SteadyState AC Power The power absorbed or delivered

Steady-State AC Power The power absorbed or delivered by any element is the product of the voltage and current associated with the element. i(t) If the current and voltage are functions of time, then so is the power: + v(t) - p(t) = v(t)i(t) Specifically, if v(t) and i(t) are sinusoids, the power, p(t), will also be a sinusoid: Using the trig identity: We can write: Spring 2001 EE 202 Steady-State AC Power 1

Example 9. 1 p(t) v(t) Spring 2001 i(t) EE 202 Steady-State AC Power 2

Average Power The average value of any periodic waveform can be found by integrating (summing) the function over one complete period and dividing by the length of the period. The average value of a sinusoid such as a sinusoidal voltage or current, then, is zero: The average value of the instantaneous power function, however, is not zero: Spring 2001 EE 202 Steady-State AC Power 3

Average Power- Continued + I=I I + V=V V - +- Z=Z Z - If Z is purely resistive Z = 0 and PAVG = VMIM/2 since cos(0) = 1. If Z is purely reactive Z = 90 and PAVG =0 since cos(90 ) =0. Reactive elements absorb (store) power during one half cycle and deliver (release) that power during the next half cycle. Spring 2001 EE 202 Steady-State AC Power 4

Maximum Average Power Transfer ZTH=ZT T + VTH = VOC V - + - ZL + VL=VL VL - PL , the power delivered to ZL , is maximized whenever ZL = Z*TH (The load impedance is equal to the complex conjugate of the Thevenin equivalent impedance). Spring 2001 EE 202 Steady-State AC Power 5

Effective (RMS) Value The effective (or rms) value of a periodic signal is the dc equivalent value. For example, household power is typically 110 V, rms. This means, that a resistive load connected to 110 V ac source would absorb the same amount of power if it were connected to a 110 V dc source. The term rms stands for root-mean-square and comes from the mathematical definition of effective value: It is easily shown that for a sinusoidal signal: So, the 110 Vrms household power is a sinusoid with a peak value of: The rms value of non-sinusoidal signals must be found by applying the definition. Spring 2001 EE 202 Steady-State AC Power 6

Effective (RMS) Values In Power Calculations The average power formulas were written in terms of the peak values of the sinusoidal voltage and current: But these can be re-written in terms of rms values: For a resistor: Spring 2001 EE 202 Steady-State AC Power 7

The Power Factor For a given load, the average ac power absorbed by the load is given by: I=Irms i P = Vrms. Irmscos ( v - i) Notice that the passive sign convention is being used: power absorbed (+) , power generated (-) + V=Vrms v ZL - The product Vrms. Irmsis called the apparent power, and is the power that would be delivered if ZL were a purely resistive load. The units for P, sometimes called the real power, is watts (W). However, apparent power is specified in volt-amps (VA) or kilovolt-amps (k. VA). The power factor (pf) is the ratio of the real power to the apparent power and is given by: Spring 2001 EE 202 Steady-State AC Power 8

The Power Factor - Continued I=Irms i P = Vrms. Irmscos ( v - i) + V=Vrms v ZL - For a purely resistive load, pf =1 since the voltage and current are in phase ( v = i ) For a purely reactive load, pf =0 since the voltage and current are 90 out of phase. The pf will range between 0 and 1 since the phase angle between the voltage and current varies between 90 . The power factor angle, ( v - I), is determined by the angle of the impedance ZL, since I = V/ZL. If ZL is inductive (ZL = R + j. X = Z + ), then the pf angle will be negative and I will lag V lagging power factor. If ZL is capacitive (ZL = R - j. X = Z - ), then the pf angle will be positive and I will lead V leading power factor. Spring 2001 EE 202 Steady-State AC Power 9

Example 9. 12 RW = 0. 08 VS IL + P=88 k. W @ 0. 707 pf lagging 480 V rms - An industrial load consumes 88 k. W at a pf of 0. 707 lagging. The voltage at the load is 480 V (rms), find the total power supplied by the source if the load is connected to the source via a transmission line that has a total resistance pf 0. 08 . P = 88 k. W = Vrms. Irms pf Irms= 88 k. W/(480 V 0. 707) Irms= 259. 3 A rms. Pwire = (259. 3 A)2 0. 08 = 5. 38 k. W Psource = 88 k. W + 5. 38 k. W = 93. 38 k. W. Suppose that the load power is maintained at 88 k. W, but that the pf is improved to 0. 9, determine the rms load current and the power delivered by the source. P = 88 k. W = Vrms. Irms pf Irms= 88 k. W/(480 V 0. 9) Irms= 203. 7 A rms. PWire= (203. 7 A)2 0. 08 = 3. 31 k. W Psource = 88 k. W + 3. 31 k. W = 91. 31 k. W. Spring 2001 EE 202 Steady-State AC Power 10

Complex Power Complex power is defined as follows: Complex power has magnitude and phase and is found by multiplying the voltage phasor with the complex conjugate of the current phasor. The expression above represents the complex power in polar form, but we can also represent it is rectangular form. P = real power (Watts) Q = reactive power (volt-amps reactive or VARs) S = P + j. Q Spring 2001 EE 202 Steady-State AC Power 11

Complex Power - Continued As stated previously, the complex power is composed of a real part (Watts) and an imaginary part (VARS). S = P + j. Q P = Re(S) and Q = Im(S) The magnitude of S, is what we have called the apparent power (VA), and is the hypotenuse of the power triangle. S z Q P Some observations can be made from the power triangle using trig identities: P = |S|cos z Q = |S|sin z |S| = [P 2+Q 2]0. 5 z = tan-1(Q/P) It can also be shown that: Spring 2001 EE 202 Steady-State AC Power 12

The Power Triangle S = P + j. Q = |S| Z (inductive load) S Q z (Lagging pf) P If the load ZL is inductive, ZL= R + j. X, then Z is positive and Q is also positive, as shown in the diagram above. This is called a lagging pf since the current lags the voltage. P S = P - j. Q = |S| - Z (capacitive load) z S -Q (Leading pf) If the load ZL is capacitive, ZL= R - j. X, then Z is negative and Q is also negative, as shown above. This is called a leading pf since the current leads the voltage. If the load ZL is purely resistive, ZL= R, then Z =0 and Q =0. In this case, P = S. Spring 2001 EE 202 Steady-State AC Power 13

Power Factor Correction The overall power factor of the typical industrial plant is lagging due to the inductive nature of the devices used: motors and transformers. It is generally economically viable to take some steps to improve (correct) the overall power factor. There at least two benefits to improving power factor: (1) the same amount of real power can be delivered at a lower current; (2) most power suppliers charge for the VAs used, not for Watts. To improve power factor, a capacitive load is added to the system to offset the inductive VARs generated by other equipment. Often, the only function of the capacitive load is to improve power factor. Spring 2001 EE 202 Steady-State AC Power 14

Power Factor Correction - continued Inew = Iold + IC Ic VS Inew Iold • Since the capacitive current leads VS by 90 , Inew = Iold + IC will be smaller in magnittde than the original current. • Also, the phase angle between the voltage and current will be closer to zero (power factor will be closer to unity). Spring 2001 EE 202 Steady-State AC Power 15

Power Factor Correction - continued Power factor improvement can also be visualized using power triangles: Pnew = Pold = P Qnew = Qold - QC Snew = P + j. Qnew = tan-1[(Qold - QC)/P] Sold QC Qold Snew old P Spring 2001 EE 202 Steady-State AC Power new 16

Example 9. 15 Plastic kayaks are manufactured using a process called rotomolding, which is illustrated in the diagram below. Molten plastic is injected into a mold, which is then spun the long axis of the kayak until the plastic cools, resulting in a hollow, one-piece craft. Suppose the induction motors used to spin the molds consume 50 k. W of power at a 0. 8 lagging pf. The power source is 220 0 , rms. Determine the rating of the capacitive reactance required to raise the pf to 0. 95 lagging. Also, calculate the new current demanded from the 220 V source. kayak mold Solution: old = cos-1(0. 8) = 36. 87 Sold = 50 k/0. 8 Sold = 62. 5 k. VA old Qold = S sin old Qold = 37. 7 k. VAR P = 50 k. W new = cos-1(0. 95) = 18. 2 Snew = 50 k/0. . 95 Snew = 52. 6 k. VA P = 50 k. W Qnew = S sin new Qnew = 16. 4 k. VAR new Qc = Qnew - Qold = 21. 1 k. VAR Iold = S/V = 62. 5 k. VA/220 V = 284 A Spring 2001 induction motor Inew = S/V = 52. 6 k. VA/220 V = 239 A EE 202 Steady-State AC Power 17