Three Phase Circuitsit Revised October 6, 008. Three-Phase Circuits 1
Preliminary Comments and a quick review of phasors. We live in the time domain. We also assume a causal (nonpredictive) world. Real-world signals are always of the form: Where f(t)i is real and, f t f t t t 0, o No one has ever measured an imaginary physical quantity.. Three-Phase Circuits
Preliminary Comments and a quick review of phasors. In some applications, such as systems whose bandwidth is narrow, or because of the nature of the excitation, the signals are well-approximated by sinusoids. Furthermore, if we assume that we wait a sufficiently long time for any transient behavior to decay to zero, then all signals (voltage, current, etc.) will take the form: f t Acos t Here the amplitude A and the phase shift are real quantities.. Three-Phase Circuits 3
Preliminary Comments and a quick review of phasors. To ease our computational burdens, classical circuit theory reveals that we may express sinusoidal signals as where cos j t A t Re Ae j j t Re Ae e j t Re Ae j A Ae is known as a phasor.. Three-Phase Circuits 4
Preliminary Comments and a quick review of phasors. In EEL 3111 you learned how to express the excitation(s) in phasor form, perform the circuit analysis required to obtain the phasor form of the quantities sought, and then return to the quantity as it would be measured by multiplying the phasor form of the answer by exp[jt] then taking the real part. We review this process with a simple example.. Three-Phase Circuits 5
Preliminary Comments and a quick review of phasors. For the circuit shown, find the voltage v(t) across the capacitor C. R L + V o sin(t) ~ C v C (t) _. Three-Phase Circuits 6
Preliminary Comments and a quick review of phasors. j jt Vosint Vocost Re Voe e j o o o V V e jv Remember: e j cos jsin. Three-Phase Circuits 7
Preliminary Comments and a quick review of phasors. R R L jl jv o ~ C 1 jc V C V C C R L C V o. Three-Phase Circuits 8
Preliminary Comments and a quick review of phasors. 1 C j j VC jvo Vo 1 R jl 1 LC jcr jc V o exp j RC 1 LC 1 1 LC RC exp j tan Vo 1 RC exp j jtan 1 LC RC 1 LC. Three-Phase Circuits 9
Preliminary Comments and a quick review of phasors. j t vc t ReV Ce Re Vo exp tan RC 1 LC 1 LC RC V 1 jt j j e RC 1 LC o 1 cos t tan 1 LC RC V o 1 sin t tan 1 LC RC RC 1 LC. Three-Phase Circuits 10
Preliminary Comments and a quick review of phasors. What is the current through the capacitor, both in phasor form and as a function of time? R R L jl I jv o ~ C 1 jc V C. Three-Phase Circuits 11
Preliminary Comments and a quick review of phasors. V o ~ I T V o I, it T j t Re Ie T R L C. Three-Phase Circuits 1
Preliminary Comments and a quick review of phasors. T R j L jc 1 1 LC jrc RC j 1 LC T C 1 1 T RC LC C T tan jc 1 1 LC RC T. Three-Phase Circuits 13
Preliminary Comments and a quick review of phasors. I V o jvo T T T o 1 LC 1 1 RC 1 LC exp j tan C jv j CV o exp j tan RC 1 LC CV RC 1 LC 1 1 RC LC RC 1 LC o 1 I exp jtan j RC. Three-Phase Circuits 14
Preliminary Comments and a quick review of phasors. it j t Re Ie cost tan RC CVo 1 1 LC RC 1 LC CV o RC 1 LC sin t tan 1 1 LC RC Review your EEL 3111 text if necessary.. Three-Phase Circuits 15
Power in AC Circuits vt ~ it L Instantaneous Power: p t v t i t. Three-Phase Circuits 16
Power in AC Circuits For sinusoidal steady state excitation v t Vcos t ~ I L L L L V V L j L I e i t I cos t L L. Three-Phase Circuits 17
Power in AC Circuits For sinusoidal steady state excitation cos cos pt vtit V t I t L cos cos V I t t V I cost costcos L sintsin L V I t L t t 1 sin t L cos cos cos sin sin L. Three-Phase Circuits 18
Power in AC Circuits For sinusoidal steady state excitation: pt V Icos tcosl sin tsinl 1 Still not very useful.. Three-Phase Circuits 19
Power in AC Circuits Time-Averaged Power: T T 1 1 Pave p t dt v t i t dt T T 0 0. Three-Phase Circuits 0
T 1 1 P ave v t i t dt, T T f 0 1 V I cos tcosl sin tsinl dt 0 1 V I cos t cosldt sin t sinldt 0 0 1 L 0 0 V I cos L cos tdt V I sin sin tdt average is zero. Three-Phase Circuits 1
Power in AC Circuits 1 ave cos L cos cos L 0 P V I tdt V I Time-Averaged Power: 1 Pave V I cosl P ave is a constant. This result is quite useful.. Three-Phase Circuits
Power in AC Circuits 1 Pave V I cosl Observation: For phasor voltage V and phasor current I compute: 1 Re * VI. Three-Phase Circuits 3
Power in AC Circuits 1 * 1 j 1 L Re Re jl VI V I e V I Re e 1 V I cosl Pave Average power is very simple to compute for the case of sinusoidal steady state circuits.. Three-Phase Circuits 4
Complex Power: 1 1 1 1 cos sinl P Q * j L S VI V I e V I L j V I P jq S : Complex Power P : Average Power Q : Reactive Power S : Apparent Power cos : Power Factor L. Three-Phase Circuits 5
Complex Power: Though the units of S, P, Q, and the apparent power are all volts times amps or watts, the term watt however is reserved for the average power. Units for the reactive power Q are termed vars, for volt-amperes-reactive, and the units for the apparent power is volt-amperes. Review your EEL 3111 text if necessary.. Three-Phase Circuits 6
Complex Power: 1 1 j 1 L S I L e I L cosl jsinl * S VI, V I 1 P I L cosl Average Power 1 Q I L sin L Reactive Power. Three-Phase Circuits 7
Complex Power: j but e L R jx cos j sin L L L L L L L L 1 P I R L :Average Power 1 Q I X L : Reactive Power. Three-Phase Circuits 8
Review your EEL 3111 text if necessary.. Three-Phase Circuits 9
What is a Three-Phase System? + _ ~ A A t Vsin t + v t V sin t 10 v t V t V V0 ~ _ B V B V10 + v t Vsint 40 C ~ _ V V40 C Convention: Note that the amplitude V is expressed as an rms value, while the peak amplitude is V. Three-Phase Circuits 30
What is a Three-Phase System? A t v v t v t B C. Three-Phase Circuits 31
Relationship among phasors: V V C 10 V V B 10 10 V A 10 V A V B 10 10 V C abc phase sequence acb phase sequence. Three-Phase Circuits 3
One Common three-phase connection the Y-connection. Line-Neutral Voltage I IA B a b Line Current V A V B n neutral V C I N I C c A Y-Connected Generator. Three-Phase Circuits 33
One Common three-phase connection the Y-connection. neutral A Balanced Y-Connected Load all impedances are equal. Three-Phase Circuits 34
One Common three-phase connection the Y-connection. V A V V C A I I I B V I N I C B A Y-Connected Generator and Balanced Load a b n c I, I, I, I,? A B C N. Three-Phase Circuits 35
V A V C I I A I B V B I N I I C a b n KCL I I : entering leaving c I I I I A B C N. Three-Phase Circuits 36
Note the three independent circuits: V A V V C I a V A A I I A I B V I N I C B b n c. Three-Phase Circuits 37
Note the three independent circuits: V A V V C I a V B A I I B I B V I N I C B b n c. Three-Phase Circuits 38
Note the three independent circuits: V A I V V C C I A I I B V I N I C V C B a b n c. Three-Phase Circuits 39
V A V B V C I N IA IB IC V V V A B C V0V10V40 V j 10 j 40 1e e 0 Do the math. A three-phase power system in which the three generators have voltages that are equal in magnitude but differ in phase by 10 degrees and in which all three loads are identical is called a balanced three-phase system. In such a system the current in the neutral is zero and the neutral wire is essentially unnecessary.. Three-Phase Circuits 40
This is the same as... V A V C I A I B V B I N I C a b n c. Three-Phase Circuits 41
... this, for a balanced system V A I A I B V B a b V C I C c. Three-Phase Circuits 4
Note also that for a balances system: V V V V A B C IA IB IC IA, B, C I L I ABC,, Line Current = Phase Current. Three-Phase Circuits 43
V an V cn V V bn Line-to-Neutral Voltages I I c I a a I b b c Line-to-Line Voltages: V V V V V ab V bc ab an bn. Three-Phase Circuits 44
Line-to-Line Voltages: V V V ab an bn V 0 V 10 1 j 10 1 cos10 sin10 V e V j 1 3 3 j 3 V 1 j V 1 1 3 V tan 3 V 30 3. Three-Phase Circuits 45
Line-to-Line Voltages: V V V 3V30 V an V cn ab an bn V ab V ca V V bn V LineLine 3 V V bn V an V ab (the line-line voltage) leads V an (the phase voltage) by 30 o V cn V bc. Three-Phase Circuits 46
Another Common three-phase connection the Delta-connection. V ca I ca ~ Possible since + _ I ab I a a V Clearly: ab I b b V LineLine I V bc bc I c c V V V ab ca bc 0 V. Three-Phase Circuits 47
Another Common three-phase connection the Delta-connection. I I c V ca I ca c I bc a ~ + _ I V bc ab b V ab I a I b R R R Since the load is resistive i the phase currents will be in-phase with the phase voltages: I I 0, I I 10, I I 40 ab bc ca. Three-Phase Circuits 48
V ca I ca c I bc a ~ + _ I V bc ab b I I I V a ab ca ab I I I b bc ab R R R I I I c ca bc. Three-Phase Circuits 49
Using very similar algebra as before we find: I I I a ab ca 3I30 or I Line 3I Phase for a balanced delta-connected generator.. Three-Phase Circuits 50
Line Currents: I ca I c I bc I a Iab Ica 3I30 I a (the line current) lags I ab (the phase current) by 30 o I ab I b I ab I bc I I a ca. Three-Phase Circuits 51
For a balanced three-phase load there is an equivalence between the and Y configuration. Y in Y in Y Y. Three-Phase Circuits 5
- Y Transformation Y in Y Y Y Y in 3 3. Three-Phase Circuits 53
- Y Transformation for a Balanced Three-Phase Load in Y Y 3 3 iny a a /3 b c b c /3 /3 These are equivalent, not the same.. Three-Phase Circuits 54
- Y Transformation for a Balanced Three-Phase Source? We ve really already done it. a a Ve Ve j40 j0 j0 Ve Y V ab ~ + _ Ve b j10 j40 Ve Y c Ve Y j10 b c V ab Ve Ve Ve j0 j0 j10 Y Y j10 3 j 3 V V 1 e Y V Y 3V 30 Y. Three-Phase Circuits 55
- Y Transformation for a Balanced Three-Phase Source. a a Ve Ve j40 0 j0 Ve j0 Y ~ + _ Ve b j10 j40 Ve Y c Ve Y j10 b c V 3V e Y j30 V j30 V Y e 3. Three-Phase Circuits 56
- Y Transformation. Source or Y a b c V ac I a b I I I c V bc V ab a b c Load or Y Regardless of the configuration of the source or load ( or Y), the - Y transformation will preserve the line currents and the line- line voltages for a balanced three-phase h system. Let s look at an example.... Three-Phase Circuits 57
Example Three impedances are connected as shown. For a balanced line-to-line voltage of 08 V, find the line current, the power factor, and the total power, reactive power, and volt-amperes.. Three-Phase Circuits 58
Example Three impedances are connected as shown. For a balanced line-to-line voltage of 08 V, find the line current, the power factor, and the total power, reactive power, and apparent power. V 08 3 I V Y. Three-Phase Circuits 59
Solution: The rms amplitude of the line-to-neutral voltage of any one phase is 08 V 10 3 thus the line current is I V 10 4 36.9 5 36.9 Y. Three-Phase Circuits 60
Power Factor : cos 36.9 0.8 lagging Powers : 08 P 3 I R 3 I cos 3 4 Y Y 69. watts 5 3 08 Q 3 I X 3 I sin 3 3 Y Y 5191.7 vars 5 3 08 08 S 3V I 3 I Y 3 865.8 va 3 5 3. Three-Phase Circuits 61
Example Three impedances are connected as shown. For a balanced line-to-line voltage of 08 V, find the line current, the power factor, and the total power, reactive power, and apparent power. I I 08 15. Three-Phase Circuits 6
Solution: The current in any one leg of the is 3 times smaller than the line current, thus: V 08 I LL 13.87 36.9 15 36.9 Power Factor : cos 36.9 0.8 lagging. Three-Phase Circuits 63
Powers: 08 P 3 I R 3 I cos 3 169. watts 15 08 Q 3 I X 3 I sin 3 95191.68 vars 15 08 S 3VLL I 308 865.8 va 15 Results from the Y-connected load: P 69. watts Q 5191.7 vars S 865.8 va. Three-Phase Circuits 64
Comparison of the results previous two examples verifies the equivalence of the Y- conversion, Y 3 We see that the line-to-line voltage, g, line current,,p power factor, total power, reactive power, and volt-amperes are equal in the two cases. Conditions viewed from the terminals a, b, and c are identical, and one cannot distinguish between the two loads from their terminal quantities.. Three-Phase Circuits 65
Why do we want Three-Phase h Systems?. Three-Phase Circuits 66
Recall the instantaneous power expression for a single phase system: pt V Icos tcos t L Note how the instantaneous power varies between zero and P max. Three-Phase Circuits 67
Instantaneous power for a three-phase system: an sin sin 10 sin 40 v t V t v t V t bn an v t V t cn v ia t t + _ sin sin 10 Isin 40 i t I t a i t I t b i t I t c ic vcn t t ib t vbn t p t v t i t v t i t v t i t an a bn b cn c. Three-Phase Circuits 68
Instantaneous power for a three-phase system: an a bn b cn c sintsint VI sint 10sint 10 sint 40sin t 40 pt v ti t v ti t v ti t 3VI cos A constant! algebra The instantaneous power supplied to a balanced three-phase load is constant for all time. This is a major advantage over single phase systems.. Three-Phase Circuits 69
Instantaneous power for a three-phase system determined via Line Line voltages: We found that, pt P3VIcos Now, for a Y-connected load, I I, andv 3V V LL thus, P 3VIcos 3 ILcos 3 VLL ILcos 3 P 3 V I cos LL L L LL. Three-Phase Circuits 70
Instantaneous power for a three-phase system determined via Line Line voltages: We found that, pt P3VIcos Now, for a -connected load, I 3, I andv V IL thus, P 3VI cos 3V LL cos 3V LLIL cos 3 L P 3V I cos Exactly the same result as for the Y-connected load. LL L LL Same p(t) = P regardless of the connection of the load. Three-Phase Circuits 71
Example For the three-phase circuit shown, determine a) the magnitude of the line currents b) the line-to-line and line-to-neutral l( (phase) voltages for the load c) the average, reactive, and apparent power consumed by the load d) the power factor of the load e) the average, reactive, and apparent power consumed by the line f) the average, reactive, and apparent power supplied by the generator g) the power factor of the generator. Three-Phase Circuits 7
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I a) the magnitude of the line currents T j 10.05 3. 10.5517.66 I 10 11.37 17.66 T I 11.37. Three-Phase Circuits 73
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I b) the line-to-line and line-to-neutral (phase) voltages for the load V L V Load Load I Load 10 j3 10.4416.7 V Load 11.37 10.55 119.5 0.96 V 3 V 07.77 volts LL Load volts. Three-Phase Circuits 74
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I c) the average, reactive, and apparent power consumed by the load P V I cos or Load Load Load V 118.7 11.37 cos 16.7 19.7 watts P I R 11.37 10 19.7 watts Load Load L. Three-Phase Circuits 75
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I c) the average, reactive, and apparent power consumed by the load Q V I or sin Load Load Load V 118.7 11.37 sin 16.7 387.83 vars Q I X 11.37 3 387.8383 vars Load Load L. Three-Phase Circuits 76
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I c) the average, reactive, and apparent power consumed by the load S V I or Load Load 118.7 11.37 1349.6 va V S I 11.37 10.44 1349.6 va Load Load L. Three-Phase Circuits 77
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I d) the power factor of the load V L cos cos 16.7 0.96 Load. Three-Phase Circuits 78
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I e) the average, reactive, and apparent power consumed by the line Line 0.0505 j0. 0.06 75.964 P I cos or Line Line Line Line V 19.8 0.06 cos 75.96 6.46 watts P I R 11.37 0.05 6.46 watts Line. Three-Phase Circuits 79 L
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I e) the average, reactive, and apparent power consumed by the line Line 0.0505 j 0. 0.06 75.964 Q I sin or Line Line Line V 19.88 0.06 cos 75.96 5.84 vars Q I X 11.37 0. 5.86 vars Line Line. Three-Phase Circuits 80 L
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I e) the average, reactive, and apparent power consumed by the line V L Line 0.05 j0. 0.0675.964 S I Line Line 19.8 0.06 6.63 va. Three-Phase Circuits 81
Solution Since this is a balanced three-phase circuit, proceed on a perphase basis. I f) the average and reactive supplied by the generator V L P P P 6.46 19.7 199.16 watts Gen Line Load Q Q Q 5.84 387.83 413.67 vars Gen Line Load. Three-Phase Circuits 8
f) the apparent power supplied by the generator S P jq Gen Gen Gen Gen Gen Gen 199.16 413.67 S P Q 1363.46 va Gen Q 1 Gen tan 17.66 P Gen Gen cos 0.95. Three-Phase Circuits 83
Since this was solved on a per-phase basis we must now multiply the powers by a factor of three. PLoad 319.7 3878.1 watts QLoad 3 387.8383 1163.49 vars SLoad 31349.6 4048.8 va P Line 3 6.46 6 19.38 watts QLine 3 5.86 77.58 vars S Line 3 6.63 79.89 va PGen 3199.16 3897.48 watts QGen 3 413. 67 141. 07 vars S 3 1363.46 409 0.38 va Gen P P P QGen QLine Q S S S Gen Line Load Load Gen Line L o ad. Three-Phase Circuits 84
What have we learned? Since the neutral conductor in a balanced three-phase circuit carries no current and hence has no voltage drop across it, the neutral conductor is considered to have zero impedance. We used a single-phase equivalent circuit, in which all quantities correspond dto those in one phase of fthe three-phase h circuit. it Conditions in the other two phases being the same (except for the 10 phase displacements in the currents and voltages), there is no need for investigating them individually. Line currents in the three-phase system are the same as in the singlephase circuit, and total three-phase power, reactive power, and voltamperes are three times the corresponding quantities in the single- phase circuit. it. Three-Phase Circuits 85