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

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

1/18 Lecture 25

Sub‐Topics:

• Current transformer(contd..)
• Auto transformer
• Rotating machines

Rotating machines

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

2/18 Lecture 25

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,

L(Y) L( )

V V

& 30

Δ

∠ =

L( )

V 1 V 3

Δ =

for turns ratio = 1

• In 3 Ф 4 wire system feeding

L( ) Δ L(Y)

V 3

• In 3‐Ф, 4 wire system feeding

balanced 3 single phase non‐linear loads i → current in the neutral loads, in→ current in the neutral wire will be finite

• Potential transformer (step down transformer) is used

hi h l AC

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

to measure high voltage AC

3/18 Lecture 25

how to measure high AC ‘I’ : MVA in 400kV line could be ≅ 600

3

600 10 I 850A × I 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 thi t h ld b i d d t f t t f ⇒ this current should be independent of status of ammeter (faulty/working) ⇒ can we use a transformer to measure current? ⇒ current transformer

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

4/18 Lecture 25

⇒ no. of turns in primary = 1 ⇒ primary AT = ‘I’

(Independent of secondary AT)

⇒ In conventional transformer, primary AT is determined by secondary AT

1 1 2 2

N I N I = ⇒ current in the secondary should be low ⇒ current in the secondary should be low ∴ N2 should be high I

2

N Desirable current in secondary ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

5/18 Lecture 25

⇒ 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 I1 is determined by I2 ⇒ In C.T, I1 and therefore N1I1 remains constant and independent of N2I2 AT used to establish ‘φ’ in core N I N I I N I = − = −

1 1 2 2 2 2

N I N I I N I = − = −

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

6/18 Lecture 25

if secondary is O.C ⇒ N2I2 = 0 d b AT used by core = I ⇒ core will be driven into saturation ⇒ this flux links N2 ⇒ 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

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

7/18 Lecture 25

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 = N1 f h d

• no. of turns in the tapped portion = N2

⇒ ‘φ’ Linked by all the turns is the same V N

1 1 2 2

V N V N ∴ = AT b l i I (N N ) N (I I ) (

l i I )

AT balance requires I1 (N1 ‐ N2) = N2 (I2‐ I1) (neglecting Io)

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

8/18 Lecture 25

1 2 2 1

N I N I ∴ = VA delivered to the load = V2I2 = V2I1 + V2(I2‐ I1)

2 1

N I

2 1 2( 2 1)

V2I1 → conductively transferred VA V2(I2‐ I1) → 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 43 48A = =

1

I 43.48A 230 = = ∴ ‘I’ in bc = 4.35 A

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

9/18 Lecture 25

∴ I2 = 47.8 A ∴ VA supplied by the source : 10 000 a 10,000 2530 230 = × a 11 10kVA a 1 = × = − VA supplied inductively V (I I ) 10kVA

a → turns ratio

VA supplied inductively = V2(I2 ‐ I1) = 10kVA VA supplied conductively = V2I1 10,000 2300 = × = 100 kVA VA supplied conductively V2I1 2300 230 = ×

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

10/18 Lecture 25

Rotating Machines Electro Mechanical Energy Conversion (EMEC)

• Transformer converts electrical power from one ‘V’ level

to another Electro Mechanical Energy Conversion (EMEC) to another

• EMEC equipment convert electrical energy into

mechanical energy & vice versa mechanical energy & vice versa

Motor Mechanical Energy El t i l E Mechanical Energy Electrical Energy Generator Mechanical Energy Electrical Energy Generator

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

11/18 Lecture 25

⇒ 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 = Blν voltage induced, e Blν 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

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

12/18 Lecture 25

In order to sustain the movement, external force Fa is applied such that F > F ⇒ ‘i' will continue to flow applied such that Fa > F ⇒ I2R = 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 I t d if V > E i li d t l f i i d Instead if V > E is applied, no external force is required to maintain the motion of the conductor ⇒ Input Electrical energy

• /p mechanical energy ⇒ Motoring action

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

13/18 Lecture 25

Basic requirements of EMEC devices: For motoring /generating action, there 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

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

14/18 Lecture 25

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 (3600) coil has completed one cycle (3600) ⇒ ‘V’ induced in the coil completes two cycles in one cycle of rotation ⇒ one complete rotation = 3600 (mech) ⇒ two complete cycles = 7200 (elect) l i l d h i l d ∴ electrical degrees ≠ mechanical degrees 10 elec= (2/P)0 mech

where ‘P’ is the number of poles

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

15/18 Lecture 25

Basic structure: stator → stationary does not move stator → stationary, does not move → normally outer frame rotor → which rotates inside the stator ⇒ generally stator and rotor are made up of high rotor → which rotates inside the stator → separated by small air gap (0.5 ‐ 1mm) ⇒ 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

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

16/18 Lecture 25

⇒ some machines have non uniform air gap → salient pole structure pole structure ⇒ consider two bar magnets pivoted at their center on the same shaft ⇒ stator→ coil ⇒ both of them are carrying ‘I’ Torque ∝ angular displacement which acts to align them rotor → coil ⇒ both of them are carrying I ⇒ two MMF sources, Fs & Fr ,

s r

⇒ create magnetic flux in the air gap between stator and rotor ⇒ similar to magnetic poles on both stator and rotor centered on their respective magnetic axis p g

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

17/18 Lecture 25

⇒ torque is produced by the tendency of two magnetic fields to align g sin

r S

F S r F

T F F ∝ ∠ i FF δ sin

S r

FF ∝ δ

1

sin sin

Sr r

F F δ = δ

1

sin

S Sr

T FF ∴ ∝ δ

2

sin sin

Sr S

F F δ = δ

2

sin

r Sr

T FF ∴ ∝ δ ⇒ for steady torque F & F should be stationary with ⇒ for steady torque FS & Fr should be stationary with respect to each other

Tue Oct 6, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

18/18 Lecture 25