Unformatted text preview:

2/5/201411Ideal TransformerProperties• High permeability of the core • No Leakage Flux• No winding resistances.• Ideal core has no reluctance.• No Core losses.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2The negative sign is the statement of the Lenz’s law stating that the polarity of the induced voltage should be such that a current produced by it produces a flux in the opposite of the original flux. This is illustrated belowFaraday’s LawIf a flux φ passes through N turns of a coil, the induced voltage in the coil is given byindde Ndtϕ= −Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.3Relationships• From Faraday’s Law• Since there is no magnetic potential drop in the ideal core,1( ) ( )( )( )s p s spss pN I t N I tI tNI t N a== =Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.4Relationships• Since• thenPower in equals power out.No power loss in ideal transformer!5Relationships• Reflected Impedance ==== 6Dot Convention• Help to determine the polarity of the voltage and direction of the current in the secondary winding.• Voltages at the dots are in phase.• When the primary current flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.2/5/20142Compare the following distribution schemes11Non-Ideal Single-Phase TransformersNon-ideal facts:• Winding resistances modeled as series resistors Rpand RsAlso called Copper Losses.• Leakage flux modeled as series inductances Xpand Xs• Core Losses (Eddy Current and Hysteresis Losses) produce heating losses modeled as a shunt resistor RCin the primary winding.•Magnetizing Current flows in the primary to establish the flux in the core. Modeled as a shunt inductance Xmin the primary winding.Mutual (M) and Leakage (L) Flux2/5/20143Excitation CurrentMagnetization Current (IM) is not sinusoidal because of the non-linear relation between the current and flux (magnetization curve)Total Excitation CurrentIn a well designed transformer Iexis small. 17The equivalent circuit for a single-phase non-ideal transformer is shown below:Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.18• The equivalent circuit may be simplified by reflectingimpedances, voltages, and currents from the secondary to the primary as shown below:• Below is the transformer model referred to Secondary Side.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2/5/2014419• Simplified equivalent circuit referred to primary side:• Simplified equivalent circuit referred to secondary side:Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Transformer Equivalent Circuit without Excitation Branch Transformer Voltage Regulation• Because a real transformer has series impedances within it, the output voltage will vary with the load even if the input voltage remains constant.• Voltage Regulation compares the no load to full load voltage. =|,||,||,|x100% =|||,||,|x 100%Transformer Phasor Diagram• ApplyingKirchhoff‘slaw0= 1+ 3451+ 673451• For a Lagging PF (I lags V)Unity PFLeading PF24Transformer Efficiency • Pout= Ps= VsIsCos(θs)• Pin= Ps+ PLosses= VsIsCos(θs)+ Pcore+ Pcu( ) 100%( )out S S Sin S S S core cuP V I CosxP V I Cos P Pθηθ•= =+ +Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2/5/20145Ex.. Given: 2.2 kVA, 440/220V transformer. Equiv. circuit parameters referred to the primary are :Req = 3Ω, Xeq = 4Ω, Rc = 2.5kΩ, Xm = 2 KΩFind : VR and η. Assume transformer is delivering rated current and voltage at full load and 0.707 pf.Solution: 1. V s,fl= 220 V <0°This is the full load secondary voltage. We must now find the secondary voltage assuming the load was removed V s,nl. This is found by adding in the voltage drop across Req + j Xeq that occurs at full load.1,9:=0=1,;:+51(=>+67=>)2. Is= 2200 VA / 220V = 10 A3. θ = cosA0.707 = 45°4. Using the primary referred circuit we get Vp,nl0,9:==> + 67=> + 1=(10<-45°)/2*(3+j4)+2(220)=464.8<0.4°So,1,9:=,=232.4<0.4=|,||,||,|x100%=P.QRR= 5.63%• To find η we need Pin and Pout1. Pin = Vp ∙ Ip cos(∅) = Re{Vp ∙Ip*}= Re{(464.8<0.4°)(QUQ.VWR.QXRR+QUQ.VWR.QYRRR+ARWQX°)*}=Re{(464.8<0.4°)(5.29<45.3°)} = 1,717 W2. Pout = Vs Is cos(∅) = |S| pf = 2200*0.707 = 1,555 W3. η = Z,[[[Z,\Z\= 90.6% 4. Note: Pin = Pout + Pcopper loss + Pcore loss= Pout +]^_`abc+de_af= 1555 +(5)23 +(464.8)2/250028Determining the Values of Components in the Transformer ModelTransformer impedances may be obtained from two tests:• Open-circuit test: to determine core losses and magnetizing reactance (Rcand Xm)• Short-circuit test: to determine equivalent Series Impedance Req. and equivalent leakage reactances, Xeq)Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.29Open-Circuit Test to Determine Rcand XmWith secondary open, Voc, Ioc, and Pocare measured on the primary side. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.30Short-Circuit Test to Determine Reqand XeqWith secondary shorted, a reduced voltage is applied to primary such that rated current flows in the primary. VSC, ISC, and PSCare measured on the primary side.1ecos Rsceqq eqscVZ PF jXI−= − = +Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.2/5/20146ReqRcXeq1 2Xm12IpVp a VsIs/a+-+------->------>Lab 31. Open circuit (oc) test data to find Rc and XmPFoc = cos(θoc) = Poc/(Voc Ioc) = 2.0 / (120.0)(0.02) = 0.8333θoc = arcos(0.833) = 33.6 deg.Y= 1/Rc – j/Xm = Ioc/Voc <- θoc = 0.02/120.0 <-33.6 = 1.6667 x 10-4<-33.6 =(1.3882 – j 0.92232) x 10-4Rc = 1/1.3882 x 10-4 = 7,204 ΩXm = 1/0.92232 x 10-4= 10,842 ΩPre- lab 3 Solution2. Short circuit


View Full Document

The Citadel ELEC 316 - Lecture notes

Download Lecture notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?