Lecture 05 Power in AC circuit

CA2627 Building Science Lecture 05 Power in AC circuit Instructor: Jiayu Chen Ph.D.

Announcement 1. Makeup Midterm 2. Midterm grade Grade 25 20 15 10 5 0 10 15 20 25 30 35 40 Grade Jiayu Chen, Ph.D. 2

Learning Objectives 1. Understand the meaning of instantaneous and average power and AC power notation. Compute average power for AC circuits. Compute the power factor of a complex load. 2. Learn complex power notation; compute apparent, real, and reactive power for complex loads. Draw the power triangle, and compute the capacitor size required to perform power factor correction on a load. Jiayu Chen, Ph.D. 3

Electricity serves as a carrier of energy to the user. Energy present in a fossil fuel or a nuclear fuel is converted to electric power to transport and readily distribute it to customers. Electric power may be transported readily with low attendant losses, and improved methods for safe handling of electric power have been developed over the past 90 years. Jiayu Chen, Ph.D. 4

http://www.youtube.com/watch?v=g17f9j1-r-k Jiayu Chen, Ph.D. 5

Instantaneous Power and Average Power We are interested in determining the power generated and absorbed in a circuit, and distinguish between several types of power, for example, instantaneous power, average power, and complex power. Instantaneous power is the power delivered at any one instant of time, which is the product of the time-domain voltage and current associated with one or more circuit elements. The instantaneous power delivered to and absorbed by this circuit element is the product of the voltage v(t) and the current i(t): p t = v t i t The unit of power is watts (W) or kw. If the circuit has an applied voltage v (t) = V m sin(ωt + θ) and an input current i(t) = I m sin ωt, The instantaneous power absorbed by the circuit is p = V m I m sin ωt + θ sin ωt Jiayu Chen, Ph.D. 6

Average power is the average value of a periodic function, which is the integral of the time function over a complete period, divided by the period. P = 1 T t0 t 0 +Tp t dt Jiayu Chen, Ph.D. 7

Recall: For a sinusoidal input, v (t) = V m sin(ωt + θ) and i(t) = I m sin ωt, the average of v(t) over one period is the effective voltage or rms voltage: V V eff = V rms = V m / 2 I I eff = I rms = I m / 2 Note: In the case of power transmission and use in the home, the voltage is said to be 220 V (or 110 V in N America, 100 V in Japan), and it is understood that these values refer to the rms or effective values of the sinusoidal voltage v(t) = 220 2 sin(ωt + θ) V In electronics or communications circuits, the voltage could be described as 10 V, which typically indicates the maximum or peak amplitude, V m. V(t) = 10 sin(ωt + θ) V Jiayu Chen, Ph.D. 8

How to calculate average power of sinusoidal inputs? For example, v t = V m cos ωt + θ V, i t = I m cos ωt + θ I p t = v t i t = V m I m cos ωt + θ V cos ωt + θ I Using the trigonometric identity p t = V mi m cos θ 2 V θ I + cos 2ωt + θ V + θ I Then we do the calculation and find the average power P = 1 T V m I m cos θ T 0 2 V θ I + cos 2ωt + θ V + θ I dt = 1 T V m I m T 0 2 cos θ V θ I dt + 1 T V m I m T 0 2 cos 2ωt + θ V + θ I dt = 1 T V m I m cos θ V θ I dt T 2 0 P = V mi m 2 cos(θ V θ I ) Jiayu Chen, Ph.D. 9

Example: Find the average power delivered to a resistor R when the current through the resistor is i(t) The current waveform repeats every T seconds and attains a maximum value of I m. Using the period from t=0 to t=t, we will have i = I m t 0 t < T T Then the instantaneous power is p = i 2 R = I m 2 t 2 R 0 t < T Then the average power is T 2 2 R P = 1 T I 2 m R T 0 T 2 t2 dt = I m P = I m 2 R 3 T 3 0 W Tt 2 dt = I m 2 R T 3 T 3 3 Jiayu Chen, Ph.D. 10

Power Factor The phase angle of the load impedance plays a very important role in the absorption of power by a load impedance. The term cos θ is referred to as the power factor (pf). The angle θ is called the power factor angle. As mentioned, it is often also the impedance angle. Note that the power factor is equal to 0 for a purely inductive or capacitive load and equal to 1 for a purely resistive load. Jiayu Chen, Ph.D. 11

Power Factor The power factor angle has different signs for inductive and capacitive circuits, but since cosθ = cos( θ), the sign of the power factor angle has no effect on the power factor, which means that the power factors of inductive and capacitive circuits cannot be distinguished mathematically. So they are distinguished by name: The power factor of an inductive circuit is called a lagging power factor. To remember this, we recall that we use voltage as the reference. For an inductive circuit, the current lags the voltage. The power factor of a capacitive circuit is called a leading power factor. For a capacitive circuit the current leads the voltage. Other form of average power formula: Substitute V = IZ into P = VIcosθ P = VI cosθ = I. Z I cosθ = I 2 Z cosθ = I 2 R where R = Z cosθ is the input resistance = Re{Z} Jiayu Chen, Ph.D. 12

Example: The instantaneous power absorbed by a circuit is p = (10 + 8 sin(377t + 40 )) Watts Find the maximum, minimum, and average absorbed powers. The maximum value occurs at those times when the sinusoidal term is a maximum. Since this term has a maximum value of 8, P max = (10 + 8) W= 18 W. The minimum value occurs when the sinusoidal term is at its minimum value of -8: P min = 10-8 = 2 W. Because the sinusoidal term has a zero average value, the average power absorbed is P = 10 + 0 = 10W. Jiayu Chen, Ph.D. 13

Example: With v = 300 cos(20t + 30 ) V applied, a circuit draws i = 15 cos(20t 25 ) A. Find the power factor and also the average, maximum, and minimum absorbed powers. P = V mi m 2 cos(θ V θ I ) The power factor of the circuit is the cosine of the power factor angle, which is the angle by which the voltage leads the current: PF = cos [30 - (-25 )] = cos 55 = 0.574. The average power absorbed is 300 15 P = 0.574 = 1.29 10 3 W = 1.29 kw 2 The maximum and minimum absorbed powers can be found from the instantaneous power which is P max = 2250 1 + cos55 W = 3.54 kw P min = 2250 1 + cos55 = 959 W Jiayu Chen, Ph.D. 14

Example: Compute the average and instantaneous power dissipated by the load R = 4Ω L = 8 mh V = 10 2 π Z = R + iwl = 4 + i3 = 5 0.644 I = V Z = 10 2 π 5 (0.644) = 2 ( 2.215) The average power can be computed from the phasor quantities: P av = 10 2 cos 0.644 = 16 W The instantaneous power is given by the expression: p t = v t i t = 2 10 sin(377t) 2 2 sin(377t 2.215) W Jiayu Chen, Ph.D. 15

Wattmeter Average power can be measured by an instrument called a wattmeter. It has two pairs of terminals: a pair of voltage terminals on the left-hand side and a pair of current terminals on the right-hand side. The wattmeter is a four terminal device. Jiayu Chen, Ph.D. 16

Reactive Power For a two-terminal circuit with an input rms voltage V and an input rms current I, the reactive power Q is: Q = VI sin θ (unit: var) The SI unit is watt. In industrial use, to distinguish it from resistive power, the unit used is voltampere reactive, the symbol for which is var. Reactive power is often referred to as vars. Volt-ampere reactive (var) is a unit used to measure reactive power in an AC electric power system. Reactive power exists in an AC circuit when the current and voltage are not in phase Jiayu Chen, Ph.D. 17

Example: A resistor in series with a capacitor absorbs 10 W when the combination is connected to a 120V, 400Hz source. If the power factor is 0.6 leading, what are the resistance and capacitance? P = VI power factor Thus, P I = = 0.1389 A V power factor = 10 120 0.6 The magnitude of the impedance is Z = V I cos 1 power factor = 120 0.1389 cos 1 0.6 = 864 53.13 = 518 i691 Ω From the real part the resistance is R = 518 Ω, and from the imaginary part and X = 1/ωC, the capacitance is C = 1 ωx = 1 2π 400 691 = 0.576 μf Jiayu Chen, Ph.D. 18

Suppose that a linear circuit with a sinusoidal input is at steady state. All the element voltages and currents will be sinusoidal, with the same frequency as the input. Such a circuit can be analyzed in the frequency domain, using phasors and impedances. Indeed, we can calculate the power generated or absorbed in a circuit or in any element of a circuit, in the frequency domain, using phasors and impedances. we can write the power in the frequency-domain representations of the element current and voltage. I ω = I m θ I and V ω = V m θ V The complex power could be defined as S = VI 2 = V mi m 2 θ V θ I = V mi m 2 cos θ V θ I + i V mi m 2 sin θ V θ I The magnitude of S is called apparent power S = V mi m 2 The complex power also could be represent as S = P + iq Jiayu Chen, Ph.D. 19

Complex Power S = P + iq where P is the average power Q is the reactive power P = V mi m 2 cos θ V θ I Q = V mi m 2 sin θ V θ I Jiayu Chen, Ph.D. 20

V m vs. V rms Jiayu Chen, Ph.D. 21

The complex power is very useful for obtaining the total complex power of several loads energized by the same source, usually in parallel. The total complex power is the sum of the individual complex powers, regardless of how the loads are connected. S = V mi m 2 cos θ V θ I + i V mi m 2 sin θ V θ I In general, apparent powers cannot be added to obtain a total apparent power, but complex powers can be added. P = VI power factor Power factor = P average power = VI apparent power Jiayu Chen, Ph.D. 22

Reactive Factor Similar to the power factor, the reactive factor is Q = VI reactive factor reactive factor = Q reactive power = VI apparent power Example: What is the reactive factor of an inductive load that has an apparent power input of 50 kva while absorbing 30 kw? We can find the angle θ of the power factor, which is θ = cos 1 P 30000 = cos 1 S 50000 = 53.1 Therefore the RF = sin53.1 = 0.8 Jiayu Chen, Ph.D. 23

Example: Two circuit element in series consume 60 var when connected to a 120 V, 60 Hz source. If the reactive factor is 0.6, what are the two components and what are their values? The two components can be found from the input impedance. The angle of this impedance is the arcsine of the reactive factor: θ = sin 1 0.6 = 36.9. The magnitude of the impedance can be found by substituting I = V/Z into Q = VI RF: Z = V2 RF Q Q = V V Z RF = 1202 0.6 60 = 144 Ω So Z = 144 36.9 = 115 + i86.4 Ω From the impedance, the inductance could be derived as L = X ω = 86.4 2π(60) = 0.229 Jiayu Chen, Ph.D. 24

Power factor correction In the consumption of electric power, a large power factor is desirable, the larger the better. The reason is that the current required to deliver a given amount of power to a load is inversely proportional to the load power factor. Although reactive power is of no practical use, it does cause an undesirable increase in the current that must be generated by the electric company, resulting in additional line losses. Thus, it is customary to try to reduce reactive power. A measure of the presence of reactive power at a load is the power factor, equal to the cosine of the angle of the load impedance. By adding a suitable reactance to the load, it is possible to attain power factors close to ideal (unity). This procedure is called power factor correction. Jiayu Chen, Ph.D. 25

Power factor correction Jiayu Chen, Ph.D. 26

Power factor correction We can represent the impedance as Z = R + ix Similarly the load impedance is Z C = R C + ix C We can Z C to absorb no average power, therefore we choose a reactive element so that Z C = ix C The impedance of the parallel combination is Z P as Z P = ZZ C Z + Z C We can write the parallel impedance as Z P = R P + ix P = Z P θ P Jiayu Chen, Ph.D. 27

Power factor correction PFC = cosθ P = cos(tan 1 X P ) R P Where PFC is the corrected power factor, the correct phase θ C = θ P, follows the algebra: Z P = R + ix ix C = R + ix + ix C Therefore, the ratio of X P to R P is X P We can also write it as Then we will have Finally, X C = 2 RX C R 2 + X + X 2 + i R2 X C + X C + X XX C C R 2 + X + X 2 C = R2 + X C + X X R P RX C X P R P = tan cos 1 PFC R 2 + X 2 Rtan(cos 1 PFC) X Z C = i ωc = ix C Jiayu Chen, Ph.D. 28

Power factor correction Further, R 2 + X 2 Rtan(cos 1 PFC) X = 1 ωc Solving for ωc gives ωc = X Rtan(cos 1 PFC) R X R 2 + X 2 = R 2 + X 2 R tan cos 1 PFC Let θ = tan 1 X, then R R ωc = R 2 + X 2 tanθ tanθ C Where θ = cos 1 (PF) and θ C = cos 1 (PFC) Jiayu Chen, Ph.D. 29

Example: A load as shown in following figure has an impedance of Z = 100 + i100 Ω. Find the parallel capacitance required to correct the power factor to (a) 0.95 lagging and (b) 1.0. Assume that the source is operating at ω = 377 rad/s. The phase angle of the impedance is θ = 45, so the original load has a lagging power factor with cosθ = cos 45 = 0.707 We wish to correct the PF to PFC=0.95 lagging. Then we can have 100 2 + 100 2 X C = 100tan(cos 1 0.95) 100 = 297.9 Ω The capacitor required is determined from 1 ωc = X C Since ω = 377 rad/s, then we will have Jiayu Chen, Ph.D. 30

Example: C = 1 = 8.9μF ωx C If we wish to correct the load to PFC = 1, we have 2 10 4 X C = 100tan(cos 1 1) 100 = 200 The capacitor required to correct the power factor to 1 is determined from C = 1 1 = = 13.3 μf ωx C 377 200 C also could be calculated by ωc = 100 2 10 4 tanθ tanθ C = 5 10 3 tan45 tan0 = 5 10 3 And C = 5 10 3 377 = 13.3 μf Jiayu Chen, Ph.D. 31

Electromagnetic Compatibility (EMC) Additional The electromagnetic energy can be best described by electric and magnetic fields, as well as voltage and current. Both electric and magnetic fields are radiated in the air, whi1e both voltage and current are propagated through circuit conductors. Units: E (V/m), H (A/m) Jiayu Chen, Ph.D. 32

Electromagnetic Compatibility (EMC) Additional Electromagnetic interference Dirty Electricity Jiayu Chen, Ph.D. 33

Electromagnetic Compatibility (EMC) Additional Examples: 1. Helicopter blades and bodies tribocharge as they swish through the air, and they don't have a convenient green/yellow wire handy. There are reports of an oil rig computer system crashing whenever a helicopter landed, due to its sudden electrical discharge into the metal decking. The Coast Guards' standards practice calls for NEVER lowering a flotation ring or sling to a person in the water; the person will automatically reach for the line! Instead, they dip the line into the water and drag it to the person. 2. Railways cover most of the country and can pass quite close to residential buildings and hence affect the equipment inside them. Additional immunity constraints are placed on the users of information technology equipment in the near vicinity. However, it is unclear how suppliers know that their users will be situated near railway systems. Jiayu Chen, Ph.D. 34

Electromagnetic Compatibility (EMC) Additional Electromagnetic Disturbances An electromagnetic disturbance is any electromagnetic phenomenon, which may degrade the performance of a device, equipment or system, or adversely affect living or inert matter. The electromagnetic disturbance can be in the nature of an electromagnetic noise, or an unwanted signal, or a change in the propagation medium itself. The electromagnetic disturbance may be classified, according to IEC, into several categories, as follows: Conducted low-frequency phenomena Radiated low-frequency phenomena Conducted high-frequency phenomena Radiated high-frequency phenomena Electrostatic discharge (ESD) phenomena Nuclear electromagnetic pulse Harmonics, interharmonics signalling systems Voltage fluctuations Voltage dips and interruptions Voltage unbalance Power frequency variations Induced low-frequency voltages DC in AC networks Jiayu Chen, Ph.D. 35

Thank You! Jiayu Chen, Ph.D. 36