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2017 Some basic conceptions and misconceptions in electrical engineering

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G Karthik Raja
G Karthik Raja

In this presentation, I discuss some basic concepts in electrical engineering that are not well known to students like the concept of Live, Neutral and Earth wiring; Active, Reactive and Apparent power and Foundation of 3-phase system. I have brought in a new perspective to understand these concepts. I also discuss some misconceptions that students face and resolve it the right way.

Engineering
1 of 63
Conceptions & Misconceptions 1. The term β€˜Electricity’? 2. The term β€˜Current Flow’. 3. The term β€˜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 β€˜Source’ & β€˜Load’. 9. Foundation of 3-Phase Power. 10. Significance of per unit system.
Misconception #1 Related to the term β€˜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.
Electricity is simply β€˜Quantity of 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
Most people use the term β€˜Current Flow’ to describe the direction of current in a circuit. While current itself means flow of charges, so the term β€˜Current Flow’ would mean β€˜Flow of Flow of Charges’ For example, we say β€˜Current Flows from A to B’ It should really be said as β€˜Current has a direction from A to B’ Misconception #2 The term β€˜Current Flow’ A B I
Misconception #3 The term β€˜Power Flow’. Most people use the term β€˜Power Flow’ which is a misnomer. Power itself means transfer of energy per unit time, so the term β€˜Power Flow’ would mean β€˜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 β€˜Power Flows from A to B’ It should really be said as β€˜Energy Transfer has a direction from A to B’ A B E
How & Why the concept of Live, Neutral and Earth came into existence? Misconception #4 (if any)
Floating Voltage (No Live & Neutral) Distribution Transformer
continued
Disadvantages of Floating Voltage
continued
How Live and Neutral eliminate the problem
Not safe to touch!
Case of equipment connected to Neutral What problem would occur?
Continued (Non Ideal Grounding)
Introduction of Separate Earth Wire Safe to Touch!
Definitions β€’ Live Wire : It is ungrounded wire of the secondary side of distribution transformer. β€’ Neutral Wire: It is the grounded wire of the secondary side of the distribution transformer and it carries the load current β€’ Earth Wire: It is a thick separate wire which is grounded and it carries leakage current, and expected to have zero voltage.
Misconception #5 Are Capacitors and Inductors linear? 𝑖 𝐢 = 𝐢 𝑑𝑣 𝐢 𝑑𝑑 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 𝑣1(t) 𝑣2(t) 𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2(t) 𝑖1 t + 𝑖2(t) 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 Current through capacitors is given by the relation 1. Capacitor as a System
𝑣 𝐢 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ)Similarly, voltage across capacitor is given by the relation 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) 𝑣1(t) + 𝑉𝐢 (𝑑 π‘œ)𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2 t + 𝑉𝐢(𝑑 π‘œ) 𝑖1 t + 𝑖2(t) 𝑣2(t) + 𝑉𝐢 (𝑑 π‘œ) 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) ∴Capacitor as a system will be linear only when its initial condition, 𝑉𝐢(𝑑 π‘œ) = 0
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 = 𝐢𝑉 2. Capacitor as an Element
When Voltage Current (V-I) characteristic is considered it is an ellipse, so Capacitor is a Non Linear element! Current Voltage 𝑖 𝐢 = 𝐢 𝑑𝑣 𝐢 𝑑𝑑
Question: If V-I characteristics of capacitor and inductor are Non-Linear then how Ohm’s Law is defined for them?
Answer: β€œCapacitor” and β€œInductor” do not obey Ohm’s Law! Generalized Statement of Ohm’s Law is β€œAt any time t, voltage across any element is in direct proportion to current through it” v(t) a i(t) BUT, β€œCapacitive Reactance” and β€œInductive Reactance” obey Ohm’s Law! 𝑽 t = 1 jω𝐢 𝑰 t , where 𝟏 π£π›šπ‚ is Capacitive Reactance, 𝑋 𝐢 𝑽 t = jω𝐿 𝑰 t ,where π’‹πŽπ‘³ is Inductive Reactance, 𝑋 𝐿 Where 𝑽 t and 𝑰 t are complex voltage and complex current respectively.
Misconception #6 Active, Reactive and 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) Γ— i(t) is positive, then p(t) is the power consumed/absorbed by the element. c) If v(t) Γ— i(t) is negative, then p(t) is the power deliveredby the element. π‘ƒπ‘œπ‘€π‘’π‘Ÿ, 𝑝(𝑑) = π‘Šπ‘œπ‘Ÿπ‘˜ π·π‘œπ‘›π‘’ 𝑖𝑛 π‘‘π‘–π‘šπ‘’ 𝑑𝑑 π‘‡π‘–π‘šπ‘’, 𝑑𝑑 = π‘·π’π’•π’†π’π’•π’Šπ’‚π’ π‘«π’Šπ’‡π’‡π’†π’“π’†π’π’„π’† π’ƒπ’†π’•π’˜π’†π’†π’ 𝑨 𝒂𝒏𝒅 𝑩 𝒂𝒕 π’•π’Šπ’Žπ’† 𝒕 Γ— π‘ͺπ’‰π’‚π’“π’ˆπ’† 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 π’‡π’“π’π’Ž 𝑨 𝒕𝒐 𝑩 π’Šπ’ π’•π’Šπ’Žπ’† 𝒅𝒕 π‘»π’Šπ’Žπ’†, 𝒅𝒕 = 𝑣(𝑑) Γ— dq 𝑑𝑑 𝑻𝒉𝒖𝒔 π’Šπ’π’”π’•π’‚π’π’•π’‚π’π’†π’π’–π’” π’‘π’π’˜π’†π’“ π’„π’π’π’”π’–π’Žπ’†π’… 𝒂𝒄𝒄. 𝒕𝒐 π’„π’π’π’—π’†π’π’•π’Šπ’π’, 𝒑(𝒕) = 𝒗(𝒕) Γ— π’Š(𝒕) watts A B
Example 1: Purely Resistive Load i(t) v(t) + - Source Purely Resistive Load R Passive Sign Convention A V 𝑣 𝑑 = π‘‰π‘š 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = π‘‰π‘š 𝑅 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = 𝐼 π‘š 𝑆𝑖𝑛 πœ”π‘‘ Phase Difference, πœ‘ = 0Β° For periodic signal, instead of defining instantaneous power, We define average power π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ, π‘ƒπ‘Žπ‘£π‘” = π‘Šπ‘œπ‘Ÿπ‘˜ π·π‘œπ‘›π‘’ 𝑖𝑛 π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑇 π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘, 𝑇
Time, t Amplitude  Voltage Waveform v(t) at 50Hz  Current Waveform i(t) at 50 HZ  Power Waveform p(t) at 100HZ  Average Power, 𝑷 π’‚π’—π’ˆ 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 π’ŽΓ—π‘° π’Ž 𝟐 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 watts
Example 2: Purely Inductive Load 𝑣 𝑑 = π‘‰π‘š 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = π‘‰π‘š πœ”πΏ 𝑆𝑖𝑛 πœ”π‘‘ βˆ’ 90Β° i 𝑑 = 𝐼 π‘š 𝑆𝑖𝑛 πœ”π‘‘ βˆ’ 90Β° Phase Difference, πœ‘ = βˆ’90Β°  Voltage Waveform v(t) at 50Hz  Current Waveform i(t) at 50 HZ  Power Waveform p(t) at 100HZ  Average Power, 𝑷 π’‚π’—π’ˆ 𝑷 π’‚π’—π’ˆ = 𝟎
Conclusion: οƒ˜ Average power consumed by a purely resistive load is 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 π’˜π’‚π’•π’•π’” οƒ˜ Average power consumed by a purely inductive or capacitive load is 𝑷 π’‚π’—π’ˆ = 𝟎 π’˜π’‚π’•π’•π’” 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.
Power Factor: It is the measure of the degree -- to which -- a given load-- matches to that of – a pure resistance.
π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 𝑃𝐿𝑂𝐴𝐷 = π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΆπ‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ 𝑏𝑦 π‘Ž 𝑔𝑖𝑣𝑒𝑛 π‘™π‘œπ‘Žπ‘‘ 𝑃𝑅 = π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΆπ‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ 𝑏𝑦 π‘Ž π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘–π‘£π‘’ π‘™π‘œπ‘Žπ‘‘ π‘“π‘œπ‘Ÿ π‘ π‘Žπ‘šπ‘’ 𝑅𝑀𝑆 π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Žπ‘›π‘‘ π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘. Generalized formula of Power Factor 𝑃. 𝐹. = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 = 517 1750 β‰… 0.3
Circuit with low power 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
i(t) v(t) + - RL Load V A Consider an R-L load Concept of Reactive and Apparent Power A Graphical Approach
𝝋 = βˆ’πŸ•πŸŽΒ° 𝝋 = 𝟎° 𝝋 = βˆ’πŸ—πŸŽΒ° 𝑰 π’Žπ’‚π’™ 𝑰 π’Žπ’‚π’™ 𝐜𝐨𝐬 πŸ•πŸŽΒ° 𝑰 π’Žπ’‚π’™ 𝐬𝐒𝐧 πŸ•πŸŽΒ°  In Phase Component of Current  Phase Quadrature Component of Current  Load Voltage  Load Current
𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 𝐜𝐨𝐬 πŸ•πŸŽΒ° 𝑽 π’Ž 𝑰 π’Ž cos πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž cos πŸ•πŸŽΒ° Power due to β€˜in phase component’ of current
Power due to β€˜phase quadrature component’ of current 𝑽 π’Ž 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’‚π’—π’ˆ = 𝟎
𝑽 π’Ž 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑻𝒉𝒖𝒔, 𝑷 π’‚π’—π’ˆ|π’—π’Šπ’“π’•π’–π’‚π’ = π‘πžπšπœπ­π’π―πž 𝐏𝐨𝐰𝐞𝐫, 𝐐 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 sin πŸ•πŸŽΒ° 𝐕𝐀𝐑 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 sin πŸ•πŸŽΒ° By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
Resultant Power 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 cos πŸ•πŸŽΒ°
𝑻𝒉𝒖𝒔, 𝑷 π’‚π’—π’ˆ|π’—π’Šπ’“π’•π’–π’‚π’ = π€π©π©πšπ«πžπ§π­ 𝐏𝐨𝐰𝐞𝐫, 𝐒 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 𝐕𝐀 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
Conclusion β€’ Only Active Power is the real & genuine power which is obtained by taking the average of the instantaneous power. β€’ β€˜Reactive Power’ and β€˜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. β€’ 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.
Power Triangle οƒ˜π‘† = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 οƒ˜π‘ƒ = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 cos πœ‘ οƒ˜π‘„ = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 sin πœ‘ P S Q 𝝋 οƒ˜ 𝑺 = 𝑷 𝟐 + 𝑸 𝟐 οƒ˜ 𝑷 = 𝑺 𝐜𝐨𝐬 𝝋 οƒ˜ 𝑸 = 𝑺 𝐬𝐒𝐧 𝝋
Misconception #7 Inductor always absorb reactive power while capacitor always deliver reactive power.
Power, p(t) Power, p(t) Inductor Capacitor time time
Misconception #8 Distinction between β€˜Source’ & β€˜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.
Element Parameter Nature Ideal Resistor 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 β€˜Increasing the Load’ means? β€˜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.
Apparent Electrical Power, 𝑆 𝑒 = 𝑉𝐼 = 𝑉2 𝑍 𝑒 = 𝐼2 𝑍 𝑒, where 𝑍 𝑒=𝑅 𝑒+j[Ο‰L + 1 𝑗2 πœ”πΆ ] Apparent Mechanical Power, 𝑆 π‘š 𝑠 = Ο„ 𝑠 Ο‰ 𝑠 = 𝜏(𝑠)2 𝑍 π‘š(𝑠) = π‘ πœ”(𝑠)2 𝑍 π‘š(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.
Foundation of 3-Phase Power
1 – Ξ¦ Generator
2 – Ξ¦ Generator
2 – Ξ¦ Generator under load
3 – Ξ¦ Generator
Total Instantaneous Power Phase B Instantaneous Power Phase Y Instantaneous Power Phase R Instantaneous Power Peak Power ERIR = Pm Er = ER ∠ 0Β° Ey = EY ∠ 120Β° Eb = EB ∠ -120Β° Ey Eb Ir Iy Ib Peak Power EY IY = Pm Peak Power EB IB = Pm Output Power = 1.5 Pm 3 – Ξ¦ Generator under load Er
The Star Connection Three Phase, 6 Wire System & Corresponding Phasor Diagram
Three Phase, 4 Wire System & its Line Currents
ELine-Line (RMS) = √3 x EPhase (RMS) Where, Eab,Ebc and Eca are line-line voltages Ean, Ebn and Ecn are phase voltages
0 EMF A B C B C A EAB EBC ECA 0 Voltage Drop The Delta Connection
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
Transformer Example The Per-Unit System
Sn kVA 1 10 100 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
Sn kVA 1 10 100 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
Distinction between Flux & Flux Linkage Open Surface Flux
Flux Linkage ψ = 𝐡. 𝑑𝑠 Open Surface having Conductor as its Contour Open Surface
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
End of Presentation.

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2017 Some basic conceptions and misconceptions in electrical engineering

  • 1.
  • 2. Conceptions & Misconceptions 1. The term β€˜Electricity’? 2. The term β€˜Current Flow’. 3. The term β€˜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 β€˜Source’ & β€˜Load’. 9. Foundation of 3-Phase Power. 10. Significance of per unit system.
  • 3. Misconception #1 Related to the term β€˜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 β€˜Quantity of 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 use the term β€˜Current Flow’ to describe the direction of current in a circuit. While current itself means flow of charges, so the term β€˜Current Flow’ would mean β€˜Flow of Flow of Charges’ For example, we say β€˜Current Flows from A to B’ It should really be said as β€˜Current has a direction from A to B’ Misconception #2 The term β€˜Current Flow’ A B I
  • 6. Misconception #3 The term β€˜Power Flow’. Most people use the term β€˜Power Flow’ which is a misnomer. Power itself means transfer of energy per unit time, so the term β€˜Power Flow’ would mean β€˜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 β€˜Power Flows from A to B’ It should really be said as β€˜Energy Transfer has a direction from A to B’ A B E
  • 7. How & Why the concept of Live, Neutral and Earth came into existence? Misconception #4 (if any)
  • 8. Floating Voltage (No Live & Neutral) Distribution Transformer
  • 12. How Live and Neutral eliminate the problem
  • 13. Not safe to touch!
  • 14. Case of equipment connected to Neutral What problem would occur?
  • 15. Continued (Non Ideal Grounding)
  • 16. Introduction of Separate Earth Wire Safe to Touch!
  • 17. Definitions β€’ Live Wire : It is ungrounded wire of the secondary side of distribution transformer. β€’ Neutral Wire: It is the grounded wire of the secondary side of the distribution transformer and it carries the load current β€’ 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 Capacitors and Inductors linear? 𝑖 𝐢 = 𝐢 𝑑𝑣 𝐢 𝑑𝑑 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 𝑣1(t) 𝑣2(t) 𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2(t) 𝑖1 t + 𝑖2(t) 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 𝑓(𝑑) = 𝐢 𝑑(𝑣 𝐢) 𝑑𝑑 Current through capacitors is given by the relation 1. Capacitor as a System
  • 19. 𝑣 𝐢 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ)Similarly, voltage across capacitor is given by the relation 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) 𝑣1(t) + 𝑉𝐢 (𝑑 π‘œ)𝑖1(t) 𝑖2(t) 𝑣1 t + 𝑣2 t + 𝑉𝐢(𝑑 π‘œ) 𝑖1 t + 𝑖2(t) 𝑣2(t) + 𝑉𝐢 (𝑑 π‘œ) 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) 𝑓 𝑑 = 1 𝐢 𝑑 π‘œ 𝑑 𝑖 𝐢 𝑑𝑑 + 𝑉𝐢(𝑑 π‘œ) ∴Capacitor as a system will be linear only when its initial condition, 𝑉𝐢(𝑑 π‘œ) = 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 = 𝐢𝑉 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 𝐢 = 𝐢 𝑑𝑣 𝐢 𝑑𝑑
  • 22. Question: If V-I characteristics of capacitor and inductor are Non-Linear then how Ohm’s Law is defined for them?
  • 23. Answer: β€œCapacitor” and β€œInductor” do not obey Ohm’s Law! Generalized Statement of Ohm’s Law is β€œAt any time t, voltage across any element is in direct proportion to current through it” v(t) a i(t) BUT, β€œCapacitive Reactance” and β€œInductive Reactance” obey Ohm’s Law! 𝑽 t = 1 jω𝐢 𝑰 t , where 𝟏 π£π›šπ‚ is Capacitive Reactance, 𝑋 𝐢 𝑽 t = jω𝐿 𝑰 t ,where π’‹πŽπ‘³ is Inductive Reactance, 𝑋 𝐿 Where 𝑽 t and 𝑰 t are complex voltage and complex current respectively.
  • 24. Misconception #6 Active, Reactive and 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) Γ— i(t) is positive, then p(t) is the power consumed/absorbed by the element. c) If v(t) Γ— i(t) is negative, then p(t) is the power deliveredby the element. π‘ƒπ‘œπ‘€π‘’π‘Ÿ, 𝑝(𝑑) = π‘Šπ‘œπ‘Ÿπ‘˜ π·π‘œπ‘›π‘’ 𝑖𝑛 π‘‘π‘–π‘šπ‘’ 𝑑𝑑 π‘‡π‘–π‘šπ‘’, 𝑑𝑑 = π‘·π’π’•π’†π’π’•π’Šπ’‚π’ π‘«π’Šπ’‡π’‡π’†π’“π’†π’π’„π’† π’ƒπ’†π’•π’˜π’†π’†π’ 𝑨 𝒂𝒏𝒅 𝑩 𝒂𝒕 π’•π’Šπ’Žπ’† 𝒕 Γ— π‘ͺπ’‰π’‚π’“π’ˆπ’† 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 π’‡π’“π’π’Ž 𝑨 𝒕𝒐 𝑩 π’Šπ’ π’•π’Šπ’Žπ’† 𝒅𝒕 π‘»π’Šπ’Žπ’†, 𝒅𝒕 = 𝑣(𝑑) Γ— dq 𝑑𝑑 𝑻𝒉𝒖𝒔 π’Šπ’π’”π’•π’‚π’π’•π’‚π’π’†π’π’–π’” π’‘π’π’˜π’†π’“ π’„π’π’π’”π’–π’Žπ’†π’… 𝒂𝒄𝒄. 𝒕𝒐 π’„π’π’π’—π’†π’π’•π’Šπ’π’, 𝒑(𝒕) = 𝒗(𝒕) Γ— π’Š(𝒕) watts A B
  • 25. Example 1: Purely Resistive Load i(t) v(t) + - Source Purely Resistive Load R Passive Sign Convention A V 𝑣 𝑑 = π‘‰π‘š 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = π‘‰π‘š 𝑅 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = 𝐼 π‘š 𝑆𝑖𝑛 πœ”π‘‘ Phase Difference, πœ‘ = 0Β° For periodic signal, instead of defining instantaneous power, We define average power π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ, π‘ƒπ‘Žπ‘£π‘” = π‘Šπ‘œπ‘Ÿπ‘˜ π·π‘œπ‘›π‘’ 𝑖𝑛 π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑇 π‘‡π‘–π‘šπ‘’ π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘, 𝑇
  • 26. Time, t Amplitude  Voltage Waveform v(t) at 50Hz  Current Waveform i(t) at 50 HZ  Power Waveform p(t) at 100HZ  Average Power, 𝑷 π’‚π’—π’ˆ 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 π’ŽΓ—π‘° π’Ž 𝟐 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 watts
  • 27. Example 2: Purely Inductive Load 𝑣 𝑑 = π‘‰π‘š 𝑆𝑖𝑛 πœ”π‘‘ i 𝑑 = π‘‰π‘š πœ”πΏ 𝑆𝑖𝑛 πœ”π‘‘ βˆ’ 90Β° i 𝑑 = 𝐼 π‘š 𝑆𝑖𝑛 πœ”π‘‘ βˆ’ 90Β° Phase Difference, πœ‘ = βˆ’90Β°  Voltage Waveform v(t) at 50Hz  Current Waveform i(t) at 50 HZ  Power Waveform p(t) at 100HZ  Average Power, 𝑷 π’‚π’—π’ˆ 𝑷 π’‚π’—π’ˆ = 𝟎
  • 28. Conclusion: οƒ˜ Average power consumed by a purely resistive load is 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 π’˜π’‚π’•π’•π’” οƒ˜ Average power consumed by a purely inductive or capacitive load is 𝑷 π’‚π’—π’ˆ = 𝟎 π’˜π’‚π’•π’•π’” 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 is the measure of the degree -- to which -- a given load-- matches to that of – a pure resistance.
  • 30. π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 𝑃𝐿𝑂𝐴𝐷 = π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΆπ‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ 𝑏𝑦 π‘Ž 𝑔𝑖𝑣𝑒𝑛 π‘™π‘œπ‘Žπ‘‘ 𝑃𝑅 = π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘ƒπ‘œπ‘€π‘’π‘Ÿ πΆπ‘œπ‘›π‘ π‘’π‘šπ‘’π‘‘ 𝑏𝑦 π‘Ž π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘–π‘£π‘’ π‘™π‘œπ‘Žπ‘‘ π‘“π‘œπ‘Ÿ π‘ π‘Žπ‘šπ‘’ 𝑅𝑀𝑆 π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘£π‘œπ‘™π‘‘π‘Žπ‘”π‘’ π‘Žπ‘›π‘‘ π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘. Generalized formula of Power Factor 𝑃. 𝐹. = 𝑃𝐿𝑂𝐴𝐷 𝑃𝑅 = 517 1750 β‰… 0.3
  • 31. Circuit with low power 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-L load Concept of Reactive and Apparent Power A Graphical Approach
  • 33. 𝝋 = βˆ’πŸ•πŸŽΒ° 𝝋 = 𝟎° 𝝋 = βˆ’πŸ—πŸŽΒ° 𝑰 π’Žπ’‚π’™ 𝑰 π’Žπ’‚π’™ 𝐜𝐨𝐬 πŸ•πŸŽΒ° 𝑰 π’Žπ’‚π’™ 𝐬𝐒𝐧 πŸ•πŸŽΒ°  In Phase Component of Current  Phase Quadrature Component of Current  Load Voltage  Load Current
  • 34. 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 𝐜𝐨𝐬 πŸ•πŸŽΒ° 𝑽 π’Ž 𝑰 π’Ž cos πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž cos πŸ•πŸŽΒ° Power due to β€˜in phase component’ of current
  • 35. Power due to β€˜phase quadrature component’ of current 𝑽 π’Ž 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’‚π’—π’ˆ = 𝟎
  • 36. 𝑽 π’Ž 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž sin πŸ•πŸŽΒ° 𝑻𝒉𝒖𝒔, 𝑷 π’‚π’—π’ˆ|π’—π’Šπ’“π’•π’–π’‚π’ = π‘πžπšπœπ­π’π―πž 𝐏𝐨𝐰𝐞𝐫, 𝐐 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 sin πŸ•πŸŽΒ° 𝐕𝐀𝐑 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 sin πŸ•πŸŽΒ° By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
  • 37. Resultant Power 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 cos πŸ•πŸŽΒ°
  • 38. 𝑻𝒉𝒖𝒔, 𝑷 π’‚π’—π’ˆ|π’—π’Šπ’“π’•π’–π’‚π’ = π€π©π©πšπ«πžπ§π­ 𝐏𝐨𝐰𝐞𝐫, 𝐒 = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 𝐕𝐀 𝑽 π’Ž 𝑰 π’Ž 𝑷 π’Ž = 𝑽 π’Ž Γ— 𝑰 π’Ž 𝑷 π’‚π’—π’ˆ = 𝑽 𝑹𝑴𝑺 Γ— 𝑰 𝑹𝑴𝑺 By manually shifting the voltage and current waveform to bring them in phase Virtual Power Waveform
  • 39. Conclusion β€’ Only Active Power is the real & genuine power which is obtained by taking the average of the instantaneous power. β€’ β€˜Reactive Power’ and β€˜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. β€’ 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 οƒ˜π‘† = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 οƒ˜π‘ƒ = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 cos πœ‘ οƒ˜π‘„ = 𝑉𝑅𝑀𝑆 Γ— 𝐼 𝑅𝑀𝑆 sin πœ‘ P S Q 𝝋 οƒ˜ 𝑺 = 𝑷 𝟐 + 𝑸 𝟐 οƒ˜ 𝑷 = 𝑺 𝐜𝐨𝐬 𝝋 οƒ˜ 𝑸 = 𝑺 𝐬𝐒𝐧 𝝋
  • 41. Misconception #7 Inductor always absorb reactive power while capacitor always deliver reactive power.
  • 43. Misconception #8 Distinction between β€˜Source’ & β€˜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 Ideal Resistor 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 β€˜Increasing the Load’ means? β€˜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, 𝑆 𝑒 = 𝑉𝐼 = 𝑉2 𝑍 𝑒 = 𝐼2 𝑍 𝑒, where 𝑍 𝑒=𝑅 𝑒+j[Ο‰L + 1 𝑗2 πœ”πΆ ] Apparent Mechanical Power, 𝑆 π‘š 𝑠 = Ο„ 𝑠 Ο‰ 𝑠 = 𝜏(𝑠)2 𝑍 π‘š(𝑠) = π‘ πœ”(𝑠)2 𝑍 π‘š(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.
  • 47. 1 – Ξ¦ Generator
  • 48. 2 – Ξ¦ Generator
  • 49. 2 – Ξ¦ Generator under load
  • 50. 3 – Ξ¦ Generator
  • 51. Total Instantaneous Power Phase B Instantaneous Power Phase Y Instantaneous Power Phase R Instantaneous Power Peak Power ERIR = Pm Er = ER ∠ 0Β° Ey = EY ∠ 120Β° Eb = EB ∠ -120Β° Ey Eb Ir Iy Ib Peak Power EY IY = Pm Peak Power EB IB = Pm Output Power = 1.5 Pm 3 – Ξ¦ Generator under load Er
  • 52. The Star Connection Three Phase, 6 Wire System & Corresponding Phasor Diagram
  • 53. Three Phase, 4 Wire 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
  • 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
  • 58. Sn kVA 1 10 100 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 1 10 100 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 = 𝐡. 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

Editor's Notes

  1. 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.
  2. 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.
  3. Generator has been loaded, its phasor diagram and instantaneous power output have been shown.