Elec467 Power Machines & Transformers Electric Machines by Hubert, Chapter 2 Topics: Exciting current, Ideal transformer, Equivalent circuits, Per Unit calculation, Open circuit/short circuit testing models

Typical physical designs for a transformer

Basic transformer action ex is the counter emf (Formula 2-3) Voltage ratio = turns ratio

Calculating transformer flux From Example 2.1 Given that a transformer 50-kVA, 200 turns, 240V, 60 Hz primary, what is the maximum flux, Φmax? Use formula 2.1 Ep = 4.44 Np f Φmax Insert knowns… 240 = 4.44·200·60· Φ Solve for the unknown Φmax = 240/(4.44·200·60)→.0045 Wb

Section 2-5—No load conditions In this diagram is a simplified view of the primary side of a transformer. The input current IP = I0 (the exciting current) during no load conditions which can be measured directly with an ampere meter. We see the eddy loss component (Ife) caused by the resistance Rfe (a fictitious resistance to account for core loss) and the magnetizing current (IM) present during no load conditions whose reactance component is jXM is in parallel. The value jXM is a fictitious reactance to account for the magnetizing current. Their vector summation gives the current input when there is no load: I0 = Ife + IM. Additional equations from formula 2-4: Ife = VT/Rfe IM = VT/jXM

Example 2.3—Calculating flux loss for no-load conditions The exciting circuit, Io , is a no-load current. Let’s calculate the exciting circuit in the primary of a transformer using data from Example 2.3: Given a 25-kVA, 2400—240 V, 60 Hz, draws 138 W at no-load condition with a .210 lagging power factor (current into a coil lags the voltage). 1st get the phase angle Ѳ…cos-1(.210) = 77.88° 2nd insert the power factor angle of 77.88° into the formula for the phase difference between the voltage and current: Ѳ = Ѳv – Ѳi Using 0° phase for Ѳv gives: 77.88° = – Ѳi therefore Ѳi = – 77.88° This value is the angle the current vector I0 makes with the vector Ife We’ll need this information to calculate the IM current but first let’s do the core-loss component that draws real power: Pcore loss = Vt ∙ Ife Insert known values: 138 = 2400 ∙ Ife Solve for the unknown: Ife = .0575 amps …giving us the amplitude for the vector Ife Utilize this value and a trig formula from the power triangle: cos Ѳi = Ife / Io Insert known values: .210 = .0575/ Io yields up Io = .2738 amps Using Io = Ife + IM we can solve for IM = Io - Ife → .2738 - .0575 = .268 amps Summary: Io = .2738 amps, Ife = .0575 amps, IM = .268 amps

Formulas used for no load calculations ΦM =the mutual flux produced by the magnetizing component of the exciting current NpIM is aka magnomotive force, F VT = IpRp + Ep Ip = (VT – Ep)/Rp Above formulas are 2-6, 2-7 & 2-8 from text, page 46.

Formulas used for calculation of quadrature components Vt = Ife ∙ Rfe multiply both sides by Ife and we get real power loss in watts equal to I2R Pcore loss = Vt ∙ Ife Vt = IM ∙ j XM reaction power loss in watts Io = Ife + IM (this is a vector formula) The vector diagram from Fig. 2-4 →

A closer look at the phasors Figure 2-4 phasor drawing is seen at the right. Ѳi is the power factor angle. When rearranged as seen on the right cosine = adj/hyp→Ife/Io giving us one more handy formula: Io = Ife/cos Ѳi or Io = Ife/power factor Io IM Ѳ Ife

Mutual flux When a load is placed on the secondary the primary side changes from just the exciting current (Io = Ife + IM) to the exciting current plus a load current, IP = I0 + IP,load (formula 2-11).

Section 2-7—Flux loss Flux leakage results in a different induced voltage in the secondary voltage than is no leakage occurred. It diminishes the efficiency of the transformer. Flux leakage is accounted for in the formulas: Φp = ΦM + Φlp formula 2-12 Φs = ΦM – Φls formula 2-13

Section 2-8—Ideal transformer The ideal transformer is represented above. The induced counter-emf voltages and the input impedance are designated with primed symbols. There is a long list of what is not included in the ideal transformer. When the turns ratio is not available use the nameplate voltage ratio. The following slide lists a number of formulas used with an ideal transformer. The effects of flux leakage and winding resistance are insignificant at no load.

Ideal transformer formulas Assuming the primary is the High Side: Input Impedance of an Ideal Transformer: since The apparent power input must equal the apparent power output:

Equivalent circuit parameters for a real transformer

Close up look at the ideal transformer from Fig. 2-8 In the ideal: Ep = 4.44Np f Φp and Es = 4.44Ns f Φs If the windings have the same polarity, the current will appear to flow into the transformer and flow out the same end as if the primary and secondary circuits were connected to together.

Equivalent circuit details An equivalent circuit of a real transformer using an ideal transformer and external components to represent what is going on internal to the transformer. Losses are represented by IPRP, Elp, IfeRfe and by the magnetizing current IMjXM.

Leakage Reactance When we include the flux leakage (formulas 2-12 &13) in the ideal’s equations we get: Ep = 4.44Np f ΦM + 4.44Np f Φlp formula 2-20 Es = 4.44Ns f ΦM - 4.44Ns f Φ ls formula 2-21

Formulas we need to know to calculate voltages around the loops Expressing formulas 2-20 & 21 in simpler form EP = E’P + Elp ES = E’S - Els VT = Ep + IpRp VT = E lp + E’p + IpRp Ip = Ife + IM + Ip load formula 2-26 Es = IsRs + Vs load E’s = Els + IsRs + Vs load formula 2-28 The sinusoidal voltage generated: E = Emaxcos(2πft) with Emax = 2πfNΦmax formula 2-30 L = N2/R formula 2-32

Equivalent impedance using referred parameters

Referring 2nd parameters to the primary side

Formulas used for Z’in Ideal transformer: Z’in = a2E’s/Is (see Fig. 2-9) Where Is = E’s/(Rs + jXls + Zload) Substituting Is into the first equation gives: Z’in = a2(Rs + jXls + Zload) → Z’in = a2Rs + a2jXls + a2Zload The impedance seen by the source is Zin = Z’in + (RP +jXlP)

How to calculate the core losses seen by the primary Neglect exciting current at near rated load conditions…the equivalent impedance is… ZeqP = Rp + a2Rs + j(Xlp + a2Xls) and, Zin, HS = ZeqP + a2Zload (note: this is not Z’in)

Equivalent impedance HS is defined for a step-down transformer Here they have taken the impedance in { Rp + a2Rs + j(Xlp + a2Xls) } from the previous slide and changed it into Zeq,HS = Req,HS + jXeq,HS by grouping the resistances and impedance together: Req,HS = RP + a2RS jXeq,HS = j(XlP + a2XlS) (note: Req,HS & jXeq,HS are the parameters obtained in the short-circuit test)

Taking the core losses to the secondary Notice that the loads when reflected thru the transformer use a2 but current and voltage relations only use “a”. Current as usual is inverted.

How to calculate the wiring and flux losses in the secondary ZeqS = Rs + Rp/a2 + j(Xls + Xlp/a2) Here the primary values are adjusted by 1/a2 which is the result of going the other direction across the turns ratio.

Transformer as a lens Think of the transformer as a telescope. Telescopes make thing look larger or smaller depending on which way you look thru it. When the high voltage is stepped down, going from primary to secondary make values smaller Thus the primary losses when taken to the secondary are divided by a2. But bringing the secondary values to the primary side makes them bigger thus Zload and the secondary losses are multiplied by a2.

Equivalent circuit for a step-down transformer—low side The secondary impedance is changed from {Rs + Rp/a2 + j(Xls + Xlp/a2)} seen in the last slide into Zeq,LS = Req,LS + jXeq,LS by grouping the resistances & impedances together: Req,LS = RLS + RHS/a2 & jXeqLS = j(XLS + XHS/a2)

Equivalent circuit for a step-up transformer can go either way also

Changing from a Step-Down to a Step-Up If we think of the inputs and output of a transformer as primary (the driving circuit) and secondary (the load) an adjustment to “a” is necessary when changing from a step-down to a step-up or visa-versa. The value of “a” is inverted because the turns ratio is inverted. We can avoid this problem by using formulas that refers to the opposite sides of the transformer in terms of voltage high side or low side.

Equivalent circuit for a step-up transformer—low side For the step-up mode the source is on the low side (smaller number of windings). The equivalent impedance is Zeq,LS = RLS+RHS/a2+ j(XlLS+ XlHS/a2) or Zeq,LS = Req,LS + jXeq,LS where Req,LS gather the real terms and jXeq,LS gathers the imaginary terms. Zin,LS = (1/a2 )*Zload,HS + Zeq,LS

Equivalent circuit for a step-up transformer—high side The equivalent impedance on the high side uses Zeq,HS = RHS + a2RLS + j(XlHS + a2XlLS). The impedance show above is Zeq,HS = Req,HS + jXeq,HS where Req,HS gathers the real terms and jXeq,HS gathers the imaginary terms.

Voltage Regulation Regulation = (Enl – Vrated)/ Vrated Enl is the voltage present at the output terminals when no load is present. Vrated is the name plate voltage. Enl = ILSZeq,LS + VLS formula 2-45 (p. 62) ILS = Papparent/VLS consider VLS as Vrated Zeq,LS = Zeq,HS/a2 formula 2-43a (p. 59)

Per Unit Parameters Use by professionals in the power industry, per unit parameters simplify voltage calculation for transmission lines. ZPU = Irated Zeq)/Vrated RPU = Irated Req)/Vrated XPU = Irated Xeq)/Vrated Zbase = Vrated/Irated also V2rated/Srated Phase angle α = tan-1(XPU/RPU)

Transformer losses and efficiency η = Pout/Pin η = Pout/(Pout + Pcore + I2Req) Pout is the apparent power (xxKVA) Pcore is the wattage reading showing the real power lost under no load conditions. This happens to be the (HS) value of the power reading taken during the open circuit test! I2Req is the wattage reading taken during the short circuit test! Therefore: η = Pout/(Pout + POC + PSC)

Open Circuit Test Open circuit test is measured on the low side with the high side open so that only the exciting current parameters are detected.

Formulas used in the Open Circuit Test Parameters measure during an open circuit test are: POC, VOC, IOC POC = VOCIfe → Ife = POC/VOC IOC = √(Ife2 +IM2) → IM = √(IOC2 – Ife2) Rfe,LS = VOC/Ife XM,LS = VOC/IM Note: By conducting an open circuit test from both sides, Rfe and XM can be determined for both primary and secondary sides. These values are used in no-load calculations in step-down and step-up transformers.

Short circuit test Short-circuit test detects the high side equivalent parameters as there is no-load seen and the exciting current is ignored.

Formulas for the short circuit test Parameters measure during a short circuit test are: PSC, VSC, ISC ISC = VSC/Zeq,HS → Zeq,HS = VSC/ISC PSC = ISC2 Req,HS → Req,HS = PSC/ISC2 Zeq,HS = √(Req,HS2 + Xeq,HS2) → Xeq,HS = √(Zeq,HS2 - Req,HS2)

Homework Assignment 2-3 2-5 2-9 2-11 2-13 2-17 a & b 2-27 2-34 a