Chapter 10 ACSS Power - PDF Free Download

Objectives: Power concepts: instantaneous power, average power, reactive power, coplex power, power factor Relationships aong power concepts the power triangle Balancing power in AC circuits Condition for axiu power transfer to a load in AC circuits

nstantaneous power: p( t) v( t) i( t) [... cos( t )][ (lots of v cos( t ) trigonoetry) cos( v i) cos( v i)cost sin( v i)sin t Notes: The first ter is a constant instantaneous power is not syetric about the tie axis! The second and third ters have a frequency that is twice the frequency of v(t) and i(t)! i...

nstantaneous power

nstantaneous power, rewritten: p( t) P where P cost Q sin t P cos( v i ) Q sin( v i ) Notes: P is the average (or real) power and has the units watts [W] P 1 T t0 T t 0 p( t) dt Q is the reactive power and has the units volt-apere-reactive [var]

ook at the expressions for P (real power) and Q (reactive power) for the three different circuit coponents R,, and C. The expressions depend on the difference between the phase angle of the voltage drop across the coponent and the phase angle of the current through the coponent.

A. Dependent on the source frequency B. Dependent on the resistance C. Equal to the voltage phase angle D. The negative of the voltage phase angle

Reeber for a resistor in the presence of an AC source, i i v i v R R 0 ) )( 0 ( The voltage and current for a resistor are inphase! Thus, for a resistor, t P P t p Q P i v i v i v i v cos ) ( 0 ) sin( ; ) cos( 0 ) sin( 1; ) cos(0 ) cos(

A. The sae as the voltage phase angle B. 90 o ore than the voltage phase angle C. 90 o less than the voltage phase angle D. Dependent on the inductance

How can you reeber the relationship between the voltage and current phase angles for inductors and capacitors? E the CE an! Huh? n the early days of electrical engineering, voltage was known as electrootive force, and was represented by the sybol e in equations! Current is represented by the letter i, as usual. is inductance, C is capacitance.

Chapter 10 ACSS Power E the CE an! E voltage is ahead of (i.e. EADS) current in an inductor, by 90 o. CE voltage is behind (i.e. AGS) current in a capacitor, by 90 o. Where did the 90 o coe fro? Reeber Oh s law for phasors: v ( v j 90 j C) 90 i i v ( 90)( v (1 C ) i 90)( ) i

A. The sae as the voltage phase angle B. 90 o ore than the voltage phase angle C. 90 o less than the voltage phase angle D. Dependent on the inductance E the CE an!

The voltage leads the current by 90 o in an inductor (E!) Therefore, t Q t p Q P i v i v i v i v sin ) ( ) sin( 0; ) cos( 1 ) sin(90 ) sin( 0; ) cos(90 ) cos(

A. The sae as the voltage phase angle B. 90 o ore than the voltage phase angle C. 90 o less than the voltage phase angle D. Dependent on the capacitance E the CE an!

The voltage lags the current by 90 o in a capacitor (CE!) Therefore, cos( ) v i cos( 90) P cos( v i ) 0; p( t) Q sin t 0; Q sin( ) v i sin( v sin( 90) 1 ) i

Suary: Resistors P > 0, Q = 0 Resistors absorb real power and have no reactive power nductors P = 0, Q > 0 nductors absorb reactive power and have no real power Capacitors P = 0, Q < 0 Capacitors generate reactive power and have no real power

Power factor (of a single coponent or a collection of coponents): pf cos( v i), 0 pf 1 This is a ter that appears in the definition of average power. When pf = 1, the coponent is purely resistive. When pf = 0, the coponent is purely reactive. To distinguish between inductive and capacitive reactance, we use the odifiers leading and lagging : When the pf is leading, the current leads the voltage; when pf is lagging, the current lags the voltage.

A. A capacitor B. An inductor C. A capacitor and a resistor D. An inductor and a resistor E the CE an!

Exaple (AP 10.1) i( t) 0cos( t 15) A v(t) 100cos( t 45) Find the average power, the reactive power, and the power factor. (0)(100) P cos( v i) cos( 45 15) 500 W (0)(100) Q sin( v i ) sin( 45 15) 866 var pf cos( ) cos( 45 15) 0.5 v i

i( t) 0cos( t 15) A v(t) 100cos( t 45) A. eading B. agging C. Can t tell fro the inforation

A. Generating P and generating Q B. Generating P and absorbing Q C. Absorbing P and generating Q D. Absorbing P and absorbing Q

Exaple (AP 10.1), continued i( t) 0cos( t 165) A v(t) 100cos( t 45) Find the average power, the reactive power, and the power factor. (0)(100) P cos( v i ) cos( 45 165) 866 W (0)(100) Q sin( v i) sin( 45 165) 500 var pf cos( ) cos( 45 165) 0.866 leading v i

A. A resistor and a capacitor B. A resistor and an inductor C. None of the above Warning trick question!

Exaple (AP 10.1), continued i( t) 0cos( t 165) A P 866 W, Q 500 var, v(t) 100cos( t 45) The circuit in the box is generating both average and reactive power. Capacitors generate reactive power, Only sources generate average power, The pf < 1, so there ust also be a resistor. Thus, the siplest circuit in the box has a source, a resistor, and a capacitor! pf 0.866 leading

Coplex power: S P jq S pf A Notes: S is apparent power pf is the power factor angle Units for both coplex power and apparent power are volt-aperes [A] The power triangle brings everything together! Q S pf pf cos pf P

Q S A. Purely resistive B. R C. RC D. Purely capacitive P pf

The coplex power in an AC circuit balances! S 0 P 0 and Q 0 Exaple: Find the phasor voltage and current for the load, and show that the power in the circuit balances. 39 40 39 j6 (500) j30 j6 34 j13 39 j6 (34 (4 j13) j3) A 34.36-3.18 5-36.87 A

Exaple: Find the phasor voltage and current for the load, and show that the power in the circuit balances. 34.36-3.18, P Q P P P P S 1 S 39 j4 j6 [(50)(5) [(50)(5) [(51)(5) [(5 39)(5) 0; 0; Q 1 Q ]cos(0 36.87) 500 W ]sin(0 36.87) 375 var ](1) 1 ](1) 5 W; [(5 4)(5) [(5 6)(5) 487.5 W; ](1) 5-36.87 A Q ](1) 1 0 Q 1 50 var 0 35 var

Exaple: Find the phasor voltage and current for the load, and show that the power in the circuit balances. Coponent P[W] Q[var] S[A] Source 500 375 500j375 1 1.5 0 1.5+j0 j4 0 50 0+j50 39 487.5 0 487.5+j0 j6 0 35 0+j35 Total 0 0 0

Review: nstantaneous power p( t) v( t) i( t) P Pcost Qsin t [W] Average (real) power P cos( v i) [W] ( P PC 0) Reactive power Q sin( v i) [var] ( QR 0) Power factor pf cos( v i), 0 pf 1 (leading, i(t) leads v(t), RC; lagging, i(t) lags v(t), R)

Review, continued: Coplex power S P jq S Apparent power S P Q pf [A] [A] The coplex power in an AC circuit balances! S 0 or P 0 and Q 0 The power triangle shows how the different types of power are related: Q P S pf pf cos pf

Q S pf pf cos pf A. Purely resistive B. R C. RC D. Can t tell fro the triangle P

P= S A. Purely resistive B. R C. RC D. Purely capacitive

An aside, using an exaple: Find an expression for the average power delivered to the resistor in the circuit shown here. R dt t T R dt R t v T dt t p T P rs T t t v T t t T t t 0 0 0 0 0 0 ) ( cos 1 1 ) ( 1 ) ( 1 The rs value of a periodic function, say f(t) with period T, is the root of the ean of the square of that function: T t t rs dt t f T F 0 0 ) ( 1

RMS, continued: Find the rs value of a cosine wavefor. v( t) cos( t ) rs 1 T t0 t0 T v cos ( t ) dt (lots ath) What use is the rs value of voltage or current in ac circuits? of v The rs value of any periodic voltage or current delivers the sae average power to a resistor as a dc voltage with the sae value rs values allow us to copare the effect of various periodic voltages to the effect of a dc voltage, so are soeties called effective values!

A. 10 B. Greater than 10 C. ess than 10

RMS, continued: Often, we use rs values in the equations for real and reactive power: P Q pf v i pf sin( ) rs pf sin( v i) rs rs rs sin( ) You ust pay attention to the stateent of probles involving power calculations you pick the correct equations for P and Q depending on whether the agnitude of the voltage (or current) is given, or the rs value of the voltage (or current) is given! v i

Exaple 10.4 A load requires a voltage of 40 rs, absorbs 8kW and has a power factor of 0.8 lagging. Find the coplex power of the load and the load ipedance. We can find the coplex power of the load by drawing the power triangle:

A. 8 kw of power B. Power factor is 0.8 C. Power factor is lagging

Exaple 10.4, continued A load requires a voltage of 40 rs, absorbs 8kW and has a power factor of 0.8 lagging. Find the coplex power of the load and the load ipedance. 1-1 cos pf cos 0.8 36.87 Q S pf 8000 W pf Q S 8000 P tan36.87 Q Q 10,000 A 6000 var 8000 6000 S 8000 j6000 A 10,00036.87 A

Z Z Chapter 10 ACSS Power Exaple 10.4, continued A load requires a voltage of 40 rs, absorbs 8kW and has a power factor of 0.8 lagging. Find the coplex power of the load and the load ipedance. How can we find the load ipedance? Start fro its definition. P 40 rs rs rs rs pf rs ( v i) 40 41.67 5.76 but what about rs pf P pf rs Z 36.87 rs? 8000 (40)(0.8) 41.67 A rs 5.7636.87 4.61 j3.46

A. An inductor and a capacitor B. A resistor and a capacitor C. An inductor and a resistor

Exaple 10.4 A load requires a voltage of 40 rs, absorbs 8kW and has a power factor of 0.8 lagging. Find the coplex power of the load and the load ipedance. 1-1 cos pf cos 0.8 36.87 Q S pf 8000 W pf Q S 8000 P tan36.87 Q Q 10,000 A 6000 var 8000 6000 S 8000 j6000 A 10,00036.87 A

A. An inductor and a capacitor B. A resistor and a capacitor C. An inductor and a resistor

Exaple 10.6 Two industrial loads each require 500 (rs) at 60 Hz. oad 1 absorbs 8 kw at 0.8 leading; load absorbs 0 ka at 0.6 lagging. The line that transits the power fro the source to the load has an ipedance of (0.05 + j0.5). To begin to analyze the circuit described, draw a block diagra.

A. n series B. n parallel

Exaple 10.6, continued Two loads each require 500 (rs) at 60 Hz. oad 1 absorbs 8 kw at 0.8 leading; load absorbs 0 ka at 0.6 lagging. The line that transits the power fro the source to the load has an ipedance of (0.05 + j0.5). To begin to analyze the circuit described, draw a block diagra.

Exaple 10.6, continued Find the power supplied by the source. 1 : 8 kw at : 0 ka 0.8 leading at 0.6 lagging Find the total power required for the loads. -6 kvar -36.87 o 8 kw 16 kvar + 53.13 o 0 ka 1 kw 10 kvar = 6.57 o 0 kw.36 ka

A. True B. False

Exaple 10.6, continued Find the power supplied by the source. S load.366.57 A Find the power lost in the line. How, since we don t know the source voltage, so can t find the voltage drop across the line ipedance? Find the line current, which is the sae as the load current. P load load line load pf load load Pload pf load load 0,000 (50)(0.89) 89.443A(rs)

Exaple 10.6, continued P Q S line line line line 89.443 A(rs) [0.05(89.443)](89.443)(1) [0.5(89.443)](89.443)(1) P line S source jq line S 400 line S load j4000 A 400 W 4000 var 0,400 j14,000 A The power copany bills for the real power it supplies to you this includes the real power you actually use, and the real power lost in the transission line.

A. Gives the power copany ore oney B. ncreases the cost of products produced by the industry C. Wars the feet of the birds who sit on the power lines D. All of the above

Exaple 10.6, continued How can we reduce the real power lost in the line? P R ( )]( )(1) line [ line line line We can t do uch about the resistance of the transission line, and usually it is pretty sall. Can we ake the line current saller? line load P load load pf load

line load P load load pf load A. The real power required by the load(s) B. The voltage drop across the load(s) C. The power factor of the load(s)

A. Maxiize the load power factor (set it to 1) B. Miniize the load power factor (set it to 0) C. Neither (A) or (B)

A. Capacitor in series with the loads B. nductor in series with the loads C. Capacitor in parallel with the loads D. nductor in parallel with the loads 10 kvar 6.57 o 0 kw.36 ka

Exaple 10.6, continued We add a capacitor in parallel with the loads because We don t want to change the voltage drop across the loads We want to cancel the positive (inductive) reactive power required by the loads We want to have no effect on the real power required by the loads 10 kvar 6.57 o 0 kw.36 ka + -10 kvar = 0 kw pf load(total ) 1

Exaple 10.6, continued The capacitor added ust generate 10 kvar of reactive power. ts voltage drop is 50 (rs). Calculate the value of capacitance. Q Z cap C cap cap cap cap 1 C cap ( 1) Qcap ( 1) cap 50 40 6.5 6.5 10,000 (50)( 1) 40 A(rs) 1 C [ (60)](6.5) 44.4F

Exaple 10.6, continued This process is called power factor correction, and is usually required by the power copanies for large industrial loads, to conserve energy there is a onetary penalty associated with net power factors less than a pre-set value, like 0.95.

P Chapter 10 ACSS Power Exaple 10.6, concluded Did this power factor correction actually reduce the power lost to the line? By how uch? line line load P load load pf load 0,000 (50)(1) [0.05(80)](80)(1) 30 W 80 A(rs) (used to be 400 W) (used to be 89.443A(rs)) Therefore, there was a 0% reduction in the real power lost to the line. This would be even ore draatic if the original power factor of the cobined loads was less than 0.89.

AP 10.6 n the figure below, oad 1 absorbs 15 ka at 0.6 lagging and oad absorbs 6 ka at 0.8 leading. Find the source voltage phasor, s. The plan of attack: S line load line load load line load j1 j1 P pf load and j1( load i ) i v pf 0 pf

AP 10.6, continued n the figure below, oad 1 absorbs 15 ka at 0.6 lagging and oad absorbs 6 ka at 0.8 leading. Find the source voltage phasor, s. f we find the total coplex power of the load, we can find the two unknowns we need the total real power of the load, and the total power factor of the load!

AP 10.6, continued n the figure below, oad 1 absorbs 15 ka at 0.6 lagging and oad absorbs 6 ka at 0.8 leading. Find the source voltage phasor, s. -3.6 kvar 1 4.8 kw kvar + -36.87 o 6 ka 53.13 o 15 ka 9 kw 8.4 kvar = 31.33 o 16,155.5 A 13.8 kw

AP 10.6, continued n the figure below, oad 1 absorbs 15 ka at 0.6 lagging and oad absorbs 6 ka at 0.8 leading. Find the source voltage phasor, s. 8.4 kvar 31.33 o 16,155.5 A 0 i load line S P load j1 line load line pf pf line 13,800 (00)cos(31.33) 0 31.33 load 80.78 A 80.78 31.33A rs rs j1(80.78 31.33) 80.7858.67 load (80.7858.67) (000) 13.8 kw rs 51.615.9 rs

A. line B. line C. line line

A. True B. False

Maxiu power transfer: Suppose we attach a load ipedance to the Thevenin equivalent of a source circuit. For what value of load ipedance will the axiu real power be delivered to the load ipedance?

A. True B. False

Maxiu power transfer, continued ) ( ) ( (1) Th Th Th Z X X R R R R R P P R To find the resistive and reactive values of load ipedance for axiu power, take the partial derivative of P with respect to R and X, set the partial derivatives to zero, and solve for R and X.

Maxiu power transfer, continued The text sketches the derivation for the condition for axiu real power to the load ipedance: * Z Z Th Note that these two ipedances are coplex nubers; the star (*) operator represents the coplex conjugate.

A. (18 j36) B. (36 j18) C. (18 + j36)

A. 6045 B. 4560 C. 6045

AP 10.7 The source current is 3 cos 5000t A(rs). Find the ipedance that should be connected to the terinals a,b for axiu real power delivered to that ipedance, and find the axiu real power delivered.

AP 10.7, continued Phasor-transfor the circuit: oc Z P Th (30)((0 4 j18 (0 pf Z Z * Th j40) j40) (48 (0 53.67 0 (0 j10) j4) (rs) 53.67 6.57 (rs) j10) 53.67 j10 0 j10 1 36 W

Objectives: Power concepts: instantaneous power, average power, reactive power, coplex power, power factor Relationships aong power concepts the power triangle Balancing power in AC circuits Power factor correction Condition for axiu power transfer to a load in AC circuits