Electrical Engineering (EE) > Short Notes for Electrical Engineering > Short Notes: Transformers

Download, print and study this document offline |

Page 1 Impact of dimensions on various parameters of Transformer KVA Rating ? (Core Dimension) 4 Voltage Rating ? (Core Dimension) 2 Current Rating ? (Core Dimension) 2 No-Load Current ? Core Dimension Core Loss ? Core Volume Induced EMF in a Transformer ? ? ? ? ?? ?? 11 22 1 1 m 2 2 m d EN dt d EN dt E (rms) 4.44fN E (rms) 4.44fN ? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. ? F is the flux in the transformer and F m is maximum value of flux. ? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other. ? Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities. Exact equivalent circuit Page 2 Impact of dimensions on various parameters of Transformer KVA Rating ? (Core Dimension) 4 Voltage Rating ? (Core Dimension) 2 Current Rating ? (Core Dimension) 2 No-Load Current ? Core Dimension Core Loss ? Core Volume Induced EMF in a Transformer ? ? ? ? ?? ?? 11 22 1 1 m 2 2 m d EN dt d EN dt E (rms) 4.44fN E (rms) 4.44fN ? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. ? F is the flux in the transformer and F m is maximum value of flux. ? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other. ? Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities. Exact equivalent circuit Exact equivalent circuit w.r.t. primary 2 2 2 1 1 1 2 2 2 2 L L 2 2 2 N N N R = R ; X = X ; Z = Z N N N ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; ? Approximately Equivalent Circuit 01 1 2 R = R R ? ? 01 1 2 X = X X ? ? Tests Conducted on a Transformer (i) Open Circuit Test o Conducted on LV side keeping HV side open circuited o Equivalent Circuit Page 3 Impact of dimensions on various parameters of Transformer KVA Rating ? (Core Dimension) 4 Voltage Rating ? (Core Dimension) 2 Current Rating ? (Core Dimension) 2 No-Load Current ? Core Dimension Core Loss ? Core Volume Induced EMF in a Transformer ? ? ? ? ?? ?? 11 22 1 1 m 2 2 m d EN dt d EN dt E (rms) 4.44fN E (rms) 4.44fN ? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. ? F is the flux in the transformer and F m is maximum value of flux. ? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other. ? Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities. Exact equivalent circuit Exact equivalent circuit w.r.t. primary 2 2 2 1 1 1 2 2 2 2 L L 2 2 2 N N N R = R ; X = X ; Z = Z N N N ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; ? Approximately Equivalent Circuit 01 1 2 R = R R ? ? 01 1 2 X = X X ? ? Tests Conducted on a Transformer (i) Open Circuit Test o Conducted on LV side keeping HV side open circuited o Equivalent Circuit o Power reading = 2 1 1 0 0 c V P = V I cos = R ? -------- (i) o Ammeter reading ? 0 I = I o 0 10 P cos = VI ? o Calculate 2 00 sin = 1 - cos ?? o 2 1 1 0 0 m V Q = V I sin = X ? ------- (ii) Calculate c R from (i) & m X from (ii) (ii) Short Circuit Test o Conducted on HV side keeping LV side short circuited o Equivalent Circuit o 01 R & 01 X are equivalent winding resistance & equivalent leakage reactor referred to HV side. o Wattmeter reading = 2 sc 01 P = I R from this equation, we can calculate 01 R o sc 01 sc V Z = I & 22 01 01 01 X = Z R ? o We obtain 01 R , 01 X & full load copper losses from this test. Losses on Transformers o Copper Loss 22 Cu 1 1 2 2 P = I R I R ? 22 1 01 2 02 = I R I R ? Where 1 I = primary current 2 I = secondary current Page 4 Impact of dimensions on various parameters of Transformer KVA Rating ? (Core Dimension) 4 Voltage Rating ? (Core Dimension) 2 Current Rating ? (Core Dimension) 2 No-Load Current ? Core Dimension Core Loss ? Core Volume Induced EMF in a Transformer ? ? ? ? ?? ?? 11 22 1 1 m 2 2 m d EN dt d EN dt E (rms) 4.44fN E (rms) 4.44fN ? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. ? F is the flux in the transformer and F m is maximum value of flux. ? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other. ? Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities. Exact equivalent circuit Exact equivalent circuit w.r.t. primary 2 2 2 1 1 1 2 2 2 2 L L 2 2 2 N N N R = R ; X = X ; Z = Z N N N ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; ? Approximately Equivalent Circuit 01 1 2 R = R R ? ? 01 1 2 X = X X ? ? Tests Conducted on a Transformer (i) Open Circuit Test o Conducted on LV side keeping HV side open circuited o Equivalent Circuit o Power reading = 2 1 1 0 0 c V P = V I cos = R ? -------- (i) o Ammeter reading ? 0 I = I o 0 10 P cos = VI ? o Calculate 2 00 sin = 1 - cos ?? o 2 1 1 0 0 m V Q = V I sin = X ? ------- (ii) Calculate c R from (i) & m X from (ii) (ii) Short Circuit Test o Conducted on HV side keeping LV side short circuited o Equivalent Circuit o 01 R & 01 X are equivalent winding resistance & equivalent leakage reactor referred to HV side. o Wattmeter reading = 2 sc 01 P = I R from this equation, we can calculate 01 R o sc 01 sc V Z = I & 22 01 01 01 X = Z R ? o We obtain 01 R , 01 X & full load copper losses from this test. Losses on Transformers o Copper Loss 22 Cu 1 1 2 2 P = I R I R ? 22 1 01 2 02 = I R I R ? Where 1 I = primary current 2 I = secondary current 1 R = primary winding resistance 2 R = secondary winding resistance 22 12 01 1 2 02 2 1 21 NN R = R R ; R = R R NN ? ? ? ? ?? ? ? ? ? ? ? ? ? o Core Loss (i) Hysteresis Loss x n n m P = K B f X = 1.6 m B = maximum value of flux density 1.6 n n m P = K B f m V B f ? V = applied voltage f = frequency ? ?? ?? ?? ?? 1.6 1.6 0.6 nhh V P = K f = K V f f If V is constant & f is increased, h P decreases (ii) Eddy Current Loss 2 2 e e m P = K B f m V B f ? 2 22 e e e V P = K f = K V f ?? ?? ?? ?? Core loss = c e n P = P P ? Page 5 Impact of dimensions on various parameters of Transformer KVA Rating ? (Core Dimension) 4 Voltage Rating ? (Core Dimension) 2 Current Rating ? (Core Dimension) 2 No-Load Current ? Core Dimension Core Loss ? Core Volume Induced EMF in a Transformer ? ? ? ? ?? ?? 11 22 1 1 m 2 2 m d EN dt d EN dt E (rms) 4.44fN E (rms) 4.44fN ? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. ? F is the flux in the transformer and F m is maximum value of flux. ? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary should be such that primary and secondary flux should oppose each other. ? Also, primary current enters the positive terminal of primary winding as primary absorbs power and secondary current leaves the positive terminal of secondary winding as secondary delivers power and this way we can mark emf polarities. Exact equivalent circuit Exact equivalent circuit w.r.t. primary 2 2 2 1 1 1 2 2 2 2 L L 2 2 2 N N N R = R ; X = X ; Z = Z N N N ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ; ? Approximately Equivalent Circuit 01 1 2 R = R R ? ? 01 1 2 X = X X ? ? Tests Conducted on a Transformer (i) Open Circuit Test o Conducted on LV side keeping HV side open circuited o Equivalent Circuit o Power reading = 2 1 1 0 0 c V P = V I cos = R ? -------- (i) o Ammeter reading ? 0 I = I o 0 10 P cos = VI ? o Calculate 2 00 sin = 1 - cos ?? o 2 1 1 0 0 m V Q = V I sin = X ? ------- (ii) Calculate c R from (i) & m X from (ii) (ii) Short Circuit Test o Conducted on HV side keeping LV side short circuited o Equivalent Circuit o 01 R & 01 X are equivalent winding resistance & equivalent leakage reactor referred to HV side. o Wattmeter reading = 2 sc 01 P = I R from this equation, we can calculate 01 R o sc 01 sc V Z = I & 22 01 01 01 X = Z R ? o We obtain 01 R , 01 X & full load copper losses from this test. Losses on Transformers o Copper Loss 22 Cu 1 1 2 2 P = I R I R ? 22 1 01 2 02 = I R I R ? Where 1 I = primary current 2 I = secondary current 1 R = primary winding resistance 2 R = secondary winding resistance 22 12 01 1 2 02 2 1 21 NN R = R R ; R = R R NN ? ? ? ? ?? ? ? ? ? ? ? ? ? o Core Loss (i) Hysteresis Loss x n n m P = K B f X = 1.6 m B = maximum value of flux density 1.6 n n m P = K B f m V B f ? V = applied voltage f = frequency ? ?? ?? ?? ?? 1.6 1.6 0.6 nhh V P = K f = K V f f If V is constant & f is increased, h P decreases (ii) Eddy Current Loss 2 2 e e m P = K B f m V B f ? 2 22 e e e V P = K f = K V f ?? ?? ?? ?? Core loss = c e n P = P P ? Efficiency ? ? ? ? 2 i Cu,FL x KVA cos = x KVA cos P x P ? ? ? ? ? X = % loading of Transformer cos = power factor ? i P = iron loss Cu,FL P = Full load copper losses KVA = Power rating of Transformer For maximum efficiency, i Cu,FL P x = P Voltage Regulation of Transformer Regulation down NL FL NL VV 100 V ? ?? Regulation up NL FL FL VV 100 V ? ?? Equivalent circuit with respect to secondary K = Transformation Ratio 2 1 N N ? No-load voltage 2 V ? Full-load voltage 2 V ? ? Approximate Voltage Regulation ? ? 2 02 2 02 2 2 I R cos X sin VR = V ? ? ?Read More

Download as PDF