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CA2627 Building Science Lecture 06 Transformers, three phase AC circuits, building power distribution Instructor: Jiayu Chen Ph.D. Learning Objectives 1. Analyze the ideal transformer; compute primary and secondary currents and voltages and turns ratios. Calculate reflected sources and impedances across ideal transformers. 2. Understand maximum power transfer. 3. Learn three-phase AC power notation; compute load currents and voltages for balanced wye and delta loads. 4. Understand the basic principles of residential electrical wiring and of electrical safety. © Jiayu Chen, Ph.D. 2 Power Distribution Electricity supply in Hong Kong Ref.: Supply rules from CLP https://www.clponline.com.hk/MyHome/CustomerService/OpenAccount/SupplyRule s/Pages/Default.aspx?lang=en Supply rules from HK Electric http://www.heh.com/NR/rdonlyres/F98E0121-03CA-4C4F-B2DAF77774833876/0/SupplyRule_en_2008.pdf © Jiayu Chen, Ph.D. 3 Power Distribution EMSD Code of Practice EMSD Code of Practice for Electricity (Wiring) Regulations, 2009 Edition (http://www.emsd.gov.hk/emsd/e_download/pps/pub/COP_E.pdf) © Jiayu Chen, Ph.D. 4 Power Distribution Electricity is supplied by the HKE or CLP in Hong Kong. The nominal values are: β’ Frequency ππ π―π ± π% β’ Phase supply o for a single phase supply, a nominal voltage of 220 volts (root mean square value) plus or minus 6% between a phase conductor and neutral conductor; o for a 3 phase supply, using a 3 phase 4 wire system, a nominal voltage of 380 volts plus or minus 6% (root mean square value) between phase conductors and 220 volts (root mean square value) plus or minus 6% between phase and neutral conductors; o for a 3 phase supply, a nominal voltage of 11,000 volts (root mean square value) plus 10% or minus 2% between phase conductors; and o for a 3 phase supply, a nominal voltage of 132,000 volts (root mean square value) plus 10% or minus 2% between phase conductors. © Jiayu Chen, Ph.D. 5 Power Distribution Beyond the generation plant, an electric power network distributes energy to several substations. This network is usually referred to as the power grid. © Jiayu Chen, Ph.D. 6 Transformers A transformer is a device that couples two AC circuits magnetically rather than through any direct conductive connection and permits a βtransformationβ of the voltage and current between one circuit and the other (for example, by matching a highvoltage, low-current AC output to a circuit requiring a low-voltage, high current source). © Jiayu Chen, Ph.D. 7 Transformers The ideal transformer consists of two coils that are coupled to each other by some magnetic medium. There is no electrical connection between the coils. The coil on the input side is termed the primary, and that on the output side the secondary. Let us assume the primary coil is wound so that it has π1 turns, while the secondary has π2 turns. We define the turns ratio N as π2 π= π1 πΌ1 π2 = ππ1 πππ πΌ2 = π An ideal transformer multiplies a sinusoidal input voltage by a factor of N and divides a sinusoidal input current by a factor of N. © Jiayu Chen, Ph.D. 8 Transformers ο§ If N is greater than 1, the output voltage is greater than the input voltage and the transformer is called a step-up transformer. ο§ If N is less than 1, then the transformer is called a step-down transformer. ο§ A transformer with N = 1 is called an isolation transformer and may perform a very useful function if one needs to electrically isolate two circuits from each other. Note: β’ Any DC currents at the primary will not appear at the secondary coil. β’ An important property of ideal transformers is conservation of power β’ We can easily verify an ideal transformer by π2 π1 = πΌ1 π1 = ππΌ2 = πΌ2 π2 = π2 π © Jiayu Chen, Ph.D. 9 Transformers Example: We require a transformer to deliver 500 mA at 24 V from a 120 V rms line source. How many turns are required in the secondary? What is the primary current? Assume we have the first coil with 3000 turns, We will have π1 π2 = π1 π2 π1 π2 24 π2 = = 3000 × = 600 π‘π’πππ π1 120 Knowing the turns, we can now compute the primary current, π1 πΌ1 = π2 πΌ2 π2 600 πΌ1 = πΌ2 = × 500 = 100 ππ΄ π1 3000 © Jiayu Chen, Ph.D. 10 Transformers In many practical circuits, the secondary is tapped at two different points, giving rise to two separate output circuits. The most common configuration is the center-tapped transformer, which splits the secondary voltage into two equal voltages. The most common occurrence of this type of transformer is found at the entry of a power line into a household, where a high-voltage primary is transformed to 240 V, and split into two 120-V lines. 110 V vs 220 V ? © Jiayu Chen, Ph.D. 11 Transformers Impedance Reflection A very common and rather general situation is that where an AC source, represented by its Thévenin equivalent, is connected to an equivalent load impedance by means of a transformer. π2 πππ πΌ1 = ππΌ2 π At the primary connection, we can have V1 V2 /N 1 π2 1 Zβ² = = = 2 = 2 ππΏ I1 ππΌ2 π πΌ2 π π1 = © Jiayu Chen, Ph.D. 12 Transformers Impedance Reflection We will have πππΆ = πππ ππ 1 πΌππΆ = ππ π And apply Thévenin equivalent π β²β² = πππΆ πππ = = π 2 ππ ππ 1 πΌππΆ ππ π Thus the load sees the source impedance multiplied by a factor of π 2, this is called impedance reflection. © Jiayu Chen, Ph.D. 13 Transformers Maximum power transfer The power transfer efficiency is defined as: πππππ π= ππ ππ’πππ Recall that the maximum power to be transferred form the source to the load, the impedance must be matched (Proof was omitted) Conclusion: We know ππ = π π + πππ ππΏ = ππβ That is π πΏ = π π ππΏ = βππ Where ππβ is the conjugate of the source impedance When the load impedance is equal to the complex conjugate of the source impedance, the load and source impedances are matched and maximum power is transferred to the load. ππΏ = ππβ 2 π π πΏ = π 2 π π ππΏ = βπ 2 ππ © Jiayu Chen, Ph.D. 14 Transformers Example: Find the transformer turns ratio and load reactance that results in maximum power transfer in the circuit. ππ = 240β 0 π; π π = 10 Ξ©; πΏπ = 0.1 π»; π πΏ = 400 Ξ©;π = 377 For the maximum power transfer, we will require π πΏ = π 2 π π Thus, π πΏ 400 π2 = = π = 40 = 6.325 π π 10 Further, to cancel the reactive power we require that ππΏ = βπ 2 ππ ππ = π × 0.1 = 37.7 And ππΏ = β40 × 37.7 = β1508 Thus the load reactance should be a capacitor with the value of 1 1 πΆ=β = = 1.76 ππΉ ππΏ π β1508 × 377 © Jiayu Chen, Ph.D. 15 Transformers A substantial amount, 20% of the power from the source, is lost in the line if no transformers are used. Since power goes into heating up the line (line has resistance), and P ~ I2R, the loss in the line can be reduced by reducing the current in the line, by using transformers. Transformers at the source step up the voltage and step down the current, keeping the power constant, before transmission. At the load, we use a step-down transformer to step down the voltage and step up the current. © Jiayu Chen, Ph.D. 16 Transformers Example: If the transformer delivers 50 A at 110 V rms with a certain resistive load, what is the power transfer efficiency between the source and the load? ππ = πππ ππ = 220 π πππ The transformer primary voltage The transformer primary current The current through the inductance πΌπ = πΌπΏ = ππ 220 = β β 90° = 11β β 90° π΄ πππ π20 20 The source current is (add phasor currents) The source power is 1 πΌ = 25 π΄ πππ π π ππ πΌπ = 2 πΌπΏ2 + πΌπ = 27.3 π΄ πππ ππ = 1Ξ© β πΌπ2 + ππ ππ = 745 + 5.5 π = 6.245 ππ Power transfer efficiency π= © Jiayu Chen, Ph.D. πππππ ππ ππ 5.5 = = = 0.88 ππ ππ’πππ ππ ππ’πππ 6.245 17 3-phase AC circuits Generation of Three Phase AC Electricity The generation of electrical power is more efficient in 3phase systems. A 3-phase system employs three balanced voltages, equal in magnitude and differing in phase by 360°/3 = 120°. http://www.fam-oud.nl/~plugsocket/Generator_3phase.html http://ele.aut.ac.ir/~wind/en/tour/wtrb/syncgen.htm http://www.launc.tased.edu.au/online/sciences/physics/3pha se/threeph.htm © Jiayu Chen, Ph.D. 18 3-phase AC circuits As the rotor rotates counterclockwise at 50 r/sec, its magnetic field cuts the armature windings, thereby inducing in them the sinusoidal voltages. These voltages have peaks at one-third of a period apart, or 120° apart, because the armature windings are displaced by 120° in space. As a result, the alternator produces three voltages of the same rms value, but with phase differences of 120°. πππβ² = ππ πππ ππ‘ πππβ² = ππ cos(ππ‘ β 120°) πππβ² = ππ cos(ππ‘ β 240°) In HK: phase voltage = 220 V line voltage = 380 V © Jiayu Chen, Ph.D. 19 3-phase AC circuits A three-phase circuit generates, distributes, and uses energy in the form of three voltages equal in magnitude and symmetric in phase. The three similar portions of a three-phase system are called phases. . Line voltage, abbreviated from line-to-line voltage Because the voltage in phase aa' reaches its maximum first, followed by that in phase bb' and then by that in phase cc', we say the phase rotation is abc. The three voltages are said to be balanced voltages because they have identical amplitude, Vm, and frequency, Ο, and are out of phase with each other by exactly 120°. πππβ² + πππβ² + πππβ² = 0 The positive phase sequence is abc; the sequence acb is called the negative phase sequence. © Jiayu Chen, Ph.D. 20 3-phase AC circuits 3-phase circuits and machines possess some unique advantages: β’ The power transmitted in a three-phase circuit is constant or independent of time rather than pulsating, as it is in a single-phase circuit. β’ Three-phase motors start and run much better than do single-phase motors. β’ The transmission of electrical power is more efficient in 3-phase systems employing three sinusoidal voltages. For example, for the addition of one more conductor, a three-phase supply provides 73% more power than a single-phase supply. With a three-phase supply the voltage between two line or phase cables is β3 = 1.73 times that between the neutral and any one of the line cables, i.e. 220 volts x 1.73 =380 volts. © Jiayu Chen, Ph.D. 21 3-phase AC circuits Y and Ξ connections Referring to the generator, there are six terminals and three voltages, va, vb and vc. We use phasor notation and assume that each phase winding provides a source voltage in series with a negligible impedance. Under these assumptions, there are two ways of interconnecting the three sources. Y connection The Y connection selects terminals a', b', and c' and connects them together as neutral. The common terminal is called the neutral terminal and is labeled n. The neutral terminal may or may not be available for connection. Balanced loads result in no current in a neutral wire, and thus it is often not needed. © Jiayu Chen, Ph.D. 22 3-phase AC circuits Ξ connection The Ξ source connection is seldom used in practice because any slight imbalance in magnitude or phase of the three-phase voltages will not result in a zero sum. The result will be a large circulating current in the generator coils that will heat the generator and depreciate the efficiency of the generator. © Jiayu Chen, Ph.D. 23 3-phase AC circuits Y-connected source The line-to-line voltage, πππ , of the Y-connected sources is πab = πa β πb π½π = ππ β 0° πππ π½π = ππ β β 120° π½ππ = ππ β ππ β0.5 β π0.866 = Vm 1.5 + i0.866 π½ππ = 3ππ β 30° And π½ππ = 3ππ β β 90° π½ππ = 3ππ β β 210° Therefore, in a Y connection, the line-to-line voltage is 3 times the phase voltage and is displaced 30° in phase. The line current is equal to the phase current. The relation can also be obtained from the sine law, π½ππ π½π = π ππ120° π ππ30° © Jiayu Chen, Ph.D. 24 3-phase AC circuits Example: The Y-connected three-phase voltage source π½π = 120β β 240° V rms. Find the line-toline voltage π½ππ Answer: 207.8β β 90° V rms © Jiayu Chen, Ph.D. 25 3-phase AC circuits Y-to-Y circuit This three-phase circuit consists of three parts: a three-phase source, a three-phase load, and a transmission line. 4-wire Y-to-Y The transmission line used to connect the source to the load consists of four wires, including a wire connecting the neutral node of the source to the neutral node of the load. 3-wire Y-to-Y The three-phase source is connected to the load using three wires, without a wire connecting the neutral node of the source to the neutral node of the load. © Jiayu Chen, Ph.D. 26 3-phase AC circuits 4-wire Y-to-Y Each impedance of the three-phase load is connected directly across a voltage source of the three-phase source. Therefore, the voltage across the impedance is known, and the line currents are easily calculated as: ππ ππ ππ πππ = , πππ = , and πππ = ππ ππ ππ The current in the wire connecting the neutral node of the source to the neutral node of the load (the neutral wire) is: ππ ππ ππ πππ§ = πππ + πππ + πππ = + + ππ ππ ππ The total average power delivered by the three-phase source to the three-phase load is calculated by adding up the average power delivered to each impedance of the load. P = PA + PB + PC Where, eg.PA is the power absorbed by ZA. Since we can calculate IaA , we can then calculate PA . Denote the phase voltages of the Y-connected source as ππ = Vp β 0° V rms, ππ = Vp β β 120° V rms, and ππ = Vp β 120° V rms, Notice that we are using effective values because the units of Vp are V rms © Jiayu Chen, Ph.D. 27 3-phase AC circuits 4-wire Y-to-Y For a balanced load, ππ΄ = ππ΅ = ππΆ = π = πβ π, the load is said to be a balanced load. The currents are: ππ Vp β 0° ππ Vp β β 120° ππ Vp β 120° πππ = = ,π = = , and πππ = = ππ πβ π ππ ππ πβ π ππ πβ π Vp Vp Vp πππ = β β ΞΈ, πππ = β β ΞΈ β 120°, and πππ = β β ΞΈ + 120° Z Z Z The line currents have equal magnitudes and differ in phase by 120°. The current in the neutral wire (which connects the neutral node of the source to the neutral node of the load) is: Vp πππ§ = πππ + πππ + πππ = 1β 0° + 1β β 120° + 1β 120° = 0 Z Vp πππ§ = πππ + πππ + πππ = 1β β π + 1β β π β 120° + 1β β π β 120° = 0 Z This shows that the neutral wire has no current. Remember that we are using eff or rms values. The average power delivered to the load is V V V V2 P = PA + PB + PC = V cos βπ + V cos βπ + V cos βπ = 3 cos(π) Z Z Z Z The average power delivered to the load can be determined by calculating PA and multiplying by 3 © Jiayu Chen, Ph.D. 28 3-phase AC circuits 3-wire Y-to-Y A 3-wire Y-Y circuit would not necessarily have the neutral current InN = 0, and the node equation at node N needs to be solved. However, it can be shown that for a balanced 3wire Y-Y circuit, the neutral current InN = 0. The balanced three-wire Y-to-Y circuit acts like the balanced four-wire Y-to-Y circuit. The line currents and the average power delivered to the load are as given above. The per-phase equivalent circuit For balanced circuits, all 3 phases are similar, and we need only calculate one line current, say IaA. The per-phase equivalent circuit consists of the voltage source and impedances in the one of the phases. The neutral nodes, n and N, are connected by a short circuit in the per-phase equivalent circuit to indicate that VnN=0 in a balanced Y-to-Y circuit. The per-phase equivalent circuit can be used to analyze either 3-wire or 4-wire balanced Y-to-Y circuits, but it can only be used for balanced circuits. © Jiayu Chen, Ph.D. 29 3-phase AC circuits The per-phase equivalent circuit The Balanced Y-to-Y Circuit © Jiayu Chen, Ph.D. 30 3-phase AC circuits Example: Determine the complex power delivered to the three-phase load of a four-wire Y-to-Y circuit. The phase voltages of the Y-connected source are π½π = 110β 0° V rms, π½π = 110β 120° V rms, and π½π = 110β 120° V rms. The load impedances are ππ¨ = ππ© = ππͺ = 50 + π80 Ξ© We need to calculate only one line current, πΌππ΄ , and the complex power ππ΄ , delivered to only one of the load impedances, ππ΄ . The power delivered to the three-phase load is 3ππ΄ . We begin by calculating πΌππ΄ as π½π 110β 0° π°ππ¨ = = = 1.16β β 58° π΄ πππ ππ¨ 50 + π80 The complex power delivered to ππ¨ is β πΊπ¨ = π°ππ¨ ππ = 1.16β β 58° β 110β 0° = 68 + π109 ππ΄ The complex power delivered to the three-phase load is 3πΊπ¨ = 204 + π325 ππ΄ © Jiayu Chen, Ph.D. 31 3-phase AC circuits Example: Determine the complex power delivered to the three-phase load of a four-wire Y-to-Y circuit. The phase voltages of the Y-connected source are π½π = 110β 0° V rms, π½π = 110β 120° V rms, and π½π = 110β 120° V rms. The load impedances are ππ¨ = 50 + π80 Ξ©, ππ© = π50 Ξ©, and ππͺ = 100 + π25 Ξ© The line currents of an unbalanced four-wire Y-to-Y circuit are calculated using, π½π 110β 0° π½π 110β β 120° π½π 110β 120° π°ππ¨ = = , π°ππ© = = , πππ π°ππͺ = = ππ¨ 50 + π80 ππ© π50 ππͺ 100 + π25 So π°ππ¨ = 1.16β β 58° π΄ πππ , π°ππ© = 1.2β 150° π΄ πππ , π°ππͺ = 1.07β 106° π΄ πππ The complex power delivered to ππ¨ is β πΊπ¨ = π°ππ¨ ππ = 1.16β β 58° β 110β 0° = 68 + π109 ππ΄ Similarly we can find πΊπ© and πΊπͺ πΊπ© = 2.2β 150° β 110β β 120° = π242 ππ΄ πΊπͺ = 1.07β 106° β 110β 120° = 114 + π28 ππ΄ The total complex power is πΊπ¨ + πΊπ© + πΊπͺ = 182 + π379 ππ΄ © Jiayu Chen, Ph.D. 32 3-phase AC circuits Y-Ξ transformations The Ξ-to-Y and Y-to-Ξ transformations convert Ξ-connected loads to equivalent Yconnected loads and vice versa. © Jiayu Chen, Ph.D. 33 3-phase AC circuits Y-Ξ Circuit We use the phasor diagram. In a Ξ connection, the line current is β3 times the phase current and is displaced -30° in phase. The line-to-line voltage is equal to the phase voltage. IA IAB Using sine law, sin120° = sin30° We will have πππ = 3 π Or IL = 3Ip © Jiayu Chen, Ph.D. 34 3-phase AC circuits Balanced 3-phase circuit We have only two possible practical configurations for three-phase circuits, Y-to-Y and Y-to-Ξ and we can convert the latter to a Y-to-Y form. Thus, a practical three-phase circuit can always be converted to the Y-to-Y circuit. Y-to-Ξ circuit Y-to-Y circuit πΞ ππ = 3 © Jiayu Chen, Ph.D. The per-phase equivalent circuit 35 3-phase AC circuits Instantaneous and Average Power in a Balanced 3-Phase Load One advantage of three-phase power is the smooth flow of energy to the load. Consider a balanced load with resistance R. Then the instantaneous power is π£ππ π‘ 2 π£ππ π‘ 2 π£ππ π‘ 2 p t = + + R R R Where vab (π‘) = VcosΟt , and the other two-phase voltage have a phase of ±120°. Furthermore, cos2 Ξ±t = (1 + cos2Ξ±)/2 Therefore, V2 p t = 1 + cos2Οt + 1 + cos2 Οt β 120° + 1 + cos2 Οt β 240° 2R 3V 2 V 2 = + [cos2Οt + cos 2Οt β 240° + cos(2Οt β 480°)] 2R 2R The bracketed term is equal to zero for all time. Hence, 3V 2 p t = 2R The instantaneous power delivered to a balanced three-phase load is a constant. © Jiayu Chen, Ph.D. 36 Earthing System - Additional Earthing System: An electric connection to the general mass of earth, whose dimensions are very large in comparison to the electrical system being considered. The terms βGroundβ and βGroundingβ are synonymous with βEarthβ and βEarthingβ and are more prevalent in some countries like North America. © Jiayu Chen, Ph.D. 37 Earthing System - Additional Earth Electrode 1 : β’ Electrical connect to earth β’ Independent earth electrode for each system/equipment β’ Maintain sufficient distance, 3.5m or twice of driven length, from on another without significant affect the potential of the others Main Earthing Terminal 6 : β’ Connect group of protective conductors to the earth electrode 1 through earth conductor 2 Protective / bonding Conductor 3 : β’ Exposed conductive parts β’ Conductive part of electrical installation - not a live part β’ Extraneous conductive parts 4 β’ Conductive part not forming part of the electrical installation © Jiayu Chen, Ph.D. 38 Earthing System - Additional Residual Current Devices (RCDs) The RCD is a circuit breaker which continuously compares the current in the phase with that in the neutral. The difference between the two will flow to earth, because it has left the supply through the phase and has not returned in the neutral. 2 1 3 The purpose of the residual current device is to monitor the residual current and to switch off the circuit quickly if it rises to a preset level. When the amount of residual current, and hence of tripping current, reaches a predetermined level, the circuit breaker trips, opening the main contacts and interrupting the circuit. © Jiayu Chen, Ph.D. 39 Lightning System - Additional Thunder Cloud β’ β’ β’ Unstable upper atmosphere before lightning generation Positive charge and negative charge of thunder cloud separated Light weight positive charge at higher altitude, heavier negative charge at the base of the cloud Lightning Discharge β’ β’ β’ β’ May be cloud to cloud or cloud to ground Faintly luminescent leader channel from cloud toward to ground into many fingers Rapid escalating electric field at ground cause streamer move upward Discharge commenced the ionized channel completed at the junction of the leader and streamer Useful lighting projection system wiki: http://www.electrical-installation.org/enwiki/General_rules_of_lightning_protection © Jiayu Chen, Ph.D. 40 Lightning System - Additional The Surge Protection Device (SPD) is a component of the electrical installation protection system. This device is connected in parallel on the power supply circuit of the loads that it has to protect It can also be used at all levels of the power supply network. This is the most commonly used and most efficient type of overvoltage protection. Note: Surge Protection Devices (SPD) are used for electric power supply networks, telephone networks, and communication and automatic control buses. © Jiayu Chen, Ph.D. 41 Thank You! © Jiayu Chen, Ph.D. 42