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 1/10 Prof. B.G.Fernandes Lecture 21 24, 2009

Sub ‐ Topics: • Eq. circuit(contd..) • O.C and S.C tests • Efficiency of Transformer • Efficiency of Transformer EE 111: Introduction to Electrical Systems Thu Sep 2/10 Prof. B.G.Fernandes Lecture 21 24, 2009

Review • Winding connected to the source is known as primary g p y winding • H.V → H.V. winding, L.V → L.V. winding g, g • I 1 ≠ 0 when I 2 = 0 V V N N E E I I • For an ideal transformer, • For an ideal transformer = = = 1 1 1 1 1 1 2 2 V N E I 2 2 2 1 and Input VA = Output VA • Eq. secondary ‘R’ referred to primary 2 ⎛ ⎛ ⎞ ⎞ N N 1 r '=r ⎜ ⎟ 2 2 ⎝ N ⎠ 2 EE 111: Introduction to Electrical Systems Thu Sep 3/10 Prof. B.G.Fernandes Lecture 21 24, 2009

• Under normal operation, V 1 ≅ V 2 ’ If V 1 = 11kV, V 2 ’ ≅ 11kV or If V 1 = 440V, V 2 ’ ≅ 440V 440V V ’ ≅ 440V If V EE 111: Introduction to Electrical Systems Thu Sep 4/10 Prof. B.G.Fernandes Lecture 21 24, 2009

eg. 100kVA, 11kV/440V, 1 ‐ φ transformer and HV winding is connected to the 11kV Transformer is fully loaded and is connected to the 11kV. Transformer is fully loaded and load P.F. is 0. 6lag what does 100kVA mean? h t d 100kVA ? Is it primary VA or secondary VA or both? ⇒ only in case of ideal transformer, primary VA = sec. VA when secondary winding is kept open y g p p ⇒ primary current is finite ⇒ this current has two components (core loss + magnetizing) EE 111: Introduction to Electrical Systems Thu Sep 5/10 Prof. B.G.Fernandes Lecture 21 24, 2009

⇒ since secondary is open → there is no load on the Transformer Transformer ⇒ The corresponding current is known as no ‐ load current ≅ 5 ‐ 10% of full load current (?) ≅ 5 10% of full load current (?) ⇒ It is the rated load that the transformer can supply at rated terminal voltage rated terminal voltage ⇒ V 2 I 2 VA ∴ = Full load current V VA VA 2 2 ≅ under this condtion, primary 'I' may not be , but it is V 1 EE 111: Introduction to Electrical Systems Thu Sep 6/10 Prof. B.G.Fernandes Lecture 21 24, 2009

what is V 2 ? when primary winding is connected to 11kV, secondary when primary winding is connected to 11kV, secondary terminal ‘V’ need not be 440V ⇒ but it is ≅ 440V ∴ using VA, V 1 and V 2 , it is possible to estimate the rated ‘I’ in the primary & secondary windings without using the eq. circuit 3 x 100 10 = = 227A at 0.6 lag rated load current on L.V. side 440 440 r' 1 = r 2 = 0.0112 Ω x' l1 = x l2 = 0.096 Ω X’ M = 44 Ω & R’ c =132 Ω EE 111: Introduction to Electrical Systems Thu Sep 7/10 Prof. B.G.Fernandes Lecture 21 24, 2009

( )( ) = ∠ + ∠− ∠ E' 440 0 227 53 0.0966 83.34 = ∠ 459.067 1.38V E = = ∠ ∠− I I' 10.43 10.43 88.7A 88.7A m jX M E = = = = ∠ ∠ I' I 3 47 1 38A 3.47 1.38A C R C + + = ∠ ∠− ' Total I I Total I ' = I I I 237.52 237.52 54A 54A 1 1 2 2 0 0 ( )( ) = ∠ + ∠− ∠ V' 459.067 1.38 237.52 54 0.0966 83.34 1 = ∠ ∠ O O 479.5 3 V 9 3 EE 111: Introduction to Electrical Systems Thu Sep 8/10 Prof. B.G.Fernandes Lecture 21 24, 2009

Observations: • winding resistance << leakage flux (??) • winding resistance << leakage flux (??) • I o is predominantly reactive • E 1 ≅ V’ 2 ≅ V 1 ’ ⇒ can we shift the magnetizing branch to supply side ? V V = − 1 1 I j o I 0 R X c M = + + ' I = I I Z R jX eq eq eq s 2 0 ∠σ = ∠ + ∠φ φ ∠θ θ & V V' 0 I' Z 1 2 2 eq ⇒ approximate eq circuit ⇒ approximate eq. circuit EE 111: Introduction to Electrical Systems Thu Sep 9/10 Prof. B.G.Fernandes Lecture 21 24, 2009

with approximate eq. circuit 479.5 = = I 10.89A % change = 3.6 M 44 479 5 479.5 = = I 3.62A % change = 4.3 c 132 lagging P.F. unity P.F. leading P.F. EE 111: Introduction to Electrical Systems Thu Sep 10/10 Prof. B.G.Fernandes Lecture 21 24, 2009