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

• Prof. B.G.Fernandes

Thu, Sep 10, 2009

1/14 Lecture 19

Sub‐Topics:

• Hysteresis & Eddy current loss(contd )
• Hysteresis & Eddy current loss(contd..)
• Loss representation in equivalent circuit

Loss representation in equivalent circuit

EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

Thu, Sep 10, 2009

2/14 Lecture 19

Review

• Value of ‘L’ can be varied by varying length(lg) of air gap
• Relationship between B‐H is non‐linear

y y g g ( g) g p ⇒ As ‘lg’ ↑, ‘L’ ↓ & leakage flux ↑

• Initial portion is almost linear

NI l ℜ i l

• Relationship between B‐H is non‐linear

⇒ μr is almost constant , NI since = φ ℜ A ℜ = μ is almost constant μr ⇒ Beyond ‘C’ core gets saturated ℜ ↑ μ ↓ ∵

r

ℜ ↑ μ ↓ ∵ ⇒ Magnetization curve is linear for air core ‘L’ i ll & l k ill b hi h

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

⇒ ‘L’ is very small & leakage will be high

3/14 Lecture 19

Hysteresis Loss: Assume ‘r’ = 0 & leakage flux = 0 E t f f t l d f ti i t l Assume r 0 & leakage flux d e = N dt φ Magnitude of voltage induced in the coil,

t2

(ei)dt = ∫ Energy transfer from source to load from time interval t1 & t2

t1

2

φ φ

φ = = φ

∫ ∫

d N idt Nid dt

1

φ

dt Now Hl i N φ = ΒΑ & = N

2

B B

H(lA) N dB N ∴ = ∫

2 1

B core B

(volume) HdB =

∫

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

1

B

N

1

B

4/14 Lecture 19

2 1

1

B core B

Energy = Power.Time =V HdB ( ) ∴ →

∫

1

B

Power.Time W/cycle V HdB ⎛ ⎞ ⎜ ⎟

∫

• ∴Energy transfer over one cycle of variation

core B

W/cycle = V HdB cycle = ⎜ ⎟ ⎝ ⎠

∫

• = Vcore * Area of hysteresis loop

‘H’ is +ve & ↑, ‘B’ is also ↑ ∴ R H S of (1) is +ve ∴ R.H.S of (1) is +ve (area of shaded portion) ‘H’ is +ve & ↓, ‘B’ is also ↓ ∴ R H S of (1) is ‐ve

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

∴ R.H.S of (1) is ve

5/14 Lecture 19

⇒ Power is fed back to source ∝ area of shaded portion ⇒ Difference in energy is dissipated as heat in the core = area of hysteresis loop ⇒ If B‐H curve is a straight line (air core), area of hysteresis loop=0 y p hysteresis loop 0

Power loss f ∴ = Vcore * Area of hysteresis loop f

Power loss = Vcore * Area of hysteresis loop * f α f α f ∵ B‐H curve is non‐linear, it is difficult to determine the area of the loop

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

area of the loop

6/14 Lecture 19

1.6 max h h

Hysteresis loss P K B f ∴ = = d d l depends on material Eddy Current Loss: ⇒ It is due to variation of flux density in the core Consider a small path in the c/s. ‘V’ will be induced in the path ⇒ Path is complete ⇒ ie will flow (V2/R is the loss) ⇒ Laminate the core ⇒ ie will flow (V /R is the loss) ⇒ ↑ ‘R’, path resistance to ↓ the loss

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

⇒ Laminate the core

7/14 Lecture 19

⇒ Use thin sheets which are insulated from each other & stack them together & stack them together

2 2

P K B f ⇒ =

max e e

P K B f ⇒ = constant ⇒ Both Ph & Pe take place in the core (iron) ⇒ Known as core loss (iron loss) ⇒ Known as core loss (iron loss) ⇒ Depend on B & f ⇒ Remain constant if B(or φ) & f are held constant ⇒ Can be called as ‘constant loss’ if φ & f are held constant

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

φ

8/14 Lecture 19

Electrical equivalent circuit: Case i: ‘r’ leakage flux core losses are neglected Case i: r , leakage flux, core losses are neglected µr is assumed to remain constant ( l K Ai ) (core loss = 0, µr = K ⇒ Air core) If µr ∞, 0, NI required to bl h h ℜ

d v e N dt φ = =

establish Ф in the core 0

dt

m

Let sin t φ = φ ω 2 ( ) φ φ

& 4 44

m

N E N E f N φ ω ∴ = φ ω = = φ

2

m m

v e N cos t N sin( t ) = = φ ω ω = φ ω ω + π

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

& 4 44 2

m m rms m

E N E . f N ∴ φ ω φ

9/14 Lecture 19

‘i' which is responsible to produce φ is in phase ith it Observations: phase with it Observations: i) Power supplied by the source = 0 2

I V

, & I lags V π ∠ = ⇒ ⇒ Ideal L 4 44 ii) V f N E = φ = 4 44

m

ii) V . f N E = φ = ⇒ φ in the core is determined by supply ‘V’ alone.

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

φ

10/14 Lecture 19

Case ii) core losses are taken into account ⇒ Source has to supply the core loss ⇒ If ‘f’ is constant , core loss depends on ‘φ’ since winding resistance, ‘r’ & leakage φ are neglected 4 44

m

E V . f N = = φm ∴Core loss depends on V ⇒ Core loss can be modeled as

2

I R V E where R = = ⇒ Core loss can be modeled as ,

C C

I R

C C

where R I = ⇒ Source also has to supply AT to produce ‘φ’ in the core ⇒ This φ induces E (= V)

m

di e L = This effect can be modeled as

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

m

e L dt This effect can be modeled as

11/14 Lecture 19

IC flows in a resistor (RC)

∴ I h t t Im responsible for producing φ ∴source I has two components

Im flows in a inductor (Lm) (reactor Xm)

Observations: ⇒ magnetizing the core Im produce ‘φ’ in the core ⇒ magnetizing the core ⇒ Lm is called as magnetizing inductance

• r X

as magnetizing reactance

• r Xm as magnetizing reactance

⇒ In case(i), θ = 900 In this case θ should 90 ( ) ( ), I V X should be I V R ⇒

• m

C

X R

• Thu, Sep

10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

( ) ( ),

m m c c

I V X should be I V R ⇒

• m

C

12/14 Lecture 19

Case iii: case ii + winding resistance + leakage flux Total flux φ + φ Total flux = φL + φm

1 s

v Ri e ∴ = + d d φ φ

1 1 L m

d d e N N dt dt φ φ = +

(??)

Flux is produced by the current flow in ‘L’

( )

N φ &

m

d L N φ = ∴Leakage inductance, Ll

1 L

N I φ = & m L N dt =

1 L S m m

E X I I X ∴ = +

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

1 L S m m

13/14 Lecture 19

Air gap = 4mm Leakage flux is more

Thu, Sep 10, 2009 EE 111: Introduction to Electrical Systems

• Prof. B.G.Fernandes

14/14 Lecture 19