Electrcal Crcuts (ECE33b SteadyState Power Analyss Anests Dounas The Unersty of Western Ontaro Faculty of Engneerng Scence
SteadyState Power Analyss (t AC crcut: The steady state oltage and current can be wrtten as: (t Z Fg : Smple AC crcut ( t cos( ω t θ ( t cos( ω t θ m, m : ampltude or maxmum alue ω: radan or angular frequency θ, θ : phase angle nstantaneous power: s the product of the oltage and current. p ( t ( t ( t
SteadyState Power Analyss nstantaneous power p ( t ( t ( t ( t cos( ω t θ ( t cos( ω t θ p ( t cos( ω t θ cos( ω t θ Trgonometrc dentty: cos φ cos φ [cos( φ φ cos( φ φ ] nstantaneous power p( t [cos( θ θ cos( ωt θ θ ]
Example : nstantaneous Power (t (t Z The crcut has the followng parameters: (t 4 cos(ωt6 and Z 3. Fnd the current and nstantaneous power as a functon of tme. Smple AC crcut The oltage n phasor notaton s 4 6 Current 4 6 Z 3 3 nstantaneous power (t cos(ωt3 A p ( t [cos( θ θ cos( ω t θ θ ] 4 * p ( t [cos( 6 3 cos( ω t 6 3 p ( t 3.46 4 cos( ω t 9 ]
Aerage Power Aerage Power: s smply the energy delered durng one full cycle P T T where: ( t ( t dt T π/ω s the perod P s n watts ( t cos( ω t θ ( t cos( ω t θ P T T ( [cos( θ θ cos( ω t θ θ ] dt P cos( θ θ θ θ : s the phase angle of the mpedance
Example : H s (t.5kω.5µf For the network shown, s ( t cos( ωt 45, where ωπ(6 s. Calculate the aerage power delered by the source and erfy that t equals the aerage power delered to the resstance. mpedance of nductor jω jπ(6 377j mpedance of capactor j/ωc j/[π(6.5e6] 768j
Example : j377 Ω Z s 5Ω j768ω Z j377 5 j768 j377. j.5656 Z j377 j377 48 5. 79.8 5.79 Z j377 48(cos( 5.79 j sn( 5.79 463 j46 53 8.3
Example : j377 Ω s 5Ω j768ω Aerage Power delered to the network by the source: P P ( Z cos( θ cos( θ θ θ where Z Z 53 8.3 ( P cos(8.3. 69 53 W
Example : j377 Ω s 5Ω j768ω Power delered to the resstance: oltage dder: 48 5.79 j377 48 5.79 S.97 43.77.97.97 * S.97 Aerage Power dsspated by the resstance.97 P cos( θ θ cos(. Z 5 68 W
Example 3: s (t j377 Ω j768ω For the network shown, s ( t cos( ωt 45, where ωπ(6 s. Calculate the aerage power delered to the load. P cos( ± 9
Example 4: j Ω 6 Ω Fnd the aerage power absorbed by total mpedance Z 6 j 6.83 45 3.53 5 Aerage power P ( (3.53 cos( 6 5 cos( θ θ P. 5 W
Example 4: Alternate solutons 6 oltage Dder j Ω ( 6 j Ω 7.7 5 Z or P Aerage power P ( cos( θ θ Z ( (7.7. 5 ( Z cos( θ θ ( (3.53. 5 W W
Example 5: Ω 6 Z 4Ω j4ω For the crcut shown, fnd the aerage power absorbed by each resstor and the total aerage power suppled. nput mpedance Z j4.88 45 4.47 4 6.57
Example 5: Ω 6 Z 4Ω j4ω Z 4.47 6 6.57.685 86.57 Aerage Power of Ω resstor P (.685 ( 7. W
Example 5: Ω 6 X 4Ω j4ω X.88 45 4.47 6.57 ( 6 7.59 4.57 Aerage Power of 4Ω resstor (7.59 P 7. 4 W
Example 5:.685 86.57 Ω 6 4Ω j4ω Aerage Power of oltage source P cos( θ θ ((.685 cos(6 86.57 P 4. 4 W (power suppled Note that the power absorbed by the resstors s equal to the power suppled by the oltage source
f supply s absorbng or generatng power f P s poste, then power s beng absorbed f P s negate, then power s beng generated
axmum Power Transfer (eew For resste load only load P To fnd alue of that yelds the maxmum power to the load Dfferentate P load w.r.t. and equate the derate to zero. ( ( ( 4 load d dp axmum power at load occurs at:
axmum Aerage Power Transfer For any load mpedance, where Z TH TH Z jx jx TH Z TH Z To fnd alue of Z that yelds the maxmum power to the load. ( X X / [ ] ( TH TH Pload ( ( TH XTH X To decrease the magntude of the denomnator select X X TH
axmum Aerage Power Transfer Ths reduces the power expresson to: Pload ( TH The aboe equaton s smlar to the purely resste case and yelds maxmum power when TH The maxmum aerage power transfer to the load occurs when: Z jx jx TH TH Z TH *
axmum Aerage Power Transfer For the case when X (.e. load s purely resste, Z TH t can be shown that maxmum power at the load occurs when: TH X TH
Example 6: Fnd Z for maxmum aerage power transfer to the load and the maxmum aerage power transferred to the load 3 45 Ω j Ω Z oad: Z j Ω axmum power at load: 3 Pload. 565 ( ( TH W
Example 7: For the crcut shown, fnd Z for maxmum aerage power transfer and the maxmum aerage power transferred to the load Ω 4 jω Z j Ω Frst step to sole ths crcut must fnd Theenn equalent crcut
Example 7: To fnd the Theenn equalent crcut at the load, we must compute the opencrcut oltage. X 4 j.73 45 4 Ω X jω j Ω TH TH TH [ 9 8 j8 6 j8 *.73 45 ] TH 8.97 7.57
Example 7: Next step s to fnd Theenn equalent mpedance jω Z TH ust short crcut all oltage sources ust open crcut all current sources Ω X j Ω Z TH j j j
Example 7: Z TH j 8.97 7.57 Z oad for maxmum power: Z j Ω axmum power at load: (8.97 Pload 45 ( ( TH W
Effecte or oot ean Square ( alues The power delered to loads depends on the type of source DC source: Power snusodal source: Power ½( A technque to compare the effecteness of dfferent perodc waefo n delerng power to a resste load s requred. By knowng the effecte current and/or oltage of perodc sgnals the power delered to the resste load can be expressed as T where eff P T ( t dt T T eff ( t dt
Example 8: Compute the alue for the snusodal waeform ( t cos( ωt θ, whch has a perod of Tπ/ω. eff T cos ( ωt θ dt / Trgonometrc dentty T eff cos(ωt θ dt cos φ cos φ / Comparng ampltude m wth n power calculaton P
Example 9: Calculate the alue for the sawtooth oltage waeform shown A For < t T ( t t T T 3T t A T T T / T A A ( t dt t dt T T 3
Power Factor Aerage power: P cos( θ θ Power factor The product s referred to as the apparent power where P θ Z cos( θ θ unts cos( θ (A θ θ phase angle of mpedance power factor angle Hgher power factor mnmzes transmsson lne losses θ θ Z Z laggng leadng pf pf (nducte (capacte Z ust label pf as leadng or laggng laggng and leadng refers to the phase of current w.r.t. oltage
Example : Fnd the power factor and draw phasor dagram for the oltage and current (a Z j and (b Z j Z (a Pf cos(arctan ( cos( 6.57. 894 (laggng Z.36 6.57.447 6.57 6.57 (b Pf.894 (leadng.447 6.57 6.57
Example : An ndustral load consumes kw at.77 pf laggng. The 6Hz lne oltage at the load s 48. The transmsson lne resstance between the power company transformer and the load s. Ω. Determne the power sangs that could be obtaned f the pf s changed to.94 laggng..ω s 48 P kw Pf.77(laggng P e3 94. 7 (pf(.77 (48 A
Example : 94.7.Ω s 48 P kw Pf.77(laggng PS P e3 (94.7. 8. 68 kw f pf s changed to.94 laggng P e3. 6 (pf(.94(48 A PS P e3 (.6. 4. 9 kw Power sang 8.68 4.9 3.77 kw
Complex Power Complex power S * f θ and θ, then Z Z θ S θ θ θ θ S cos( θ θ j sn( θ θ S P jq where P cos( θ θ Q sn( θ θ where S P Q Apparent power (A P Q real power (W reacte power (A
Complex Power: Phasor Dagrams Assume θ m θ e mpedance Dagram Z Z θ jx Z θ m jx θ e
Complex Power: Phasor Dagrams Power Trangle P Q X cos θ sn θ S P Q / ( m jq S θ P Power factor cosθ P/( cos(sn [Q/( ] cos(tan [Q/P] e Note: f θ s poste f θ s negate nducte network Capacte network Complex power, lke energy s consered. The total complex power generated s equal to sum of complex power delered to each nddual load
Example : An ndustral load requres 4kW at.84 pf laggng. The load oltage s at 6Hz. The transmsson lne mpedance s.j.5ω. Determne the real and the reacte power losses n the lne and the real and reacte power requred at the nput to the transmsson lne. j.5ω.ω s P 4kW Pf.84(laggng
Example : j.5ω.ω s P 4kW Pf.84(laggng P 4 S 47 69 pf.84 A pf.84 cos θ cosθ θ 3.86 Complex power at the load S 47 69 3.86 4 j5 837 S * 47 69 3.86 * 6.45 3.86
Example : j.5ω.ω s Complex power losses n the lne P 4kW Pf.84(laggng Slne Z (6.45 (. j.5 4 685 j 73 A Plne 4 685 W Qlne 73 AS Complex power at the generator S S S S S S lne 44 685 j37 55 A ( 4 j5 837 ( 4 685 j 73 PS 44 685 W Qlne 37 55 AS
Example 3: Complete the followng table: P n kw Q n kas Apparent Power (ka Power factor mpedance (Ohms oltage (absolute alue Current (absolute alue 5.8.89 lag 3. 6.6 6 87.95 lag 3 5.84 5 Apparent Power ( 5 /.8 ka Power factor /.8.89 lag Current P/(cosθ e3/(6*.89 87 A mpedance / 6/87 3. Ω Phase of mpedance tan (5/ 6.6
Example 3: Exercse: Complete table and erfy solutons P n kw Q n kas Apparent Power (ka Power factor mpedance (Ohms oltage (absolute alue Current (absolute alue 5.8.89 lag 3. 6.6 6 87.4 3.75.95 lag. 8. 3 7 3 3.9 lag 5.84 6 5
Power Factor Correcton ost ndustral loads hae laggng power factor (eg. nducton motors Can we mproe the power factor by usng a capacte load n parallel. ncreases the power factor. n addton, ths does not change the real power T C Generator ndustral load wth laggng pf C
Power Factor Correcton T C Generator ndustral load wth laggng pf C S S C S T jm jm jm S θ Q S T θ T Q C P e Q C e P e
Power Factor Correcton T C Generator ndustral load wth laggng pf C Complex Power S old P old jq old S old θ old S S cap new cap cap jq S S old S cap S new 9 θ new S cap * C Z * j ω C
Example 4: A small machne shop s suppled. 6 Hz, and has a load of kw and.8 laggng power factor. What alue of capactor must be connected n parallel to produce unty power factor. P kw Pf.8(laggng pf cosθ old.8 θ old 36.87 tanθ old Q old /P old Q old ( tan(36.87 6 5 ars Q cap Q new Q old ωc 6 5 (π6c( C 94µf
Example 5: 4 6 Hz P 48kW Pf.6(laggng P 4kW Pf.96(leadng For the network shown aboe calculate the followng:. The total real power, reacte power, apparent power.. The magntude of the total plant current. 3. The power factor of the combned loads. 4. How many ars of capactors must be connected n parallel to the plant power factor to.9 lag. 5. What s the alue of the capactance.
Example 5: 4 6 Hz P 48kW Pf.6(laggng P 4kW Pf.96(leadng. The total real power, reacte power, apparent power. oad oad Total Power eal power P 48 kw Apparent power S P /pf 48k/.6 8 kars eacte power Q (8 48 / 64 ka eal power P 4 kw Apparent power S P /pf 4k/.96 5 kars eacte power Q (5 4 / 7 ka P T 48k4k 7 kw Q T 64k7k 57 kars S T (7 57 / 9.8 ka
Example 5:. The magntude of the total plant current. T S T / 9 8/4 38.5 A 3. The power factor of the combned loads. Pf P T /S T 7/9.8.7843 lag 4. How many ars of capactors must be connected parallel to the plant power factor to.9 lag. pf cosθ new.9 θ new 5.84 tanθ new Q new /P T Q new (7 tan(5.84 34 87 ars Q cap Q new Q old 34 87 57 9 ars 5. What s the alue of the capactance. Q cap ωc C 9/(π*6*4 9 µf
SnglePhase ThreeWre Crcuts Typcal house hold connecton a a neutral n b n b aa n N b B Z Z A N B a n b Z lne Z n Z lne Z Z A N B Z et an nb aa /Z bb /Z nn ( aa bb (/Z /Z AB
Summary nstantaneous power: s the product of the oltage and current as a functon of tme. p ( t ( t ( t Aerage power: s obtaned by aeragng the nstantaneous power, s the aerage rate at whch energy s absorbed or suppled. For snusodal sgnals: Power at resste load: P P cos( θ θ snce m and m are n phase. Power at reacte load: P cos( ± 9
Summary axmum aerage power transfer: Z jx jx TH TH TH the load mpedance s equal to the complex conjugate of the Theenn equalent mpedance of the network Z * Z TH Z S or effecte alue of perodc waeform: Used to measure the effecteness of perodc waeform sgnals n delerng power to a resste load. The effecte alue s of a perodc wae form s found by determnng the rootmeansquare alue of the waeform For snusodal functon s the ampltude dded by For sawtooth functon s the ampltude dded by 3
Summary Power factor: s the rato of the aerage power to the apparent power leadng laggng phase of current leads the oltage phase of current lags the oltage aggng power factor can be corrected by placng a capactor n parallel wth the load. Complex power: S * where θ, θ S P jq where P real or aerage power (W Q magnary or reacte power (A