Download to read offline
![Apparent Electrical Power,
𝑆 𝑒 = 𝑉𝐼 =
𝑉2
𝑍 𝑒
= 𝐼2 𝑍 𝑒,
where 𝑍 𝑒=𝑅 𝑒+j[ωL +
1
𝑗2 𝜔𝐶
]
Apparent Mechanical Power,
𝑆 𝑚 𝑠 = τ 𝑠 ω 𝑠 =
𝜏(𝑠)2
𝑍 𝑚(𝑠)
= 𝑠𝜔(𝑠)2 𝑍 𝑚(𝑠)
where, Zm s = B s + s[ J(s) +
𝐾(𝑠)
𝑠2 ]
Increase in Load means:
1. Decrease in Electrical Impedance, if applied voltage is kept constant.
2. Increase in Electrical Impedance, if current is kept constant.
3. Decreasein Mechanical Impedance, if applied torque is kept constant.
4. Increase in Mechanical Impedance, if angular velocity is kept constant.](https://image.slidesharecdn.com/2017-somebasicconceptionsandmisconceptionsinelectricalengineering-180306052721/85/2017-Some-basic-conceptions-and-misconceptions-in-electrical-engineering-45-320.jpg)
Uploaded byG Karthik Raja
2017 Some basic conceptions and misconceptions in electrical engineering
The document discusses various concepts and misconceptions related to electricity, such as the definitions of electricity, current flow, and power flow. It addresses common misunderstandings regarding capacitors and inductors, reactive and apparent power, and the distinction between sources and loads in electrical systems. The document also outlines the foundations of 3-phase power and the significance of the per unit system.
More Related Content
Similar to 2017 Some basic conceptions and misconceptions in electrical engineering
2017 Some basic conceptions and misconceptions in electrical engineering
- 2. Conceptions & Misconceptions 1.The term ‘Electricity’? 2. The term ‘Current Flow’. 3. The term ‘Power Flow’. 4. Concept of live, neutral and earth terminals. 5. Are capacitor and inductor linear, and do obey Ohm’s law? 6. Active, Reactive and Apparent Power. 7. Inductor always absorb reactive power while capacitor always deliver reactive power. 8. Distinction between ‘Source’ & ‘Load’. 9. Foundation of 3-Phase Power. 10. Significance of per unit system.
- 3. Misconception #1 Related tothe term ‘Electricity’ What comes to our Mind? 1. Electricity is a form of Energy. 2. Electricity is the flow of electrons. 3. Electricity is made of electrons. 4. Electricity is weightless. 5. Electricity flows nearly at the speed of light. 6. Batteries and generators create electricity. 7. Electric companies sell electricity. 8. Flow of Electricity results in flow of Electrical Energy 9. Cost of Electricity / Electricity Bill.
- 4. Electricity is simply ‘Quantityof Electric Charge’ Its SI unit is Coulomb Reference: The National Institute of Standards and Technology http://physics.nist.gov/cuu/Units/units.html Table 3 Reality
- 5. Most people usethe term ‘Current Flow’ to describe the direction of current in a circuit. While current itself means flow of charges, so the term ‘Current Flow’ would mean ‘Flow of Flow of Charges’ For example, we say ‘Current Flows from A to B’ It should really be said as ‘Current has a direction from A to B’ Misconception #2 The term ‘Current Flow’ A B I
- 6. Misconception #3 The term‘Power Flow’. Most people use the term ‘Power Flow’ which is a misnomer. Power itself means transfer of energy per unit time, so the term ‘Power Flow’ would mean ‘Flow of Transfer of Energy’ IT IS THE ENERGY WHICH IS TRANSFERRED, NOT POWER! SO THE TERM POWER FLOW IS COMPLETELY WRONG For example, we say ‘Power Flows from A to B’ It should really be said as ‘Energy Transfer has a direction from A to B’ A B E
- 7. How & Whythe concept of Live, Neutral and Earth came into existence? Misconception #4 (if any)
- 8. Floating Voltage (NoLive & Neutral) Distribution Transformer
- 9.
- 10. Disadvantages of FloatingVoltage
- 11.
- 12. How Live andNeutral eliminate the problem
- 13. Not safe totouch!
- 14. Case of equipmentconnected to Neutral What problem would occur?
- 15. Continued (Non IdealGrounding)
- 16. Introduction of SeparateEarth Wire Safe to Touch!
- 17. Definitions • Live Wire: It is ungrounded wire of the secondary side of distribution transformer. • Neutral Wire: It is the grounded wire of the secondary side of the distribution transformer and it carries the load current • Earth Wire: It is a thick separate wire which is grounded and it carries leakage current, and expected to have zero voltage.
- 18. Misconception #5 Are Capacitorsand Inductors linear? 𝑖 𝐶 = 𝐶 𝑑𝑣 𝐶 𝑑𝑡 𝑓(𝑡) = 𝐶 𝑑(𝑣 𝐶) 𝑑𝑡 𝑣1(t) 𝑣2(t) 𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2(t) 𝑖1 t + 𝑖2(t) 𝑓(𝑡) = 𝐶 𝑑(𝑣 𝐶) 𝑑𝑡 𝑓(𝑡) = 𝐶 𝑑(𝑣 𝐶) 𝑑𝑡 Current through capacitors is given by the relation 1. Capacitor as a System
- 19. 𝑣 𝐶 = 1 𝐶 𝑡𝑜 𝑡 𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜)Similarly, voltage across capacitor is given by the relation 𝑓 𝑡 = 1 𝐶 𝑡 𝑜 𝑡 𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜) 𝑣1(t) + 𝑉𝐶 (𝑡 𝑜)𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2 t + 𝑉𝐶(𝑡 𝑜) 𝑖1 t + 𝑖2(t) 𝑣2(t) + 𝑉𝐶 (𝑡 𝑜) 𝑓 𝑡 = 1 𝐶 𝑡 𝑜 𝑡 𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜) 𝑓 𝑡 = 1 𝐶 𝑡 𝑜 𝑡 𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜) ∴Capacitor as a system will be linear only when its initial condition, 𝑉𝐶(𝑡 𝑜) = 0
- 20. When Charge-Voltage (Q-V)characteristic is considered it is a straight line passing through origin, so it is a Linear element! Charge Voltage 𝑄 = 𝐶𝑉 2. Capacitor as an Element
- 21. When Voltage Current(V-I) characteristic is considered it is an ellipse, so Capacitor is a Non Linear element! Current Voltage 𝑖 𝐶 = 𝐶 𝑑𝑣 𝐶 𝑑𝑡
- 22. Question: If V-I characteristicsof capacitor and inductor are Non-Linear then how Ohm’s Law is defined for them?
- 23. Answer: “Capacitor” and “Inductor”do not obey Ohm’s Law! Generalized Statement of Ohm’s Law is “At any time t, voltage across any element is in direct proportion to current through it” v(t) a i(t) BUT, “Capacitive Reactance” and “Inductive Reactance” obey Ohm’s Law! 𝑽 t = 1 jω𝐶 𝑰 t , where 𝟏 𝐣𝛚𝐂 is Capacitive Reactance, 𝑋 𝐶 𝑽 t = jω𝐿 𝑰 t ,where 𝒋𝝎𝑳 is Inductive Reactance, 𝑋 𝐿 Where 𝑽 t and 𝑰 t are complex voltage and complex current respectively.
- 24. Misconception #6 Active, Reactiveand Apparent Power. How instantaneous power is measured? + - i(t) v(t) a) First we assign sign for voltage & current by making use of passive sign convention. b) If v(t) × i(t) is positive, then p(t) is the power consumed/absorbed by the element. c) If v(t) × i(t) is negative, then p(t) is the power deliveredby the element. 𝑃𝑜𝑤𝑒𝑟, 𝑝(𝑡) = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑑𝑡 𝑇𝑖𝑚𝑒, 𝑑𝑡 = 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝑨 𝒂𝒏𝒅 𝑩 𝒂𝒕 𝒕𝒊𝒎𝒆 𝒕 × 𝑪𝒉𝒂𝒓𝒈𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 𝒇𝒓𝒐𝒎 𝑨 𝒕𝒐 𝑩 𝒊𝒏 𝒕𝒊𝒎𝒆 𝒅𝒕 𝑻𝒊𝒎𝒆, 𝒅𝒕 = 𝑣(𝑡) × dq 𝑑𝑡 𝑻𝒉𝒖𝒔 𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒂𝒄𝒄. 𝒕𝒐 𝒄𝒐𝒏𝒗𝒆𝒏𝒕𝒊𝒐𝒏, 𝒑(𝒕) = 𝒗(𝒕) × 𝒊(𝒕) watts A B
- 25. Example 1: PurelyResistive Load i(t) v(t) + - Source Purely Resistive Load R Passive Sign Convention A V 𝑣 𝑡 = 𝑉𝑚 𝑆𝑖𝑛 𝜔𝑡 i 𝑡 = 𝑉𝑚 𝑅 𝑆𝑖𝑛 𝜔𝑡 i 𝑡 = 𝐼 𝑚 𝑆𝑖𝑛 𝜔𝑡 Phase Difference, 𝜑 = 0° For periodic signal, instead of defining instantaneous power, We define average power 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟, 𝑃𝑎𝑣𝑔 = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝑖𝑛 𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑 𝑇 𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇
- 26. Time, t Amplitude VoltageWaveform v(t) at 50Hz Current Waveform i(t) at 50 HZ Power Waveform p(t) at 100HZ Average Power, 𝑷 𝒂𝒗𝒈 𝑽 𝒎 𝑰 𝒎 𝑷 𝒎 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 𝑷 𝒂𝒗𝒈 = 𝑽 𝒎×𝑰 𝒎 𝟐 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 watts
- 27. Example 2: PurelyInductive Load 𝑣 𝑡 = 𝑉𝑚 𝑆𝑖𝑛 𝜔𝑡 i 𝑡 = 𝑉𝑚 𝜔𝐿 𝑆𝑖𝑛 𝜔𝑡 − 90° i 𝑡 = 𝐼 𝑚 𝑆𝑖𝑛 𝜔𝑡 − 90° Phase Difference, 𝜑 = −90° Voltage Waveform v(t) at 50Hz Current Waveform i(t) at 50 HZ Power Waveform p(t) at 100HZ Average Power, 𝑷 𝒂𝒗𝒈 𝑷 𝒂𝒗𝒈 = 𝟎
- 28. Conclusion: Average powerconsumed by a purely resistive load is 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝒘𝒂𝒕𝒕𝒔 Average power consumed by a purely inductive or capacitive load is 𝑷 𝒂𝒗𝒈 = 𝟎 𝒘𝒂𝒕𝒕𝒔 So why bother the presence of Inductor or Capacitor in load circuit? Reasons: 1. For same values of RMS voltage and RMS current magnitude, their active power output is less than that of a purely resistive load 2. For same active power load, they draw more RMS current than purely resistive load 3. This excess current increases copper loss in the transmission line and increases current and thermal rating of all components in the power system. For these reasons there is need to define some quantity which can account the effect of presence of reactive elements in the load.
- 29. Power Factor: It isthe measure of the degree -- to which -- a given load-- matches to that of – a pure resistance.
- 30. 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 𝑃𝐿𝑂𝐴𝐷= 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟 𝐶𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑙𝑜𝑎𝑑 𝑃𝑅 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟 𝐶𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑎 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 𝑓𝑜𝑟 𝑠𝑎𝑚𝑒 𝑅𝑀𝑆 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡. Generalized formula of Power Factor 𝑃. 𝐹. = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 = 517 1750 ≅ 0.3
- 31. Circuit with lowpower factor draws more current from the supply causing more copper loss in the transmission line and increases current and thermal rating of all components in the power system
- 32. i(t) v(t) + - RL Load V A Consider an R-Lload Concept of Reactive and Apparent Power A Graphical Approach
- 33. 𝝋 = −𝟕𝟎° 𝝋= 𝟎° 𝝋 = −𝟗𝟎° 𝑰 𝒎𝒂𝒙 𝑰 𝒎𝒂𝒙 𝐜𝐨𝐬 𝟕𝟎° 𝑰 𝒎𝒂𝒙 𝐬𝐢𝐧 𝟕𝟎° In Phase Component of Current Phase Quadrature Component of Current Load Voltage Load Current
- 34. 𝑷 𝒂𝒗𝒈 =𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝐜𝐨𝐬 𝟕𝟎° 𝑽 𝒎 𝑰 𝒎 cos 𝟕𝟎° 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 cos 𝟕𝟎° Power due to ‘in phase component’ of current
- 35. Power due to‘phase quadrature component’ of current 𝑽 𝒎 𝑰 𝒎 sin 𝟕𝟎° 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 sin 𝟕𝟎° 𝑷 𝒂𝒗𝒈 = 𝟎
- 36. 𝑽 𝒎 𝑰 𝒎sin 𝟕𝟎° 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 sin 𝟕𝟎° 𝑻𝒉𝒖𝒔, 𝑷 𝒂𝒗𝒈|𝒗𝒊𝒓𝒕𝒖𝒂𝒍 = 𝐑𝐞𝐚𝐜𝐭𝐢𝐯𝐞 𝐏𝐨𝐰𝐞𝐫, 𝐐 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 sin 𝟕𝟎° 𝐕𝐀𝐑 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 sin 𝟕𝟎° By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
- 37. Resultant Power 𝑽 𝒎 𝑰𝒎 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 cos 𝟕𝟎°
- 38. 𝑻𝒉𝒖𝒔, 𝑷 𝒂𝒗𝒈|𝒗𝒊𝒓𝒕𝒖𝒂𝒍= 𝐀𝐩𝐩𝐚𝐫𝐞𝐧𝐭 𝐏𝐨𝐰𝐞𝐫, 𝐒 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝐕𝐀 𝑽 𝒎 𝑰 𝒎 𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
- 39. Conclusion • Only ActivePower is the real & genuine power which is obtained by taking the average of the instantaneous power. • ‘Reactive Power’ and ‘Apparent Power’ are not real as they are obtained by manually shifting the waveform to 0 phase difference and calculating the virtual power by taking average of the virtual power waveform. • Since Q & S power quantities are obtained by shifting they are totally independent quantities, they have no physical relationship with active power P. But they posses a mathematical relationship.
- 40. Power Triangle 𝑆 =𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆 𝑃 = 𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆 cos 𝜑 𝑄 = 𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆 sin 𝜑 P S Q 𝝋 𝑺 = 𝑷 𝟐 + 𝑸 𝟐 𝑷 = 𝑺 𝐜𝐨𝐬 𝝋 𝑸 = 𝑺 𝐬𝐢𝐧 𝝋
- 41. Misconception #7 Inductor alwaysabsorb reactive power while capacitor always deliver reactive power.
- 42.
- 43. Misconception #8 Distinction between‘Source’ & ‘Load’. How to tell which is source and which is load? At any time t, The one which deliver power is source. The one which absorb power is load.
- 44. Element Parameter Nature IdealResistor Resistance, R Always a Load Inductor Inductance, L Load/Source Capacitor Capacitance, C Load/Source Ind. Voltage/Current Source Voltage/Current Source/Load Dep. Voltage/Current Source Voltage/Current Source/Load Ideal Transformer Turns Ratio Neither load nor source Element Parameter Nature Ideal Viscous Damper Coefficientof viscous friction, B Always a Load Spring Spring Constant, K Load/Source Mass/Moment of Inertia Mass, M/Moment of Inertia, J Load/Source Force/Torque Force, F/ Torque 𝜏 Source/Load Ideal Lever/Gear Teeth Ratio Neither load nor source Now what exactly the term ‘Increasing the Load’ means? ‘Increasing the Load’ means varying R, L, C in Electrical Domain and varying B, K, J in Mechanical Domain such that apparent power consumed by the system is increased.
- 45. Apparent Electrical Power, 𝑆𝑒 = 𝑉𝐼 = 𝑉2 𝑍 𝑒 = 𝐼2 𝑍 𝑒, where 𝑍 𝑒=𝑅 𝑒+j[ωL + 1 𝑗2 𝜔𝐶 ] Apparent Mechanical Power, 𝑆 𝑚 𝑠 = τ 𝑠 ω 𝑠 = 𝜏(𝑠)2 𝑍 𝑚(𝑠) = 𝑠𝜔(𝑠)2 𝑍 𝑚(𝑠) where, Zm s = B s + s[ J(s) + 𝐾(𝑠) 𝑠2 ] Increase in Load means: 1. Decrease in Electrical Impedance, if applied voltage is kept constant. 2. Increase in Electrical Impedance, if current is kept constant. 3. Decreasein Mechanical Impedance, if applied torque is kept constant. 4. Increase in Mechanical Impedance, if angular velocity is kept constant.
- 46.
- 47. 1 – ΦGenerator
- 48. 2 – ΦGenerator
- 49. 2 – ΦGenerator under load
- 50. 3 – ΦGenerator
- 51. Total Instantaneous Power PhaseB Instantaneous Power Phase Y Instantaneous Power Phase R Instantaneous Power Peak Power ERIR = Pm Er = ER ∠ 0° Ey = EY ∠ 120° Eb = EB ∠ -120° Ey Eb Ir Iy Ib Peak Power EY IY = Pm Peak Power EB IB = Pm Output Power = 1.5 Pm 3 – Φ Generator under load Er
- 52. The Star Connection ThreePhase, 6 Wire System & Corresponding Phasor Diagram
- 53. Three Phase, 4Wire System & its Line Currents
- 54. ELine-Line (RMS) =√3 x EPhase (RMS) Where, Eab,Ebc and Eca are line-line voltages Ean, Ebn and Ecn are phase voltages
- 55.
- 56. A C B 0 EMF 0 Voltage Drop ILine (RMS) =√3 x IPhase (RMS) Where, IA, IB and IC are line current IAB, IBC and ICA are phase current
- 57.
- 58. Sn kVA 110 100 1000 400000 Enp V 2400 2400 12470 69000 13800 Ens V 460 347 600 6900 424000 Inp A 0.417 4.17 8.02 14.5 29000 Ins A 2.17 28.8 167 145 943 R1 Ω 58.0 5.16 11.6 27.2 0.0003 R2 Ω 1.9 0.095 0.024 0.25 0.354 Xf1 Ω 32 4.3 39 151 0.028 Xf2 Ω 1.16 0.09 0.09 1.5 27 Xm Ω 200000 29000 150000 505000 460 Rm Ω 400000 51000 220000 432000 317 Io A 0.0134 0.0952 0.101 0.210 52.9 ACTUAL TRANSFORMER VALUES
- 59. Sn kVA 110 100 1000 400000 Enp V 2400 2400 12470 69000 13800 Ens V 460 347 600 6900 424000 Inp A 0.417 4.17 8.02 14.5 29000 Ins A 2.17 28.8 167 145 943 Znp Ω 5760 576 1555 4761 0.4761 Zns Ω 211.6 12.0 3.60 47.61 449.4 R1 (pu) – 0.0101 0.0090 0.0075 0.0057 0.00071 R2 (pu) – 0.0090 0.0079 0.0067 0.0053 0.00079 Xf1 (pu) – 0.0056 0.0075 0.0251 0.0317 0.0588 Xf2 (pu) – 0.0055 0.0075 0.0250 0.0315 0.0601 Xm (pu) – 34.7 50.3 96.5 106 966 Rm (pu) – 69.4 88.5 141.5 90.7 666 Io (pu) – 0.032 0.023 0.013 0.015 0.0018 PER-UNIT TRANSFORMER VALUES
- 60. Distinction between Flux& Flux Linkage Open Surface Flux
- 61. Flux Linkage ψ =𝐵. 𝑑𝑠 Open Surface having Conductor as its Contour Open Surface
- 62. References 1. Electrical Machines,Drives and Power System, Theodore Wildi 2. The National Institute of Standards and Technology 3. Power System Analysis, Grainger & Stevenson 4. Basic Electrical Technology, NPTEL 5. National Instruments
- 63.
Editor's Notes
- #48 The average power is one half the peak power. The electric power is derived from the mechanical power provided by the turbine driving the generator. As a result , the turbine must deliver its mechanical energy in pulses, to match the pulsed electrical output. This set up mechanical vibrations whose frequency is twice the electrical frequency. Consequently, the generator will vibrate and tend to be noisy.
- #49 In two phase generator, the armature winding are placed such that the phase diff. b/w the induced emf of the windings is 90 degrees.
- #50 Generator has been loaded, its phasor diagram and instantaneous power output have been shown.