7. Transformers 7.1 What does one look like? 7.2 How does it work? 7.3 What is the model? 7.4 How does one find the parameters? 7.5 How are transformers used? 7.6 What's missing?

7.1 What does one look like? Primary winding V1, I

1, N

1. Secondary winding V

2, I

2, N

2. HV & LV Core Flux Φ links 1 and 2

7.2 How does it work? Simplified Approach- IDEAL Transformer: Flux, EMF: AC Voltage, V1 applied to primary, secondary OC. Small AC current, I

0, flows

Φ. E

1 = N

1dΦ/dt = 4.44 f N

1Φ V1. E

2= N

2 dΦ/dt = 4.44 f N

2Φ V

2. V1/V2 = N1 / N2 = k

A3 Electrical Power

Transformer action Balance of Ampere Turns: Load R2. I

2 flows decrease in Φ E

1. BUT E

1 V

1 so I

1 ,such that N

1I

1 = N

2 I

2 I1/ I2 = N2 / N1 = 1/k Effective load on primary: R'2: R'2 = V1/I2 = kV2/(I2/k) = k2 V2/I2. R'2 = k2R2

7.3 What is the model? Primary R1 X1 Xm

or B RI or G Secondary X

2 R2

Full equivalent circuit Refer primary to secondary (can also refer to secondary) E1= E'2.

A3 Electrical Power

Approximate Equivalent Circuit

If R1,X

1 X

M,R

I then move parallel branch to

left.

7.4 How does one find the parameters? Open circuit test (LV) Short Circuit test (HV)

R P

I2

Z VI

X Z2

R2

RI V

2

P

IX I

2 VR

I 2

X ! VI

X

7.5 How are transformers used?" Ratings" Performance

parameters" Three Phase connections" Special Transformers

A3 Electrical Power

Ratings# Rating is NOT necessarily operating condition. # Main (Nameplate)$ Voltage$ VA# Secondary$ Frequency$ % Impedance =% Z = Vp/Isc

IRated

IS C & 100

Performance parameters# Eff iciency

# % Regulation:

' ( POut

PI n

( POut)

POut * P

Cu * PFe +' ( P

out)P

out * I2R * P

Fe +'max , P

Cu

( PFe

,unity pf

VN L - V

L

VN L . 100

Three Phase conn ections/ 3 single phase transformers or 1 three phase/ Connections:0 Star-star0 Delta - delta0 Star-delta, delta-star/ Always work in PHASE quantities. Remember 1 3

A3 Electrical Power

Special Transformers2 Power transformer2 Distribution2 Welding

Example2 A 500 kVA, 3 Phase 22kV, 11kV star-star transformer has a no load current of 1.1A. It has a maximum eff iciency of 96% when operating at rated load, unity power factor. The short circuit current is 8 times rated current.2 Determine:3 the parameters of the approximate equivalent circuit.3 the regulation when supplying 80% rated current at 0.8 pf

lag

3 V*p= 22k/ 4 3 =12.7 kV I*p= S/3V

P= 500/12.7 = 39.4A3 Approximate Circuit 5 POut + 2Pfe= POut/η 6 Pfe= 0.5(1/η -1)POut= 10 kW5 RI= 3Vp

2 /Pfe = 3 127102 /(10000) = 48.5 kΩ5 IR=Vp/RI= 0.26A5 XM= Vp/ 7 ( I0

2-IR2) = 12700/ 7 ( 1.12-0.262) = 11.9 kΩ .5 R= Pcu/3Ip

2 = 10000/(3 39.42) = 2.1Ω .5 Z = Vp/(8 Ip) = 12700/(8 39.4) = 40.3 Ω5 X = 7 (Z2-R2) = 40.2Ω

Solution

A3 Electrical Power

Solution part 28 Solve for V'2: V1= 12.7 kV I= 0.8 9 39.4 = 31.5A8 V1 ∠α = V'2∠ 0+ I∠−36.8 (R+jX)8 Imaginary: V1 sin α = IR sin -36.8 + IX cos -36.88 α = sin-1((-39.6+1014)/12700) = 4.8°8 Real: V2 cos α = V'2 + IR cos -36.8 - IX sin-36.88 V'2 = V2 cos α − IR cos -36.8 + IX sin-36.8 = 11.9 kV 8 Regulation: (12.7-11.9)/12.7 = 6.3%

7.6 What's missing?8 Alternative connections8 Saturation8 Inrush currents8 Interwinding capacitance8 Zig-zag windings etc.

A3 Electrical Power