Transcript Today’s Topics - Department of Electrical Engineering

Today’s Topics
1- The per unit system
2-Transformer Voltage Regulation
3- Transformer Efficiency
4- Transformer Taps
5- Transformer Rating and Related Problems
The Per Unit System
In power systems there are so many different elements such as
Motors, Generators and Transformers with very different sizes
and nominal values.
To be able to compare the performances of a big and a small
element, per unit system is used.
In per unit system, each electric quantity is
measured as a decimal fraction of some base level
Per Unit Value =
Actual value
Base value
e.g. in a synchronous generator with 13.8 Kv as its nominal voltage,
Instead of saying the voltage is 12.42 Kv, we say the voltage is 0.9 p.u.
How Are the Base Values Defined
For an electric element, we have : Power, Voltage, Current
and Impedance
Usually, the nominal apparent power (S) and nominal voltage (V)
are taken as the base values for power and voltage
The base values for the current and impedance can be calculated
Sbase ,Vbase
I base
Sbase

Vbase
Z base
Vbase V 2base


I base
Sbase
Example 1:
A load takes 20 A at a power factor 0.9 lead
when connected to a 600 Voltage. If the base
voltage and apparent power are 600 and 6000
VA, determine the variables and impedance in
per unit system
600
Z 
 30
20
Sbase  6000,Vbase  600
I base
Sbase

 10
Vbase
V  1 .0
Z base
Vbase

 60
I base
Z  0 .5
I  2 .0
Ideal Transformers in Per Unit Values
vp
vs

Np
Ns

ip
Ns 1


is N p 
S p  Vp .I p  Vs .I s  Ss  S
Sbase  V p . I p  Vs . I s
V p ,base
Vs ,base
vp
vs
1
ip
is
1
Ignore the transformer
Example: Given the transformer in the previous example
rated at 20 kVA, 8000/240, calculate the equivalent circuit,
with per unit values
Sbase  20000
I p ,base
Sbase

 2 .5
V p ,base
Z p ,base
V 2 p ,base

 3200
Sbase
V p ,base  8000
Vs ,base  240
I s ,base
Z p ,base
Sbase

 83.33
V p ,base
V 2 p ,base

 2.88
Sbase
The equivalent circuit
in per unit values
Referred to the primary
or secondary
Transformer Voltage Regulation
Fact: As the load current is increased, the voltage (usually) drops.
Transformer voltage regulation is defined as:
VR 
Vs ,nl  Vs , fl
Vs , fl
Other useful formulas
VR 
VP / a  Vs , fl
Vs , fl
VR 
Vs ,nl  Vs , fl
VR 
Vs , fl
VP  Vs , fl
Vs , fl
Prefer small VR
For an ideal transformer : VR=0
Note: If there were only resistive loads, the VR would
not be much large in real transformers, but because loads
normally have different power factors, VR may be
considerably large in real transformers.
VR in real transformers
VR 
VP / a  Vs , fl
Vs , fl
V P / a  V s  Req.I s  jX eq.I s
V P / a  V s  Req.I s  jX eq.I s
At unity power factor
Resistive load
At lagging power factor
Inductive load
At leading power factor
Capacitive load
Transformer Efficiency
Pout

Pin
Pout
Pout


Ploss  Pout PCu  Pcore  Pout
Vs . I s . cos

PCu  Pcore  Vs . I s . cos
Is 2
PCu  ReqP .( )
a
2
 ReqP . I P
Pcore
2
P
V

Rc
PCu  ReqS . I
Pcore
2
P
V

Rc
2
S
Transformer Taps
The voltage in a distribution line is not constant.
It may be 1.05 p.u. at generator terminal and 0.95 at the
load side.
Depending on the place the transformer is used, we may
need to adjust the transformer ratio to get similar load
voltage. That’s why we need Tap
In Tap Changing
Under Load (TCUL)
transformer, the tap
changes automatically
to keep the output
voltage constant
Transformer Rating and Related Problems
Transformers have three major ratings:
Voltage, Apparent power (current) and frequency
The parameters of equivalent circuit are given in p.u.
As long as a transformer is working under rated
conditions, there should be (is) no problem
What should we consider if we want to use a
transformer under de-rating conditions?
Voltage rating
The insulation has been designed to stand up to a particular voltage
10% increase in voltage in the linear part of B-H curve would
result in 10% increase in magnetization current, but in the
saturation part, the magnetization current may be unacceptable
Frequency rating
Exactly the same problem.
If a 60 Hz transformer is going to be used in a 50 Hz system,
the applied voltage should be reduced by a factor 5/6.
Vm
e  Vm . cos t   
. sin t
N .
Apparent power (current) rating
Each coil has limit for current magnitude and also the higher
current results in higher power loss (which is converted to heat)
and may damage the insulation. This has a dramatic effect on
transformer life.
If the voltage rating has been reduced, the apparent power
should be reduced by the same factor.
Current Inrush
If a transformer is switched to a system, there may be
some current with large magnitude (Current Inrush), in the
first few milliseconds after the switch is closed.
The reason behind the current inrush is transient
Transients
Transient in electrical systems
is the severe changes in
current and voltage after a
switch is closed or opened
Vm
e  Vm . cos t   
. sin t
N .