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 Thu, Sep 1/14 Prof. B.G.Fernandes Lecture 19 10, 2009

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 Thu, Sep 2/14 Prof. B.G.Fernandes Lecture 19 10, 2009

Review • Value of ‘L’ can be varied by varying length(l g ) of air gap y y g g ( g ) g p ⇒ As ‘l g ’ ↑ , ‘L’ ↓ & leakage flux ↑ • Relationship between B ‐ H is non ‐ linear • Relationship between B ‐ H is non ‐ linear • Initial portion is almost linear l NI NI ℜ = μ ℜ φ ℜ i is almost constant l since = , A ⇒ μ r is almost constant μ r ⇒ Beyond ‘C’ core gets saturated ℜ ↑ ℜ ↑ μ μ ↓ ↓ ∵ ∵ r ⇒ Magnetization curve is linear for air core ⇒ ‘L’ is very small & leakage will be high ‘L’ i ll & l k ill b hi h EE 111: Introduction to Electrical Systems Thu, Sep 3/14 Prof. B.G.Fernandes Lecture 19 10, 2009

Hysteresis Loss: Assume ‘r’ = 0 & leakage flux = 0 Assume r 0 & leakage flux 0 φ d e = N dt Magnitude of voltage induced in the coil, E Energy transfer from source to load from time interval t f f t l d f ti i t l t2 t 1 & t 2 = ∫ (ei)dt t1 φ φ d 2 = = φ ∫ ∫ N idt Nid dt dt φ φ 1 Hl φ = ΒΑ & = N i Now N B B 2 H(lA) = 2 ∴ = ∫ ∫ (volume) HdB N dB core N N B B B B 1 1 1 EE 111: Introduction to Electrical Systems Thu, Sep 4/14 Prof. B.G.Fernandes Lecture 19 10, 2009

B 2 ∴ → ∫ Energy = Power.Time =V HdB ( ) 1 core B B 1 1 ∴ Energy transfer over one cycle of variation ⎛ ⎞ Power.Time = ∫ ∫ � � W/cycle W/cycle = ⎜ ⎜ ⎟ ⎟ V V HdB HdB core ⎝ ⎠ cycle B = V core * 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 ∴ R.H.S of (1) is ve EE 111: Introduction to Electrical Systems Thu, Sep 5/14 Prof. B.G.Fernandes Lecture 19 10, 2009

⇒ Power is fed back to source ∝ area of shaded portion ⇒ Difference in energy is dissipated as heat in the core = area of hysteresis loop y p ⇒ If B ‐ H curve is a straight line (air core), area of hysteresis loop=0 hysteresis loop 0 Power loss = V core * Area of hysteresis loop ∴ f f Power loss = V core * 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 area of the loop EE 111: Introduction to Electrical Systems Thu, Sep 6/14 Prof. B.G.Fernandes Lecture 19 10, 2009

∴ = = 1.6 Hysteresis loss P K B f h h max d depends on material d l 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 ⇒ i e will flow (V 2 /R is the loss) ⇒ i e will flow (V /R is the loss) ⇒ ↑ ‘R’, path resistance to ↓ the loss ⇒ Laminate the core ⇒ Laminate the core EE 111: Introduction to Electrical Systems Thu, Sep 7/14 Prof. B.G.Fernandes Lecture 19 10, 2009

⇒ Use thin sheets which are insulated from each other & stack them together & stack them together ⇒ ⇒ = = 2 2 P P K B K B f f e e max constant ⇒ Both P h & P e 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 φ EE 111: Introduction to Electrical Systems Thu, Sep 8/14 Prof. B.G.Fernandes Lecture 19 10, 2009

Electrical equivalent circuit : Case i: ‘r’ leakage flux core losses are neglected Case i: r , leakage flux, core losses are neglected µ r is assumed to remain constant (core loss = 0, µ r = K ⇒ Air core) ( l 0 K Ai ) ℜ ∞ , 0, NI required to If µ r establish Ф in the core 0 bl h h φ d = = v e N dt dt φ = φ ω Let sin t m = = φ ω φ ω = φ ω φ ω + π v e N cos t N sin( t ( ) ) 2 2 m m φ ω N ∴ ∴ = φ ω φ ω = = φ φ m E E N N E E . f f N N & & 4 44 4 44 m m rms m 2 EE 111: Introduction to Electrical Systems Thu, Sep 9/14 Prof. B.G.Fernandes Lecture 19 10, 2009

‘i' which is responsible to produce φ is in phase with it phase ith it Observations: Observations: i) Power supplied by the source = 0 π ∠ = I , & I lags V V 2 ⇒ ⇒ Ideal L = = φ φ = = ii) V ii) V . f f N N E E 4 44 4 44 m ⇒ φ in the core is determined by supply ‘V’ alone. φ EE 111: Introduction to Electrical Systems Thu, Sep 10/14 Prof. B.G.Fernandes Lecture 19 10, 2009

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 = = φ m E V . f N 4 44 m ∴ Core loss depends on V = V E = = ⇒ Core loss can be modeled as ⇒ Core loss can be modeled as 2 where R where R I R I R , C C C I C ⇒ Source also has to supply AT to produce ‘ φ ’ in the core ⇒ This φ induces E (= V) di = m This effect can be modeled as This effect can be modeled as e e L L m dt EE 111: Introduction to Electrical Systems Thu, Sep 11/14 Prof. B.G.Fernandes Lecture 19 10, 2009

I m responsible for producing φ I C flows in a resistor (R C ) ∴ ∴ source I has two components I h t t I m flows in a inductor (L m ) (reactor X m ) Observations: produce ‘ φ ’ in the core I m ⇒ magnetizing the core ⇒ magnetizing the core ⇒ L m is called as magnetizing inductance or X or X m as magnetizing reactance as magnetizing reactance ⇒ In case(i), θ = 90 0 In this case θ should 90 � ⇒ ⇒ � � X R I I V X V X should be should be I V R I V R ( ( ) ) ( ( ), ), m m C C m m c c EE 111: Introduction to Electrical Systems Thu, Sep 12/14 Prof. B.G.Fernandes Lecture 19 10, 2009

Case iii: case ii + winding resistance + leakage flux Total flux = φ L + φ m φ + φ Total flux ∴ = + v Ri e s 1 φ φ φ φ d d d d = + L m e N N 1 1 dt dt (??) ( ) Flux is produced by the current flow in ‘L’ N φ φ N ∴ Leakage inductance, L l = L 1 I N φ d = = m L L N dt & m & ∴ = + E X I I X L S L S m m m m 1 1 EE 111: Introduction to Electrical Systems Thu, Sep 13/14 Prof. B.G.Fernandes Lecture 19 10, 2009

Air gap = 4mm Leakage flux is more EE 111: Introduction to Electrical Systems Thu, Sep 14/14 Prof. B.G.Fernandes Lecture 19 10, 2009