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ENERGY CONVERSION ONE (Course 25741) Chapter Two TRANSFORMERS …continued

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Transformer Voltage Regulation and Efficiency Output Voltage of Transformer Varies with Load Due to Voltage Drop on Series Impedance of Transformer Equivalent Model Full Load Regulation Parameter, compares output no-load Voltage with its Full Load Voltage: V.R. = At no load V S = V P / a thus : V.R.= in per unit: V.R. = For Ideal Transformer V.R.=0

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Transformer Voltage Regulation and Efficiency The transformer phasor diagram To determine the voltage regulation of a transformer: The voltage drops should be determined In below a Transformer equivalent circuit referred to the secondary side shown:

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Transformer Voltage Regulation and Efficiency since current which flow in magnetizing branch is small can be ignored Assuming secondary phasor voltage as reference V S with an angle of 0 Writing the KVL equation: From this equation the phasor diagram can be shown: At lagging power factor:

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Transformer Voltage Regulation and Efficiency If power factor is unity, V S is lower than V P so V.R. > 0 V.R. is smaller for lagging P.F. With a leading P.F., V S is larger V P V.R.<0 P.F. =1 P.F. leading

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Transformer Voltage Regulation and Efficiency Table Summarize possible Value for V.R. vs Load P.F.: Since transformer usually operate at lagging P.F., a simplified method is introduced Lagging P.F.V P / a > V S V.R. > 0 Unity P.F. V P / a > V S V.R. >0 (smaller) Leading P.F.V S > V P / aV.R. < 0

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Transformer Voltage Regulation and Efficiency Simplified Voltage Regulation Calculation For lagging loads: the vertical components related to voltage drop on R eq & X eq partially cancel each other angle of V P /a very small

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Transformer Voltage Regulation and Efficiency Transformer Efficiency (as applied to motors, generators and motors) Losses in Transformer: 1- Copper I²R losses 2- Core Hysteresis losses 3- Core Eddy current losses Transformer efficiency may be determined as follows:

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Transformer Voltage Regulation and Efficiency Example: A 15kVA, 2300/230 V transformer tested to determine 1- its excitation branch components, 2- its series impedances, and 3- its voltage regulation Following data taken from the primary side of the transformer: Open Circuit TestShort Circuit Test V OC =2300 VV SC =47 V I OC =0.21AI SC =6 A P OC = 50 WP SC = 160 W

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Transformer Voltage Regulation and Efficiency (a) Find the equivalent circuit referred to H.V. side (b) Find the equivalent circuit referred to L. V. side (c) Calculate the full-load voltage regulation at 0.8 lagging PF, 1.0 PF, and at 0.8 leading PF (d) Find the efficiency at full load with PF 0.8 lagging SOLUTION: Open circuit impedance angle is: Excitation admittance is:

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Transformer Voltage Regulation and Efficiency Impedance of excitation branch referred to primary: Short Circuit Impedance angle: Equivalent series Impedance: Req=4.45 Ω, Xeq=6.45 Ω

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Transformer Voltage Regulation and Efficiency The equivalent circuits shown below:

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Transformer Voltage Regulation and Efficiency (b) To find eq. cct. Referred to L.V. side, impedances divided by a²=N P /N S =10 R C =1050 Ω, X M =110 Ω R eq = Ω, X eq = Ω (c) full load current on secondary side: I S,rated =S rated / V S,rated =15000/230 =65.2 A To determine V.R., V P / a is needed V P /a = V S + R eq I S + j X eq I S, and: I S =65.2/_-36.9 A, at PF=0.8 lagging

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Transformer Voltage Regulation and Efficiency Therefore: V P / a = V.R.=( )/230 x 100 %=2.1 % for 0.8 lagging At PF=0.8 leading I S =65.2/_36.9 A V P / a =

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Transformer Voltage Regulation and Efficiency V.R. = ( )/230 x 100%= % At PF=1.0, I S = 65.2 /_0 A V P /a= V.R. = ( )/230 x 100% = 1.28 % for PF=1

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Transformer Voltage Regulation and Efficiency Example: Phasor Diagrams …

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Transformer Voltage Regulation and Efficiency (d) to plot V.R. as a function of load is by repeating the calculations of part c for many different loads using MATLAB

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Transformer Voltage Regulation and Efficiency (e) Efficiency of Transformer: - Copper losses: P Cu =(I S )²R eq =(65.2)² (0.0445)=189 W - Core losses: P Core = (V P /a)² / R C = (234.85)² / 1050=52.5 W output power: P out =V S I S cosθ=230x65.2xcos36.9=12000 W η= V S I S cosθ / [P Cu +P Core +V S I S cosθ] x 100%= 12000/ [ ] = %

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Efficiency of Distribution Transformers

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Energy Losses in Electrical Energy Systems The total electrical energy use per annum of the world is estimated as 13,934 TeraWatthours [TWh] (1 TWh = 10^9 kWh) it is further estimated [2] that the losses in all of the worlds electrical distribution systems total about 1215 TWh or about 8.8% of the total electrical energy consumed. About 30-35% of these losses are generated in the Transformers in the Distribution systems. Studies estimate that some 40-80% of these transformer losses are potentially saveable by increasing transformer efficiencies, i.e TWh.

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Electrical Energy Losses in Distribution Networks

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Transformer Taps & Voltage Regulation Distribution Transformers have a series taps in windings which permit small changes in turn ratio of transformer after leaving factory A typical distribution transformer has four taps in addition to nominal setting, each has a 2.5% of full load voltage with the adjacent tap This provides possibility for voltage adjustment below or above nominal setting by 5%

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Transformer Taps & Voltage Regulation Example: A 500 kVA, 13200/480 V distribution transformer has 4, 2.5 % taps on primary winding. What are voltage ratios? Five possible voltage ratings are: +5% tap 13860/480 V +2.5% tap 13530/480 V Nominal rating 13200/480 V -2.5% tap 12870/480 V -5% tap 12540/480 V

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Transformer Taps & Voltage Regulation Taps on transformer permit transformer to be adjusted in field to accommodate variations in tap voltages While this tap can not be changed when power is applied to transformer Some times voltage varies widely with load, i.e. when high line impedance exist between generators & particular load; while normal loads should be supplied by an essentially constant voltage One solution is using special transformer called: tap changing under load transformer A voltage regulator is a tap changing under load transformer with built-in voltage sensing circuitry that automatically changes taps to preserve system voltage constant

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AUTO TRANSFORMER some occasions it is desirable to change voltage level only by a small amount i.e. may need to increase voltage from 110 to 120 V or from 13.2 to 13.8 kV This may be due to small increase in voltage drop that occur in a power system with long lines In such cases it is very expensive to hire a two full winding transformer, however a special transformer called: auto-transformer can be used

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AUTO TRANSFORMER Diagram of a step-up auto-transformer shown in figure below: C: common, SE: series

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AUTO TRANSFORMER A step-down auto-transformer : I H =I SE I L =I SE +I C

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AUTO TRANSFORMER In step-up autotransformer: V C / V SE = N C / N SE (1) N C I C = N SE I SE (2) voltages in coils are related to terminal voltages as follows: V L =V C (3) V H =V C +V SE (4) current in coils are related to terminal currents: I L =I C +I SE (5) I H =I SE (6)

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AUTO TRANSFORMER Voltage & Current Relations in Autotransformer V H =V C +V SE since V C /V SE =N C /N SE V H =V C + N SE /N C. V C Noting that: V L =V C V H =V L + N SE /N C. V L = (N SE +N C )/N C. V L V L / V H = N C / (N SE +N C ) (7) Current relations: I L =I C +I SE employing Eq.(2) I C =(N SE / N C )I SE I L = (N SE / N C )I SE + I SE, since I SE =I H I L = (N SE / N C )I H +I H = (N SE + N C )/N C. I H I L / I H = (N SE + N C )/N C (8)

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AUTO TRANSFORMER Apparent Power Rating Advantage of Autotransformer Note : not all power transferring from primary to secondary in autotransformer pass through windings Therefore if a conventional transformer be reconnected as an autotransformer, it can handle much more power than its original rating The input apparent power to the step-up autotransformer is : S in =V L I L And the output apparent power is: S out =V H I H

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AUTO TRANSFORMER And : S in =S out =S IO Apparent power of transformer windings: S W = V C I C =V SE I SE This apparent power can be reformulated: S W = V C I C =V L (I L -I H ) =V L I L -V L I H employing Eq.(8) S W = V L I L -V L I L N C /(N SE +N C ) =V L I L [(N SE +N C )-N C ] /(N SE +N C )=S IO N SE /(N SE +N C ) S IO / S W = (N SE +N C ) / N SE (9)

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AUTO TRANSFORMER Eq.(9); describes apparent power rating advantage of autotransformer over a conventional transformer – smaller the series winding the greater the advantage Example one: A 5000 kVA autotransformer connecting a 110 kV system to a 138 kV system has an N C /N SE of 110/28 for this autotransformer actual winding rating is: S W =S IO N SE /(N SE +N C )=5000 x 28/ (28+110)=1015 kVA Example Two: A 100 VA 120/12 V transformer is connected as a step-up autotransformer, and primary voltage of 120 applied to transformer.

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AUTO TRANSFORMER (a) what is the secondary voltage of transformer (b) what is its maximum voltampere rating in this mode of operation (c) determine the rating advantage of this autotransformer connection over transformers rating of conventional 120/12 V operation Solution: N C /N SE = 120/12 (or 10:1) (a) using Eq.(7),V H = (12+120)/120 x 120 = 132 V (b) maximum VA rating 100 VA I SE,max =100/12=8.33 A

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AUTO TRANSFORMER S out =V S I S =V H I H = 132 x 8.33 = 1100 VA = S in (c) rating advantage: S IO /S W =(N SE +N C )/N SE =(12+120)/12=11 or: S IO /S W = 1100/100 = 11 It is not normally possible to reconnect an ordinary transformer as an autotransformer due to the fact that insulation of L.V. side may not withstand full output voltage of autotransformer connection Common practice: to use autotransformer when two voltages fairly close Also used as variable transformers, where L.V. tap moves up & down the winding Disadvantage: direct physical connection between primary & secondary circuits, and electrical isolation of two sides is lost

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AUTO TRANSFORMER Internal Impedance of an Autotransformer Another disadvantage: effective per unit impedance of an autotransformer w.r.t. the related conventional transformer is the reciprocal of power advantage This is a disadvantage where the series impedance is required to limit current flows during power system faults (S.C.)

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AUTO TRANSFORMER Example three: A transformer rated 1000 kVA, 12/1.2 kV, 60 Hz when used as a two winding conventional transformer and its series resistance & reactance are 1 and 8 percent per unit It is used as a 13.2/12 kV autotransformer (a) what is now the transformers rating ? (b) what is the transformers series impedance in per unit?

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AUTO TRANSFORMER Solution: (a) N C /N SE = 12/1.2 (or 10:1) the voltage ratio of autotransformer is 13.2/12 kV & VA rating : S IO =(1+10)/1 x 1000 kVA=11000 kVA (b) transformers impedance in per-unit when connected as conventional transformer: Z eq = j 0.08 pu Power advantage of autotransformer is 11, so its per unit impedance would be: Z eq =(0.01+j0.08)/11= j pu

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Example of Variable Auto-Transformer

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