Example 1 A 9375 kVA, 13,800 kV, 60 Hz, two pole, Y-connected synchronous generator is delivering rated current at rated voltage and unity PF. Find the armature resistance and synchronous reactance given that the filed excitation voltage is 11935.44 V and leads the terminal voltage by an angle 47.96

°.

Example 2 A cylindrical rotor machine is delivering active power of 0.8 p.u. and reactive power of 0.6 p.u. at a terminal voltage of 1 p.u. If the power angle is 20 °, compute the excitation voltage and the machine’s synchronous reactance.

Announcement  Quiz I : Next Tuesday (April 5, 2011) at 12:00 in H11  Assignment I: will be posted today and due Next Tuesday (April 5, 2011) at 12:00 in H11

Power Transformers

Transformation ratio Primary (supply) Secondary (Load)

Transformers at no load

I F Ic Q c E 1 Im f I f I F Ic Im The no load current I f is needed to supply the no load losses and to magnetize the transformer core.

E 1

Transformer losses

 The transformer losses are divided into electrical losses (copper losses) and Magnetic losses (Iron losses).

 Copper losses in both the primary and secondary windings.

I

1 2

R

1 

I

2 2

R

2  Magnetic losses, these losses are divided into eddy current losses and hysteresis losses.

P mag



P eddy



P hysterises



V

1

I m

Loaded Transformer

Z 2 ’ is the load impedance referred to the primary

Equivalent circuit

V 1 : Primary voltage (supply) I 1 : Primary current.

V 2 : Secondary voltage (load) I 2: : Secondary current

Exact Circuit

Approximate Circuit

(a) (b) The no load current ranges from 1% to 3% of the full load current.

Therefore, the circuit can be simplified to circuit (b).

Phasor Diagram

V

1 

V

2 ' 

I

' 2 (

R eq



jX eq

)

Performance Measures

 The percent regulation  The transformer efficiency

Voltage Regulation

VR



V

1 

V

2 ' 

I

' 2

R eq

cos f 

I

' 2

X eq

sin f

Efficiency

 The efficiency of the transformer is the ratio of output (secondary) power to the input (primary) power. Formally the efficiency is η: Where,  

P

2

P

1 P 1 : The input power (Primary) = V 1 I 1 cos f 1 P 2 : The output power (Secondary) = V 2 I 2 cos f 2

P

1 

P

2 

P L

Where, P L is the power loss in the transformer = P copper + P iron  

V

2 '

I

' 2

V

2 ' cos f 2

I

' 2  cos f 2

I

' 2 2

R eq



P iron

Example

A 100-kVA, 400/2000 V, single-phase transformer has the following parameters

R

1 = 0.01

X

1 = 0.03 ohms

R X

2 = 0.25 ohms 2 = 0.75 ohms The transformer supplies a load of 90 kVA at 2000 V and 0.8 PF lagging.   Calculate the primary voltage and current using the simplest equivalent circuit.

Find also the V.R. and efficiency for the transformer

Solution

Voltage Regulation

VR



V

1 

V

2 ' 

I

' 2

R eq

cos f 

I

' 2

X eq

sin f

VR

 225  0 .

02  0 .

8  225  0 .

06  0 .

6

VR

 11 .

7

V

%

VR

 11 .

7 411 .

96  2 .

84 %

Efficiency