Introduction to Electrical Systems Course Code: EE 111 Course Code: - PowerPoint PPT Presentation

Introduction to Electrical Systems Course Code: EE 111 Course Code: EE 111 Department: Electrical Engineering Department: Electrical Engineering Instructor Name: B G Fernandes Instructor Name: B.G. Fernandes E ‐ mail id: bgf @ee iitb ac in E ‐ mail id: bgf @ee.iitb.ac.in EE 111: Introduction to Electrical Systems Tue Oct 1/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Sub ‐ Topics: • Current transformer(contd..) • Auto transformer • Rotating machines Rotating machines EE 111: Introduction to Electrical Systems Tue Oct 2/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Review • If Δ /Y or Y/ Δ connections are used in 3 ‐ φ transformer • If Δ /Y or Y/ Δ connections are used in 3 ‐ φ transformer, V Δ = 1 ∠ = V 0 & 30 L( ) L(Y) for turns ratio = 1 V V V 3 3 Δ Δ L( L( ) ) L(Y) • In 3 Ф 4 wire system feeding • In 3 ‐Ф , 4 wire system feeding balanced 3 single phase non ‐ linear loads i → current in the neutral loads, i n → current in the neutral wire will be finite • Potential transformer (step down transformer) is used to measure high voltage AC hi h l AC EE 111: Introduction to Electrical Systems Tue Oct 3/18 Prof. B.G.Fernandes Lecture 25 6, 2009

how to measure high AC ‘I’ : MVA in 400kV line could be ≅ 600 × 3 600 10 ∴ = ≅ I I 850A 850A × 3 400 ammeter should be connected in series and this meter is ammeter should be connected in series and this meter is placed in control room ⇒ this current should be independent of status of thi t h ld b i d d t f t t f ammeter (faulty/working) ⇒ can we use a transformer to measure current? ⇒ current transformer EE 111: Introduction to Electrical Systems Tue Oct 4/18 Prof. B.G.Fernandes Lecture 25 6, 2009

⇒ no. of turns in primary = 1 ⇒ primary AT = ‘I’ (Independent of secondary AT) ⇒ In conventional transformer, primary AT is determined by secondary AT = N I N I 1 1 2 2 ⇒ current in the secondary should be low ⇒ current in the secondary should be low ∴ N 2 should be high ⎛ ⎞ I = ⎜ N ⎟ 2 ⎝ ⎠ Desirable current in secondary EE 111: Introduction to Electrical Systems Tue Oct 5/18 Prof. B.G.Fernandes Lecture 25 6, 2009

⇒ ammeter is used to measure the current in secondary ⇒ equivalent to shorted secondary ⇒ In conventional transformer short circuit in ⇒ In conventional transformer, short circuit in secondary side is avoided (transformer is protected against S.C) (transformer is protected against S.C) ⇒ because I 1 is determined by I 2 ⇒ In C.T, I 1 and therefore N 1 I 1 remains constant and independent of N 2 I 2 AT used to establish ‘ φ ’ in core = = − − = − = − N I N I N I N I I I N I N I 1 1 2 2 2 2 EE 111: Introduction to Electrical Systems Tue Oct 6/18 Prof. B.G.Fernandes Lecture 25 6, 2009

if secondary is O.C ⇒ N 2 I 2 = 0 AT used by core = I d b ⇒ core will be driven into saturation ⇒ this flux links N 2 ⇒ high ‘V’ will be induced in secondary ⇒ high V will be induced in secondary ⇒ shock hazard/ affects insulation ⇒ ensure low ‘R’ path in secondary ⇒ either connect ammeter or S C while disconnecting ⇒ either connect ammeter or S.C while disconnecting the ammeter EE 111: Introduction to Electrical Systems Tue Oct 7/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Auto transformer In transformer two sides are completely isolated In transformer, two sides are completely isolated In I.I.T.B. incomer is 22kV. This ‘V’ is stepped down to 440V ⇒ ‘V’ levels in two windings is high ⇒ V levels in two windings is high ⇒ If the voltage levels aren’t very high, isolation may not be required ⇒ use auto transformer total no. of turns = N 1 no. of turns in the tapped portion = N 2 f h d ⇒ ‘ φ ’ Linked by all the turns is the same V V N N ∴ = 1 1 V N 2 2 AT balance requires I 1 (N 1 ‐ N 2 ) = N 2 (I 2 ‐ I 1 ) ( neglecting I o ) AT b l i I (N N ) N (I I ) ( I ) l i EE 111: Introduction to Electrical Systems Tue Oct 8/18 Prof. B.G.Fernandes Lecture 25 6, 2009

N I ∴ = 1 2 N N I I 2 2 1 1 VA delivered to the load = V 2 I 2 = V 2 I 1 + V 2 (I 2 ‐ I 1 ) 2 ( 2 1 ) 2 1 V 2 I 1 → conductively transferred VA V 2 (I 2 ‐ I 1 ) → inductively transferred VA 2 ( 2 1 ) y Prob:10kVA, 2300/230 V, 2 winding transformer is connected as an auto transformer. Current in ‘ab’ (L.T.) winding is equal to the rated current of L.T. winding 10,000 = = = = I I 43.48A 43 48A 1 230 ∴ ‘I’ in bc = 4.35 A EE 111: Introduction to Electrical Systems Tue Oct 9/18 Prof. B.G.Fernandes Lecture 25 6, 2009

∴ I 2 = 47.8 A ∴ VA supplied by the source : 10 000 10,000 = × 2530 230 a a = × = a → turns ratio 11 10kVA a 1 − VA supplied inductively VA supplied inductively = V 2 (I 2 ‐ I 1 ) = 10kVA V (I I ) 10kVA 10,000 = = × × VA supplied conductively = V 2 I 1 VA supplied conductively V 2 I 1 2300 2300 230 = 100 kVA EE 111: Introduction to Electrical Systems Tue Oct 10/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Rotating Machines Electro Mechanical Energy Conversion (EMEC) Electro Mechanical Energy Conversion (EMEC) • Transformer converts electrical power from one ‘V’ level to another to another • EMEC equipment convert electrical energy into mechanical energy & vice versa mechanical energy & vice versa Motor Mechanical Energy Mechanical Energy Electrical Energy El t i l E Generator Generator Electrical Energy Mechanical Energy EE 111: Introduction to Electrical Systems Tue Oct 11/18 Prof. B.G.Fernandes Lecture 25 6, 2009

⇒ In both systems, there is electrical system & mechanical system mechanical system ⇒ coupled by a magnetic field Elementary concepts: Elementary concepts: conductor of length ‘l’ is moving with a uniform velocity of ‘ ν ’ with a uniform velocity of ν If ‘B’ is the magnetic field, voltage induced, e = Bl ν voltage induced, e Bl ν If external ‘R’ is connected, ‘ i ’ will flow ⇒ current carrying conductor placed in a magnetic field ⇒ current carrying conductor placed in a magnetic field experiences a force, F = Bil ⇒ This ‘F’ opposes the movement of conductor ⇒ This F opposes the movement of conductor EE 111: Introduction to Electrical Systems Tue Oct 12/18 Prof. B.G.Fernandes Lecture 25 6, 2009

In order to sustain the movement, external force F a is applied such that F a > F applied such that F > F ⇒ ‘i' will continue to flow ⇒ I 2 R = power is available at the o/p ⇒ I R = power is available at the o/p ⇒ this power is generated from mechanical input input → generator action Instead if V > E is applied, no external force is required I t d if V > E i li d t l f i i d to maintain the motion of the conductor ⇒ Input Electrical energy o/p mechanical energy ⇒ Motoring action EE 111: Introduction to Electrical Systems Tue Oct 13/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Basic requirements of EMEC devices: For motoring /generating action, t here has to be a field produced by a set of coils → field coil ⇒ this flux induces ‘V’ or ‘I’ in another coil which is rotating in the magnetic field known as armature ⇒ do we need to have conductor carrying ‘I’ to experience a force ? p In rotating machine ‘V’ induced in the conductor is sinusoidal, irrespective of nature of field flux , p ∴ If dc is required, some arrangement must be made to convert ac to dc convert ac to dc EE 111: Introduction to Electrical Systems Tue Oct 14/18 Prof. B.G.Fernandes Lecture 25 6, 2009

consider a coil rotating in a magnetic field, ⇒ V induced is sinusoidal ⇒ ‘V’ induced is sinusoidal ⇒ In one rotation ‘V’ induced in the coil has completed one cycle (360 0 ) coil has completed one cycle (360 0 ) ⇒ ‘V’ induced in the coil completes two cycles in one cycle of rotation ⇒ one complete rotation = 360 0 (mech) ⇒ two complete cycles = 720 0 (elect) ∴ electrical degrees ≠ mechanical degrees l i l d h i l d 1 0 elec= (2/P) 0 mech where ‘P’ is the number of poles EE 111: Introduction to Electrical Systems Tue Oct 15/18 Prof. B.G.Fernandes Lecture 25 6, 2009

Basic structure: stator → stationary does not move stator → stationary, does not move → normally outer frame rotor → which rotates inside the stator rotor → which rotates inside the stator → separated by small air gap (0.5 ‐ 1mm) ⇒ generally stator and rotor are made up of high ⇒ generally stator and rotor are made up of high permeability ferromagnetic material ⇒ length of the air gap is kept as small as possible so ⇒ length of the air gap is kept as small as possible so that AT required to establish φ in the air gap is small ⇒ stator bore & rotor are perfectly round ⇒ stator bore & rotor are perfectly round ⇒ air gap is uniform & ∴ reluctance ⇒ non salient pole machines ⇒ non ‐ salient pole machines EE 111: Introduction to Electrical Systems Tue Oct 16/18 Prof. B.G.Fernandes Lecture 25 6, 2009