Chapter 4 POWER IN AC CIRCUITS AND POWER FACTOR CORRECTION EE-425 Electrical Circuits II Engr. Sarah Jane F. Fruelda, REE, RME POWER IN AC CIRCUITS Real Power, P Real power, also known as true or active power, performs the “real work” within an electrical circuit. Real power, measured in watts, defines the power consumed by the resistive part of a circuit. Then real power, P in an AC circuit is the same as power in a DC circuit. As resistances do not produce any phasor difference (phase shift) between voltage and current waveforms, all the useful power is delivered directly to the resistance and converted to heat, light and work. Then the power consumed by a resistance is real power which is fundamentally the circuit average power. POWER IN AC CIRCUITS Reactive Power, Q Reactive power, sometimes called wattles power, is the power consumed in an an AC circuit that does not perform any useful work but has a big effect on the phase shift between the voltage and current waveforms. It is linked to the reactance produced by inductors and capacitors and counteracts the effects of real power. Unlike real power which does all the work, reactive power takes power away away from a circuit due to the creation and reduction of both inductive magnetic fields and capacitive electrostatic fields, thereby making it harder for the true power to supply power directly to a circuit or load. POWER IN AC CIRCUITS Apparent Power, S There is a mathematical relationship between the real power (P), and the reactive power power (Q), which is called the complex power. The product of the rms voltage applied to an AC circuit and the rms current flowing into that circuit is called the “volt-ampere product” (VA) given the symbol S and the magnitude is known generally as apparent power. This complex power is not equal to the algebraic sum of the real and reactive powers added together, but is instead the vector sum of P and Q given in volt-amps (VA). It is the complex power that is represented by the power triangle. The rms value of the volt-ampere product is known more commonly as the apparent power as, apparently this is the total power consumed by a circuit even though the real power that does the work is a lot less. AC POWER CALCULATIONS USING MAGNITUDE ONLY E2 P = EIcosθ = = I2 R = R E2 Q = EIsinθ = = I2 X = X S2 − P2 E2 S = EI = = I2 Z = Z where: E R X S 2 − Q2 (𝑊) (𝑉𝐴𝑅 − 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑜𝑟 𝑖𝑛𝑑𝑐𝑢𝑡𝑖𝑣𝑒) P 2 + Q2 (𝑉𝐴) = voltage in rms volts I = current in rms amperes = resistance in Z = impedance in ohms ohms = reactance (inductive or capacitive) in ohms AC POWER CALCULATIONS USING COMPLEX QUANTITIES Current Conjugate S = EI ∗ Note: If θ and Q are positive (+), p.f. is lagging (inductive) If θ and Q are negative (-), p.f. is leading (capacitive) Voltage Conjugate S = E∗I Note: If θ and Q are negative (-), p.f. is lagging (inductive) If θ and Q are positive (+), p.f. is leading (capacitive) EXAMPLE 5 1. Given a circuit with an impedance of 6 + j8 ohms and an applied phasor voltage of 100⎿-30° V. Determine the power triangle. 2. Determine the power components of a combination of three individuals loads specified as follows: Load A: 250 VA, 0.5 lagging pf Load B: 180 W, 0.8 leading pf Load C: 300 VA, 100 VAR (inductive) SOLUTION: SOLUTION: EXAMPLE 6 1. An ac circuit is rated 160kVA at a lagging power factor of 0.75. Determine real and reactive powers. 2. A 50 microfarad capacitor is connected in series with a 50-ohm resistor. What is the power if the source voltage is V = 100sin(377t+30)? 3. A load of 20+j35 ohms is connected across a 220 volts source. Determine the VARS. 4. A 25-ohm resistor connected in series with a coil of 50 ohm resistance and 150mH inductance. What is the power factor of the circuit? ANYONE? POWER FACTOR Power factor is the ratio between the useful (true) power (kW) to the total (apparent) power (kVA) consumed by an item of a.c. electrical equipment or a complete electrical installation. It is a measure of how efficiently electrical power is converted into useful work output. The ideal power factor is unity, or one. Anything less than one means that extra power is required to achieve the actual task at hand. WHY IMPROVED POWER FACTOR? The benefits that can be achieved by applying the correct power correction are: Environmental benefit. Reduction of power consumption due to improved energy efficiency. Reduced power consumption means less greenhouse gas emission and fossil fuel depletion by power stations. Reduction of electricity bills Extra kVA available from the existing supply Reduction of I2R losses in transformers and distribution equipment Reduction of voltage drop in long cables Extended equipment life – reduced electrical burden on cables and electrical component. POWER FACTOR CORRECTION Power factor correction is the process of increasing the power factor without altering the voltage or current to the original load. Power factor correction is the process of compensating for the lagging current by creating a leading current by connecting capacitors to the supply. It can be achieved by the addition of capacitors in parallel with the connected motor or lighting circuits and can be applied at the equipment, distribution board or at the origin of installation. POWER FACTOR CORRECTION Illustration of Power Factor Correction using a bank of Static Condenser (Capacitors) connected across the load POWER FACTOR CORRECTION Illustration of Power Factor Correction using a Synchronous Condenser (a synchronous motor operated at a leading power factor) connected across the load POWER FACTOR CORRECTION where: Sold = old apparent power, when power factor is not yet corrected Snew = new apparent power, after power factor is corrected Ssyn = apparent power rating of the synchronous motor needed to correct the low power factor Qc = reactive power rating of the static condensers needed to correct the low power factor Pold = old active power before correction Pnew = new active power after correction Psyn = active power rating of the synchronous motor cos θold = power factor before correction cos θnew = power factor after correction cos θsyn = power factor of the synchronous motor EXAMPLE 7 An existing industrial plant has an average load of 900kW at 0.6 pf lagging. To raise the overall power factor to 0.92, a synchronous motor driving a DC generator is to installed. If the input of the synchronous motor will be at 250 kW, calculate its KVA input rating. SOLUTIO N: EXAMPLE 8 A load of 10,000 kVA, 80% pf lagging is connected to a 13,200 volts line. How much capacitive reactive power is needed to correct the power factor to 0.97 lagging? ANYONE? THANK YOU!
