Transcript Power Transformers

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
Ic
IF
E1
Qc
Im
f
The no load current If is needed to supply the no load
losses and to magnetize the transformer core.
E1
If
IF
Ic Im
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.
I12 R1  I 22 R2

Magnetic losses, these losses are divided into eddy current
losses and hysteresis losses.
Pmag  Peddy  Physterises  V1I m
Loaded Transformer
Z2’ is the load impedance
referred to the primary
Equivalent circuit
V1: Primary voltage (supply)
I1 : Primary current.
V2: Secondary voltage (load)
I2: : 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
'
V1  V2
'
 I 2 ( Req
 jX eq )
Performance Measures

The percent regulation

The transformer efficiency
Voltage Regulation
VR  V1  V2'  I 2' Req cosf  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,
P1 : The input power (Primary) = V1I1 cosf1
P2 : The output power (Secondary) = V2I2 cosf2
P1  P2  PL
Where,
PL is the power loss in the transformer = Pcopper + Piron

V2' I 2' cosf2
V2' I 2' cosf2  I 2'2 Req  Piron
P2
P1
Example
A 100-kVA, 400/2000 V, single-phase transformer has the following
parameters
R1 = 0.01
R2 = 0.25 ohms
X1 = 0.03 ohms
X2 = 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  V1  V2

'
I 2 Req cosf
'
 I 2 X eq sin f
VR  225  0.02  0.8  225  0.06  0.6
VR  11.7V
11.7
%VR 
 2.84%
411.96
Efficiency