltages are of the form e re (t)= ˆe cos η cos ωt (5.18a) e im (t)=−ˆe sin η sin ωt (5.18b) (5.18c) The introductionof these two voltage componentsallows usto rewrite thepower equation p e = iu e as p e = ie re  p e re + ie im  p e im (5.19) which shows that the instantaneous power expression is again defined by two terms, which with the aid of equations (5.3), (5.18) can also be written as p e re =ˆe ˆ i cos η cos 2 ωt (5.20a) p e im = −ˆe ˆ i sin η cos ωtsin ωt (5.20b) 128 FUNDAMENTALS OF ELECTRICAL DRIVES Equation (5.20) can also be rewritten in the form given below p e re = ˆe ˆ i 2 cos η    P e + ˆe ˆ i 2 cos η cos 2ωt (5.21a) p e im = ˆe ˆ i 2 sin η    Q e (−sin 2ωt) (5.21b) Expressions (5.21) clearly show the real and reactive power contributions P e , Q e respectively, which are associated with the voltage source u e . It is at this stage helpful to return to the numerical example given for the supply side. For example, if we set ˆe = √ 2/2V, η = − π 4 rad, R =0.5Ω, ωL =  √ 3+0.5  Ω and ˆ i =1A, then a simple phasor analysis of this cir- cuit learns that the voltage across the current supply source will be equal to u =2cos(ωt + π/3), which is the waveform used for the supply example that corresponds to the power/energy figure 5.3. Use of these circuit parameters with equations (5.14), (5.16) and (5.21) leads to the power waveforms and ‘energy’ surfaces as given in figure 5.5. The axis scaling has purposely been chosen to match that of figure 5.3. The energy levels linked with this circuit are again shown, where ‘green’ implies an energy flow into the circuit element and ‘red’ out of an element. For example, in the resistance, energy flow is always into this component, given that this energy is dissipated as heat. The average value of the waveform p R represents the real power P R =0.25W dissipated in the resistance given our choice of parameter values. For the inductance we observe a pulsating energy flow, which corresponds to a reactive power value of Q L =1.116VA . Finally, we need to consider the voltage source u e which has two power components, where thefirst p e re has an energy flow whichmust be unidirectional and (for this example) into this element. The average power level with the present parameter values is equal to P e =0.25W. A reactive power term is also evident, which in this case was arbitrarily chosen to be capacitive, which corresponds with a negative reactive value of Q e = −0.25VA. The total real power which is absorbed by the circuit is equal to P R + P e =0.25 + 0.25 = 0.5W, which corresponds to the power P =0.5W (supplied by the current source), as shown in figure 5.3. The reactive component sumisequal to Q L +Q e =1.116−0.25 = 0.866VA , which corresponds to the reactive power level of the circuit as shown in fig- ure 5.3. From this analysiswe canalsoobserve wheretherealandreactive power ends up in the circuit. The real power supplied by the source is equally divided between the resistance and voltage source u e . Furthermore, in this example, Concept of Real and Reactive Power 129 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 0 1 time (s) p R VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 0 1 time (s) p L VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 0 1 time (s) p e re VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 0 1 time (s) p e im VA P R =0.25 Q L =1.116 P e =0.25 Q e =−0.25 Figure 5.5. Power plot for R, L, e circuit, component side the analysis shows the total reactive power Q is predominantly linked with that of the inductance. However, a small capacitive component (-0.25/0.866) (0.866VA represents the total reactive power level) is also linked with the volt- age source u e . 5.3 Power in three-phase systems The approach used to explain the concept of power in single phase systems is extended in this section to three-phase systems. Use will be made of the material presented in chapter 4 in particular with respect to the use of space vectors. It is helpful to assume a three-phase supply current source of the form i R = ˆ i cos (ωt) (5.22a) i S = ˆ i cos (ωt − γ) (5.22b) i T = ˆ i cos (ωt − 2γ) (5.22c) The phase voltages which appear across the respective supply current sources may be written as 130 FUNDAMENTALS OF ELECTRICAL DRIVES u R =ˆu cos (ωt + ρ) (5.23a) u S =ˆu cos (ωt + ρ − γ) (5.23b) u T =ˆu cos (ωt + ρ − 2γ) (5.23c) The corresponding space vector representation (power invariant notation, see equation (4.26)) of the supply waveforms is in this case given as u =  3 2 ˆue jρ e jωt (5.24a)  i =  3 2 ˆ ie jωt (5.24b) where ˆu, ˆ i represent the peak values of the three phase sinusoidal variables. The three-phase circuit configuration as described for the single phase is again used here, which means that each phase consists of a series network in the form of a resistance R, inductance L and voltage source u e which for the three phases 1, 2, 3 is now of the form u e1 =ˆe cos (ωt + η) (5.25a) u e2 =ˆe cos (ωt + η −γ) (5.25b) u e3 =ˆe cos (ωt + η −2γ) (5.25c) which can also be presented in its space vector form u e =  3 2 ˆee jη e jωt (5.26) It is instructive to represent the three-phase network as considered here in its space vector form according to the approach outlined in chapter 4. We will assume for simplicity a ‘star’ connected circuit, given that phase variables are under these conditions equal to supply variables (there is no zero sequence voltage component in this case u 0 =0). The resultant space vector based circuit diagram is of the form shown in figure 5.6. On the basis of figure 5.6 we will examine the power/energy concepts. A suitable starting point is the total Figure 5.6. Space vector rep- resentation of the three-phase circuit [...]... respectively The R-L-ue circuit model is implemented in a single sub-module 141 Concept of Real and Reactive Power SCOPE1 u c 159 7 .57 3u SCOPE3 u L 9.998 SCOPE4 i -9 89.810u u s 10.000 MUL u u L 9.998 B_S u c 159 7 .57 3u + C2 101.32uF p SCOPE5 s -9 898.102u -9 89.810u u L3 100mH 10 s i10.000 - MUL L i9.998 p SCOPE6 L -9 896 .52 3u -9 89.810u MUL u i -9 89.810u c i 159 7 .57 3u p SCOPE7 c -1 58 1.294n -9 89.810u ground... obtained Use of equations (5. 38), (5. 39), (5. 40) leads to PR QL Pe Qe = i (i)∗ R = i (i)∗ ωL = {e (i)∗ } = {e (i)∗ } (5. 45) (5. 46) (5. 47) (5. 48) 5. 5 Tutorials for Chapter 5 5 .5. 1 Tutorial 1 This tutorial is concerned with the energy flow within a resonant circuit connected to a battery source ub = 10V via a switch S which is closed at time t = 0 The resonant period of the circuit, as shown in figure 5. 12 is... (taken from figure 5. 19) P = 293W 142 FUNDAMENTALS OF ELECTRICAL DRIVES 400 u i*10 e 300 200 u (V), i (A), e (V) 100 0 −100 −200 −300 −400 0 0.01 0.02 Figure 5. 18 0.03 time (s) 0.04 0. 05 0.06 Simulink:results u, i, ue 3000 p in p e 250 0 2000 pin (VA), pe (VA) 150 0 1000 50 0 0 50 0 −1000 − 150 0 −2000 0 0.01 0.02 Figure 5. 19 0.03 time (s) 0.04 Simulink: results pin , pe 0. 05 0.06 143 Concept of Real and Reactive... the inductance An example of an m-file implementation for this tutorial is given below: m-file Tutorial 1, chapter 5 %Tutorial 1, chapter 5 close all L=100e-3; C=101.32e-6; Ub=10; T=pi*sqrt(L*C); i=dat(:,4); uc=dat(:,3); ul=dat(:,2); %%%%%%%%%%%%%%%%%%%%%%%% t=[dat(: ,5) ; T]; subplot(3,1,1) plot(dat(: ,5) ,i,’b’); xlabel(’time (s)’) ylabel(’i(A)’) grid subplot(3,1,2) plot(dat(: ,5) ,ul,’r’); xlabel(’time(s)’)... 2V, ρ = π rad, ω = 100rad/s i ˆ 3 1 .5 Q =0.866 R pRim VA 1 0 .5 0 −0 .5 −1 0 0.002 0.004 0.006 0.008 0.01 time (s) 0.012 0.014 0.016 0.018 0.02 0.01 time (s) 0.012 0.014 0.016 0.018 0.02 0.016 0.018 0.02 1 .5 QS=0.866 pSim VA 1 0 .5 0 −0 .5 −1 0 0.002 0.004 0.006 0.008 1 .5 QT=0.866 pTim VA 1 0 .5 0 −0 .5 −1 0 0.002 0.004 0.006 Figure 5. 10 0.008 0.01 time (s) 0.012 0.014 Three-phase ‘reactive’ power plot The... Table 5. 1 Real and reactive value for circuit example Variable Real power resistance Reactive power inductance Real power source ue Reactive power source ue Value PR QL Pe Qe 0. 75 W 3.348 VA 0. 75 W -0 . 75 VA According to table 5. 1 the total real power utilized by the circuit elements is equal to PR + Pe = 0. 75 + 0. 75 = 1.5W which is precisely the power level shown in figure 5. 8 The reactive power of the... phase and orthogonal to the current phasor, not to the voltage phasor 144 FUNDAMENTALS OF ELECTRICAL DRIVES m-file Tutorial 3, chapter 5 %Tutorial 3, chapter 5 %we set E=100; eta =-7 *pi/6 assume we don’t know these values %determine real reactive components on the basis of measured %current close all R=10; % resistance L=100e-3; % inductance w=100*pi; % frequency in rad/s %%%%%%%%%%%%%%%% t=dat(:,4);... example of an m-file for this problem is given below: m-file Tutorial 4, chapter 5 %Tutorial 4, chapter 5 %%%%eta=-pi/3 R=10; L=100e-3; % phase resistance % phase inductance(H) Concept of Real and Reactive Power 147 w=100*pi; % supply frequency (rad/s) X=w*L; % load reactance C=sqrt(2/3); % space vector constant power invariant U=220; % supply phase voltage RMS E=100; % EMF phase voltage RMS rho_e=-pi/3;... 0.1 0 0 0.001 0.002 0.003 0.004 0.0 05 time (s) 0.006 0.007 0.008 0.009 0.01 0 0.001 0.002 0.003 0.004 0.0 05 time (s) 0.006 0.007 0.008 0.009 0.01 0 0.001 0.002 0.003 0.004 0.0 05 time (s) 0.006 0.007 0.008 0.009 0.01 10 ul (V) 5 0 5 −10 20 uc (V) 15 10 5 0 Figure 5. 14 Simulink: results i (t), uL (t), uc (t) be studied with the aid of the instantaneous power for each of the elements The power for the... calculation of the real power of for example the R-L-ue circuit is defined by equation (5. 30) Substitution of equation (5. 42) leads to the phasor based form namely P = {u (i)∗ } (5. 43) 137 Concept of Real and Reactive Power A similar approach for conversion from space vector to phasor format can also be carried out for the reactive power equation (5. 36) which gives Q= {u (i)∗ } (5. 44) Similarly the real . 0.02 −1 −0 .5 0 0 .5 1 1 .5 time (s) p re VA 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 −1 −0 .5 0 0 .5 1 1 .5 time (s) p im VA P=0 .5 P=0 .5 Q=0.866 Figure 5. 3. Power plot for R-L-u e circuit,. (5. 39), (5. 40) leads to P R = i (i) ∗ R (5. 45) Q L = i (i) ∗ ωL (5. 46) P e = {e (i) ∗ } (5. 47) Q e = {e (i) ∗ } (5. 48) 5. 5 Tutorials for Chapter 5 5 .5. 1 Tutorial 1 This tutorial is concerned. in terms of two components which with the aid of equations (5. 3), (5. 7) can also be written as p re =ˆu ˆ i cos ρ cos 2 ωt (5. 10a) p im = −ˆu ˆ i sin ρ cos ωt sin ωt (5. 10b) 124 FUNDAMENTALS OF ELECTRICAL