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
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 %